unorderable types: list() > int()
Peter Otten
__peter__ at web.de
Fri Oct 13 03:31:12 EDT 2017
Andrew Z wrote:
> Hello,
> pos = {"CLown":10,"BArbie":20}
> I want to return integer (10) for the keyword that starts with "CL"
>
>
> cl_ = [v for k, v in pos.items() if k.startswith('CL')]
> cl_pos = cl_[0]
> if cl_pos > 0:
>
> blah..
>
>
> There are 2 issues with the above:
> a. ugly - cl_pos = cl_ [0] . I was thinking something like
There's another issue: what do you want to do if there is more than one
match or no match at all?
If you are OK with a ValueError in both cases you can rewrite
> cl_ = [v for k, v in pos.items() if k.startswith('CL')][0]
[match] = (v for k, v in pos.items() if k.startswith("CL"))
If the values are orderable you can standardise on the minimum or maximum
match = min(v for k, v in pos.items() if k.startswith("CL"))
or if you always want an arbitrary match you can pick the first one
explicitly:
match = next(v for k, v in pos.items() if k.startswith("CL"))
This will raise a StopIteration if there are no matches.
> but that is too much for the eyes - in 2 weeks i'll rewrite this into
> something i can understand without banging against the desk.
> So what can be a better approach here ?
>
>
> b. in "cl_pos > 0:, cl_pos apparently is still a list. Though the run in a
> python console has no issues and report cl_pos as int.
>
>
> Appreciate.
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