Lazily transpose a list in Python

Question

So, I have an iterable of 3-tuples, generated lazily. I'm trying to figure out how to turn this into 3 iterables, consisting of the first, second, and third elements of the tuples, respectively. However, I wish this to be done lazily. So, for example, I wish `[(1, 2, 3), (4, 5, 6), (7, 8, 9)]` to be turned into `[1, 4, 7]`, `[2, 5, 8]`, `[3, 6, 9]`. (Except I want iterables not lists.) The standard `zip(*data)` idiom doesn't work, because the argument unpacking expands the entire iterable. (You can verify this by noting that `zip(*((x, x+1, x+2) for x in itertools.count(step=3)))` hangs.) The best I've come up with thus far is the following: ``` def transpose(iterable_of_three_tuples): teed = itertools.tee(iterable_of_three_tuples, 3) return map(lambda e: e[0], teed[0]), map(lambda e: e[1], teed[1]), map(lambda e: e[2], teed[2]) ``` This seems to work. But it hardly seems like clean code. And it does a lot of what seems to be unnecessary work.

Accepted Answer

Your `transpose` is pretty much exactly what you need.

With any solution you'd choose, you'd have to buffer the unused values (e.g. to get to the 7, you have to read 1-6, and store them in memory for when the other iterables ask for them). `tee` already does exactly that kind of buffering, so there's no need implementing it yourself.

The only other (minor) thing is that I'd write it slightly differently, avoiding the `map` and `lambda`s:

```
def transpose(iterable_of_three_tuples):
    teed = itertools.tee(iterable_of_three_tuples, 3)
    return ( e[0] for e in teed[0] ),  ( e[1] for e in teed[1] ),  ( e[2] for e in teed[2] )
```

Lazily transpose a list in Python

Solution

Problem