python3 import idlelib.PyShell fails
Hi, I have a strange error. When I try import idlelib.PyShell from Python3.3 it fails with Python 3.3.2+ (3.3:68ff68f9a0d5+, Jun 30 2013, 12:59:15) [GCC 4.7.3] on linux Type "help", "copyright", "credits" or "license" for more information. >>> import idlelib.PyShell Traceback (most recent call last): File "", line 1, in File "/usr/lib64/python3.3/idlelib/PyShell.py", line 29, in from idlelib.EditorWindow import EditorWindow, fixwordbreaks File "/usr/lib64/python3.3/idlelib/EditorWindow.py", line 11, in import webbrowser File "/usr/lib64/python3.3/webbrowser.py", line 7, in import shlex File "./shlex.py", line 56 print 'shlex: reading from %s, line %d' \ ^ SyntaxError: invalid syntax Looking at shlex.py from /usr/lib64/python3.3 there is no print statement in line 56. But, looking at /usr/lib/python2.7/shlex.py there is this print statement in line 56. Why does idlelib.PyShell when imported from Python3 import shlex from Python2 ? What's going on here? Many thanks for a hint, Helmut. -- http://mail.python.org/mailman/listinfo/python-list
Re: python3 import idlelib.PyShell fails
On Sun, 30 Jun 2013 13:20:24 +0200, Peter Otten wrote: Thanks a lot! Helmut. -- http://mail.python.org/mailman/listinfo/python-list
How to make this faster
Hi,
I have coded a simple algorithm to solve a Sudoku (probably not the first one).
Unfortunately, it takes 13 seconds for a difficult problem which is more than
75 times slower
than the same algorithm coded in C++.
Is this to be expected or could I have made my Python version faster ***
without *** sacrificing readability.
Profiling shows that the function find_good_cell is called (only) 45267 times
and this take 12.9 seconds
CPU time (on a 3.2 GHz machine)
For your pleasure I include the full code below.
Many thanks for a hint,
Helmut
#!/usr/bin/python3
import numpy as np
Grid= np.zeros((9,9),dtype=int)
Row_Digits = np.asarray(np.zeros((9,10)),dtype=bool)
Col_Digits = np.asarray(np.zeros((9,10)),dtype=bool)
Sqr_Digits = np.asarray(np.zeros((9,10)),dtype=bool)
def Read_Sudoku(Input) :
r= -1
R_Cells= 81
for line in Input :
line= line.strip()
if len(line) == 0 or line[0] == '#' : continue
r+= 1
for (c,ch) in enumerate(line) :
if ch == '_' : continue
if not ch in "123456789_" :
raise ValueError("invalid character {0} in input line
{1}".format(c,line))
Sq_No= (r//3)*3+c//3
d= int(ch)
Grid[r,c]= d
Row_Digits[r,d]= True
Col_Digits[c,d]= True
Sqr_Digits[Sq_No,d]= True
R_Cells-= 1
return R_Cells
def Print_Grid() :
Sep = "+---+---+---#---+---+---#---+---+---+"
SepK= "#"
print(Sep)
for r in range(9) :
print('|',end='')
for c in range(9) :
d= Grid[r,c]
print(" {0} {1}".format( str(d) if d>0 else ' ',
'#' if (c+1)%3==0 and c>0 and c<8 else '|' ),
end='')
print("\n{}".format(SepK if (r+1)%3==0 and r>0 and r<8 else Sep))
def find_good_cell() :
Best= None
minPoss= 10
for r in range(9) :
for c in range(9) :
if Grid[r,c] > 0 : continue
Sq_No= (r//3)*3+c//3
Possibilities= 0
for d in range(1,10) :
if Row_Digits[r,d] or Col_Digits[c,d] or Sqr_Digits[Sq_No,d] : continue
Possibilities+= 1
if ( Possibilities < minPoss ) :
minPoss= Possibilities
Best= (r,c)
if minPoss == 0 : Best=(-1,-1)
return Best
def Solve(R_Cells) :
if R_Cells == 0 :
print("\n\n++ S o l u t i o n ++\n")
Print_Grid()
return True
r,c= find_good_cell()
if r < 0 : return False
Sq_No= (r//3)*3+c//3
for d in range(1,10) :
if Row_Digits[r,d] or Col_Digits[c,d] or Sqr_Digits[Sq_No,d] : continue
# put d into Grid
Grid[r,c]= d
Row_Digits[r,d]= True
Col_Digits[c,d]= True
Sqr_Digits[Sq_No,d]= True
Success= Solve(R_Cells-1)
# remove d again
Grid[r,c]= 0
Row_Digits[r,d]= False
Col_Digits[c,d]= False
Sqr_Digits[Sq_No,d]= False
if Success :
Zuege.append((d,r,c))
return True
return False
from io import StringIO
Problem='''
_
_3_85
__1_2
___5_7___
__4___1__
_9___
5__73
__2_1
4___9
'''
Input= StringIO(Problem)
from time import process_time
R_Cells= Read_Sudoku(Input)
Print_Grid()
Zuege=[]
Start= process_time()
# import cProfile
# cProfile.run("Success = Solve(R_Cells)")
Success = Solve(R_Cells)
Stop= process_time()
print("after {} seconds:".format(Stop-Start))
if Success :
print("\nZuege:")
n=0
for Z in reversed(Zuege) :
print("{0} -> ({1},{2})\t".format(Z[0],Z[1]+1,Z[2]+1),end='')
n+= 1
if n%5 == 0 :
print()
print()
--
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Re: How to make this faster
On Fri, 05 Jul 2013 10:38:35 +0100, Fábio Santos wrote: > [Skipping to bottleneck] > >> def find_good_cell() : > > In this function you are accessing global variables a lot of times. Since > accessing globals takes much more time than accessing locals, I advise you > to assign them to local names, and use them. I've tried to use local "references" like G= Grid and using G instead of Grid below and similarly for Row_Digits, Col_Digits and Sqr_Digits but it had no noticeable effect. > >> Best= None >> minPoss= 10 >> for r in range(9) : >> for c in range(9) : >> if Grid[r,c] > 0 : continue >> Sq_No= (r//3)*3+c//3 >> Possibilities= 0 >> for d in range(1,10) : > > On this condition (below) you can try to check which condition is True more > often and put that condition first in order to take advantage of short > circuiting and minimize array access. Unfortunately, that's unpredictable. > >> if Row_Digits[r,d] or Col_Digits[c,d] or Sqr_Digits[Sq_No,d] : > continue Many thanks, Helmut -- http://mail.python.org/mailman/listinfo/python-list
Re: How to make this faster
On Fri, 05 Jul 2013 11:13:33 +0100, Oscar Benjamin wrote: > My one comment is that you're not really making the most out of numpy > arrays. Numpy's ndarrays are efficient when each line of Python code > is triggering a large number of numerical computations performed over > the array. Because of their N-dimensional nature and the fact that > they are in some sense second class citizens in CPython they are often > not as good as lists for this kind of looping and indexing. > > I would actually expect this program to run faster with ordinary > Python lists and lists of lists. It means that you need to change e.g. > Grid[r, c] to Grid[r][c] but really I think that the indexing syntax > is all you're getting out of numpy here. > Thanks Oscar, that was a big improvement, indeed. Using lists of lists instead of numpy arrays made the code more than twice as fast (13 seconds down to 6 seconds) Since I don't do any numerical stuff with the arrays, Numpy doesn't seem to be a good choice. I think this is an argument to add real arrays to Python. I even tried to use dictionaries instead of Numpy arrays. This version is a bit slower then the lists of lists version (7.2 seconds instead of 6 second) but still much faster than the Numpy array solution. Helmut -- http://mail.python.org/mailman/listinfo/python-list
Re: How to make this faster
On Fri, 05 Jul 2013 14:41:23 +0100, Oscar Benjamin wrote: > On 5 July 2013 11:53, Helmut Jarausch wrote: >> I even tried to use dictionaries instead of Numpy arrays. This version is a >> bit >> slower then the lists of lists version (7.2 seconds instead of 6 second) but >> still >> much faster than the Numpy array solution. > > When you switched to dictionaries did you take advantage of the > sparseness by iterating over dictionary keys instead of indices? This > is the kind of thing that I meant when I said that in Python it's > often easier to implement a better algorithm than in C. What I mean is > that if Grid is a dict so that Grid[(r, c)] is the entry at row r and > column c (if it exists) then you can change a loop like: > > for r in range(9): > for c in range(9): > if Grid[r, c] > 0: continue > # do stuff > > so that it looks like: > > for r, c in Grid: > # do stuff > > If the grid is sparsely occupied then this could be a significant improvement. > > > Oscar This gives a big speedup. Now, the time is gone down to 1.73 seconds in comparison to original 13 seconds or the 7 seconds for the first version above. Many thanks, it seems hard to optimize a Python program, Helmut -- http://mail.python.org/mailman/listinfo/python-list
Re: How to make this faster
On Fri, 05 Jul 2013 12:02:21 +, Steven D'Aprano wrote: > On Fri, 05 Jul 2013 10:53:35 +, Helmut Jarausch wrote: > >> Since I don't do any numerical stuff with the arrays, Numpy doesn't seem >> to be a good choice. I think this is an argument to add real arrays to >> Python. > > Guido's time machine strikes again: > > import array > > > By the way, I'm not exactly sure how you go from "I don't do numerical > calculations on numpy arrays" to "therefore Python should have arrays". I should have been more clear. I meant multi-dimensional arrays (2D, at least) Numpy is fine if I do math with matrices (without loops in python). Given that I don't like to use the old FORTRAN way (when "dynamic" arrays are passed to functions) of indexing a 2-d array I would need a MACRO or an INLINED function in Python or something like a META-compiler phase transforming def access2d(v,i,j,dim1) : # doesn't work on the l.h.s. return v[i*dim1+j] access2d(v,i,j,dim1) = 7# at compile time, please to v[i*dim1+j]= 7 # this, by itself, is considered ugly (the FORTRAN way) Thanks for the discussion, Helmut -- http://mail.python.org/mailman/listinfo/python-list
Re: How to make this faster
On Fri, 05 Jul 2013 13:44:57 +0100, Fábio Santos wrote: May I suggest you avoid range and use enumerate(the_array) instead? It might be faster. How does this work? Given Grid= [[0 for j in range(9)] for i in range(9)] for (r,c,val) in (Grid) : Helmut -- http://mail.python.org/mailman/listinfo/python-list
Re: How to make this faster
On Fri, 05 Jul 2013 15:45:25 +0100, Oscar Benjamin wrote:
> Presumably then you're now down to the innermost loop as a bottle-neck:
>
> Possibilities= 0
> for d in range(1,10) :
> if Row_Digits[r,d] or Col_Digits[c,d] or Sqr_Digits[Sq_No,d] :
> continue
> Possibilities+= 1
>
> If you make it so that e.g. Row_Digits[r] is a set of indices rather
> than a list of bools then you can do this with something like
>
> Possibilities = len(Row_Digits[r] | Col_Digits[c] | Sqr_Digits[Sq_No])
>
> or perhaps
>
> Possibilities = len(set.union(Row_Digits[r], Col_Digits[c],
> Sqr_Digits[Sq_No]))
>
> which I would expect to be a little faster than looping over range
> since the loop is then performed under the hood by the builtin
> set-type.
>
> It just takes practice.
indeed
> It's a little less obvious in Python than in
> low-level languages where the bottlenecks will be and which operations
> are faster/slower but optimisation always involves a certain amount of
> trial and error anyway.
>
>
> Oscar
I've tried the following version
def find_good_cell() :
Best= None
minPoss= 10
for r,c in Grid :
if Grid[(r,c)] > 0 : continue
Sq_No= (r//3)*3+c//3
Possibilities= 9-len(Row_Digits[r] | Col_Digits[c] | Sqr_Digits[Sq_No])
if ( Possibilities < minPoss ) :
minPoss= Possibilities
Best= (r,c)
if minPoss == 0 : Best=(-1,-1)
return Best
All_digits= set((1,2,3,4,5,6,7,8,9))
def Solve(R_Cells) :
if R_Cells == 0 :
print("\n\n++ S o l u t i o n ++\n")
Print_Grid()
return True
r,c= find_good_cell()
if r < 0 : return False
Sq_No= (r//3)*3+c//3
for d in All_digits - (Row_Digits[r] | Col_Digits[c] | Sqr_Digits[Sq_No]) :
# put d into Grid
Grid[(r,c)]= d
Row_Digits[r].add(d)
Col_Digits[c].add(d)
Sqr_Digits[Sq_No].add(d)
Success= Solve(R_Cells-1)
# remove d again
Grid[(r,c)]= 0
Row_Digits[r].remove(d)
Col_Digits[c].remove(d)
Sqr_Digits[Sq_No].remove(d)
if Success :
Zuege.append((d,r,c))
return True
return False
which turns out to be as fast as the previous "dictionary only version".
Probably, set.remove is a bit slow
Thanks,
Helmut
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Re: How to make this faster
On Fri, 05 Jul 2013 16:18:41 +0100, Fábio Santos wrote: > On 5 Jul 2013 15:59, "Helmut Jarausch" wrote: >> >> On Fri, 05 Jul 2013 13:44:57 +0100, Fábio Santos wrote: >> May I suggest you avoid range and use enumerate(the_array) instead? It >> might be faster. >> >> How does this work? >> >> Given >> >> Grid= [[0 for j in range(9)] for i in range(9)] >> >> for (r,c,val) in (Grid) : >> >> Helmut > > for r, row_lst in enumerate(Grid): > for c, val in enumerate(row_lst): This is only slightly faster. I assume the creation of the temporary lists "row_list" is a bit expensive. Taking 5.4 seconds it's much slower than the current champion ( 0.79 seconds ) Thanks, Helmut. -- http://mail.python.org/mailman/listinfo/python-list
Re: How to make this faster
On Fri, 05 Jul 2013 16:38:43 +0100, Oscar Benjamin wrote:
> On 5 July 2013 16:17, Helmut Jarausch wrote:
>>
>> I've tried the following version
>>
>> def find_good_cell() :
>> Best= None
>> minPoss= 10
>> for r,c in Grid :
>> if Grid[(r,c)] > 0 : continue
>
> Sorry, I think what I meant was that you should have a structure
> called e.g. Remaining which is the set of all (r, c) pairs that you
> want to loop over here. Then there's no need to check on each
> iteration whether or not Grid[(r, c)] > 0. When I said "sparse" I
> meant that you don't need to set keys in Grid unless you actually have
> a value there so the test "Grid[(r, c)] > 0" would look like "(r, c)
> in Grid". Remaining is the set of all (r, c) pairs not in Grid that
> you update incrementally with .add() and .remove().
