I would want to do something like:
>>> lst = [1, 2, 3, 4, 5]
>>> lst.find(lambda x: x % 2 == 0)
2
>>> lst.findall(lambda x: x % 2 == 0)
[2, 4]
Is there anything nearing such behavior in Python's standard libraries?
I know it's very easy to roll-your-own here, but I'm looking for a more standard way.
You can use the filter method:
>>> lst = [1, 2, 3, 4, 5]
>>> filter(lambda x: x % 2 == 0, lst)
[2, 4]
or a list comprehension:
>>> lst = [1, 2, 3, 4, 5]
>>> [x for x in lst if x %2 == 0]
[2, 4]
to find a single element, you could try:
>>> next(x for x in lst if x % 2 == 0)
2
Though that would throw an exception if nothing matches, so you'd probably want to wrap it in a try/catch. The () brackets make this a generator expression rather than a list comprehension.
Personally though I'd just use the regular filter/comprehension and take the first element (if there is one).
These raise an exception if nothing is found
filter(lambda x: x % 2 == 0, lst)[0]
[x for x in lst if x %2 == 0][0]
These return empty lists
filter(lambda x: x % 2 == 0, lst)[:1]
[x for x in lst if x %2 == 0][:1]
Generators and list comprehensions are more pythonic than chainable functions.
>>> lst = [i for i in range(1, 6)]
>>> lst
[1, 2, 3, 4, 5]
>>> gen = (x for x in lst if x % 10 == 0)
>>> next(gen, 'not_found')
'not_found'
>>> [x for x in gen]
[]
For example, I use it like this sometimes:
>>> n = next((x for x in lst if x % 10 == 0), None)
>>> if n is None:
... print('Not found')
...
Not found
Otherwise, you can define your utility function oneliners like this:
>>> find = lambda fun, lst: next((x for x in lst if fun(x)), None)
>>> find(lambda x: x % 10 == 0, lst)
>>> find(lambda x: x % 5 == 0, lst)
5
>>> findall = lambda fun, lst: [x for x in lst if fun(x)]
>>> findall(lambda x: x % 5 == 0, lst)
[5]
