using method as function in python

Question

Say I have a list of strings, and I want to filter out all non-upper case strings. Is there a simpler way than doing filter(lambda x: x.isupper(), list)?

• CromTheDestroyer: @The Communist Duck: why is boilerplate bad? It's not a big deal, but it is annoying.
• Eric Wilson: You could define a function to do this, but it will still take one line to invoke it.
• S.Lott: @CromTheDestroyer: What part is "boilerplate"? Can you be more specific on what you object to?
• CromTheDestroyer: @S.Lott: the lambda is. In my head I was contrasting it to the case where a function is passed in. Cosmologicon's answer is what I was looking for.
• S.Lott: @CromTheDestroyer: The lambda is as much boilerplate as the word filter. Only in the most trivial of cases could is be considered "boilerplate". lambda x: x.isupper() and x not in ('Z','z') would demonstrate that the lambda is essential syntax.

Answer 1

While I would prefer a list comprehension, this seems to be what you're looking for:

filter(str.isupper, list)

• Jess: If the function already works as-is, why bother with the comprehension?
• Cosmologicon: You're right, this is a perfectly fine use of filter. I hardly ever use it is all. I just like comprehensions.
• senderle: Clearly you're using list as a metasyntactic variable here, but if you weren't, I would have to point out that using list this way masks a built-in ;).

Answer 2

uppers = [s for s in list if s.isupper()]

• Steven Rumbalski: I think this would be easier to read with s as the variable name instead of i.
• CromTheDestroyer: I'm sorry, I should have clarified that the question was about passing methods into functions as functions, not solving this trivial problem.
• manji: ok, no problem. (if you re-read your question you will see that it's nearly impossible to understand that).