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:

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:

So 3,4 and 1 are the digits of the number you want to create? — Devesh Kumar Singh, Apr 25 '19 at 14:08
Possible duplicate of [Get the cartesian product of a series of lists?](https://stackoverflow.com/questions/533905/get-the-cartesian-product-of-a-series-of-lists) — mkrieger1, Apr 25 '19 at 14:08
one thing i want to tell you is that printing stuff slows computation time. — EHM, Apr 25 '19 at 14:11
@Hilea I know, that's just for debugging in my case at this point of my code. But thanks! — kevinunger, Apr 25 '19 at 14:51

score 5 · Answer 1 · answered Apr 25 '19 at 14:12

5

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)

answered Apr 25 '19 at 14:12

Error - Syntactical Remorse

7,468
4
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That's perfect for me! Thanks! – kevinunger Apr 25 '19 at 15:12
If it answers your question don't forget to mark it as an answer! – Error - Syntactical Remorse Apr 25 '19 at 15:27

jdehesa · Accepted Answer · 2019-04-25T14:32:14.727

3

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)
# ...

edited Apr 25 '19 at 14:32

answered Apr 25 '19 at 14:10

jdehesa

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Fixed the bit about not producing negative numbers (didn't read that at first). – jdehesa Apr 25 '19 at 14:33

Devesh Kumar Singh · Answer 3 · 2019-04-25T14:39:07.430

1

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.

edited Apr 25 '19 at 14:39

answered Apr 25 '19 at 14:15

Devesh Kumar Singh

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I know he didn't explicitly say it but what if I used a negative number in this solution? – Error - Syntactical Remorse Apr 25 '19 at 14:24
Updated @Error-SyntacticalRemorse! Can you check and upvote if it looks good? – Devesh Kumar Singh Apr 25 '19 at 14:39

How can I simplifiy this python iteration?

3 Answers3