如何计算列表中“None”的出现次数？

Question

I'm trying to count things that are not None , but I want False and numeric zeros to be accepted too. Reversed logic: I want to count everything except what it's been explicitly declared as None .

Example

Just the 5th element it's not included in the count:

>>> list = ['hey', 'what', 0, False, None, 14]
>>> print(magic_count(list))
5

I know this isn't Python normal behavior, but how can I override Python's behavior?

What I've tried

So far I founded people suggesting that a if a is not None else "too bad" , but it does not work.

I've also tried isinstance , but with no luck.

Answer 1

Just use sum checking if each object is not None which will be True or False so 1 or 0.

lst = ['hey','what',0,False,None,14]
print(sum(x is not None for x in lst))

Or using filter with python2:

print(len(filter(lambda x: x is not None, lst))) # py3 -> tuple(filter(lambda x: x is not None, lst))

With python3 there is None.__ne__() which will only ignore None's and filter without the need for a lambda.

sum(1 for _ in filter(None.__ne__, lst))

The advantage of sum is it lazily evaluates an element at a time instead of creating a full list of values.

On a side note avoid using list as a variable name as it shadows the python list .

Answer 2

Two ways:

One, with a list expression

len([x for x in lst if x is not None])

Two, count the Nones and subtract them from the length:

len(lst) - lst.count(None)

Answer 3

lst = ['hey','what',0,False,None,14]
print sum(1 for i in lst if i != None)

Answer 4

You could use Counter from collections .

from collections import Counter

my_list = ['foo', 'bar', 'foo', None, None]

resulted_counter = Counter(my_list) # {'foo': 2, 'bar': 1, None: 2}

resulted_counter[None] # 2

Answer 5

I recently released a library containing a function iteration_utilities.count_items (ok, actually 3 because I also use the helpers is_None and is_not_None ) for that purpose:

>>> from iteration_utilities import count_items, is_not_None, is_None
>>> lst = ['hey', 'what', 0, False, None, 14]
>>> count_items(lst, pred=is_not_None)  # number of items that are not None
5

>>> count_items(lst, pred=is_None)      # number of items that are None
1

Answer 6

Use numpy

import numpy as np

list = np.array(['hey', 'what', 0, False, None, 14])
print(sum(list != None))

Answer 7

I needed to make sure that only one param was send on each call. Therefore at least 2 of the variables must be None and this worked for me.

a_list = [param_1, param_2, param_3] 
count = a_list.count(None) 
if count < 2:
    raise error

Answer 8

im pretty sure that using the length of the list minus the number of Nones here sould be the best option for performance and simplicity

>>> list = ['hey', 'what', 0, False, None, 14]
>>> print(len(list) - list.count(None))
5