How to count the number of occurrences of `None` in a list?
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.
Answers:
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.
lst = ['hey','what',0,False,None,14]
print sum(1 for i in lst if i != None)
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)
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
Use numpy
import numpy as np
list = np.array(['hey', 'what', 0, False, None, 14])
print(sum(list != None))
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
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
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
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.
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.
lst = ['hey','what',0,False,None,14]
print sum(1 for i in lst if i != None)
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)
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
Use numpy
import numpy as np
list = np.array(['hey', 'what', 0, False, None, 14])
print(sum(list != None))
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
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
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