Python 如何计算列表中“无”的出现次数？
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How to count the number of occurrences of `None` in a list?
提问by Hrabal
I'm trying to count things that are not None, but I want Falseand numeric zeros to be accepted too. Reversed logic: I want to count everything except what it's been explicitly declared as None.
我正在尝试计算不是的东西None，但我也希望False数字零也被接受。反向逻辑：我想计算除明确声明为None.
Example
例子
Just the 5th element it's not included in the count:
只有第 5 个元素不包含在计数中：
>>> 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?
我知道这不是 Python 的正常行为，但是如何覆盖 Python 的行为？
What I've tried
我试过的
So far I founded people suggesting that a if a is not None else "too bad", but it does not work.
到目前为止，我发现有人建议这样做a if a is not None else "too bad"，但它不起作用。
I've also tried isinstance, but with no luck.
我也试过isinstance，但没有运气。
采纳答案by Padraic Cunningham
Just use sumchecking if each object is not Nonewhich will be Trueor Falseso 1 or 0.
只是使用sum检查是否每个对象is not None，这将是True或False所以1或0。
lst = ['hey','what',0,False,None,14]
print(sum(x is not None for x in lst))
Or using filterwith python2:
或filter与 python2 一起使用：
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.
使用 python3， None.__ne__()它只会忽略 None 和过滤器，而无需 lambda。
sum(1 for _ in filter(None.__ne__, lst))
The advantage of sumis it lazily evaluates an element at a time instead of creating a full list of values.
的优点sum是它一次懒惰地评估一个元素，而不是创建一个完整的值列表。
On a side note avoid using listas a variable name as it shadows the python list.
在旁注中避免list用作变量名，因为它会影响 python list。
回答by kvorobiev
lst = ['hey','what',0,False,None,14]
print sum(1 for i in lst if i != None)
回答by RemcoGerlich
Two ways:
两种方式：
One, with a list expression
一、用列表表达式
len([x for x in lst if x is not None])
Two, count the Nones and subtract them from the length:
二、数数 Nones 并从长度中减去它们：
len(lst) - lst.count(None)
回答by MSeifert
I recently released a library containing a function iteration_utilities.count_items(ok, actually 3 because I also use the helpers is_Noneand is_not_None) for that purpose:
我最近发布了一个包含一个函数的库iteration_utilities.count_items（好吧，实际上是 3，因为我也使用了 helpersis_None和is_not_None）用于这个目的：
>>> 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
回答by Rémi Baudoux
Use numpy
使用 numpy
import numpy as np
list = np.array(['hey', 'what', 0, False, None, 14])
print(sum(list != None))
回答by Vikto
You could use Counterfrom collections.
您可以使用Counterfrom 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
