unorderable types: list() > int()

[Peter Otten]

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.