On Tuesday, March 11, 2014 9:25:29 AM UTC-7, John Gordon wrote:
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>
>
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> Why do you say that 'key=lambda x: x.name.lower' is the correct form? That
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> returns the str.lower() function object, which is a silly thing to sort
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> on. Surely you want to sort on the *result* of that function, w
A comprehensive and educational answer, Peter. Thank you.
Josh
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Josh English wrote:
> I am running into a strange behavior using the sorted function in Python
> print list(sorted(all_the_stuff, key=lambda x: x.name.lower))
> print list(sorted(all_the_stuff, key=lambda x: x.name.lower()))
Let's simplify your example some more:
>>> items = ["B", "2", "a"]
>>>
In <058a4a9e-7893-44ef-97c0-999a3589e...@googlegroups.com> Josh English
writes:
> print list(sorted(all_the_stuff, key=lambda x: x.name.lower))
> print list(sorted(all_the_stuff, key=lambda x: x.name.lower()))
> # END
> The output is:
> [Thing d, Thing f, Thing 2, Thing a, Thing b, Thing C]
>
On 03/11/2014 09:13 AM, Josh English wrote:
I am running into a strange behavior using the sorted function in Python 2.7.
The key parameter is not behaving as the docs say it does:
Here is a snippet of code, simplified from my full program:
#begin code
class Thing(object):
def __init__(se
On Tue, Mar 11, 2014 at 9:13 AM, Josh English wrote:
> print list(sorted(all_the_stuff, key=lambda x: x.name.lower))
>
In this case, the key being sorted on is the function object x.name.lower,
not the result of the call.
It might make more sense if you break the lambda out into a separate def
s
I am running into a strange behavior using the sorted function in Python 2.7.
The key parameter is not behaving as the docs say it does:
Here is a snippet of code, simplified from my full program:
#begin code
class Thing(object):
def __init__(self, name):
self.name = name
def __