>
> Then this
>
>for r,c in Grid :
> if Grid[(r,c)] > 0 : continue
>
> becomes
>
> for r, c in Remaining:
>
>> Sq_No= (r//3)*3+c//3
>> Possibilities= 9-len(Row_Digits[r] | Col_Digits[c] | Sqr_Digits[Sq_No])
>> if ( Possibilities < minPoss ) :
>> minPoss= Possibilities
>> Best= (r,c)
>>
>> if minPoss == 0 : Best=(-1,-1)
>> return Best
>>
>> All_digits= set((1,2,3,4,5,6,7,8,9))
>
> All_digits= set(range(1, 10))
>
> or
>
> All_digits = {1,2,3,4,5,6,7,8,9}
>
>>
>> def Solve(R_Cells) :
>> if R_Cells == 0 :
>> print("\n\n++ S o l u t i o n ++\n")
>> Print_Grid()
>> return True
>>
>> r,c= find_good_cell()
>> if r < 0 : return False
>> Sq_No= (r//3)*3+c//3
>>
>> for d in All_digits - (Row_Digits[r] | Col_Digits[c] | Sqr_Digits[Sq_No]) :
>> # put d into Grid
>> Grid[(r,c)]= d
>> Row_Digits[r].add(d)
>> Col_Digits[c].add(d)
>> Sqr_Digits[Sq_No].add(d)
>>
>> Success= Solve(R_Cells-1)
>>
>> # remove d again
>> Grid[(r,c)]= 0
>> Row_Digits[r].remove(d)
>> Col_Digits[c].remove(d)
>> Sqr_Digits[Sq_No].remove(d)
>>
>> if Success :
>> Zuege.append((d,r,c))
>> return True
>>
>> return False
>>
>> which turns out to be as fast as the previous "dictionary only version".
>> Probably, set.remove is a bit slow
>
> No it's not and you're not using it in your innermost loops anyway.
> Probably the loop I referred to isn't your bottleneck.
>
I've tried your suggestion, but unless I've misunderstood your suggestion it's
a bit slower
than the solution above.
The solution above take 0.79 seconds (mean of 100 calls) while the following
version
take 1.05 seconds (mean of 100 calls):
Grid = {(i,j):0 for i in range(9) for j in range(9)}
Remaining= {(i,j) for i in range(9) for j in range(9)}
Row_Digits = [set() for r in range(9)]
Col_Digits = [set() for c in range(9)]
Sqr_Digits = [set() for s in range(9)]
remove some pairs from Remaining for the initial set of the given Sudoku
def find_good_cell() :
Best= None
minPoss= 10
for r,c in Remaining :
Sq_No= (r//3)*3+c//3
Possibilities= 9-len(Row_Digits[r] | Col_Digits[c] | Sqr_Digits[Sq_No])
if ( Possibilities < minPoss ) :
minPoss= Possibilities
Best= (r,c)
if minPoss == 0 : Best=(-1,-1)
return Best
All_digits= set(range(1,10))
def Solve(R_Cells) :
if R_Cells == 0 :
print("\n\n++ S o l u t i o n ++\n")
Print_Grid()
return True
r,c= find_good_cell()
if r < 0 : return False
Sq_No= (r//3)*3+c//3
for d in All_digits - (Row_Digits[r] | Col_Digits[c] | Sqr_Digits[Sq_No]) :
# put d into Grid
Grid[(r,c)]= d
Remaining.remove((r,c))
Row_Digits[r].add(d)
Col_Digits[c].add(d)
Sqr_Digits[Sq_No].add(d)
Success= Solve(R_Cells-1)
# remove d again
Grid[(r,c)]= 0
Remaining.add((r,c))
Row_Digits[r].remove(d)
Col_Digits[c].remove(d)
Sqr_Digits[Sq_No].remove(d)
if Success :
Zuege.append((d,r,c))
return True
return False
--
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Re: How to make this faster
On Fri, 05 Jul 2013 16:50:41 +, Steven D'Aprano wrote: > On Fri, 05 Jul 2013 16:07:03 +, Helmut Jarausch wrote: > >> The solution above take 0.79 seconds (mean of 100 calls) while the >> following version take 1.05 seconds (mean of 100 calls): > > 1) How are you timing the calls? I've put the work horse "Solve" into a loop executing 100 times. That's on an otherwise idle Linux machine. > > 2) Don't use the mean, that's the wrong statistic when you are measuring > something where the error is always one sided. You want the minimum, not > the mean. Here you assume time measuring itself is without error - I doubt that. If the timing version, which executes function "Solve" one hundred times, runs about 80-100 seconds without a significant variation, then taking the mean is mathematically correct. I can't take the minimum since I don't measure the time a single call takes. > > When you measure the time taken for a piece of code, the number you get > is made up of two components: > > > 1) the actual time the code would have taken, if there were no random > fluctuations due to other processes, etc.; and > > 2) random errors due to switching to other processes, etc. > > Both of these are unknown; you only know the total. But obviously the > random errors are always positive. They can never be negative, and you > can never measure a time which is less than the fastest your code could > run. > > (If your anti-virus software starts scanning in the middle of the trial, > it can only make your code take more time to run, never less.) > > So the only way to minimize the error is to pick the minimum time, not > the average. The average just gives you: > > - some unknown "true" time, plus some unknown error, somewhere > between the smallest error and the biggest error; > > whereas the minimum gives you: > > - some unknown "true" time, plus the smallest error yet seen. -- http://mail.python.org/mailman/listinfo/python-list
Re: How to make this faster
On Fri, 05 Jul 2013 17:25:54 +0100, MRAB wrote:
> For comparison, here's my solution:
Your solution is very fast, indeed.
It takes 0.04 seconds (mean of 1000 runs) restoring "grid"
in between.
But that's a different algorithm which is IMHO more difficult to understand.
Many thanks,
Helmut
>
> from collections import Counter
>
> problem = '''
> _
> _3_85
> __1_2
> ___5_7___
> __4___1__
> _9___
> 5__73
> __2_1
> 4___9
> '''
>
> # Build the grid.
> digits = "123456789"
>
> grid = []
>
> for row in problem.splitlines():
>if not row:
> continue
>
>new_row = []
>
>for cell in row:
> if cell.isdigit():
>new_row.append({cell})
> else:
>new_row.append(set(digits))
>
>grid.append(new_row)
>
> # Solve the grid.
> changed = True
> while changed:
>changed = False
>
># Look for cells that contain only one digit.
>for r in range(9):
> for c in range(9):
>if len(grid[r][c]) == 1:
> digit = list(grid[r][c])[0]
>
> # Remove from other cells in same row.
> for c2 in range(9):
>if c2 != c and digit in grid[r][c2]:
> grid[r][c2].remove(digit)
> changed = True
>
> # Remove from other cells in same column.
> for r2 in range(9):
>if r2 != r and digit in grid[r2][c]:
> grid[r2][c].remove(digit)
> changed = True
>
> # Remove from other cells in the same block of 9.
> start_row = r - r % 3
> start_column = c - c % 3
> for r2 in range(start_row, start_row + 3):
>for c2 in range(start_column, start_column + 3):
> if (r2, c2) != (r, c) and digit in grid[r2][c2]:
>grid[r2][c2].remove(digit)
>changed = True
>
># Look for digits that occur in only one cell in a row.
>for r in range(9):
> counts = Counter()
> for c in range(9):
>counts += Counter(grid[r][c])
>
> unique = {digit for digit, times in counts.items() if times == 1}
>
> for c in range(9):
>if len(grid[r][c]) > 1 and len(grid[r][c] & unique) == 1:
> grid[r][c] &= unique
> changed = True
>
># Look for digits that occur in only one cell in a column.
>for c in range(9):
> counts = Counter()
> for r in range(9):
>counts += Counter(grid[r][c])
>
> unique = {digit for digit, times in counts.items() if times == 1}
>
> for r in range(9):
>if len(grid[r][c]) > 1 and len(grid[r][c] & unique) == 1:
> grid[r][c] &= unique
> changed = True
>
># Look for digits that occur in only one cell in a block of 9.
>for start_row in range(0, 9, 3):
> for start_column in range(0, 9, 3):
>counts = Counter()
>for r in range(start_row, start_row + 3):
> for c in range(start_column, start_column + 3):
>counts += Counter(grid[r][c])
>
>unique = {digit for digit, times in counts.items() if times == 1}
>
>for r in range(start_row, start_row + 3):
> for c in range(start_column, start_column + 3):
>if len(grid[r][c]) > 1 and len(grid[r][c] & unique) == 1:
> grid[r][c] &= unique
> changed = True
>
> # Display the solution.
> for row in grid:
>for cell in row:
> print("".join(sorted(cell)), end=" ")
>
>print()
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Re: How to make this faster
On Sat, 06 Jul 2013 03:05:30 +, Steven D'Aprano wrote: > That doesn't explain how you time it, only that you have a loop executing > 100 times. Are you using time.time, or time.clock? (I trust you're not > measuring times by hand with a stop watch.) > > I expect you're probably doing something like this: > > start = time.time() I have been using time.process_time >> If the timing version, which executes function "Solve" one hundred >> times, runs about 80-100 seconds without a significant variation, then >> taking the mean is mathematically correct. > For longer running code, like this, you might also like this: > http://code.activestate.com/recipes/577896/ Thanks for the pointer. > If the best you can say is it takes "80-100 seconds", that's pretty > significant variation, of the order of 20%. That's not a variation of a SINGLE variant. One variant takes 80 seconds and the other variant to be compared with takes 100 seconds. > > In this case, with times of the order of a second per loop, it may be > reasonable to say "in this specific case, the error is too small to care > about", or "I just don't care about the error, since it will be about the > same for different variations of my solve function". But in that case, > why bother looping 100 times? > > >> I can't take the minimum >> since I don't measure the time a single call takes. > > Then perhaps you should. Many thanks, Helmut -- http://mail.python.org/mailman/listinfo/python-list
Procedure to request adding a module to the standard library - or initiating a vote on it
Hi, I'd like to request adding the module http://pypi.python.org/pypi/regex to Python's standard library in the (near) future or to even replace the current 're' module by it. Personally I'm in need for fuzzy regular expressions and I don't see how to do this easily and efficiently without this module. For a long term project I also need some "guarantee" that this functionality will exist in future. So, is there a (formal) procedure for such a request or for initiating some sort of vote on it? I know there is a "Benevolent Dictator" for Python. Should I try to contact him personally? Many thanks for a hint, Helmut Jarausch RWTH Aachen University Germany -- http://mail.python.org/mailman/listinfo/python-list
Re: Procedure to request adding a module to the standard library - or initiating a vote on it
On Tue, 07 Aug 2012 13:15:29 +0200, Peter Otten wrote: > I don't think that will help. From PEP 408: > > """ > As part of the same announcement, Guido explicitly accepted Matthew > Barnett's 'regex' module [4] as a provisional addition to the standard > library for Python 3.3 (using the 'regex' name, rather than as a drop-in > replacement for the existing 're' module). > """ What is a "provisional addition"? Python-3.3 (20120708) doesn't have such a module. Many thanks, Helmut. -- http://mail.python.org/mailman/listinfo/python-list
print(....,file=sys.stderr) buffered?
Hi,
for tracing purposes I have added some print outs like
print('+++ before calling foo',file=sys.stderr)
x=foo(..)
print('--- after calling foo',
and within 'foo'
print('>>> entering foo ...',file=sys.stderr)
Now, when executing this, I always get
+++ before calling foo
--- after calling foo
>>> entering foo ...
When outputting to stderr from C/C++ it's guaranteed that the different
outputs appear in the same order as they have been generated.
Is this guarantee no more valid in Python 3.2 ?
Many thanks for a comment,
Helmut.
(That's a single-threaded application)
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Re: print(....,file=sys.stderr) buffered?
On Mon, 13 Aug 2012 15:43:31 +, Grant Edwards wrote:
> On 2012-08-13, Helmut Jarausch wrote:
>> Hi,
>>
>> for tracing purposes I have added some print outs like
>>
>> print('+++ before calling foo',file=sys.stderr)
>> x=foo(..)
>> print('--- after calling foo',
Sorry, this is a cut'n paste error. I did use
print('--- after calling foo',file=sys.stderr)
>>
>> and within 'foo'
>> print('>>> entering foo ...',file=sys.stderr)
>>
>> Now, when executing this, I always get
>>
>> +++ before calling foo --- after calling foo
>>>>> entering foo ...
>>
>> When outputting to stderr from C/C++ it's guaranteed that the different
>> outputs appear in the same order as they have been generated.
>
> You're not printing to stderr in the second print() call -- you're
> printing to stdout. The two file objects have separate buffers and may
> even be using two different buffering modes (e.g. line vs. block).
> You can't interleave writes to stderr and stdout and assume order is
> preserved unless you take specific steps (such as forcing them both to
> be unbuffered or flushing them at certain points).
>
>> Is this guarantee no more valid in Python 3.2 ?
>
> If you write to stderr all three times, it should work the way you want
> it to.
It seems it doesn't do so in my case.
Thanks,
Helmut.
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email with a non-ascii charset in Python3 ?
Hi,
I'm sorry to ask such a FAQ but still I couldn't find an answer - neither in
the docs nor the web.
What's wrong with the following script?
Many thanks for a hint,
Helmut.
