Dmitry,

 

Efficiency of several kinds is hotly debated and sometimes it depends a lot on 
what is done within loops.

 

Many suggest a mild speed up of some comprehensions over loops but the loops 
are not gone but somewhat hidden and perhaps some aspects are faster for having 
been written in C carefully and not interpreted.

 

Comprehensions (and there are other versions that generate dictionaries and 
tuples and sets) may also be sped up a bit for other reasons like your fairly 
expensive APPPEND that has to keep finding the end o f a growing list and is 
not done the same way in a comprehension.

 

If you do a search, you find many opinions including on using functional 
programming techniques such as map/reduce. There are also 

 

Your particular case is interesting because it just makes all combination of 
three variables. Some languages, like R, have functions that do this for you, 
like expand.grd. Python has many modules, like itertools that do things 
including combinations but perhaps not designed for your case. 

 

Here is a version of your scenario:

 

import itertools

a = range(3)

b = range(4)

c = range(5)

 

list(itertools.product(a,b,c))

 

The result comes as tuples but as you are moving the result into numpy, does it 
matter:

 

>>> list(itertools.product(a,b,c))

[(0, 0, 0), (0, 0, 1), (0, 0, 2), (0, 0, 3), (0, 0, 4), (0, 1, 0), (0, 1, 1), 
(0, 1, 2), (0, 1, 3), (0, 1, 4), (0, 2, 0), (0, 2, 1), (0, 2, 2), (0, 2, 3), 
(0, 2, 4), (0, 3, 0), (0, 3, 1), (0, 3, 2), (0, 3, 3), (0, 3, 4), (1, 0, 0), 
(1, 0, 1), (1, 0, 2), (1, 0, 3), (1, 0, 4), (1, 1, 0), (1, 1, 1), (1, 1, 2), 
(1, 1, 3), (1, 1, 4), (1, 2, 0), (1, 2, 1), (1, 2, 2), (1, 2, 3), (1, 2, 4), 
(1, 3, 0), (1, 3, 1), (1, 3, 2), (1, 3, 3), (1, 3, 4), (2, 0, 0), (2, 0, 1), 
(2, 0, 2), (2, 0, 3), (2, 0, 4), (2, 1, 0), (2, 1, 1), (2, 1, 2), (2, 1, 3), 
(2, 1, 4), (2, 2, 0), (2, 2, 1), (2, 2, 2), (2, 2, 3), (2, 2, 4), (2, 3, 0), 
(2, 3, 1), (2, 3, 2), (2, 3, 3), (2, 3, 4)]

 

Or a atd easier to read pretty printed:

 

>>> import pprint

>>> pprint.pprint(list(itertools.product(a,b,c)))

[(0, 0, 0),

(0, 0, 1),

(0, 0, 2),

(0, 0, 3),

(0, 0, 4),

(0, 1, 0),

(0, 1, 1),

(0, 1, 2),

(0, 1, 3),

(0, 1, 4),

(0, 2, 0),

(0, 2, 1),

(0, 2, 2),

(0, 2, 3),

(0, 2, 4),

(0, 3, 0),

(0, 3, 1),

(0, 3, 2),

(0, 3, 3),

(0, 3, 4),

(1, 0, 0),

(1, 0, 1),

(1, 0, 2),

(1, 0, 3),

(1, 0, 4),

(1, 1, 0),

(1, 1, 1),

(1, 1, 2),

(1, 1, 3),

(1, 1, 4),

(1, 2, 0),

(1, 2, 1),

(1, 2, 2),

(1, 2, 3),

(1, 2, 4),

(1, 3, 0),

(1, 3, 1),

(1, 3, 2),

(1, 3, 3),

(1, 3, 4),

(2, 0, 0),

(2, 0, 1),

(2, 0, 2),

(2, 0, 3),

(2, 0, 4),

(2, 1, 0),

(2, 1, 1),

(2, 1, 2),

(2, 1, 3),

(2, 1, 4),

(2, 2, 0),

(2, 2, 1),

(2, 2, 2),

(2, 2, 3),

(2, 2, 4),

(2, 3, 0),

(2, 3, 1),

(2, 3, 2),

(2, 3, 3),

(2, 3, 4)]

 

I think that is close enough to what you want but is it faster? You can try a 
benchmarking method on alternatives.

