How to do C++ style(indexed) nested loops in python?

Question

What is the equivalent of the following in python?

for (i=0; i<n; i++)
    for (j=i+1; j<n; j++)
        //do stuff with A[i], A[j]

Or in some sense, the following. It should also remove the element from A at the completion of each round of the loop.

for a in A:
    for a' in A/{a}: #i.e. rest of the elements of A
        #do something with a,a'
    #remove a from A

Is there a pythonic way of doing this without using enumerate()?

Edits:

Sorry for the bad description.

In the first example, I mean to use i & j only as indices. Their values do not matter. Its just a rough c++ equivalent of the latter.
The outer loop is executed n times. The inner loop is executed (n-1), (n-2)...0 times for each iteration of the outer loop.

Maybe this might help (pseudocode):

function next_iteration(list):
    head = first element
    tail = remaining elements #list
    each element in tail interacts with head one by one
    next_iteration(tail)

PS: All code samples above are pseudocodes. I'm trying to express something that is still a bit vague in my mind.

@katrielalex -- not explicitly, but conceptually they could both represent nested loops over the same sequence, where the inner loop starts immediately after the current element in the outer loop. — Dave Costa
– Dave Costa, Commented Mar 20, 2012 at 13:41
The second example begins the inner loop at the start and only skips element a. — hochl
– hochl, Commented Mar 20, 2012 at 13:46
It need not be the 'next' element in the inner loop. The inner loop works on the 'remaining elements'. The next iteration is also on the remaining elements. Please see next_iteration to get a better idea. — Karthick
– Karthick, Commented Mar 20, 2012 at 14:16
Both katrielalex and hochl are correct I guess. I just have to use itertools.combinations instead of itertools.permutations. Is it possible to mark both as correct? — Karthick
– Karthick, Commented Mar 20, 2012 at 14:32

Katriel · Accepted Answer · 2012-03-20 21:47:13Z

4

I intepret what you're asking as

How can I iterate over all pairs of distinct elements of a container?

Answer:

>>> x = {1,2,3}
>>> import itertools
>>> for a, b in itertools.permutations(x, 2):
...     print a, b
... 
1 2
1 3
2 1
2 3
3 1
3 2

EDIT: If you don't want both (a,b) and (b,a), just use itertools.combinations instead.

edited Mar 20, 2012 at 21:47

answered Mar 20, 2012 at 13:45

Katriel

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2 Comments

Karthick Over a year ago

All pairs of distinct elements, without considering the order of elements in each pair. how does this change to accomodate that?

Katriel Over a year ago

@Karthick use .combinations instead. See the docs.

hochl · Accepted Answer · 2012-03-20 14:36:47Z

3

Since your two questions are different, here is solution for your second problem:

for i in xrange(len(A)):
    for j in xrange(len(A)):
        if i != j:
            do_stuff(A[i], A[j])

or using itertools (I think using the included batteries is very pythonic!):

import itertools

for a, b in itertools.permutations(A, 2):
    do_stuff(a, b)

This applies do_stuff to all combinations of 2 different elements from A. I you want to store the result just use:

[do_stuff(a, b) for a, b in itertools.permutations(A, 2)]

edited Mar 20, 2012 at 14:36

answered Mar 20, 2012 at 13:35

hochl

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1 Comment

Karthick Over a year ago

Hi, almost there. :) Can you make it work in such a way that (a,b) and (b,a) are the same? That is, order doesn't matter.

MByD · Accepted Answer · 2012-03-20 13:31:09Z

0

How about:

for i in range(0,n):
    for j in range (i+1,n):
    # do stuff

answered Mar 20, 2012 at 13:31

MByD

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Comments

Grigor Gevorgyan · Accepted Answer · 2012-03-20 13:31:22Z

0

for i in range(0,n):
   for j in range(i+1,n):
   # do stuff

answered Mar 20, 2012 at 13:31

Grigor Gevorgyan

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Comments

tbd · Accepted Answer · 2012-03-20 13:36:06Z

0

Still can't leave comments.. but basically what the other two posts said - but get in the habit of using xrange instead of range.

for i in xrange(0,n):
   for j in xrange(i+1,n):
   # do stuff

answered Mar 20, 2012 at 13:36

tbd

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3 Comments

rplnt Over a year ago

range() returns generator in python 3.x

tbd Over a year ago

It does, but the op never specified version so I tend to assume < 3.x :]

Ethan Furman Over a year ago

With 2.x on the way out and 3.x on the way in it is better to get in the habit of using range.

Dave Costa · Accepted Answer · 2012-03-20 13:39:18Z

0

You could make the inner loop directly over a slice. Not saying this is any better, but it is another approach.

for i in range(0,len(x)):
  a = x[i]
  for b in x[i+1:]:
    print a, b

answered Mar 20, 2012 at 13:39

Dave Costa

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Comments

Burhan Khalid · Accepted Answer · 2012-03-20 13:43:30Z

0

Another way to approach this is - if n is an sequence that provides the iterable interface, then in Python you can simplify your code by iterating over the object directly:

for i in n:
   for some_var in n[n.index(i):]: # rest of items
     # do something

I hope I understood your loop correctly, because as others have stated - they don't do the same thing.

answered Mar 20, 2012 at 13:43

Burhan Khalid

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Comments

stranac · Accepted Answer · 2012-03-20 13:46:05Z

0

For the first one of your questions, as already mentioned in other answers:

for i in xrange(n):
    for j in xrange(i+1, n):
        # do stuff with A[i] and A[j]

For the second one:

for i, a in enumerate(A):
    for b in A[i+1:]:
        # do stuff with a and b

answered Mar 20, 2012 at 13:46

stranac

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Comments

Ethan Furman · Accepted Answer · 2012-03-20 21:55:20Z

0

Your psuedocode almost has it:

function next_iteration(list):
    head = first element
    tail = remaining elements #list
    each element in tail interacts with head one by one
    next_iteration(tail)

Python code:

def next_iteration(lst):
    head, tail = lst[0], lst[1:]
    for item in tail:
        print(head, item)
    if tail:
        next_iteration(tail)

Which, when tried with next_iteration([1, 2, 3]), prints:

1 2
1 3
2 3

answered Mar 20, 2012 at 21:55

Ethan Furman

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Comments

octopusgrabbus · Accepted Answer · 2012-07-18 17:48:53Z

0

You can use xrange to generate values for i and j respectively as show below:

for i in xrange(0, n):
   for j in xrange(i + 1, n):
       # do stuff

edited Jul 18, 2012 at 17:48

octopusgrabbus

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answered Mar 20, 2012 at 13:36

warvariuc

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Comments

Avid Coder · Accepted Answer · 2014-09-06 01:15:09Z

0

In the first for-loop, enumerate() walks through the array and makes the index,value of each element available to the second for-loop. In the second loop, range() makes j = i+1 --> len(a) available. At this point you'd have exactly what you need which is i & j to do your operation.

>>> a = [1,2,3,4]
>>> array_len = len(a)
>>> for i,v in enumerate(a):
...   for j in range(i+1, array_len):
...     print a[i], a[j]
...
1 2
1 3
1 4
2 3
2 4
3 4
>>>

answered Sep 6, 2014 at 1:15

Avid Coder

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