#!/usr/bin/python3
#_*_ coding: latin1 _*_
import smtplib
from email.message import Message
import datetime
msg= Message()
msg.set_charset('latin-1')
msg['Subject'] = "*** Email Test ***"
msg['From'] = "email_tes...@numa-sv.igpm.rwth-aachen.de"
msg['To'] = "jarau...@igpm.rwth-aachen.de"
msg['Date'] = datetime.datetime.utcnow().strftime('%m/%d/%Y %I:%M:%S %p')
server= smtplib.SMTP("igpm.igpm.rwth-aachen.de")
msg.set_payload("Gedanken über einen Test","iso-8859-1")
## I have tried msg.set_payload("Gedanken über einen
Test".encode("iso-8859-1"),"iso-8859-1")
## which fails, as well
server.send_message(msg)
Traceback (most recent call last):
File "./Test_EMail_Py3.py", line 17, in
server.send_message(msg)
File "/usr/lib64/python3.2/smtplib.py", line 812, in send_message
g.flatten(msg_copy, linesep='\r\n')
File "/usr/lib64/python3.2/email/generator.py", line 91, in flatten
self._write(msg)
File "/usr/lib64/python3.2/email/generator.py", line 137, in _write
self._dispatch(msg)
File "/usr/lib64/python3.2/email/generator.py", line 163, in _dispatch
meth(msg)
File "/usr/lib64/python3.2/email/generator.py", line 396, in _handle_text
super(BytesGenerator,self)._handle_text(msg)
File "/usr/lib64/python3.2/email/generator.py", line 201, in _handle_text
self.write(payload)
File "/usr/lib64/python3.2/email/generator.py", line 357, in write
self._fp.write(s.encode('ascii', 'surrogateescape'))
UnicodeEncodeError: 'ascii' codec can't encode character '\xfc' in position
9: ordinal not in range(128)
server.quit()
This is Python 3.2.4 (GIT 20120805)
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Re: email with a non-ascii charset in Python3 ?
On Wed, 15 Aug 2012 14:48:40 +0200, Christian Heimes wrote:
> Am 15.08.2012 14:16, schrieb Helmut Jarausch:
>> Hi,
>>
>> I'm sorry to ask such a FAQ but still I couldn't find an answer -
>> neither in the docs nor the web.
>>
>> What's wrong with the following script?
>>
>> Many thanks for a hint,
>> Helmut.
>>
>> #!/usr/bin/python3 #_*_ coding: latin1 _*_
>>
>> import smtplib from email.message import Message import datetime
>>
>> msg= Message()
>> msg.set_charset('latin-1')
>> msg['Subject'] = "*** Email Test ***"
>> msg['From'] = "email_tes...@numa-sv.igpm.rwth-aachen.de"
>> msg['To'] = "jarau...@igpm.rwth-aachen.de"
>> msg['Date'] = datetime.datetime.utcnow().strftime('%m/%d/%Y %I:%M:%S
>> %p')
>>
>> server= smtplib.SMTP("igpm.igpm.rwth-aachen.de")
>> msg.set_payload("Gedanken über einen Test","iso-8859-1")
>
> You mustn't combine set_charset() with set_payload() with a charset.
> That results into invalid output:
>
>>>> msg = Message()
>>>> msg.set_payload("Gedanken über einen Test", "iso-8859-1")
>>>> msg.as_string()
> 'MIME-Version: 1.0\nContent-Type: text/plain;
> charset="iso-8859-1"\nContent-Transfer-Encoding:
> quoted-printable\n\nGedanken =FCber einen Test'
>
>>>> msg2 = Message()
>>>> msg2.set_charset("iso-8859-1")
>>>> msg2.set_payload("Gedanken über einen Test", "iso-8859-1")
>>>> msg2.as_string()
> 'MIME-Version: 1.0\nContent-Type: text/plain;
> charset="iso-8859-1"\nContent-Transfer-Encoding:
> quoted-printable\n\nGedanken über einen Test'
>
Thanks!
Just, one mustn't use
server.send_message(msg.as_string())
But what if msg['From'] contains a non-ASCII character?
I wonder what the usage of msg.set_charset('latin-1') is.
With
msg.set_charset('latin-1')
msg.set_payload("Gedanken über einen Test") # is accepted BUT
server.send_message(msg)
gives
Traceback (most recent call last):
File "Test_EMail_Py3_2.py", line 21, in
server.send_message(msg)
File "/usr/lib64/python3.2/smtplib.py", line 812, in send_message
g.flatten(msg_copy, linesep='\r\n')
File "/usr/lib64/python3.2/email/generator.py", line 91, in flatten
self._write(msg)
File "/usr/lib64/python3.2/email/generator.py", line 137, in _write
self._dispatch(msg)
File "/usr/lib64/python3.2/email/generator.py", line 163, in _dispatch
meth(msg)
File "/usr/lib64/python3.2/email/generator.py", line 396, in
_handle_text
super(BytesGenerator,self)._handle_text(msg)
File "/usr/lib64/python3.2/email/generator.py", line 201, in
_handle_text
self.write(payload)
File "/usr/lib64/python3.2/email/generator.py", line 357, in write
self._fp.write(s.encode('ascii', 'surrogateescape'))
UnicodeEncodeError: 'ascii' codec can't encode character '\xfc' in
position 9: ordinal not in range(128)
Helmut.
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Python3.3 email policy date field
Hi,
in response to a bug report I got the follow helpful comments from R. David
Murray.
Many thanks to him. (Unfortunately, I don't know his email, so I can write him
directly)
To generate an email (with non-ascii letters)
R. David Murray wrote:
>>> But even better, so will this:
>>> m = Message(policy=policy.SMTP)
>>> m['From'] = "Günter Weiße "
This works, but now I cannot add a date field
Trying
m['Date'] = datetime.datetime.utcnow().strftime('%m/%d/%Y %I:%M:%S %p')
I get
Traceback (most recent call last):
File "Test_EMail_Py3_4.py", line 23, in
msg['Date'] = datetime.datetime.utcnow().strftime('%m/%d/%Y %I:%M:%S %p')
File "/usr/lib64/python3.3/email/message.py", line 359, in __setitem__
self._headers.append(self.policy.header_store_parse(name, val))
File "/usr/lib64/python3.3/email/policy.py", line 119, in header_store_parse
return (name, self.header_factory(name, value))
File "/usr/lib64/python3.3/email/headerregistry.py", line 583, in __call__
return self[name](name, value)
File "/usr/lib64/python3.3/email/headerregistry.py", line 194, in __new__
cls.parse(value, kwds)
File "/usr/lib64/python3.3/email/headerregistry.py", line 300, in parse
value = utils.parsedate_to_datetime(value)
File "/usr/lib64/python3.3/email/utils.py", line 243, in parsedate_to_datetime
*dtuple, tz = __parsedate_tz(data)
TypeError: 'NoneType' object is not iterable
Many thanks for a hint,
Helmut.
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Re: Python3.3 email policy date field
On Thu, 23 Aug 2012 12:36:01 +0100, MRAB wrote: > From what I've tried, it looks like the date can't be a string: > > >>> m['Date'] = datetime.datetime.utcnow() > >>> m['Date'] > 'Thu, 23 Aug 2012 11:33:20 -' Many thanks - it's even easier! Waiting for Python 3.3 to become standard! Helmut. -- http://mail.python.org/mailman/listinfo/python-list
exec with partial globals
Hi,
I'd like to give the user the ability to enter code which may only rebind
a given set of names but not all ones.
This does NOT work
A=1
B=2
Code=compile('A=7','','exec')
exec(Code,{'A':0})
print("I've got A={}".format(A)) # prints 1
How can 'filter' the gobal namespace such that modifying 'A' is allowed
but any attempt to modify 'B' should give an exception.
Many thanks for a hint,
Helmut.
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Re: exec with partial globals
On Tue, 30 Oct 2012 08:33:38 -0400, Dave Angel wrote:
> On 10/30/2012 08:00 AM, Helmut Jarausch wrote:
>> Hi,
>>
>> I'd like to give the user the ability to enter code which may only rebind
>> a given set of names but not all ones.
>> This does NOT work
>> A=1
>> B=2
>> Code=compile('A=7','','exec')
>> exec(Code,{'A':0})
>> print("I've got A={}".format(A)) # prints 1
>>
>>
>> How can 'filter' the gobal namespace such that modifying 'A' is allowed
>> but any attempt to modify 'B' should give an exception.
>>
>>
>> Many thanks for a hint,
>> Helmut.
>
> A=1
> B=2
> Code=compile('A=7','','exec')
> vars = {'A':A}
> exec(Code, vars)
> A = vars["A"]
> print("I've got A={}".format(A)) # prints 1
>
> That now prints "I've got A=7"
>
> More generally, you could write a loop, copying globals into vars, and
> another one, copying them back.
>
> No idea what you're really after; this is one of the more dangerous
> things to try.
>
> Although you can constrain the globals seen by the code, that code can
> still use builtins, do imports, delete files, etc.
>
> Further, if your user is clever enough, he can do:
>
> Code=compile('A=7; print("howdy"); import __main__;
> __main__.B=42','','exec')
>
> What are you really trying to permit him to do? Initialize some
> variables for you? How about an .ini file ?
Many thanks Chris, many thanks to Dave.
First, in my application only trusted people will use it.
Here my example. I'd like to update a spreadsheet by data given by another
spreadsheet. I like to allow to modify only certain columns of the first
spreadsheet. The update formula and possible conditions are entered at
run time and the available fields are only known once the spreadsheets
have been read.
Given spreadsheet S (Source) and D (Destination) as objects (wrapping a
dictionary) a possible (legal) input would be
D.price= D.price-S.discount
No other fields of 'D' should be modifiable.
Again, I don't need to take actions against a malicious user,
just take care about a typo or mistake
Thanks again,
Helmut.
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Python3.3 str() bug?
Hi, probably I'm missing something. Using str(Arg) works just fine if Arg is a list. But str([],encoding='latin-1') gives the error TypeError: coercing to str: need bytes, bytearray or buffer-like object, list found If this isn't a bug how can I use str(Arg,encoding='latin-1') in general. Do I need to flatten any data structure which is normally excepted by str() ? Many thanks for a hint, Helmut. -- http://mail.python.org/mailman/listinfo/python-list
Re: Python3.3 str() bug?
On Fri, 09 Nov 2012 10:37:11 +0100, Stefan Behnel wrote: > Helmut Jarausch, 09.11.2012 10:18: >> probably I'm missing something. >> >> Using str(Arg) works just fine if Arg is a list. >> But >> str([],encoding='latin-1') >> >> gives the error >> TypeError: coercing to str: need bytes, bytearray or buffer-like object, >>list found >> >> If this isn't a bug how can I use str(Arg,encoding='latin-1') in general. >> Do I need to flatten any data structure which is normally excepted by str() ? > > Funny idea to call this a bug in Python. What your code is asking for is to > decode the object you pass in using the "latin-1" encoding. Since a list is > not something that is "encoded", let alone in latin-1, you get an error, > and actually a rather clear one. > > Note that this is not specific to Python3.3 or even 3.x. It's the same > thing in Py2 when you call the equivalent unicode() function. > For me it's not funny, at all. Whenever Python3 encounters a bytestring it needs an encoding to convert it to a string. If I feed a list of bytestrings or a list of list of bytestrings to 'str' , etc, it should use the encoding for each bytestring component of the given data structure. How can I convert a data strucure of arbitrarily complex nature, which contains bytestrings somewhere, to a string? This problem has arisen while converting a working Python2 script to Python3.3. Since Python2 doesn't have bytestrings it just works. Tell me how to convert str(obj) from Python2 to Python3 if obj is an arbitrarily complex data structure containing bytestrings somewhere which have to be converted to strings with a given encoding? Helmut. -- http://mail.python.org/mailman/listinfo/python-list
Re: Python3.3 str() bug?
On Fri, 09 Nov 2012 23:22:04 +1100, Chris Angelico wrote: > On Fri, Nov 9, 2012 at 10:08 PM, Helmut Jarausch > wrote: >> For me it's not funny, at all. > > His description "funny" was in reference to the fact that you > described this as a bug. This is a heavily-used mature language; bugs > as fundamental as you imply are unlikely to exist (consequences of > design decisions there will be, but not outright bugs, usually); > extraordinary claims require extraordinary evidence. Just for the record. I first discovered a real bug with Python3 when using os.walk on a file system containing non-ascii characters in file names. I encountered a very strange behavior (I still would call it a bug) when trying to put non-ascii characters in email headers. This has only been solved satisfactorily in Python3.3. > >> Whenever Python3 encounters a bytestring it needs an encoding to convert it >> to >> a string. If I feed a list of bytestrings or a list of list of bytestrings to >> 'str' , etc, it should use the encoding for each bytestring component of the >> given data structure. >> >> How can I convert a data strucure of arbitrarily complex nature, which >> contains >> bytestrings somewhere, to a string? > > Okay, now we're getting somewhere. > > What you really should be doing is not transforming the whole > structure, but explicitly transforming each part inside it. I > recommend you stop fighting the language and start thinking about your > data as either *bytes* or *characters* and using the appropriate data > types (bytes or str) everywhere. You'll then find that it makes > perfect sense to explicitly translate (en/decode) from one to another, > but it doesn't make sense to encode a list in UTF-8 or decode a > dictionary from Latin-1. > >> This problem has arisen while converting a working Python2 script to >> Python3.3. >> Since Python2 doesn't have bytestrings it just works. > > Actually it does; it just calls them "str". And there's a Unicode > string type, called "unicode", which is (more or less) the thing that > Python 3 calls "str". > > You may be able to do some kind of recursive cast that, in one sweep > of your data structure, encodes all str objects into bytes using a > given encoding (or the reverse thereof). But I don't think this is the > best way to do things. Thanks, but in my case the (complex) object is returned via ctypes from the aspell library. I still think that a standard function in Python3 which is able to 'stringify' objects should take an encoding parameter. Thanks, Helmut. -- http://mail.python.org/mailman/listinfo/python-list
python3.3 - tk_setPalette bug?
