Python - Iteration over an unknown number of variables

user_16909173 0
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Say I have a list range_list which length is not know a priori. For example say range_list = [2,5,4,10] I would like to iterate in the following manner:

for i in range(range_list[0]):
   for j in range(range_list[1]):
        for k in range(range_list[2]):
            for l in range(range_list[3]):

However since the length of range_list isn't fixed, I cannot nest a finite number of for loops like above.

I've tried using (unsuccessfully) the zip function in the following manner:

range_list_2
for i in range(len(range_list)):
        range_list_2.append(range(range_list[i]))

for i in zip(*range_list_2):
   print(i)

As a result I would be expecting

(0,0,0,0)
(1,0,0,0)
(0,1,0,0)
...
iteration python loops
1 Answers

You can use itertools.product():

 |  product(*iterables, repeat=1) --> product object
 |
 |  Cartesian product of input iterables.  Equivalent to nested for-loops.
 |
 |  For example, product(A, B) returns the same as:  ((x,y) for x in A for y in B).
 |  The leftmost iterators are in the outermost for-loop, so the output tuples
 |  cycle in a manner similar to an odometer (with the rightmost element changing
 |  on every iteration).

For example:

from itertools import product

range_list = [2, 4, 3]
for i in product(*map(range, range_list)):
    print(i)

(0, 0, 0)
(0, 0, 1)
(0, 0, 2)
(0, 1, 0)
(0, 1, 1)
(0, 1, 2)
(0, 2, 0)
(0, 2, 1)
(0, 2, 2)
(0, 3, 0)
(0, 3, 1)
(0, 3, 2)
(1, 0, 0)
(1, 0, 1)
(1, 0, 2)
(1, 1, 0)
(1, 1, 1)
(1, 1, 2)
(1, 2, 0)
(1, 2, 1)
(1, 2, 2)
(1, 3, 0)
(1, 3, 1)
(1, 3, 2)
user_15450260
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