What is a good way to do countif in Python
Question:
I want to count how many members of an iterable meet a given condition. I’d like to do it in a way that is clear and simple and preferably reasonably optimal.
My current best ideas are:
sum(meets_condition(x) for x in my_list)
and
len([x for x in my_list if meets_condition(x)])
The first one being iterator based is presumably faster for big lists. And it’s the same form as you’d use for testing any and all. However it depends on the fact that int(True) == 1, which is somewhat ugly.
The second one seems easier to read to me, but it is different from the any and all forms.
Does anyone have any better suggestions? is there a library function somewhere that I am missing?
Answers:
The first one
sum(meets_condition(x) for x in my_list)
looks perfectly readable and pythonic to me.
If you prefer the second approach I’d go for
len(filter(meets_condition, my_list))
Yet another way could be:
map(meets_condition, my_list).count(True)
The iterator based approach is just fine. There are some slight modifications that can emphasize the fact that you are counting:
sum(1 if meets_condition(x) else 0 for x in my_list)
# or
sum(1 for x in my_list if meets_condition(x))
And as always, if the intent isn’t apparent from the code, encapsulate it in descriptively named function:
def count_matching(condition, seq):
"""Returns the amount of items in seq that return true from condition"""
return sum(1 for item in seq if condition(item))
count_matching(meets_condition, my_list)
countif for a list
#counting if a number or string is in a list
my_list=[1,2,3,2,3,1,1,1,1,1, "dave" , "dave"]
one=sum(1 for item in my_list if item==(1))
two=sum(1 for item in my_list if item==(2))
three=sum(1 for item in my_list if item==(3))
dave=sum(1 for item in my_list if item==("dave"))
print("number of one's in my_list > " , one)
print("number of two's in my_list > " , two)
print("number of three's in my_list > " , three)
print("number of dave's in my_list > " , dave)
I know the question above is very well taken care of already, but if you are new in the python world and happen to be here because you searched for the simple keyword “Countif python” (and google suggested this page as the most relevant page on its list), the following method is very simple to code and run.
You can simply use a boolean value and as Gordon said, it depends on the fact that True==1;
First, create a boolean series using the column’s condition and then add up the values.
series = boolian condition
sum(series)
Example:
We’d like to know how many students got (GPA = A) and how many received a mark over 80 in Math.
df = pd.DataFrame({'ID': [1,2,3,4,5],
'Math':[87,83,95,75,60],
'GPA': ['A','B','A','B','C']})
Answer:
GPA = df.GPA == 'A'
sum(GPA)
Out: 2
Math_Mark = df.Math > 80
sum(Math_Mark)
Out: 3
P.s: I added this answer because google lists this page as the most relevant page when you search for “countif python”. No need to give negative, if most of you think this answer shouldn’t be here, I’ll remove it.
I want to count how many members of an iterable meet a given condition. I’d like to do it in a way that is clear and simple and preferably reasonably optimal.
My current best ideas are:
sum(meets_condition(x) for x in my_list)
and
len([x for x in my_list if meets_condition(x)])
The first one being iterator based is presumably faster for big lists. And it’s the same form as you’d use for testing any and all. However it depends on the fact that int(True) == 1, which is somewhat ugly.
The second one seems easier to read to me, but it is different from the any and all forms.
Does anyone have any better suggestions? is there a library function somewhere that I am missing?
The first one
sum(meets_condition(x) for x in my_list)
looks perfectly readable and pythonic to me.
If you prefer the second approach I’d go for
len(filter(meets_condition, my_list))
Yet another way could be:
map(meets_condition, my_list).count(True)
The iterator based approach is just fine. There are some slight modifications that can emphasize the fact that you are counting:
sum(1 if meets_condition(x) else 0 for x in my_list)
# or
sum(1 for x in my_list if meets_condition(x))
And as always, if the intent isn’t apparent from the code, encapsulate it in descriptively named function:
def count_matching(condition, seq):
"""Returns the amount of items in seq that return true from condition"""
return sum(1 for item in seq if condition(item))
count_matching(meets_condition, my_list)
countif for a list
#counting if a number or string is in a list
my_list=[1,2,3,2,3,1,1,1,1,1, "dave" , "dave"]
one=sum(1 for item in my_list if item==(1))
two=sum(1 for item in my_list if item==(2))
three=sum(1 for item in my_list if item==(3))
dave=sum(1 for item in my_list if item==("dave"))
print("number of one's in my_list > " , one)
print("number of two's in my_list > " , two)
print("number of three's in my_list > " , three)
print("number of dave's in my_list > " , dave)
I know the question above is very well taken care of already, but if you are new in the python world and happen to be here because you searched for the simple keyword “Countif python” (and google suggested this page as the most relevant page on its list), the following method is very simple to code and run.
You can simply use a boolean value and as Gordon said, it depends on the fact that True==1;
First, create a boolean series using the column’s condition and then add up the values.
series = boolian condition
sum(series)
Example:
We’d like to know how many students got (GPA = A) and how many received a mark over 80 in Math.
df = pd.DataFrame({'ID': [1,2,3,4,5],
'Math':[87,83,95,75,60],
'GPA': ['A','B','A','B','C']})
Answer:
GPA = df.GPA == 'A'
sum(GPA)
Out: 2
Math_Mark = df.Math > 80
sum(Math_Mark)
Out: 3
P.s: I added this answer because google lists this page as the most relevant page when you search for “countif python”. No need to give negative, if most of you think this answer shouldn’t be here, I’ll remove it.