Creating continues nested for loop

Question

Is it possible to write a function that adds more for loops to a nested loop. So lets say I have a function for_loops(3).

    def for_loops(n):
        a = []
        for add_v in range(n):
            a.append("AB")

        for i in range(len(a[0])):
            for j in range(len(a[1])):
                for k in range(len(a[2])):
                    print(a[0][i]+" "+a[1][j]+" "+a[2][k])
   for_loops(3)

then for_loops(4)

    def for_loops(n):
        a = []
        for add_v in range(n):
            a.append("AB")

        for i in range(len(a[0])):
            for j in range(len(a[1])):
                for k in range(len(a[2])):
                    for l in range(len(a[3])):
                        print(a[0][i]+" "+a[1][j]+" "+a[2][k]+" "+a[3][l])
   for_loops(4)

You might want to look into doing something else. Either recursion, or just not having such crazily nested lists. There's not really a good way to do this in exactly this way. — Morgan Thrapp
– Morgan Thrapp, Commented Jul 13, 2016 at 19:03
Probably you could use itertools.product docs.python.org/2.7/library/itertools.html#itertools.product But I agree with Morgan Thrapp, this approach seems like it probably isn't ideal. — juanpa.arrivillaga
– juanpa.arrivillaga, Commented Jul 13, 2016 at 19:09

Abhijit · Accepted Answer · 2016-07-13 19:33:45Z

3

You can achieve a similar result using itertools.product which generates a Cartesian product

def for_loops(n):
    from itertools import product
    for row in product("AB", repeat = n):
        print " ".join(row)

The way it works is, given a set, it will generate the Cartesian product of n cardinality i.e.

(A, B) X (A, B) X (A, B) .... n

For n = 4, this would be (A, B) X (A, B) X (A, B) X (A, B)
For n = 3, this would be (A, B) X (A, B) X (A, B)

And example run with n = 4 is shown below, which matches with the result you have from the 4 nested loop

>>> for_loops(4)
A A A A
A A A B
A A B A
A A B B
A B A A
A B A B
A B B A
A B B B
B A A A
B A A B
B A B A
B A B B
B B A A
B B A B
B B B A
B B B B

Edit 1

Thanx, this does exactly what I want it to do but I'm trying to do it without using any built in function. –

If you do not want to use any builtin function, the following might be an alternate approach for you

def for_loops(n):
    for i in range(2**n):
        print " ".join("AB"[int(j)] for j in "{:04b}".format(i))

edited Jul 13, 2016 at 19:33

answered Jul 13, 2016 at 19:11

Abhijit

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1 Comment

Pierre Over a year ago

Thanx, this does exactly what I want it to do but I'm trying to do it without using any built in function.

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