Hi, AFAIK, this should work: import tkinter as Tk root= Tk.Tk() root.tk_setPalette(background = 'AntiqueWhite1', foreground = 'blue') but python-3.3:0e4574595674+ gives Traceback (most recent call last): File "Matr_Select.py", line 174, in root.tk_setPalette(background = 'AntiqueWhite1', foreground = 'blue') File "/usr/lib64/python3.3/tkinter/__init__.py", line 390, in tk_setPalette + _flatten(args) + _flatten(kw.items())) TypeError: argument must be sequence What am I missing? Thanks, Helmut. -- http://mail.python.org/mailman/listinfo/python-list
email.message.Message - as_string fails
Hi, I'm trying to filter an mbox file by removing some messages. For that I use Parser= FeedParser(policy=policy.SMTP) and 'feed' any lines to it. If the mbox file contains a white line followed by '^From ', I do Msg= Parser.close() (lateron I delete the Parser and create a new one by Parser= FeedParser(policy=policy.SMTP) ) I can access parts of the message by Msg['Message-ID'], e.g. but even for the very first message, trying to print it or convert it to a string by MsgStr=Msg.as_string(unixfrom=True) lets Python (3.3.1_pre20121209) die with Traceback (most recent call last): File "Email_Parse.py", line 35, in MsgStr=Msg.as_string(unixfrom=True) File "/usr/lib64/python3.3/email/message.py", line 151, in as_string g.flatten(self, unixfrom=unixfrom) File "/usr/lib64/python3.3/email/generator.py", line 112, in flatten self._write(msg) File "/usr/lib64/python3.3/email/generator.py", line 171, in _write self._write_headers(msg) File "/usr/lib64/python3.3/email/generator.py", line 198, in _write_headers self.write(self.policy.fold(h, v)) File "/usr/lib64/python3.3/email/policy.py", line 153, in fold return self._fold(name, value, refold_binary=True) File "/usr/lib64/python3.3/email/policy.py", line 176, in _fold (len(lines[0])+len(name)+2 > maxlen or IndexError: list index out of range What am I missing? Many thanks for a hint, Helmut. -- http://mail.python.org/mailman/listinfo/python-list
Re: email.message.Message - as_string fails
On Fri, 28 Dec 2012 20:57:46 -0500, Terry Reedy wrote: > On 12/28/2012 7:22 AM, Helmut Jarausch wrote: >> Hi, >> >> I'm trying to filter an mbox file by removing some messages. >> For that I use Parser= FeedParser(policy=policy.SMTP) >> and 'feed' any lines to it. >> If the mbox file contains a white line followed by '^From ', >> I do >> >> Msg= Parser.close() >> >> (lateron I delete the Parser and create a new one by Parser= >> FeedParser(policy=policy.SMTP) >> ) >> >> I can access parts of the message by Msg['Message-ID'], e.g. >> but even for the very first message, trying to print it or convert it >> to a string by MsgStr=Msg.as_string(unixfrom=True) >> >> lets Python (3.3.1_pre20121209) die with >> >> Traceback (most recent call last): >>File "Email_Parse.py", line 35, in >> MsgStr=Msg.as_string(unixfrom=True) >>File "/usr/lib64/python3.3/email/message.py", line 151, in as_string >> g.flatten(self, unixfrom=unixfrom) >>File "/usr/lib64/python3.3/email/generator.py", line 112, in flatten >> self._write(msg) >>File "/usr/lib64/python3.3/email/generator.py", line 171, in _write >> self._write_headers(msg) >>File "/usr/lib64/python3.3/email/generator.py", line 198, in >>_write_headers >> self.write(self.policy.fold(h, v)) >>File "/usr/lib64/python3.3/email/policy.py", line 153, in fold >> return self._fold(name, value, refold_binary=True) >>File "/usr/lib64/python3.3/email/policy.py", line 176, in _fold >> (len(lines[0])+len(name)+2 > maxlen or >> IndexError: list index out of range > > The only list index visible is 0 in lines[0]. If this raises, lines is > empty. You could trace back to see where lines is defined. I suspect it > is all or part of the Msg you started with. > > I believe that some of email was rewritten for 3.3, so it is possible > that you found a bug based on an untrue assumption. It is also possible > that you missed a limitation in the doc, or tripped over an intended but > not written limitation. So I hope you do the tracing, so if doc or code > need a fix, a tracker issue can be opened. Thanks Terry, I've debugged it and it smells like a bug. I have created http://bugs.python.org/issue16811 Helmut. -- http://mail.python.org/mailman/listinfo/python-list
re: ignore case only for a part of the regex?
Hi, is there a means to specify that 'ignore-case' should only apply to a part of a regex? E.g. the regex should match Msg-id:, Msg-Id, ... but not msg-id: and so on. I've tried the pattern r'^Msg-(?:(?i)id):' but (?i) makes the whole pattern ignoring case. In my simple case I could say r'Msg-[Ii][Dd]:' but that's a bit clumsy. Is there a more elegant way? Is there a way to compose a pattern from subpatterns which are compiled with different flags? Many thanks for a hint, Helmut. -- http://mail.python.org/mailman/listinfo/python-list
numpy.genfromtxt with Python3 - howto
Hi
I have a machine with a non-UTF8 local.
I can't figure out how to make numpy.genfromtxt work
I pipe some ascii data into the following script but get this
bytes to str hell.
Many thanks for a hint,
Helmut.
#!/usr/bin/python3
import numpy as np
import io
import sys
inpstream = io.open(sys.stdin.fileno(), "r", encoding='latin1')
data = np.genfromtxt(inpstream)
'''
Traceback (most recent call last):
File "SimpleMatInp.py", line 8, in
data = np.genfromtxt(inpstream)
File "/usr/lib64/python3.2/site-packages/numpy/lib/npyio.py", line
1274, in genfromtxt
first_values = split_line(first_line)
File "/usr/lib64/python3.2/site-packages/numpy/lib/_iotools.py", line
206, in _delimited_splitter
line = line.split(self.comments)[0].strip(asbytes(" \r\n"))
TypeError: Can't convert 'bytes' object to str implicitly
'''
print(data)
print(data.dtype)
print(data.shape)
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Re: Is it possible to open a dbf
Miklós P wrote: Paul Rubin wrote: John Fabiani <[EMAIL PROTECTED]> writes: I'm wondering if there is a module available that will open a dbf So far (more than a minute) I have discovered a reader only. So if you have a URL or a search string it would be very helpful. TIA John Yes, "dBase Python" yields only some code for reading dBase ... and lots of enquires about such a thing... I've been using http://www.fiby.at/dbfpy/ without any problems including writing/modifying dbf files. -- Helmut Jarausch Lehrstuhl fuer Numerische Mathematik RWTH - Aachen University D 52056 Aachen, Germany -- http://mail.python.org/mailman/listinfo/python-list
Re: regex question
Felix Schwarz wrote:
> Hi all,
>
> I'm experiencing problems with a regular expression and I can't figure
> out which words I use when googling. I read the python documentation for
> the re module multiple times now but still no idea what I'm doing wrong.
>
> What I want to do:
> - Extract all digits (\d) in a string.
> - Digits are separated by space (\w)
>
> What my program does:
> - It extracts only the last digit.
>
> Here is my program:
> import re
> line = ' 123'
> regex = '^' + '(?:\s+(\d))*' + '$'
> match = re.match(regex, line)
> print "lastindex is: ",match.lastindex
> print "matches: ",match.group(1)
>
>
> Obviously I do not understand how (?:\s+(\d))* works in conjunction with
> ^ and $.
>
I am sure what you like to do.
What about
regex= re.compile('\s+\d')
print regex.findall(line)
--
Helmut Jarausch
Lehrstuhl fuer Numerische Mathematik
RWTH - Aachen University
D 52056 Aachen, Germany
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BaseHTTPServer and priviledge separation?
Hi, to use a port below 1000 on a Unix system one needs root priviledges. But it's dangerous to execute all of a script under those priviledges. Therefore I'd like to drop the root priviledges as soon as possible. (How) is this possible? Many thanks for a hint, Helmut Jarausch Lehrstuhl fuer Numerische Mathematik RWTH - Aachen University D 52056 Aachen, Germany -- http://mail.python.org/mailman/listinfo/python-list
extracting a heapq in a for loop - there must be more elegant solution
Hi, I'd like to extracted elements from a heapq in a for loop. I feel my solution below is much too complicated. How to do it more elegantly? I know I could use a while loop but I don't like it. Many thanks for some lessons in Python. Here is my clumsy solution from heapq import heappush, heappop # heappop raises IndexError if heap is empty H=[] for N in 'H','C','W','I' : heappush(H,N) # how to avoid / simplify the following function def in_sequence(H) : try : while True : N= heappop(H) yield N except IndexError : raise StopIteration # and here the application: for N in in_sequence(H) : print(N) -- https://mail.python.org/mailman/listinfo/python-list
Re: extracting a heapq in a for loop - there must be more elegant solution
On Tue, 03 Dec 2013 04:40:26 -0800, rusi wrote: > On Tuesday, December 3, 2013 5:48:59 PM UTC+5:30, Helmut Jarausch wrote: >> Hi, >> >> I'd like to extracted elements from a heapq in a for loop. >> I feel my solution below is much too complicated. >> How to do it more elegantly? >> I know I could use a while loop but I don't like it. > > How about > > def in_sequence(h): > for i in range(len(h)): > yield heapq.heappop(h) > > Yeah its a bit fiddly: > 1. i needed for for but not used > 2. The range in the for loop -- normally a python 'smell' > > If python3 > > def ins3(h): >yield from (heapq.heappop(h) for i in range(len(h))) Many thanks, I'm using Python3 anyway! Helmut -- https://mail.python.org/mailman/listinfo/python-list
Re: extracting a heapq in a for loop - there must be more elegant solution
On Tue, 03 Dec 2013 13:38:58 +0100, Peter Otten wrote: > Helmut Jarausch wrote: > >> Hi, >> >> I'd like to extracted elements from a heapq in a for loop. >> I feel my solution below is much too complicated. >> How to do it more elegantly? >> I know I could use a while loop but I don't like it. >> >> Many thanks for some lessons in Python. >> >> Here is my clumsy solution >> >> from heapq import heappush, heappop >> # heappop raises IndexError if heap is empty >> >> H=[] >> for N in 'H','C','W','I' : >> heappush(H,N) > > H = ["H", "C", "W", "I"] > heapq.heapify(H) > > But see below. > >> # how to avoid / simplify the following function >> >> def in_sequence(H) : >> try : >> while True : >> N= heappop(H) >> yield N >> except IndexError : >> raise StopIteration >> >> # and here the application: >> >> for N in in_sequence(H) : >> print(N) > > If you are iterating over the complete heap I see no advantage over a sorted > list. So > > for N in sorted(H): > print(N) > > If H is huge use H.sort() instead of sorted() to save memory. > If you need only a few items use heapq.nsmallest(). Many thanks! In my real application the data which is pushed onto the heap will be extracted from a small file which is executed several thousands times. So, I thought, I could keep the CPU a bit busy while the OS is doing file I/O. Of course, I could have appended all these strings to a list which is sorted at the end. -- https://mail.python.org/mailman/listinfo/python-list
Re: extracting a heapq in a for loop - there must be more elegant solution
On Tue, 03 Dec 2013 13:06:05 +, Duncan Booth wrote: > Helmut Jarausch wrote: > >> Hi, >> >> I'd like to extracted elements from a heapq in a for loop. >> I feel my solution below is much too complicated. >> How to do it more elegantly? >> I know I could use a while loop but I don't like it. >> >> Many thanks for some lessons in Python. >> >> Here is my clumsy solution >> >> from heapq import heappush, heappop >> # heappop raises IndexError if heap is empty >> >> H=[] >> for N in 'H','C','W','I' : >> heappush(H,N) >> >> # how to avoid / simplify the following function >> >> def in_sequence(H) : >> try : >> while True : >> N= heappop(H) >> yield N >> except IndexError : >> raise StopIteration >> >> # and here the application: >> >> for N in in_sequence(H) : >> print(N) >> > > If all you want to do is pull all of the elements out of the heap in > order, you would probably be better off just doing: > > for N in sorted(H): > print(N) > > Heaps are mostly useful if you want only some of the elements, or if you > are continually producing more elements while also processing the > smallest ones. > > However, if you really wnt to do this: > > for N in iter(lambda: heappop(H) if H else None, None): > print(N) > > will work so long as H cannot contain None. If it can just replace both > occurences of None with some other sentinel: > > sentinel = object() > for N in iter(lambda: heappop(H) if H else sentinel, sentinel): > print(N) > > > Alternatively your 'in_sequence' function would look better without the > exception handling: > > def in_sequence(H) : > while H: > yield heappop(H) Many thanks! And as noted in another reply, I try to overlap CPU time with file I/O since I have to open / read / close a file for each line that gets pushed onto the heap. Helmut -- https://mail.python.org/mailman/listinfo/python-list
Re: extracting a heapq in a for loop - there must be more elegant solution
On Tue, 03 Dec 2013 15:56:11 +0200, Jussi Piitulainen wrote: > Helmut Jarausch writes: > ... >> I know I could use a while loop but I don't like it. > ... >> from heapq import heappush, heappop >> # heappop raises IndexError if heap is empty > ... >> # how to avoid / simplify the following function >> >> def in_sequence(H) : >> try : >> while True : >> N= heappop(H) >> yield N >> except IndexError : >> raise StopIteration > > That seems equivalent to this: > > def in_sequence(H): > while H: yield heappop(H) Many thanks, that's something I'd hoped for. > But I don't like the side-effect. I'd change the name to something > that indicates the draining of the heap - heapaerobic? - or consider > sorted(H) instead as others suggested. Since I fill the heap once and empty it afterwards I regard the side-effect clearly visible. Helmut -- https://mail.python.org/mailman/listinfo/python-list
Re: extracting a heapq in a for loop - there must be more elegant solution
On Wed, 04 Dec 2013 08:13:03 +1100, Cameron Simpson wrote: > On 03Dec2013 12:18, Helmut Jarausch wrote: >> I'd like to extracted elements from a heapq in a for loop. >> I feel my solution below is much too complicated. >> How to do it more elegantly? > > I can't believe nobody has mentioned PriorityQueue. > > A PriorityQueue (from the queue module in python 3 and the Queue > module in python 2) is essentially just a regular Queue using a > heapq for the storage. > > Example (untested): > > from Queue import PriorityQueue > > PQ = PriorityQueue() > for item in [1,2,3,7,6,5,9,4]: > PQ.put(item) > > while not PQ.empty(): > item = PQ.get() > ... do stuff with item ... > > I iterate over Queues so often that I have a personal class called > a QueueIterator which is a wrapper for a Queue or PriorityQueue > which is iterable, and an IterablePriorityQueue factory function. > Example: > > from cs.queues import IterablePriorityQueue > > IPQ = IterablePriorityQueue() > for item in [1,2,3,7,6,5,9,4]: > IPQ.put(item) > > for item in IPQ: > ... do stuff with item ... > > Cheers, Many thanks! I think you QueueIterator would be a nice addition to Python's library. Helmut -- https://mail.python.org/mailman/listinfo/python-list
smart splitting - how to
Hi, I'd like to read several strings by using 'input'. These strings are separated by white space but I'd like to allow for some quoting, e.g. "Guido van" Rossum should be split into 2 strings only Now, a simple split doesn't work since it splits the quoted text as well. Is there a simple way to do so? It would be nice if it could handle embedded quotes which are escaped by a backslash, too. Is there something simpler then a sophisticated regular expression or even a parser? Many thanks for a hint, Helmut -- https://mail.python.org/mailman/listinfo/python-list
Re: smart splitting - how to
On Fri, 13 Dec 2013 11:39:57 +, Chris Angelico and Robert Kern wrote: > On 2013-12-13 11:28, Helmut Jarausch wrote: >> Hi, >> >> I'd like to read several strings by using 'input'. >> These strings are separated by white space but I'd like to allow for >> some quoting, e.g. >> >> "Guido van" Rossum >> >> should be split into 2 strings only >> >> Now, a simple split doesn't work since it splits the quoted text as well. >> Is there a simple way to do so? >> It would be nice if it could handle embedded quotes which are escaped >> by a backslash, too. >> >> Is there something simpler then a sophisticated regular expression >> or even a parser? > > http://docs.python.org/3.3/library/shlex > > > [~] > |1> import shlex > > [~] > |2> shlex.split(r'"Guido \"van\" Rossum" invented Python') > ['Guido "van" Rossum', 'invented', 'Python'] Many thanks, that works perfectly and is so simple. Helmut -- https://mail.python.org/mailman/listinfo/python-list
Python3 - temporarily change the file encoding
Hi,
my locale is en_US.iso88591
But now I'd like to process a restructuredtext file which is encoded in utf-8.
