Filter a Python list by predicate

Question

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          


        

Answer 1

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]

Answer 2

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]