On Thursday, March 2, 2017 at 2:57:02 PM UTC-5, Jussi Piitulainen wrote:
> Peter Otten <__pete...@web.de> writes:
>
> > Andrew Zyman wrote:
> >
> >> On Thursday, March 2, 2017 at 11:27:34 AM UTC-5, Peter Otten wrote:
> >>> Andrew Zyman wrote:
> >>> .....
> >>> .....
> >>> > End result:
> >>> > ll =[ [a,1], [b,2], [c,3], [blah, 1000, 'new value'] ]
> >>>
> >>> >>> outer = [["a", 1], ["b", 2], ["c", 3], ["blah", 1000]]
> >>> >>> for inner in outer:
> >>> ... if inner[0] == "blah":
> >>> ... inner.append("new value")
> >>
> >> thank you. this will do.
> >> Just curious, is the above loop can be done in a one-liner?
> >
> > Ah, that newbie obsession ;)
> >
> >>>> outer = [["a", 1], ["b", 2], ["c", 3], ["blah", 1000]]
> >>>> [inner + ["new value"] if inner[0] == "blah" else inner for inner in
> > outer]
> > [['a', 1], ['b', 2], ['c', 3], ['blah', 1000, 'new value']]
> >
> > Note that there is a technical difference to be aware of -- matching
> > lists are replaced rather than modified.
>
> I take it you are too sane, or too kind, to suggest the obvious
> solution:
>
> >>> outer = [["a", 1], ["b", 2], ["c", 3], ["blah", 1000]]
> >>> [inner.append("new value") for inner in outer if inner[0] == "blah"]
> [None]
> >>> outer
> [['a', 1], ['b', 2], ['c', 3], ['blah', 1000, 'new value']]
>
> [snip]
Arh!!! this is it :)
I'm sure i'll regret this line of code in 2 weeks - after i successfully forget
what i wanted to achieve :)
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