rst2html has
#!/usr/bin/python3.3
# $Id: rst2html.py 4564 2006-05-21 20:44:42Z wiemann $
# Author: David Goodger
# Copyright: This module has been placed in the public domain.
"""
A minimal front end to the Docutils Publisher, producing HTML.
"""
try:
import locale
locale.setlocale(locale.LC_ALL, '')
except:
pass
from docutils.core import publish_cmdline, default_description
description = ('Generates (X)HTML documents from standalone reStructuredText '
'sources. ' + default_description)
publish_cmdline(writer_name='html', description=description)
--
Even if I comment out the part containing 'import locale' the utf-8 encoding of
my
rst-file is not recognized.
How can I change this so that rst2html use utf-8 when reading and writing files.
Of course, I don't want to change the docutils package for that.
Many thanks for a hint,
Helmut
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__iadd__ for a subclass of array - howto
Hi,
I'd like to subclass array.array and implement operators like __iadd__
How can this be accomplished.
I'tried
from array import array
class Vec(array) :
def __new__(cls,Vinit) :
return array.__new__(cls,'d',Vinit)
def __init__(self,*args) :
self.N = len(self)
def __str__(self) :
out=[ "{}".format(self.__getitem__(i)) for i in range(0,self.N) ]
return ",".join(out)
def __iadd__(self,Op) :
#for i,x in enumerate(self) :
# x+= Op[i] # this doesn't update self
for i in range(0,self.N) :
(self.__getitem__(i))+= Op[i] # __getitem__ doesn't return a "reference"
(like in C++)
return self
Many thanks for a hint,
Helmut
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Python3 exec locals - this must be a FAQ
Hi,
I've tried but didn't find an answer on the net.
The exec function in Python modifies a copy of locals() only.
How can I transfer changes in that local copy to the locals of my function
** without ** knowing the names of these variables.
E.g. I have a lot of local names.
Doing
_locals= locals()
expr=compile(input('statements assigning to some local variables '),
'user input','exec')
exec(expr,globals(),_locals)
How can I "copy" the new values within _locals to my current locals.
If I knew that Var1 has changed I could say
Var1 = _locals['Var1'] but what to do in general?
Many thanks for a hint,
Helmut.
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Re: Python3 exec locals - this must be a FAQ
On Tue, 12 Feb 2013 08:27:41 -0500, Dave Angel wrote:
> On 02/12/2013 06:46 AM, Helmut Jarausch wrote:
>> Hi,
>>
>> I've tried but didn't find an answer on the net.
>>
>> The exec function in Python modifies a copy of locals() only.
>> How can I transfer changes in that local copy to the locals of my
>> function ** without ** knowing the names of these variables.
>>
>> E.g. I have a lot of local names.
>>
>> Doing
>>
>> _locals= locals()
>
> This doesn't copy everything. But perhaps you know that and you're just
> testing us.
No, I didn't know. And I'm bit surprised since this is recommend
several times, e.g. in "Python Essential Reference, 4th ed" by
David Beazley.
>
>> expr=compile(input('statements assigning to some local variables '),
>> 'user input','exec')
>> exec(expr,globals(),_locals)
>>
>> How can I "copy" the new values within _locals to my current locals.
>>
>> If I knew that Var1 has changed I could say Var1 = _locals['Var1']
>> but what to do in general?
>
> locals()["Var1"] = _locals["Var1"] will set the same Var1 local.
Thanks for this hint which surprises me again since I thought
locals() by itself is a copy only.
>
> So you might write a loop on _locals.
>
> But beware if someone has deleted one of the "variables" it may not do
> what you'd like. You cannot necessarily add back a local with the above
> syntax.
Does this mean that adding something completely new won't work?
Many thanks,
Helmut.
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raw format string in string format method?
Hi,
I'd like to print a string with the string format method which uses
{0}, ...
Unfortunately, the string contains TeX commands which use lots of
braces. Therefore I would have to double all these braces just for the
format method which makes the string hardly readable.
Is there anything like a "raw" format string and any other means
to circumvent this?
Many thanks for a hint,
Helmut.
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Re: raw format string in string format method?
On Fri, 01 Mar 2013 01:22:48 +1100, Chris Angelico wrote:
> On Fri, Mar 1, 2013 at 1:11 AM, Helmut Jarausch
> wrote:
>> Hi,
>>
>> I'd like to print a string with the string format method which uses
>> {0}, ...
>>
>> Unfortunately, the string contains TeX commands which use lots of
>> braces. Therefore I would have to double all these braces just for the
>> format method which makes the string hardly readable.
>>
>> Is there anything like a "raw" format string and any other means to
>> circumvent this?
>
> You could use a different string formatting function, such as
> percent-formatting:
>
> "Hello, {0}, this is %s" % some_string
>
> The {0} will be output literally, and the %s will be replaced by the
> string. Braces are ignored, percent signs are significant.
>
> ChrisA
Originally I had used percent-formatting
But isn't it deprecated in Python 3.X ?
Thanks,
Helmut.
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new.instancemethod - how to port to Python3
Hi, I'm trying to port a class to Python3.3 which contains class Foo : def to_binary(self, *varargs, **keys): self.to_binary = new.instancemethod(to_binary, self, self.__class__) # Finally call it manually return apply(self.to_binary, varargs, keys) The last line has been transformed to return self.to_binary(*varargs, **keys) by 2to3 But how to transform the line with new.instancemethod. I've seen examples where new.instancemethod(to_binary, ) is replaced by to_binay but this doesn't work here since to_binary isn't known. If I simply delete it, I get an infinite recursion. So, what's a working transcript of this code? Many thanks for a hint, Helmut. P.S. Is there collection of examples of necessary transformations to Python3 which are not managed by 2to3 ? -- http://mail.python.org/mailman/listinfo/python-list
Re: new.instancemethod - how to port to Python3
On Sun, 07 Apr 2013 11:41:46 +0100, Arnaud Delobelle wrote: > On 7 April 2013 10:50, Helmut Jarausch wrote: >> Hi, >> >> I'm trying to port a class to Python3.3 which contains >> >> class Foo : >> >> def to_binary(self, *varargs, **keys): >> >> >> >> self.to_binary = new.instancemethod(to_binary, self, >> self.__class__) > > `self` isn't bound in this lexical scope (the class scope). Or is your > indentation incorrect? However, if this statement was within the > `to_binary` method, then `to_binary` wouldn't be bound so I can't make a > reasonable guess. > >> # Finally call it manually return apply(self.to_binary, varargs, >> keys) >> > [...] Sorry, the excerpt wasn't complete (it's not my code): Here a more complete excerpt: class Foo : def to_binary(self, *varargs, **keys): code= ... args= ... # Add function header code = 'def to_binary(self, %s):\n' % ', '.join(args) + code exec(code) self.to_binary = new.instancemethod(to_binary, self, self.__class__) return self.to_binary(*varargs, **keys) -- http://mail.python.org/mailman/listinfo/python-list
Re: new.instancemethod - how to port to Python3
On Sun, 07 Apr 2013 11:07:07 +, Steven D'Aprano wrote:
> On Sun, 07 Apr 2013 10:54:46 +, Helmut Jarausch wrote:
>
>> class Foo :
>>
>> def to_binary(self, *varargs, **keys):
>>
>>code= ...
>>args= ...
>># Add function header
>>code = 'def to_binary(self, %s):\n' % ', '.join(args) + code
>>exec(code)
>>self.to_binary = new.instancemethod(to_binary, self,
>>self.__class__)
>>return self.to_binary(*varargs, **keys)
>
>
> Self-modifying code! Yuck!
>
> Nevertheless, try changing the line with new.instancemethod to this:
>
> self.to_binary = types.MethodType(to_binary, self)
>
>
> and see if it works. If it complains about the call to exec, then change
> that part of the code to this:
Thanks, that worked
>
> ns = {}
> exec(code, ns)
I just had to replace this with
exec(code,globals(),ns)
> to_binary = ns['to_binary']
> self.to_binary = types.MethodType(to_binary, self)
Helmut
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Re: new.instancemethod - how to port to Python3
On Sun, 07 Apr 2013 10:52:11 +, Steven D'Aprano wrote:
> On Sun, 07 Apr 2013 09:50:35 +, Helmut Jarausch wrote:
>
>> Hi,
>>
>> I'm trying to port a class to Python3.3 which contains
>>
>> class Foo :
>>
>> def to_binary(self, *varargs, **keys):
>>
>>
>> self.to_binary = new.instancemethod(to_binary, self, self.__class__)
>> # Finally call it manually
>> return apply(self.to_binary, varargs, keys)
>
>
> I do not understand this code. Can you give a short example that actually
> works please?
>
> As written, your code has a class that defines a to_binary method. Then,
> *outside* of the method, in the class definition, you refer to "self",
> and use a return statement. This is not possible -- self does not exist,
> and return gives a SyntaxError.
>
> But, if the indentation is wrong, it looks like you are trying to get the
> to_binary method to replace itself with a global to_binary function. I do
> not understand this.
Sorry, the first excerpt was to too short.
As noted in the reply to Arnaud, a better excerpt is:
class Foo :
def to_binary(self, *varargs, **keys):
code= ...
args= ...
# Add function header
code = 'def to_binary(self, %s):\n' % ', '.join(args) + code
exec(code)
self.to_binary = new.instancemethod(to_binary, self, self.__class__)
return self.to_binary(*varargs, **keys)
If you'd like to see the full code (not by me)
please see the Python package http://python-xlib.sourceforge.net/
file Xlib/protocol/rq.py method to_binary in class Struct
>
> I cannot answer your question, because I don't understand it. But perhaps
> this will help:
>
>
> # === Python 2 version ===
> class Spam:
> pass
>
> x = Spam()
>
> def method(self, arg):
> return {arg: self}
>
> import new
> x.method = new.instancemethod(method, x, x.__class__)
>
> x.method("hello")
> => returns {'hello': <__main__.Spam instance at 0xa18bb4c>}
>
>
> Here is the Python 3 version:
>
>
> # === Python 3 ===
> class Spam:
> pass
>
> x = Spam()
>
> def method(self, arg):
> return {arg: self}
>
> import types
> x.method = types.MethodType(method, x)
>
> x.method("hello")
> => returns {'hello': <__main__.Spam object at 0xb7bc59ac>}
>
>
>
> Does this help?
Yes,
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NNTPlib::xover problem
Hi
I try to regularly extract recent news from some newsgroups.
If News is an NNTP object I try
(Response,Articles)= News.xover(str(int(Last)+1),'1000')
where 'Last' is the (previously saved) number of the last
article read.
If there are no new articles I get an Exception
Traceback (most recent call last):
File "/home/jarausch/Python_My/News", line 36, in -toplevel-
(Response,Articles)= News.xover(str(int(Last)+1),'1000')
File "/usr/local/lib/python2.4/nntplib.py", line 479, in xover
resp, lines = self.longcmd('XOVER ' + start + '-' + end, file)
File "/usr/local/lib/python2.4/nntplib.py", line 265, in longcmd
return self.getlongresp(file)
File "/usr/local/lib/python2.4/nntplib.py", line 236, in getlongresp
resp = self.getresp()
File "/usr/local/lib/python2.4/nntplib.py", line 219, in getresp
raise NNTPTemporaryError(resp)
NNTPTemporaryError: 420 No such article
I would have expected to get an empty 'Response' or the value None
for 'Articles'.
What am I missing?
(This is Python 2.4.3)
Many thanks for a hint,
Helmut Jarausch
Lehrstuhl fuer Numerische Mathematik
RWTH - Aachen University
D 52056 Aachen, Germany
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Re: NNTPlib::xover problem
Michiel Sikma wrote:
> Hi Helmut,
>
> I guess it simply raises an exception in case there are no articles;
> this may not be what you expected, but it would seem that this is the
> way it operates. You should try catching the exception to plan out a
> course of action in case no articles are present.
Thanks,
though the name of the exception 'nntplib.NNTPTemporaryError'
sound 'temporary'
Helmut.
> Op 7-aug-2006, om 12:50 heeft Helmut Jarausch het volgende geschreven:
>
>> Hi
>>
>> I try to regularly extract recent news from some newsgroups.