 

 

 

 

From: Popov, Dmitry Yu <dpo...@anl.gov> 
Sent: Friday, July 12, 2024 11:10 PM
To: avi.e.gr...@gmail.com; 'Popov, Dmitry Yu via Python-list' 
<python-list@python.org>; oscar.j.benja...@gmail.com
Subject: Re: Relatively prime integers in NumPy

 

Thank you very much. List comprehensions make code much more concise indeed. Do 
list comprehensions also improve the speed of calculations?

  _____  

From: avi.e.gr...@gmail.com <mailto:avi.e.gr...@gmail.com>  
<avi.e.gr...@gmail.com <mailto:avi.e.gr...@gmail.com> >
Sent: Friday, July 12, 2024 6:57 PM
To: Popov, Dmitry Yu <dpo...@anl.gov <mailto:dpo...@anl.gov> >; 'Popov, Dmitry 
Yu via Python-list' <python-list@python.org <mailto:python-list@python.org> >; 
oscar.j.benja...@gmail.com <mailto:oscar.j.benja...@gmail.com>  
<oscar.j.benja...@gmail.com <mailto:oscar.j.benja...@gmail.com> >
Subject: RE: Relatively prime integers in NumPy 

 

Dmitry, I clearly did not understand what you wanted earlier as you had not 
made clear that in your example, you already had progressed to some level where 
you had the data and were now doing a second step. So, I hesitate to say much 
until 

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Dmitry,

 

I clearly did not understand what you wanted earlier as you had not made clear 
that in your example, you already had progressed to some level where you had 
the data and were now doing a second step. So, I hesitate to say much until 
either nobody else addressed the issue (as clearly some have) or you explain 
well enough.

 Ditr

I am guessing you have programming experience in other languages and are not as 
“pythonic” as some. The code you show may not be quite how others might do it. 
Some may write mch of your code as a single line of python using a list 
comprehension such as:

 

hkl_list = [ [h, k, l] for SOMETHING in RANGE  for SOMETHING2  in RANGE2 for 
SOMETHING3 in RANGE3]         

 

Where h, k. l come from the somethings.

 

Back to the real world.

 

 

From: Popov, Dmitry Yu <dpo...@anl.gov <mailto:dpo...@anl.gov> > 
Sent: Friday, July 12, 2024 1:13 PM
To: avi.e.gr...@gmail.com <mailto:avi.e.gr...@gmail.com> ; 'Popov, Dmitry Yu 
via Python-list' <python-list@python.org <mailto:python-list@python.org> >; 
oscar.j.benja...@gmail.com <mailto:oscar.j.benja...@gmail.com> ; Popov, Dmitry 
Yu <dpo...@anl.gov <mailto:dpo...@anl.gov> >
Subject: Re: Relatively prime integers in NumPy

 

Thank you very much, Oscar. 

 

Using the following code looks like a much better solution than my current 
Python code indeed.

np.gcd.reduce(np.transpose(a)) 
or 
np.gcd.reduce(a,1)  
 
The next question is how I can generate ndarray of h,k,l indices. This can be 
easily done from a Python list by using the following code.
 
import numpy as np
hkl_list=[]
for h in range(0, max_h):
      for k in range(0, max_k):
            for l in range(0, max_l):
                  hkl_local=[]
                  hkl_local.append(h)
                  hkl_local.append(k)
                  hkl_local.append(l)
                  hkl_list.append(hkl_local)
hkl=np.array(hkl_list, dtype=np.int64)
This code will generate a two-dimensional ndarray of h,k,l indices but is it 
possible to make a faster routine with NumPy? 
 
