interleaving dictionary values

Roberto Bonvallet Roberto.Bonvallet at cern.ch
Tue Nov 21 16:18:12 CET 2006


j1o1h1n at gmail.com wrote:
[...]
> I had in mind something like this:
> 
>>>> interleave([1, 2, 3], [4,5], [7, 8, 9])
> [1, 4, 7, 2, 5, 8, 3, 9]

[...]
> But instead I (finally worked out I could) do this:
> 
>  apply(map, tuple([None] + d.values()))

apply is deprecated, it should be map(None, *d.values())

> So... my bit of code becomes:
> 
>  filter(None, flatten(map(None, apply(map, tuple([None] +
> d.values())))))
> 
> It fits on one line, but it feels far more evil than I set out to be.
> The brackets at the end are bad for my epilepsy.
> 
> Surely there is there some nice builtin function I have missed?

You have missed the "it doesn't need to be a one-liner" motto :)

>>> def interleave(*args):
...     result = []
...     for items in map(None, *args):
...         result.extend([x for x in items if x is not None])
...     return result
...
>>> interleave([1, 2, 3], [4,5], [7, 8, 9])
[1, 4, 7, 2, 5, 8, 3, 9]

Cheers,
-- 
Roberto Bonvallet



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