>> If News is an NNTP object I try
>> (Response,Articles)= News.xover(str(int(Last)+1),'1000')
>> where 'Last' is the (previously saved) number of the last
>> article read.
>> If there are no new articles I get an Exception
>>
>> Traceback (most recent call last):
>>File "/home/jarausch/Python_My/News", line 36, in -toplevel-
>> (Response,Articles)= News.xover(str(int(Last)+1),'1000')
>>File "/usr/local/lib/python2.4/nntplib.py", line 479, in xover
>> resp, lines = self.longcmd('XOVER ' + start + '-' + end, file)
>>File "/usr/local/lib/python2.4/nntplib.py", line 265, in longcmd
>> return self.getlongresp(file)
>>File "/usr/local/lib/python2.4/nntplib.py", line 236, in getlongresp
>> resp = self.getresp()
>>File "/usr/local/lib/python2.4/nntplib.py", line 219, in getresp
>> raise NNTPTemporaryError(resp)
>> NNTPTemporaryError: 420 No such article
>>
>>
>> I would have expected to get an empty 'Response' or the value None
>> for 'Articles'.
>>
>> What am I missing?
>>
>> (This is Python 2.4.3)
>>
>> Many thanks for a hint,
>>
>> Helmut Jarausch
>>
>> Lehrstuhl fuer Numerische Mathematik
>> RWTH - Aachen University
>> D 52056 Aachen, Germany
>> --http://mail.python.org/mailman/listinfo/python-list
>
--
Helmut Jarausch
Lehrstuhl fuer Numerische Mathematik
RWTH - Aachen University
D 52056 Aachen, Germany
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Re: hide python code !
John Machin wrote: > Bayazee wrote: >> hi >> can we hide a python code ? >> if i want to write a commercial software can i hide my source code from > [1] >> users access ? >> we can conver it to pyc but this file can decompiled ... so ...!! >> do you have any idea about this ...? >> >> --- >> First Iranian Open Source Community : www.python.ir > ^^[2] > > > [1] and [2] don't seem to be compatible. I suppose all of you who have commented about this, are sitting in the >>> free world <<< . But there are countries (like .ir) where the government has totally different ideas of 'freedom'. So taking the freedom to write something can be very dangerous at times. Fortunately most of those guys which intercept every email and check every web server are not so smart to reverse engineer everything in a short time since they have to check thousands of pieces of information each day. Let's make their work a bit harder! -- http://mail.python.org/mailman/listinfo/python-list
Python in a nutshell - new edition ?
Hi, is there a new edition of "Python in a Nutshell" covering Python 2.5 coming soon? Many thanks, Helmut Jarausch Lehrstuhl fuer Numerische Mathematik RWTH - Aachen University D 52056 Aachen, Germany -- http://mail.python.org/mailman/listinfo/python-list
How to terminate a main script?
Hi, I'm still looking for an elegant and clear means to terminate the main script in Python. Unfortunately, Python doesn't allow a 'return' instruction in the main script. Using sys.exit(0) produces an error message which looks dangerous to an uninitiated user. The same is true for an exception. And setting a 'flag' and testing at several place for 'fall through' is ugly and error-prone. So what is a good choice? Many thanks for a hint, Helmut Jarausch Lehrstuhl fuer Numerische Mathematik RWTH - Aachen University D 52056 Aachen, Germany -- http://mail.python.org/mailman/listinfo/python-list
Re: How to terminate a main script?
Fredrik Lundh wrote: > Helmut Jarausch wrote: > >> Using sys.exit(0) produces an error >> message which looks dangerous to an >> uninitiated user. > > sys.exit(0) doesn't print anything at all. Yes, sorry, I was trying in in 'idle' There you get Traceback (most recent call last): File "", line 1, in -toplevel- sys.exit(0) SystemExit: 0 -- Helmut Jarausch Lehrstuhl fuer Numerische Mathematik RWTH - Aachen University D 52056 Aachen, Germany -- http://mail.python.org/mailman/listinfo/python-list
upgrade 2.4 -> 2.5 HowTo
Hi, what has to be done for upgrading from Python 2.4 to 2.5? - How can I find out which packages (in addition to the core packages) have been installed up to now - Can I just copy /usr/local/lib/python2.4/site-packages to /usr/local/lib/python2.5 ? Many thanks for a hint, Helmut Jarausch Lehrstuhl fuer Numerische Mathematik RWTH - Aachen University D 52056 Aachen, Germany -- http://mail.python.org/mailman/listinfo/python-list
Re: upgrade 2.4 -> 2.5 HowTo
Sibylle Koczian wrote: > Helmut Jarausch schrieb: >> Hi, >> >> what has to be done for upgrading from Python 2.4 to 2.5? >> >> - How can I find out which packages (in addition to the core packages) >> have been >> installed up to now >> - Can I just copy /usr/local/lib/python2.4/site-packages to >> /usr/local/lib/python2.5 ? >> >> Many thanks for a hint, >> > > Which OS? > > Windows: control panel / software will list all python packages which > came with separate installers. > > Linux: the package manager of your distribution should tell you what's > installed. > It's Linux. Thanks, but I don't have a package manager. I'm working with Linux-From-Scratch. -- Helmut Jarausch Lehrstuhl fuer Numerische Mathematik RWTH - Aachen University D 52056 Aachen, Germany -- http://mail.python.org/mailman/listinfo/python-list
Re: Test for number?
Tim Williams wrote:
> On 04/09/06, Dr. Pastor <[EMAIL PROTECTED]> wrote:
>> In the following code I would like to ascertain
>> that x has/is a number. What the simplest TEST should be?
>> (Could not find good example yet.)
>> ---
>> x=raw_input('\nType a number from 1 to 20')
>> if TEST :
>> Do_A
>> else:
>> Do_B
>> ---
>
> Something simple like the following might help,
>
>>>> def numtest():
> ... x=raw_input('\nType a number from 1 to 20')
> ... try:
> ... x=int(x)
> ... print x, "is a number between 1 & 20 "
> ... except:
> ... print x, "is not a number"
> ...
>>>> numtest() # enter 1
> 1 is a number between 1 & 20
>>>> numtest() # enter "f"
> f is not a number
>
> its not a final solution though, think input = -2, 5.5 or 21
>
One step further
try:
eval(x+'0')
--
Helmut Jarausch
Lehrstuhl fuer Numerische Mathematik
RWTH - Aachen University
D 52056 Aachen, Germany
--
http://mail.python.org/mailman/listinfo/python-list
Re: mutable numeric type
[EMAIL PROTECTED] wrote: > Way to go. > Try doing this. > x = MutableNumeric(42) ^^ where is this defined? > y = x > x += 42 > print y > -- Helmut Jarausch Lehrstuhl fuer Numerische Mathematik RWTH - Aachen University D 52056 Aachen, Germany -- http://mail.python.org/mailman/listinfo/python-list
python2.5 frameobject - how to upgrade?
Hi,
I'd like to install a package ('rekall') which uses
frame->f_nlocals
which is no longer contained in frameobject.h
What's the recommended way to upgrade such an application?
Many thanks for a hint,
Helmut Jarausch
Lehrstuhl fuer Numerische Mathematik
RWTH - Aachen University
D 52056 Aachen, Germany
--
http://mail.python.org/mailman/listinfo/python-list
Re: python2.5 frameobject - how to upgrade?
[EMAIL PROTECTED] wrote:
> Helmut> I'd like to install a package ('rekall') which uses
> Helmut> frame->f_nlocals which is no longer contained in frameobject.h.
>
> Helmut> What's the recommended way to upgrade such an application?
>
> I suspect PySequence_Length(frame->f_locals) will do the trick.
Yes, thanks, that fixed it,
Happy New Year to you.
A personal question: Have you converted from Perl to Python, as well?
--
Helmut Jarausch
Lehrstuhl fuer Numerische Mathematik
RWTH - Aachen University
D 52056 Aachen, Germany
--
http://mail.python.org/mailman/listinfo/python-list
Re: File processing - is Python suitable?
ferrad wrote: > I have not used Python before, but believe it may be what I need. > > I have large text files containing text, numbers, and junk. I want to > delete large chunks process other bits, etc, much like I'd do in an > editor, but want to do it automatically. I have a set of generic > rules that my fingers follow to process these files, which all follow > a similar template. > > Question: can I translate these types of rules into programmatical > constructs that Python can use to process these files? Can Python do > the trick? I think that's one of the great strength of Python. Just some pointers http://gnosis.cx/TPiP/ http://www.egenix.com/products/python/mxBase/mxTextTools/ -- Helmut Jarausch Lehrstuhl fuer Numerische Mathematik RWTH - Aachen University D 52056 Aachen, Germany -- http://mail.python.org/mailman/listinfo/python-list
restructuredtext latin1 encoding (FAQ?)
Hi, I did try to find something on the net but I couldn't find anything useful. Is there a way to write documents with the reST tools using the Latin1 encoding? Many thanks for a hint, Helmut Jarausch Lehrstuhl fuer Numerische Mathematik RWTH - Aachen University D 52056 Aachen, Germany -- http://mail.python.org/mailman/listinfo/python-list
Re: restructuredtext latin1 encoding (FAQ?)
Marc 'BlackJack' Rintsch wrote: > On Tue, 03 Jul 2007 12:12:04 +0200, Helmut Jarausch wrote: > >> Is there a way to write documents with the reST tools using the Latin1 >> encoding? > > Yes of course there is. Just write the documents in Latin-1 encoding. > Take a look at the charset command line options if Latin-1 is not the > default encoding used by your operating system. > Many thanks! Is there also a way to tell it dugui.py ? Helmut. -- Helmut Jarausch Lehrstuhl fuer Numerische Mathematik RWTH - Aachen University D 52056 Aachen, Germany -- http://mail.python.org/mailman/listinfo/python-list
wxPython Cannot convert from the charset 'latin-1'
Hi, I'm trying to teach "dugui.py" (a tiny GUI for restructuredtext written in wxPython) to accept files in isolatin-1 encoding. It displays e.g. German umlauts correctly but then I get a popup window saying Python Error Cannot convert from the charset 'latin-1'! How can I find out where this did come from. Running it under pdb isn't helpful either. Many thanks for a hint, Helmut Jarausch Lehrstuhl fuer Numerische Mathematik RWTH - Aachen University D 52056 Aachen, Germany -- http://mail.python.org/mailman/listinfo/python-list
ReSTedit ported to Linux?
Hi, does anybody know if ReSTedit (originally for Mac OS X) has been ported to Linux? Many thanks for a hint, Helmut Jarausch Lehrstuhl fuer Numerische Mathematik RWTH - Aachen University D 52056 Aachen, Germany -- http://mail.python.org/mailman/listinfo/python-list
Re: Simple search and Display system, no need for db?
flit wrote: > Hello All, > > I was discussing with a friend and we get in a hot debate. > > We have homepage that act like a name list. > > The user goes there: > > - Choose the building -- Department and the user receive an html table > with all names . > > So the user gives two inputs and receive the names. > > Currently we are using this in mysql, This is the reason of the > debate. > > If our list will never change and we are supposing that will be true > for a long time, is it not too much use an mysql? > > Which kind of structure can we use in python? in web? > What about using a simple dictionary in Python. You can use the module pickle (or cpickle) to dump it to disk and load it next time. Furthermore you can easily write the whole web server in Python, e.g. I like http://karrigell.sourceforge.net/ -- Helmut Jarausch Lehrstuhl fuer Numerische Mathematik RWTH - Aachen University D 52056 Aachen, Germany -- http://mail.python.org/mailman/listinfo/python-list
Re: pattern match !
[EMAIL PROTECTED] wrote: > Extract the application name with version from an RPM string like > hpsmh-1.1.1.2-0-RHEL3-Linux.RPM, i require to extract hpsmh-1.1.1.2-0 > from above string. Sometimes the RPM string may be hpsmh-1.1.1.2-RHEL3- > Linux.RPM. > Have a try with import re P=re.compile(r'(\w+(?:[-.]\d+)+)-RHEL3-Linux\.RPM') S="hpsmh-1.1.1.2-0-RHEL3-Linux.RPM" PO= P.match(S) if PO : print PO.group(1) -- Helmut Jarausch Lehrstuhl fuer Numerische Mathematik RWTH - Aachen University D 52056 Aachen, Germany -- http://mail.python.org/mailman/listinfo/python-list
split a string of space separated substrings - elegant solution?
Hi,
I'm looking for an elegant solution to the following (quite common)
problem:
Given a string of substrings separated by white space,
split this into tuple/list of elements.
The problem are quoted substrings like
abc "xy z" "1 2 3" "a \" x"
should be split into ('abc','xy z','1 2 3','a " x')
For that, one probably has to protect white space between
quotes, then split by white space and finally converted the
'protected white space' to normal white space again.
Is there an elegant solution - perhaps without using a lexer
and something else. With regular expressions alone it seems
clumsy.
Many thanks for a hint,
Helmut Jarausch
Lehrstuhl fuer Numerische Mathematik
RWTH - Aachen University
D 52056 Aachen, Germany
--
http://mail.python.org/mailman/listinfo/python-list
Re: split a string of space separated substrings - elegant solution?
