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listA = [1,2,3,4,5,6]
listB = listA[:]
output = []
for a in listA:	
	for b in listB:
		if b != a:
            //function here
			output.append([a,b])
print(output)

(Sep-11-2019, 06:53 AM)3Pinter Wrote: [ -> ]Going to explain in simple language since googling this didn't really help me (unsure how to name this problem).

I have a list which items I want to check with eachother. Each combination triggers a function.
So a list with [1,2,3] should run a function for pair [1,2], for pair [1,3] and for pair [2,3]. Duplicates / themselves aren't needed [1,1] [2,1] [2,2] [3,1] [3,2]

My list might become huge (600K (if that is huge)) so I 'Yield' seems interesting too.

Below is my soup. Horrible, not logical, returns duplicates (I know ).

listA = [1,2,3,4,5,6]
listB = listA[:]
output = []
for a in listA:	
	for b in listB:
		if b != a:
            //function here
			output.append([a,b])
print(output)

listA = [1,2,3,4,5,6]
listB = listA[:]
output = []
for a in listA: 
    for b in listB:
        if b != a:
            # function here
            output.append([a,b])
print(output)

Output:
[[1, 2], [1, 3], [1, 4], [1, 5], [1, 6], [2, 1], [2, 3], [2, 4], [2, 5], [2, 6], [3, 1], [3, 2], [3, 4], [3, 5], [3, 6], [4, 1], [4, 2], [4, 3], [4, 5], [4, 6], [5, 1], [5, 2], [5, 3], [5, 4], [5, 6], [6, 1], [6, 2], [6, 3], [6, 4], [6, 5]]

(Sep-11-2019, 07:20 AM)3Pinter Wrote: [ -> ]It does leave out the 'themselves'-ones but not the duplicate matches.

In my case [1,6] will return the same result (if doing a function) as [6,1]

listA = [1,2,3,4,5,6]
listB = listA[:]
output = []
for a in listA: 
    for b in listB:
        if b != a and b > a:
            # function here
            output.append([a,b])
print(output)

Output:
[[1, 2], [1, 3], [1, 4], [1, 5], [1, 6], [2, 3], [2, 4], [2, 5], [2, 6], [3, 4], [3, 5], [3, 6], [4, 5], [4, 6], [5, 6]]

foo =  [1, 2, 3, 4, 5, 6]

for idx, num1 in enumerate(foo[:-1:], start=1):
    for num2 in foo[idx:]:
        print(num1, num2)

print( '=' * 10)
# better use itertools.combinations
import itertools
for num1, num2 in itertools.combinations(foo, 2):
    print(num1, num2)

(Sep-11-2019, 07:35 AM)3Pinter Wrote: [ -> ]Ah look at that. Nice.

Now the 600K items in a list part.
I'm by no means a programmer and do this purely because I'm an enthousiast. So I can 'see' potential inefficient parts / runtime issues. Could you guys share your thoughts on it:

1. I'm creating a duplicate list of listA to check items.
Back to the 600K list ... that would be 2x 600K list.
To my own logic I would use first item in list comparing it with the rest of the list, then the second etc.

2. Would Yield be of any help?

(Sep-11-2019, 07:35 AM)3Pinter Wrote: [ -> ]So I can 'see' potential inefficient parts / runtime issues. Could you guys share your thoughts on it:
# Sorry Buran, I saw your post of posting my own reply. I will look at yours now

Quote:The module standardizes a core set of fast, memory efficient tools that are useful by themselves or in combination. Together, they form an “iterator algebra” making it possible to construct specialized tools succinctly and efficiently in pure Python.