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

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.

Answer 1

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.

Answer 2

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)]

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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.

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

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
>>>