Many thanks to all of you! It's amazing how many elegant solutions there are in Python. -- Helmut Jarausch Lehrstuhl fuer Numerische Mathematik RWTH - Aachen University D 52056 Aachen, Germany -- http://mail.python.org/mailman/listinfo/python-list
decorators - more than just syntactic sugar
Hi, are decorators more than just syntactic sugar in python 2.x and what about python 3k ? How can I find out the predefined decorators? Many thanks for your help, Helmut Jarausch Lehrstuhl fuer Numerische Mathematik RWTH - Aachen University D 52056 Aachen, Germany -- http://mail.python.org/mailman/listinfo/python-list
Re: Drawing a graph
Ghirai wrote: > Hello list, > > I need to draw a graph, 2 axes, 2D, nothing fancy. > One of the axes is time, the other one is a series of integers. > > I don't care much about the output format. > > Are there any specialized libraries for this, or should i use PIL? > What about Gnuplot.py http://gnuplot-py.sourceforge.net/ -- Helmut Jarausch Lehrstuhl fuer Numerische Mathematik RWTH - Aachen University D 52056 Aachen, Germany -- http://mail.python.org/mailman/listinfo/python-list
sendmail a long message
Hi, to send a possibly long email I have seen a solution which does os.popen to an external sendmail program and then writes the message into that pipe. I wonder if it possible to use smtplib.SMTP.sendmail but this requires building the complete body as one long string in memory before calling this. Is there an alternative solution, e.g. where smtplib.SMTP.sendmail calls a generator. Many thanks for a hint, Helmut Jarausch Lehrstuhl fuer Numerische Mathematik RWTH - Aachen University D 52056 Aachen, Germany -- http://mail.python.org/mailman/listinfo/python-list
Re: sendmail a long message
Laszlo Nagy wrote: > Helmut Jarausch wrote: >> Hi, >> >> to send a possibly long email I have seen a solution >> which does os.popen to an external sendmail program and >> then writes the message into that pipe. >> >> I wonder if it possible to use smtplib.SMTP.sendmail >> but this requires building the complete body as one long >> string in memory before calling this. >> >> Is there an alternative solution, e.g. where >> smtplib.SMTP.sendmail calls a generator. >> > using socket.connect, and feed the message body into it in smaller chunks. > > Of course, you need to create the message body first, but it was not > your question - I suppose you already have the message stored on disk? > Thanks, it's a mail filter, the message is read in smaller chunks from sys.stdin . Helmut. -- Helmut Jarausch Lehrstuhl fuer Numerische Mathematik RWTH - Aachen University D 52056 Aachen, Germany -- http://mail.python.org/mailman/listinfo/python-list
asynchronous timer events - how to?
Hi, I'm using a web server (Karrigell) which is based on the asyncore module. I'd like to be able to checkpoint some data (e.g. pickled dictionaries) to disk from time to time. For that I would need to setup a timer which calls a Python object/function when its time interval has expired. While this function is running I need access to the variables of the server. Can this be done in a simple way? Many thanks for a hint, Helmut Jarausch Lehrstuhl fuer Numerische Mathematik RWTH - Aachen University D 52056 Aachen, Germany -- http://mail.python.org/mailman/listinfo/python-list
Re: asynchronous timer events - how to?
Fredrik Lundh wrote: > Helmut Jarausch wrote: > >> I'm using a web server (Karrigell) which is based on the asyncore module. >> I'd like to be able to checkpoint some data (e.g. pickled >> dictionaries) to disk >> from time to time. >> For that I would need to setup a timer which calls a Python >> object/function when its time interval has expired. While this >> function is running I need access to the variables of the server. > > the usual way to do that with asyncore is to add an outer control loop > that calls asyncore.loop with a count argument (or poll directly), and > checks a "task queue" at regular intervals. > > but in your case, it's probably easier to set up a cron job that does > a "wget" against your server with a "secret" URL, and have the server do > the checkpointing whenever that URL is fetched. Thanks for this ingenious idea, Helmut. -- Helmut Jarausch Lehrstuhl fuer Numerische Mathematik RWTH - Aachen University D 52056 Aachen, Germany -- http://mail.python.org/mailman/listinfo/python-list
module finalizer - is there such a beast?
Hi, when a module gets imported the statements not contained in function definitions or classes are executed. This can be thought of an initializer for the module. But how can I get control when the module gets unloaded either by Python's gc, Python's exit or by a module reload. Many thanks for a hint, Helmut Jarausch Lehrstuhl fuer Numerische Mathematik RWTH - Aachen University D 52056 Aachen, Germany -- http://mail.python.org/mailman/listinfo/python-list
common problem - elegant solution sought
Hi,
I'm looking for an elegant solution of the following tiny but common problem.
I have a list of tuples (Unique_ID,Date) both of which are strings.
I want to delete the tuple (element) with a given Unique_ID, but
I don't known the corresponding Date.
My straight forward solution is a bit lengthy, e.g.
L=[("a","070501"),("b","080115"),("c","071231")]
pos=-1
found=-1
for (Key,Date) in L :
pos+= 1
if Key == "b" :
found= pos
break
if found >= 0 :
del L[found]
print L
Most probably there are much more elegant solutions.
Unfortunately, the index-list-method doesn't take an
additional function argument for the comparisons.
Many thanks for your hints,
Helmut Jarausch
Lehrstuhl fuer Numerische Mathematik
RWTH - Aachen University
D 52056 Aachen, Germany
--
http://mail.python.org/mailman/listinfo/python-list
Re: common problem - elegant solution sought
Thanks to you all for your help. The solution to regenerate the list skipping the one to be deleted is fine for me since my lists are of moderate size and the operation is infrequent. Helmut. -- Helmut Jarausch Lehrstuhl fuer Numerische Mathematik RWTH - Aachen University D 52056 Aachen, Germany -- http://mail.python.org/mailman/listinfo/python-list
Re: common problem - elegant solution sought
Neil Cerutti wrote:
> On Jan 15, 2008 5:33 AM, Helmut Jarausch <[EMAIL PROTECTED]> wrote:
>> Hi,
>>
>> I'm looking for an elegant solution of the following tiny but common problem.
>>
>> I have a list of tuples (Unique_ID,Date) both of which are strings.
>> I want to delete the tuple (element) with a given Unique_ID, but
>> I don't known the corresponding Date.
>>
>> My straight forward solution is a bit lengthy, e.g.
>>
>> L=[("a","070501"),("b","080115"),("c","071231")]
>
> If the data is truly sorted by Unique_ID, then binary search may be
> feasible (though not actually recommended for such small list).
>
> import bisect
>
> i = bisect.bisect_left(L, ('b', '00'))
> if i[0] == 'b':
> del L[i]
>
Thanks,
unfortunately, they are sorted on 'Date'
Helmut.
--
Helmut Jarausch
Lehrstuhl fuer Numerische Mathematik
RWTH - Aachen University
D 52056 Aachen, Germany
--
http://mail.python.org/mailman/listinfo/python-list
Benchmark [was Re: common problem - elegant solution sought]
Again, many thanks to all who provide their solution.
I have timed these (though on my old P3(0.9GHz)) - see below
Helmut.
Helmut Jarausch wrote:
> Hi,
>
> I'm looking for an elegant solution of the following tiny but common
> problem.
>
> I have a list of tuples (Unique_ID,Date) both of which are strings.
> I want to delete the tuple (element) with a given Unique_ID, but
> I don't known the corresponding Date.
>
#!/usr/bin/python
import random
import timeit
Lorg=[]
def genList(L) :
for f in range(ord('A'),ord('z')+1) :
for s in range(ord('A'),ord('z')+1) :
L.append((chr(f)+chr(s),str(random.randrange(0,100
genList(Lorg)
Times= 1000
T0= timeit.Timer('L=list(Lorg)','from __main__ import Lorg').timeit(Times)
print T0
SetUp1=r'''from __main__ import Lorg
def del_by_key(L,key) :
d= dict(L)
del d[key]
L[:]= d.items()
'''
SetUp2=r'''from __main__ import Lorg
def del_by_key(L,key) :
d= dict(L)
t= (key,d[key])
L.remove(t)
'''
SetUp3=r'''from __main__ import Lorg
def del_by_key(L,key) :
index= [k for k,val in L]
pos = index.index(key)
del L[pos]
'''
SetUp4=r'''from __main__ import Lorg
def del_by_key(L,key) :
for pos, (k,d) in enumerate(L):
if k == key :
del L[pos]
break
'''
SetUp5=r'''from __main__ import Lorg
def del_by_key(L,key) :
L[:]= [(k,d) for (k,d) in L if k !=key]
'''
SetUp6=r'''from __main__ import Lorg
class Tester(object) :
def __init__(self,key) :
self.key= key
def __eq__(self,other) :
return other[0] == self.key
def del_by_key(L,key) :
del L[L.index(Tester(key))]
'''
print '*** ready ***'
T= timeit.Timer("L=list(Lorg);del_by_key(L,'Zz')",SetUp1).timeit(Times)
print "Method 1 :",T-T0
T= timeit.Timer("L=list(Lorg);del_by_key(L,'Zz')",SetUp2).timeit(Times)
print "Method 2 :",T-T0
T= timeit.Timer("L=list(Lorg);del_by_key(L,'Zz')",SetUp3).timeit(Times)
print "Method 3 :",T-T0
T= timeit.Timer("L=list(Lorg);del_by_key(L,'Zz')",SetUp4).timeit(Times)
print "Method 4 :",T-T0
T= timeit.Timer("L=list(Lorg);del_by_key(L,'Zz')",SetUp5).timeit(Times)
print "Method 5 :",T-T0
T= timeit.Timer("L=list(Lorg);del_by_key(L,'Zz')",SetUp6).timeit(Times)
print "Method 6 :",T-T0
# Results on an old P3 (0.9 GHz)
# *** ready ***
# Method 1 : 10.9850928783
# Method 2 : 5.96455168724
# Method 3 : 3.97821164131
# Method 4 : 1.66151881218
# Method 5 : 8.90886187553
# Method 6 : 6.2503888607
The clear winner is
def del_by_key(L,key) :
for pos, (k,d) in enumerate(L):
if k == key :
del L[pos]
break
--
Helmut Jarausch
Lehrstuhl fuer Numerische Mathematik
RWTH - Aachen University
D 52056 Aachen, Germany
--
http://mail.python.org/mailman/listinfo/python-list
Re: Benchmark [was Re: common problem - elegant solution sought]
Paul Rubin wrote: > Helmut Jarausch <[EMAIL PROTECTED]> writes: >> def del_by_key(L,key) : >>for pos, (k,d) in enumerate(L): >> if k == key : >>del L[pos] >>break > > This looks very dangerous, mutating L while iterating over it. No, as Bruno Desthuilliers has pointed out, because one breaks out of the loop immediately. Helmut. -- Helmut Jarausch Lehrstuhl fuer Numerische Mathematik RWTH - Aachen University D 52056 Aachen, Germany -- http://mail.python.org/mailman/listinfo/python-list
Re: super, decorators and gettattribute
Michele Simionato wrote: > I really need to publish this one day or another, since these > questions > about super keeps coming out: > > http://www.phyast.pitt.edu/~micheles/python/super.html Unfortunately the links [2], [3] and [4] are not given, Helmut. -- Helmut Jarausch Lehrstuhl fuer Numerische Mathematik RWTH - Aachen University D 52056 Aachen, Germany -- http://mail.python.org/mailman/listinfo/python-list
ctypes CDLL - which paths are searched?
Hi, how can I specify the paths to be searched for a dynamic library to be loaded by ctypes' CDLL class on a Linux system. Do I have to set os.environment['LD_LIBRARY_PATH'] ? Many thanks for a hint, Helmut. -- Helmut Jarausch Lehrstuhl fuer Numerische Mathematik RWTH - Aachen University D 52056 Aachen, Germany -- http://mail.python.org/mailman/listinfo/python-list
Re: ctypes CDLL - which paths are searched?
Thomas Heller wrote: > Helmut Jarausch schrieb: >> Hi, >> >> how can I specify the paths to be searched for a dynamic library >> to be loaded by ctypes' CDLL class on a Linux system. >> >> Do I have to set os.environment['LD_LIBRARY_PATH'] ? >> > > ctypes passes the argument given to CDLL(path) straight to > the dlopen(3) call, so your system documentation should tell you. > Thanks, but then it's difficult to use CDLL. Setting os.environ['LD_LIBRARY_PATH'] within the script which calls CDLL is too late. What other methods are possible rather than put an explicit export LD_LIBRARY_PATH=... before running the script, if I don't want to put the dynamic library into a standard system library. Many thanks, Helmut. -- Helmut Jarausch Lehrstuhl fuer Numerische Mathematik RWTH - Aachen University D 52056 Aachen, Germany -- http://mail.python.org/mailman/listinfo/python-list
Re: Removing objects
[EMAIL PROTECTED] wrote:
> I am writing a game, and it must keep a list of objects. I've been
> representing this as a list, but I need an object to be able to remove
> itself. It doesn't know it's own index. If I tried to make each object
> keep track of it's own index, it would be invalidated when any object
> with a lower index was deleted. The error was that when I called
> list.remove(self), it just removed the first thing in hte list with
> the same type as what I wanted, rather than the object I wanted. The
> objects have no identifying charachteristics, other than thier
> location in memory
>
> So my question: How do I look something up in a list by it's location
> in memory? does python even support pointers?
>
> Is there a better way?
You could use a doubly linked list.
The class 'deque' from collections is such a list
but I don't known how to delete a specific element without
this ugly (expensive?) rotate method.
Here is my solution in pure Python
#!/usr/bin/python
import re
class Queue(object):
__slots__= 'Head', 'Tail', '__iter__','NNd'
def __init__(self):
self.Head= None
self.Tail= None
self.NNd= None
def append(self,N):
if self.Head == None:
self.Head= N
self.Tail= self
N._enqueue_after(self.Tail)
self.Tail= N
def dequeue(self,N):
Father, Son= N._dequeue()
if self.Tail == N:
self.Tail= Father
if self.Head == N:
self.Head= Son
if self.Head == None: self.Tail= None
def enqueue_after(self,N,Father):
N._enqueue_after(Father)
if self.Tail == Father:
self.Tail= N
def enqueue_before(self,N,Son):
N._enqueue_before(Son)
if self.Head == Son:
self.Head= N
def __iter__(self):
next= self.Head # allows to dequeue the current element in a loop
while True:
current= next
if current == None: raise StopIteration
next= current._next()
yield current
class Node(object): # partially application specific
__slots__= 'next', 'NNd','PNd','key','data'
def __init__(self,Key,Data,P=None,N=None): # P = Previous N = Next
if P != None: P.NNd= self
if N != None: N.PNd= self
self.PNd= P
self.NNd= N
self.key= Key
self.data= Data
def _enqueue_after(self,Father):
self.NNd= Father.NNd
self.PNd= Father
Father.NNd= self
def _enqueue_before(self,Son):
self.NNd= Son
self.PNd= Son.PNd
Son.PNd= self
def _dequeue(self):
if self.PNd != None:
self.PNd.NNd= self.NNd
if self.NNd != None:
self.NNd.PNd= self.PNd
Father= self.PNd
Son=self.NNd
self.PNd= self.NNd= None
return Father,Son
def _next(self):
return self.NNd
def __str__(self): # highly application specific
return '(%s:%s)' % (self.key,self.data)
# some tests
MyQ= Queue()
MyQ.append( Node('HJ','3949') )
print MyQ.Head,MyQ.Tail,MyQ.Head.key
MyQ.append( Node('CJ','10149') )
print MyQ.Head,MyQ.Tail,MyQ.Head.key
N= MyQ.Head
print "queue: ",N.key,"->"
N= N.NNd
print "next: ",N.key,"->"
N= N.NNd
if N != None:
print "next: ",N.key,"->"
else:
print "no next:"
for N in MyQ:
print "loop->",N
print N.key,N.data
MyQ.dequeue(MyQ.Tail)
print "--- after dequeue"
print "Head: ",MyQ.Head," Tail: ",MyQ.Tail
for N in MyQ:
print "loop2->",N
print N.key,N.data
MyQ.dequeue(MyQ.Tail)
print "after second dequeue"
print "Head: ",MyQ.Head," Tail: ",MyQ.Tail
for N in MyQ:
print "loop3->",N
print N.key,N.data
ENJOY,
Helmut.
