wow i dint know that a single statement like that would make such a
difference. Thanks you very much. that really improves the performance
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Can some one help me improve this block of code...this jus converts the
list of data into tokens based on the range it falls into...but it
takes a long time.Can someone tell me what can i change to improve
it...
def Tkz(tk,data):
no_of_bins = 10
tkns = []
dmax = ma
how do I convert
b is a string b = '(1,2,3,4)' to b = (1,2,3,4)
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I tried to compile it from source. But it dint work.It was looking for
python2.3 .But I want to install it on pyrthon 2.4
PLatform you mean I am using RedHat 9.0. is that what you were
referring?
can you point me to the source you are referring. I used the sourse
from this link...
http://www.scipy
Did anyone try to install Scipy package on python2.4 linux version. I
see only python2.3 version of scipy released. When I try to install I
get an dependency warning saying scipy cannot find python2.3.
Can someone point me to python2.4 version of scipy and help me install.
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Actually slicing the way you suggested improved it to some extent. I
did profile on this and I observed that it reduced the number of calls
for __get_item__ and improved the timing to some extent. Which was
useful to some extent.
Thanks again.
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What is the fastest way to code this particular block of code below..
I used numeric for this currently and I thought it should be really
fast..
But, for large sets of data (bx and vbox) it takes a long time and I
would like to improve.
vbox = array(m) (size: 1000x1000 approx)
for b in bx:
vbo
Hi I would like to know how to resize an Image without using python
Imaging library.
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oh yes its the same case. even [0,4,9,2,0] as a set [2,6] and may be
not [2,7]. Its not that you are wrong its jus that I was not clear.
Sorry about that.
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I get a syntax error in :
py> [(min((abs(p - v), v) for v in valleys + [0] if v < p)[1],
... p,
... min((abs(p - v), v) for v in valleys if v > p)[1])
... for p in peaks]
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Hi Kent,
Thanks for that. But We are considering [..., 0, 101, 0, 0, 0, 0, 0] ->
[13,18] .In fact if you look at the list, the histogram ends at 15 that
is [0,101,0] --> [13,15]. Dont you think so.
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Hi Steve!
I am not sure if I was clear with my previous post .Ok let me rephrase
it .
Assume the values list is the
content of a histogram. Then we see that
values = [ 0, 72, 2, 4, 9, 2, 0, 0, 42, 26, 0, 282,
23, 0, 101, 0, 0, 0, 0, 0]
1 is repeated 72 times, 3 -> 4 t
what if we do something like this. Assume the values list is the
content of a histogram. Then we see that
values = [ 0, 72, 0, 4, 9, 2, 0, 0, 42, 26, 0, 282,
23, 0, 101, 0, 0, 0, 0, 0]
1 is repeated 72 times, 3 -> 4 times and so on. That is the index
would be the value
Thanks for that. My version of python does'nt find "groupby". I am
using python 2.3.2. Is there a way I could do it with out using groupby
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If this is the list.
values = [ 0, 72, 0, 4, 9, 2, 0, 0, 42, 26, 0, 282,
23, 0, 101, 0, 0, 0, 0, 0]
as we can see there are peaks in the list.that is 0,72,0 is a
group(triangle) with peak 72.then 0, 4, 9, 2, 0, 0 with peak
9 and 0, 42, 26, 0 with 42 and s
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