Python sort with key=lambda gets TypeError when trying to call x.lower()

Question

I am trying to sort a list of objects in Python 3.4 based on the value of the data attribute of each object. If I use:

db[count].sort(key=lambda a: a.data)

everything works fine. However, I want the sort to be case insensitive so I use

db[count].sort(key=lambda a: a.data.lower)

but then I get:

db[count].sort(key=lambda a: a.data.lower)

TypeError: unorderable types: builtin_function_or_method() < builtin_function_or_method()

Any ideas?

Sorry sussed it. I needed db[count].sort(key=lambda a: a.data.lower()) — user3595536
– user3595536, Commented May 2, 2014 at 8:50

Grapsus · Accepted Answer · 2014-05-02 08:50:51Z

4

key has to be a callable that returns a value to be sorted. In your case it returns another callable a.data.lower. You need to call lower in order to get the value, so the correct form is:

db[count].sort(key=lambda a: a.data.lower())

answered May 2, 2014 at 8:50

Grapsus

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tpatja · Accepted Answer · 2014-05-02 08:49:35Z

4

You are passing a reference to the lower method instead of calling it.

Try this:

db[count].sort(key=lambda a: a.data.lower())

answered May 2, 2014 at 8:49

tpatja

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