--
Helmut Jarausch
Lehrstuhl fuer Numerische Mathematik
RWTH - Aachen University
D 52056 Aachen, Germany
--
http://mail.python.org/mailman/listinfo/python-list
Re: global/local variables
Tim Rau wrote: > I'm sorry: I forgot to say what my problem was. Python seems to think > that nextID is a local, and complains that it can't find it. THis is > not the full text of the function, just the level that is causing > errors. the lack of : on the if is a transcription error. > Traceback (most recent call last): > File "C:\Documents and Settings\Owner\My Documents\NIm's code\sandbox > \sandbox.py", line 248, in > thing.step() > File "C:\Documents and Settings\Owner\My Documents\NIm's code\sandbox > \sandbox.py", line 112, in step > allThings[len(allThings)-1].id = nextID > UnboundLocalError: local variable 'nextID' referenced before > assignment But in the next line in the same function you say nextID+= 1 Since there is an assignment to nextID in the same function without a global statement, you tell Python, that nextID is a local variable. Note: Assignment (anywhere in a function) is a local definition unless there is a global statement in this function > I want to know why it says 'local variable' it never had a problem > when just allThings was in there. > > as for the objection to global variable: If I didn't make them global, > I'd make them part of a singleton class called gameVars that would get > passed everywhere. It's unlikely that they'll get mixed in with > anyhting else, as they fulfill a unique function. Also, I think it's > more convenient, and I am, after all, my own employer when it comes to > programming. -- Helmut Jarausch Lehrstuhl fuer Numerische Mathematik RWTH - Aachen University D 52056 Aachen, Germany -- http://mail.python.org/mailman/listinfo/python-list
Tkinter - incremental input ?
Hi, I don't want to reinvent the wheel but I cannot find it so far. Many editors have a so-called incremental search feature. As you type characters, elements of a set of strings which fit so far are displayed or at least, the first one of these is displayed. Now I want to do something similar in Tkinter - an Entry widget which displays possible 'completions' e.g. given the list of names (...,'Hardy','Helmut',..) As soon as enter the character 'H' the string 'Hardy' would be displayed in the Entry widget - but the cursor is still at position 2 (given 'H' is a position 1) Furthermore, as soon as I enter 'e', it would change the text to 'Helmut', and so on. While I can bind '' to a callback, I haven't figured out how to get (and later on set) the cursor within the Entry widget. In other words I need to know at which character position the last character was entered. Currently I can only see the brute force method: keeping track of all cursor positioning means like , , the '<-' and '->' keys and mouse clicks. Is there an easier method? Many thanks for a hint or even a pointer to an example, Helmut. -- Helmut Jarausch Lehrstuhl fuer Numerische Mathematik RWTH - Aachen University D 52056 Aachen, Germany -- http://mail.python.org/mailman/listinfo/python-list
Re: Trouble loading dll via ctypes
subopt inTheVicinityOf geemail.com wrote:
> I'm trying to load a dll via ctypes by doing this:
>
> cdll.LoadLibrary('/path/to/mylib.so')
>
> But i'm getting this:
>
> /path/to/mylib.so: cannot open shared object file: No such file or
> directory What am i doing wrong?
>
> The dll in question is in a directory mounted via NSF, but no part of
> the path/filename is a symlink. When i try code from the docs:
>
> cdll.LoadLibrary('libc.so.6')
>
> ,then all is fine.
>
> I've also tried to set my LD_LIBRARY_PATH before starting Python,
> checking it via os.environ, then doing the cdll.LoadLibrary(...), but
> that fails with the same complaint. What am i doing wrong?
>
I vaguely remember you need execute permissions for a dynamic library to load.
Permissions on an NFS mounted directory are different, especial for user 'root'.
Check if can execute some executable in that NFS path.
Helmut.
--
Helmut Jarausch
Lehrstuhl fuer Numerische Mathematik
RWTH - Aachen University
D 52056 Aachen, Germany
--
http://mail.python.org/mailman/listinfo/python-list
Re: Sine Wave Curve Fit Question
Iain Mackay wrote: > Python Folks > > I'm a newbie to Python and am looking for a library / function that can help > me fit a 1D data vector to a sine wave. I know the frequency of the wave, > so its really only phase and amplitude information I need. > > I can't find anything in the most widely known libraries (they seem to be > strong on polynomial fitting, but not, apparently, on trig functions) and I > wondered if any one here had recommendations? > > Something that implemented IEEE 1057 , or similar, would be perfect. Let's do a bit math first. Your model is A*sin(omega*t+alpha) where A and alpha are sought. Let T=(t_1,...,t_N)' and Y=(y_1,..,y_N)' your measurements (t_i,y_i) ( ' denotes transposition ) First, A*sin(omega*t+alpha) = A*cos(alpha)*sin(omega*t) + A*sin(alpha)*cos(omega*t) = B*sin(omega*t) + D*cos(omega*t) by setting B=A*cos(alpha) and D=A*sin(alpha) Once, you have B and D, tan(alpha)= D/B A=sqrt(B^2+D^2) Then in vector notation S=sin(omega*T) C=cos(omega*T) you get the 2x2 system for B and D : (S'*S) * B + (S'*C) * D= S'*Y (S'*C) * B + (C'*C) * D= C'*Y where S'*C is the scalar product of the vectors S and C and similarly. Now, for Python, to handle vectors and scalar products efficiently, have a look at numpy. -- Helmut Jarausch Lehrstuhl fuer Numerische Mathematik RWTH - Aachen University D 52056 Aachen, Germany -- http://mail.python.org/mailman/listinfo/python-list
Re: Sine Wave Curve Fit Question
Mikael Olofsson wrote: > Helmut Jarausch wrote: >> Your model is A*sin(omega*t+alpha) where A and alpha are sought. >> Let T=(t_1,...,t_N)' and Y=(y_1,..,y_N)' your measurements (t_i,y_i) >> ( ' denotes transposition ) >> >> First, A*sin(omega*t+alpha) = >> A*cos(alpha)*sin(omega*t) + A*sin(alpha)*cos(omega*t) = >> B*sin(omega*t) + D*cos(omega*t) >> >> by setting B=A*cos(alpha) and D=A*sin(alpha) >> >> Once, you have B and D, tan(alpha)= D/B A=sqrt(B^2+D^2) > > This is all very true, but the equation tan(alpha)=D/B may fool you. You're right: standard Python's math library missing the function arctan2. But you can get it from numpy or numarray. In case of doubt just compute the residuum with different possibilities . > This may lead you to believe that alpha=arctan(D/B) is a solution, which > is not always the case. The point (B,D) may be in any of the four > quadrants of the plane. Assuming B!=0, the solutions to this equation > fall into the two classes > > alpha = arctan(D/B) + 2*k*pi > > and > > alpha = arctan(D/B) + (2*k+1)*pi, > > where k is an integer. The sign of B tells you which class gives you the > solution. If B is positive, the solutions are those in the first class. > If B is negative, the solutions are instead those in the second class. > Whithin the correct class, you may of course choose any alternative. > > Then we have the case B=0. Then the sign of D determines alpha. If D is > positive, we have alpha=pi/2, and if D is negative, we have alpha=-pi/2. > > Last if both B and D are zero, any alpha will do. > > /MiO -- Helmut Jarausch Lehrstuhl fuer Numerische Mathematik RWTH - Aachen University D 52056 Aachen, Germany -- http://mail.python.org/mailman/listinfo/python-list
helper function in a class' namespace
Hi,
the following code works fine
def Helper(M) :
return 'H>'+M
class A(object):
def __init__(self,Msg) :
print Helper(Msg)
B=A('ZZ')
but the Helper function is in the module's namespace.
I'd like to put it into class A's namespace.
Note, the Helper function doesn't need access to any instance attributes.
The first variant
class A(object):
def Helper(M) :
return 'H>'+M
def __init__(self,Msg) :
print Helper(Msg)
doesn't work since Python is looking for a global function Helper (why?)
The alternative
class A(object):
def Helper(M) :
return 'H>'+M
def __init__(self,Msg) :
print A.Helper(Msg)
doesn't work either since now the first argument to A.Helper must be 'A'.
So, isn't it possible to have a helper function in the namespace of a class
without (the overhead of) passing a 'self' or a 'cls' parameter?
Probably I'm hurt by my C++ history.
Many thanks for your help,
Helmut.
--
Helmut Jarausch
Lehrstuhl fuer Numerische Mathematik
RWTH - Aachen University
D 52056 Aachen, Germany
--
http://mail.python.org/mailman/listinfo/python-list
Re: a trick with lists ?
Sébastien Vincent I've found some class on the Net which takes basically this form : > > ## > class Foo: > def __init__(self): > self.tasks = [] >... > > def method1(self): > tasks = [] > while True: > ... > append/pop elements into/from tasks > ... > if condition : break > > self.tasks[:] = tasks > return > ## > > What I do not fully understand is the line "self.tasks[:] = tasks". Why does > the guy who coded this did not write it as "self.tasks = tasks"? What is the > use of the "[:]" trick ? > I've just run into this difference myself. As several others have pointed out, assignment to self.task[:] modifies this list in place. Here my example showing a striking difference class MyClass(object) : def shorten_list(self,outer_list) : ll=len(outer_list) if ll > 0 : outer_list[:]= outer_list[:ll-1] mylist=[1,2,3] MyClass().shorten_list(mylist) print mylist // this prints [1, 2] (as expected) class MyClass2(object) : def shorten_list(self,outer_list) : ll=len(outer_list) if ll > 0 : outer_list= outer_list[:ll-1] mylist=[1,2,3] MyClass2().shorten_list(mylist) print mylist # this prints [1, 2, 3] The shortened list outer_list[:ll-1] has been assigned (bound in Python terms) to the LOCAL reference (to a list) 'outer_list' -- Helmut Jarausch Lehrstuhl fuer Numerische Mathematik RWTH - Aachen University D 52056 Aachen, Germany -- http://mail.python.org/mailman/listinfo/python-list
Tkinter equiv for setPalette
Hi, I am to convert an old Perl-Tk script to Python. It starts by my $MW= new MainWindow; $MW->setPalette(background => 'AntiqueWhite1', foreground => 'blue'); Is there an equivalent for Tkinter? How can I set default colors for background and foreground for the whole application (root window and its children) Many thanks for a hint, Helmut. -- Helmut Jarausch Lehrstuhl fuer Numerische Mathematik RWTH - Aachen University D 52056 Aachen, Germany -- http://mail.python.org/mailman/listinfo/python-list
Tkinter - tk_focusNext broken for years?
Hi,
I'd like to bind the '' event to a function
which does some calculations (like a database lookup)
and then invoke tk_focusNext.
I get an exception as many others have reported earlier, see
http://bugs.python.org/issue799428
There has even a patch been suggested. Why hasn't it been fixed -
I haven't seen a fix.
The problem can be reproduced by the following tiny script
I'm using Python 2.5.1 .
Should I fixed the Tkinter source myself or is there a workaround?
Many thanks for your help,
Helmut.
---
import Tkinter as Tk
def Proc_Enter(Event) :
print "Enter - normally lookup a database"
Event.widget.tk_focusNext()
root= Tk.Tk()
Tk.Label(root,text='Name').grid(column=0,row=0,sticky=Tk.W)
Nm_input= Tk.StringVar()
Nm_entry= Tk.Entry(root,textvariable=Nm_input)
Nm_entry.bind('',Proc_Enter)
Nm_entry.grid(column=1,row=0,sticky=Tk.W)
Tk.Label(root,text='City').grid(column=0,row=1,sticky=Tk.W)
City_input= Tk.StringVar()
City_entry= Tk.Entry(root,textvariable=City_input)
City_entry.bind('',Proc_Enter)
City_entry.grid(column=1,row=1,sticky=Tk.W)
Tk.mainloop()
---
>>>>>>>>>>>>
Exception in Tkinter callback
Traceback (most recent call last):
File "/usr/lib/python2.5/lib-tk/Tkinter.py", line 1403, in __call__
return self.func(*args)
File "FocusBug.py", line 6, in Proc_Enter
Event.widget.tk_focusNext()
File "/usr/lib/python2.5/lib-tk/Tkinter.py", line 479, in tk_focusNext
return self._nametowidget(name)
File "/usr/lib/python2.5/lib-tk/Tkinter.py", line 1064, in nametowidget
if name[0] == '.':
TypeError: '_tkinter.Tcl_Obj' object is unsubscriptable
---
--
Helmut Jarausch
Lehrstuhl fuer Numerische Mathematik
RWTH - Aachen University
D 52056 Aachen, Germany
--
http://mail.python.org/mailman/listinfo/python-list
Re: Why the result same as before modifing py files
zaley wrote: > In my C++ program ,python is embeded . I import One module(py file) > and execute some functions .Then I modify py files in order to modify > func parameters when process is still running .I import the module and > execute functions again.But I find the result is same as before 'import' is cached. If do want Python to re-import a just call reload(http://mail.python.org/mailman/listinfo/python-list
