CODEWITHSUNDEEP
pythonlistloopscombinations
user52931
user52931

Reputation: 3

Nested For Loops To Fill An Array

I'm learning Python's nested loops but couldn't use nested loops to fill a list with the combinations. So far I have

for i in range(3):
   for j in range(3):
       paths = [i,j]
       print paths

When this prints I get:

[0, 0]
[0, 1]
[0, 2]
[1, 0]
[1, 1]
[1, 2]
[2, 0]
[2, 1]
[2, 2]

Which is a list of the possible combinations but if I try an extra for-loop to index them it doesn't seem to work and only gives the first element. I'm trying to get an array in the form:

Array = [[0,0], [0,1]... [2,2]]

where I can index said array and get Array[0] == [0,0] and so on.

Any hints on how to do this? I've tried for-loops and using the append function neither giving me the results I wanted.

Upvotes: 0

Views: 21958

Answers (4)

Reut Sharabani
Reut Sharabani

Reputation: 31339

I like list comprehension!

print [[x, y] for x in range(3) for y in range(3)]

But it creates three lists (the iterated two stay in memory) in python 2.7 which is the common one nowdays.

If you're using big values try this:

print [[x, y] for x in xrange(3) for y in xrange(3)]

which uses Generators :)

Upvotes: 4

user2555451
user2555451

Reputation:

The easiest way to do this is to use itertools.product:

>>> from itertools import product
>>> combos = [list(x) for x in product(range(3), repeat=2)]
>>> combos
[[0, 0], [0, 1], [0, 2], [1, 0], [1, 1], [1, 2], [2, 0], [2, 1], [2, 2]]
>>> combos[0]
[0, 0]
>>> combos[1]
[0, 1]
>>> combos[8]
[2, 2]
>>>

Note that this solution is also very flexible. For example, if you want to find the combinations of the numbers one through five taken three at a time, all you need to do is change the numbers given to range and repeat:

combos = [list(x) for x in product(range(6), repeat=3)]
#                                        ^          ^

Upvotes: 1

Simeon Visser
Simeon Visser

Reputation: 122326

You can append the lists to the result:

result = []
for i in range(3):
   for j in range(3):
       paths = [i,j]
       result.append(paths)
       print paths

print result

This says: create a result list [] and append the paths value to it at each step. You'll then have a list of lists and you can access the first element as result[0] (which is indeed [0, 0]).

Upvotes: 2

BartoszKP
BartoszKP

Reputation: 35891

You can achieve it using itertools:

>>> import itertools
>>> list(itertools.product(range(3), range(3)))
[(0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (1, 2), (2, 0), (2, 1), (2, 2)]

If you really need lists instead of tuples:

>>> import itertools
>>> list(itertools.product(range(3), range(3)))
>>> [list(t) for t in itertools.product(range(3), range(3))]
[0, 0], [0, 1], [0, 2], [1, 0], [1, 1], [1, 2], [2, 0], [2, 1], [2, 2]]

Upvotes: 0

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