How can I simplifiy this python iteration?

Question

I want to get all possible combinations of three (or more) numbers. The numbers themselves need to be in the range of +-1. The range is to find 'similar numbers' - for example the number 3 needs to be iterated as 2,3,4. E.g I have:

    num1 = 3 
    num2 = 4
    num3 = 1

So in this example I want all combinations for these three numbers and every number +-1. (E.g. 341, 241, 441; 351, 331, ... ). So for the example numbers I should get 27 combinations.

First idea was to use 3 for-loops in python like this:

    num1 = 3
    num2 = 4
    num3 = 1

    def getSimilar(num1,num2,num3):

        num1 = n1 - 2

        for i in range (3):
            num1 = num1 + 1
            num2 = n2 - 2

            for j in range(3):

                num2 = num2 + 1
                num3 = n3 - 2

                for k in range(3):
                    num3 = num3 + 1
                    print(num1,num2,num3)

Output I get:

    2 3 0
    2 3 1
    2 3 2
    2 4 0
    2 4 1
    2 4 2
    2 5 0
    2 5 1
    2 5 2
    3 3 0
    3 3 1
    3 3 2
    3 4 0
    3 4 1
    3 4 2
    3 5 0
    3 5 1
    3 5 2
    4 3 0
    4 3 1
    4 3 2
    4 4 0
    4 4 1
    4 4 2
    4 5 0
    4 5 1
    4 5 2

Is there a smarter and faster way to do this in python instead of using 3 for loops? The order of the output does not matter. A small problem I additionally have: If one number is 0, I need it to only iterate to 0 and 1, not to -1. So the output for 0; 4; 1 should be:

    0 4 1
    0 4 2
    0 4 0

    0 3 1
    0 3 2
    0 3 0

    0 5 1
    0 5 2
    0 5 0

    1 4 1
    1 4 2
    1 4 0

    1 3 1
    1 3 2
    1 3 0

    1 5 1
    1 5 2
    1 5 0

Answer 1

Here is a solution that handles your edge case:

from itertools import product

nums = [0, 4, 1]
options = [[x - 1, x, x + 1] for x in nums]
result = [similar for similar in product(*options) if all(x >= 0 for x in similar)]
for x in result:
    print(x)

Output:

(0, 3, 0)
(0, 3, 1)
(0, 3, 2)
(0, 4, 0)
(0, 4, 1)
(0, 4, 2)
(0, 5, 0)
(0, 5, 1)
(0, 5, 2)
(1, 3, 0)
(1, 3, 1)
(1, 3, 2)
(1, 4, 0)
(1, 4, 1)
(1, 4, 2)
(1, 5, 0)
(1, 5, 1)
(1, 5, 2)

Answer 2

You can do that like this:

from itertools import product

def getSimilar(*nums):
    return product(*(range(max(n - 1, 0), n + 2) for n in nums))

num1 = 3
num2 = 4
num3 = 1
for comb in getSimilar(num1, num2, num3):
    print(comb)

# (2, 3, 0)
# (2, 3, 1)
# (2, 3, 2)
# (2, 4, 0)
# ...

Answer 3

Start by creating the list of valid digits via list comprehension, then use itertools.product to create the possible combinations

from itertools import product
digits = [3,4,0,-1]

#Generate all possible digits
all_digits = [ (k-1,k,k+1) for k in digits]

#Valid digits, ensuring not to consider negative digits
valid_digits = [digits for digits in all_digits if all(x >= 0 for x in digits)]

#Create the iterator of all possible numbers
nums = product(*valid_digits)

#Print the combinations
for num in nums:
    print(*num)

The output will look like.

2 3
2 4
2 5
3 3
3 4
3 5
4 3
4 4
4 5