Regards,
Dmitry
 
 
 
  _____  


From: Python-list <python-list-bounces+dpopov=anl....@python.org 
<mailto:python-list-bounces+dpopov=anl....@python.org> > on behalf of Popov, 
Dmitry Yu via Python-list <python-list@python.org 
<mailto:python-list@python.org> >
Sent: Thursday, July 11, 2024 2:25 PM
To: avi.e.gr...@gmail.com <mailto:avi.e.gr...@gmail.com>  
<avi.e.gr...@gmail.com <mailto:avi.e.gr...@gmail.com> >; 'Popov, Dmitry Yu via 
Python-list' <python-list@python.org <mailto:python-list@python.org> >
Subject: Re: Relatively prime integers in NumPy 

 

Thank you for your interest. My explanation is too concise indeed, sorry. So 
far, I have used Python code with three enclosed 'for' loops for this purpose 
which is pretty time consuming. I'm trying to develop a NumPy based code to 
make this 

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Thank you for your interest. My explanation is too concise indeed, sorry. So 
far, I have used Python code with three enclosed 'for' loops for this purpose 
which is pretty time consuming. I'm trying to develop a NumPy based code to 
make this procedure faster. This routine is kind of 'heart' of the algorithm to 
index of X-ray Laue diffraction patterns. In our group we have to process huge 
amount of such patterns. They are collected at a synchrotron radiation 
facility. Faster indexation routine would help a lot.
 
This is the code I'm currently using. Any prompts how to implement it in NumPy 
would be highly appreciated.
 
for h in range(0, max_h):
      for k in range(0, max_k):
            for l in range(0, max_l):
                  chvec=1
                  maxmult=2
                  if h > 1:                     
                        maxmult=h
                  if k > 1:
                        maxmult=k
                  if l > 1:
                        maxmult=l
                  if h > 1:
                        if maxmult > h:
                              maxmult=h
                  if k > 1:
                        if maxmult > k:
                              maxmult=k
                  if l > 1:
                        if maxmult > l:
                              maxmult=l
                  maxmult=maxmult+1
                  for innen in range(2, maxmult):
                        if h in range(0, (max_h+1), innen):
                              if k in range(0, (max_k+1), innen):
                                    if l in range(0, (max_l+1), innen):
                                          chvec=0
                  if chvec==1:
                        # Only relatively prime integers h,k,l pass to this 
block of the code
 
 
________________________________
From: avi.e.gr...@gmail.com <mailto:avi.e.gr...@gmail.com>  
<avi.e.gr...@gmail.com <mailto:avi.e.gr...@gmail.com> >
Sent: Thursday, July 11, 2024 1:22 PM
To: Popov, Dmitry Yu <dpo...@anl.gov <mailto:dpo...@anl.gov> >; 'Popov, Dmitry 
Yu via Python-list' <python-list@python.org <mailto:python-list@python.org> >
Subject: RE: Relatively prime integers in NumPy
 
Дмитрий, You may think you explained what you wanted but I do not see what 
result you expect from your examples. Your request is a bit too esoteric to be 
a great candidate for being built into a module like numpy for general purpose 
se but
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Дмитрий,
 
You may think you explained what you wanted but I do not see what result you
expect from your examples.
 
Your request is a bit too esoteric to be a great candidate for being built
into a module like numpy for general purpose se but I can imagine it could
be available in modules build on top of numpy.
 
Is there a reason you cannot solve this mostly outside numpy?
 
It looks like you could use numpy to select the numbers you want to compare,
then call one of many methods you can easily search for to see  how to use
python to make some list or other data structure for divisors of each number
involved and then use standard methods to compare the lists and exact common
divisors. If needed, you could then put the results back into your original
data structure using numpy albeit the number of matches can vary.
 
Maybe a better explanation is needed as I cannot see what your latter words
about -1 and 1 are about. Perhaps someone else knows.
 
 
 
 
-----Original Message-----
From: Python-list <python-list-bounces+avi.e.gross=gmail....@python.org 
<mailto:python-list-bounces+avi.e.gross=gmail....@python.org> > On
Behalf Of Popov, Dmitry Yu via Python-list
Sent: Monday, July 8, 2024 3:10 PM
To: Popov, Dmitry Yu via Python-list <python-list@python.org 
<mailto:python-list@python.org> >
Subject: Relatively prime integers in NumPy
 
Dear Sirs.
 
Does NumPy provide a simple mechanism to identify relatively prime integers,
i.e. integers which don't have a common factor other than +1 or -1? For
example, in case of this array:
[[1,5,8],
  [2,4,8],
  [3,3,9]]
I can imagine a function which would return array of common factors along
axis 0: [1,2,3]. Those triples of numbers along axis 1 with the factor of1
or -1 would be relatively prime integers.
 
Regards,
Dmitry Popov
 
Argonne, IL
USA
 
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