# How to compare each item in a list with the rest, only once?

*Solution:*

Of course this will generate each pair twice as each `for`

loop will go through every item of the list.

You could use some itertools magic here to generate all possible combinations:

```
import itertools
for a, b in itertools.combinations(mylist, 2):
compare(a, b)
```

`itertools.combinations`

will pair each element with each other element in the iterable, but only once.

You could still write this using index-based item access, equivalent to what you are used to, using nested `for`

loops:

```
for i in range(len(mylist)):
for j in range(i + 1, len(mylist)):
compare(mylist[i], mylist[j])
```

Of course this may not look as nice and pythonic but sometimes this is still the easiest and most comprehensible solution, so you should not shy away from solving problems like that.

Use `itertools.combinations(mylist, 2)`

```
mylist = range(5)
for x,y in itertools.combinations(mylist, 2):
print x,y
0 1
0 2
0 3
0 4
1 2
1 3
1 4
2 3
2 4
3 4
```

I think using enumerate on the outer loop and using the index to slice the list on the inner loop is pretty Pythonic:

```
for index, this in enumerate(mylist):
for that in mylist[index+1:]:
compare(this, that)
```