## Generate lists with elements less or equal to the given list

number of elements less than or equal to a given number in a given array

count smaller elements on left side in a given array

compare values within list python

python list

one to one correspondence in list python

accessing list of lists python

python select from list

What is the most pythonic way to solve the following problem?

Given a list A, find all lists B, such that for i in range(len(A)): B[i] <= A[i]. Example of what I expect:

#Input A = [1,2,0] #Output B = [[0,0,0], [1,0,0], [1,1,0], [1,2,0], [0,1,0], [0,2,0]]

Thanks in advance!

You can use `itertools.product`

to do this easily

>>> from itertools import product >>> A [1, 2, 0] >>> B = list(product(*[list(range(e+1)) for e in A])) >>> B [(0, 0, 0), (0, 1, 0), (0, 2, 0), (1, 0, 0), (1, 1, 0), (1, 2, 0)] >>>

If you want the o/p as list of list, convert the tuples to list

>>> B = [list(e) for e in B] >>> B [[0, 0, 0], [0, 1, 0], [0, 2, 0], [1, 0, 0], [1, 1, 0], [1, 2, 0]] >>>

If you dont wan't to use `itertools.product`

, you can have your custom implementation of `product`

>>> B = [[]]; >>> for t in [range(e+1) for e in A]: ... B = [x+[y] for x in B for y in t] ... >>> B [[0, 0, 0], [0, 1, 0], [0, 2, 0], [1, 0, 0], [1, 1, 0], [1, 2, 0]] >>>

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Here is a quite smart solution without using `itertools`

but using recursive function call. It works whatever the size of the input list:

A = [1,2,0] def compute(A, B=[], n=0): for i in range(A[n]+1): A[n] = i if A not in B: B.append(A[:]) if n+1 < len(A): compute(A, B, n+1) # recursive call here return sorted(B) B = compute(A) print(B) # [[0, 0, 0], [0, 1, 0], [0, 2, 0], [1, 0, 0], [1, 1, 0], [1, 2, 0]]

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I guess you could first generate all unique combinations of these numbers:

>>> import itertools >>> A = [1,2,0] >>> combs = set(itertools.combinations(itertools.chain(*(range(col+1) for col in A)), len(A))) >>> combs {(0, 1, 1), (0, 2, 0), (1, 1, 0), (1, 2, 0), (0, 0, 2), (0, 1, 2), (1, 0, 0), (0, 0, 1), (1, 0, 1), (1, 1, 2), (0, 0, 0), (0, 1, 0), (1, 0, 2)}

Then pick the ones that follow your condition:

>>> sorted(comb for comb in combs if all(comb[i] <= x for i, x in enumerate(A))) [(0, 0, 0), (0, 1, 0), (0, 2, 0), (1, 0, 0), (1, 1, 0), (1, 2, 0)]

But this is alot more inefficient than using `itertools.product`

, since it has to generate all the combinations beforehand, and sort at the end to retain order.

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This also works:

A = [1,2,0] answers = [[] for element in A] answers[0] = [[number,] for number in range(A[0]+1)] for i in range(len(A)-1): for possibility in answers[i]: for valid_number in range(A[i+1]+1): answers[i+1].append(possibility + [valid_number, ]) your_answer = answers[-1]

It could probably be written in a more compact way.

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##### Comments

- Sounds like homework.
- Great! The itertools was in my list. Unfortunatelly, I'm not allowed to use any side libraries. Maybe there are straightforward solutions without it? Thanks!
- I can't see it working for an arbitrary len(A). Am I missing something? Is is possible to extend your second solution to any len(A)?
- Updated the answer
- @Sanitha, looks awesome! Maybe you know, why does the itertools implementation work much faster than the custom implementation? Thanks!