Python count if greater than

I have a list of numbers and I want to get the number of times a number appears in a list that meets a certain criteria. I can use a list comprehension (or a list comprehension in a function) but I am wondering if someone has a shorter way.

# list of numbers
j=[4,5,6,7,1,3,7,5]
#list comprehension of values of j > 5
x = [i for i in j if i>5]
#value of x
len(x)

#or function version
def length_of_list(list_of_numbers, number):
     x = [i for i in list_of_numbers if j > number]
     return len(x)
length_of_list(j, 5)

is there an even more condensed version?

asked May 10, 2012 at 23:01

You could do something like this:

>>> j = [4, 5, 6, 7, 1, 3, 7, 5]
>>> sum(i > 5 for i in j)
3

It might initially seem strange to add True to True this way, but I don't think it's unpythonic; after all, bool is a subclass of int in all versions since 2.3:

>>> issubclass(bool, int)
True

answered May 10, 2012 at 23:03

senderlesenderle

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5

You can create a smaller intermediate result like this:

>>> j = [4, 5, 6, 7, 1, 3, 7, 5]
>>> len([1 for i in j if i > 5])
3

answered May 10, 2012 at 23:08

Greg HewgillGreg Hewgill

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1

if you are otherwise using numpy, you can save a few strokes, but i dont think it gets much faster/compact than senderle's answer.

import numpy as np
j = np.array(j)
sum(j > i)

answered May 10, 2012 at 23:08

ludaavicsludaavics

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A (somewhat) different way:

reduce(lambda acc, x: acc + (1 if x > 5 else 0), j, 0)

answered May 10, 2012 at 23:15

If you are using NumPy (as in ludaavic's answer), for large arrays you'll probably want to use NumPy's sum function rather than Python's builtin sum for a significant speedup -- e.g., a >1000x speedup for 10 million element arrays on my laptop:

>>> import numpy as np
>>> ten_million = 10 * 1000 * 1000
>>> x, y = (np.random.randn(ten_million) for _ in range(2))
>>> %timeit sum(x > y)  # time Python builtin sum function
1 loops, best of 3: 24.3 s per loop
>>> %timeit (x > y).sum()  # wow, that was really slow! time NumPy sum method
10 loops, best of 3: 18.7 ms per loop
>>> %timeit np.sum(x > y)  # time NumPy sum function
10 loops, best of 3: 18.8 ms per loop

(above uses IPython's %timeit "magic" for timing)

answered Dec 7, 2015 at 11:18

Different way of counting by using bisect module:

>>> from bisect import bisect
>>> j = [4, 5, 6, 7, 1, 3, 7, 5]
>>> j.sort()
>>> b = 5
>>> index = bisect(j,b) #Find that index value
>>> print len(j)-index
3

answered Nov 8, 2016 at 8:08

ShashankShashank

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I'll add a map and filter version because why not.

sum(map(lambda x:x>5, j))
sum(1 for _ in filter(lambda x:x>5, j))

answered Sep 10, 2019 at 16:42

You can do like this using function:

l = [34,56,78,2,3,5,6,8,45,6]  
print ("The list : " + str(l))   
def count_greater30(l):  
    count = 0  
    for i in l:  
        if i > 30:  
            count = count + 1.  
    return count
print("Count greater than 30 is : " + str(count)).  
count_greater30(l)

Joseph Budin

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answered Oct 1, 2020 at 17:06

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