How to find the length of a "filter" object in python

>>> n = [1,2,3,4]


>>> filter(lambda x:x>3,n)
<filter object at 0x0000000002FDBBA8>


>>> len(filter(lambda x:x>3,n))
Traceback (most recent call last):
File "<pyshell#3>", line 1, in <module>
len(filter(lambda x:x>3,n))
TypeError: object of type 'filter' has no len()

I could not get the length of the list I got. So I tried saving it to a variable, like this...

>>> l = filter(lambda x:x>3,n)
>>> len(l)
Traceback (most recent call last):
File "<pyshell#5>", line 1, in <module>
len(l)
TypeError: object of type 'filter' has no len()

Instead of using a loop, is there any way to get the length of this?

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最佳答案

You have to iterate through the filter object somehow. One way is to convert it to a list:

l = list(filter(lambda x: x > 3, n))


len(l)  # <--

But that might defeat the point of using filter() in the first place, since you could do this more easily with a list comprehension:

l = [x for x in n if x > 3]

Again, len(l) will return the length.

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The docs for python 3 say it returns an iterator

"Construct an iterator from those elements of iterable for which function returns true."

In python 2 it returned a list: see here. You will need to iterate the filter object to find its length.

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This is an old question, but I think this question needs an answer using the map-reduce ideology. So here:

from functools import reduce


def ilen(iterable):
return reduce(lambda sum, element: sum + 1, iterable, 0)


ilen(filter(lambda x: x > 3, n))

This is especially good if n doesn't fit in the computer memory.

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Generally, filter and reduce are not pythonic.

@arshajii metioned this solution:

len([x for x in n if x > 3])

This is quite simple, but is not describing what you exactly want to do, and it makes a list that may use some additional memory. A better solution is using sum with generator:

sum(1 for x in n if x > 3)

(See more about generator here: https://www.python.org/dev/peps/pep-0289/#rationale)

However, sum with generator is actually slower in most cases because of the implementation (tested in CPython 3.6.4):

In [1]: %timeit len([1 for x in range(10000000)])
356 ms ± 17.4 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)


In [2]: %timeit sum(1 for x in range(10000000))
676 ms ± 7.05 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
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Converting a filter to a list will take extra memory, which may not be acceptable for large amounts of data. You can find length of the filter object without converting it to a list:

sum(1 for _ in filter(lambda x: x > 3, n))
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Can be solved using more-itertools:

from more_itertools import ilen
f = filter(lambda x:x>3,n)
return ilen(f)

Note that ilen consumes the filter (you won't be able to use it afterwards - it will return an empty list)


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