CODEWITHSUNDEEP
pythoncombinationspython-itertools
user9893356
user9893356

Reputation: 37

All combinations of n items with lenght l, where n < l

There are many examples how to get all combinations of n items of length l, if l <= n. But how about get all combinations of n items (f.i. digits) of lenght l, where n < l, for instance:

n = 3 ([0,1,2])
l = 5
sequence to get:
00000
00001
00002
00010
...
22222

It is not a problem to implement it, but I suppose it is possible to do using itertools. I can not find how.

Upvotes: 0

Views: 59

Answers (2)

Shipof123
Shipof123

Reputation: 233

Since this set is essentially just a ternary number system, we can use a ternary function from here

def ternary (n):
    if n == 0:
        return '0'
    nums = []
    while n:
        n, r = divmod(n, 3)
        nums.append(str(r))
    return ''.join(reversed(nums))

And you can treat this as an array, and iterate to it through a for loop that goes something like this:

for i in range(3**5):
    print(ternary(i))

Upvotes: 0

ssh
ssh

Reputation: 167

From itertools documentation :

product(range(2), repeat=3) --> 000 001 010 011 100 101 110 111

So for your case the code will be 

itertools.product(range(3), repeat=5)

Upvotes: 1

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