# python – How to count the occurrence of certain item in an ndarray?

## The Question :

446 people think this question is useful

In Python, I have an ndarray y that is printed as array([0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 1])

I’m trying to count how many 0s and how many 1s are there in this array.

But when I type y.count(0) or y.count(1), it says

numpy.ndarray object has no attribute count

What should I do?

• Can’t you use sum and length function, since you only have aces and zeros?
• In this case, it is also possible to simply use numpy.count_nonzero.

719 people think this answer is useful
a = numpy.array([0, 3, 0, 1, 0, 1, 2, 1, 0, 0, 0, 0, 1, 3, 4])
unique, counts = numpy.unique(a, return_counts=True)
dict(zip(unique, counts))

# {0: 7, 1: 4, 2: 1, 3: 2, 4: 1}



Non-numpy way:

import collections, numpy
a = numpy.array([0, 3, 0, 1, 0, 1, 2, 1, 0, 0, 0, 0, 1, 3, 4])
collections.Counter(a)

# Counter({0: 7, 1: 4, 3: 2, 2: 1, 4: 1})



296 people think this answer is useful

What about using numpy.count_nonzero, something like

>>> import numpy as np
>>> y = np.array([1, 2, 2, 2, 2, 0, 2, 3, 3, 3, 0, 0, 2, 2, 0])

>>> np.count_nonzero(y == 1)
1
>>> np.count_nonzero(y == 2)
7
>>> np.count_nonzero(y == 3)
3



154 people think this answer is useful

Personally, I’d go for: (y == 0).sum() and (y == 1).sum()

E.g.

import numpy as np
y = np.array([0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 1])
num_zeros = (y == 0).sum()
num_ones = (y == 1).sum()



46 people think this answer is useful

For your case you could also look into numpy.bincount

In : a = np.array([0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 1])

In : np.bincount(a)
Out: array([8, 4])  #count of zeros is at index 0 : 8
#count of ones is at index 1 : 4



23 people think this answer is useful

Convert your array y to list l and then do l.count(1) and l.count(0)

>>> y = numpy.array([0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 1])
>>> l = list(y)
>>> l.count(1)
4
>>> l.count(0)
8



21 people think this answer is useful
y = np.array([0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 1])



If you know that they are just 0 and 1:

np.sum(y)



gives you the number of ones. np.sum(1-y) gives the zeroes.

For slight generality, if you want to count 0 and not zero (but possibly 2 or 3):

np.count_nonzero(y)



gives the number of nonzero.

But if you need something more complicated, I don’t think numpy will provide a nice count option. In that case, go to collections:

import collections
collections.Counter(y)
> Counter({0: 8, 1: 4})



This behaves like a dict

collections.Counter(y)
> 8



14 people think this answer is useful

If you know exactly which number you’re looking for, you can use the following;

lst = np.array([1,1,2,3,3,6,6,6,3,2,1])
(lst == 2).sum()



returns how many times 2 is occurred in your array.

9 people think this answer is useful

Honestly I find it easiest to convert to a pandas Series or DataFrame:

import pandas as pd
import numpy as np

df = pd.DataFrame({'data':np.array([0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 1])})
print df['data'].value_counts()



Or this nice one-liner suggested by Robert Muil:

pd.Series([0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 1]).value_counts()



9 people think this answer is useful

No one suggested to use numpy.bincount(input, minlength) with minlength = np.size(input), but it seems to be a good solution, and definitely the fastest:

In : choices = np.random.randint(0, 100, 10000)

In : %timeit [ np.sum(choices == k) for k in range(min(choices), max(choices)+1) ]
100 loops, best of 3: 2.67 ms per loop

In : %timeit np.unique(choices, return_counts=True)
1000 loops, best of 3: 388 µs per loop

In : %timeit np.bincount(choices, minlength=np.size(choices))
100000 loops, best of 3: 16.3 µs per loop



That’s a crazy speedup between numpy.unique(x, return_counts=True) and numpy.bincount(x, minlength=np.max(x)) !

8 people think this answer is useful

What about len(y[y==0]) and len(y[y==1]) ?

6 people think this answer is useful

I’d use np.where:

how_many_0 = len(np.where(a==0.))
how_many_1 = len(np.where(a==1.))



6 people think this answer is useful

y.tolist().count(val)

with val 0 or 1

Since a python list has a native function count, converting to list before using that function is a simple solution.

5 people think this answer is useful

Yet another simple solution might be to use numpy.count_nonzero():

import numpy as np
y = np.array([0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 1])
y_nonzero_num = np.count_nonzero(y==1)
y_zero_num = np.count_nonzero(y==0)
y_nonzero_num
4
y_zero_num
8



Don’t let the name mislead you, if you use it with the boolean just like in the example, it will do the trick.

5 people think this answer is useful

To count the number of occurrences, you can use np.unique(array, return_counts=True):

In : boo = np.array([0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 1])

# use bool value True or equivalently 1
In : uniq, cnts = np.unique(boo, return_counts=1)
In : uniq
Out: array([0, 1])   #unique elements in input array are: 0, 1

In : cnts
Out: array([8, 4])   # 0 occurs 8 times, 1 occurs 4 times



4 people think this answer is useful

take advantage of the methods offered by a Series:

>>> import pandas as pd
>>> y = [0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 1]
>>> pd.Series(y).value_counts()
0    8
1    4
dtype: int64



2 people think this answer is useful

A general and simple answer would be:

numpy.sum(MyArray==x)   # sum of a binary list of the occurence of x (=0 or 1) in MyArray



which would result into this full code as exemple

import numpy
MyArray=numpy.array([0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 1])  # array we want to search in
x=0   # the value I want to count (can be iterator, in a list, etc.)
numpy.sum(MyArray==0)   # sum of a binary list of the occurence of x in MyArray



Now if MyArray is in multiple dimensions and you want to count the occurence of a distribution of values in line (= pattern hereafter)

MyArray=numpy.array([[6, 1],[4, 5],[0, 7],[5, 1],[2, 5],[1, 2],[3, 2],[0, 2],[2, 5],[5, 1],[3, 0]])
x=numpy.array([5,1])   # the value I want to count (can be iterator, in a list, etc.)
temp = numpy.ascontiguousarray(MyArray).view(numpy.dtype((numpy.void, MyArray.dtype.itemsize * MyArray.shape)))  # convert the 2d-array into an array of analyzable patterns
xt=numpy.ascontiguousarray(x).view(numpy.dtype((numpy.void, x.dtype.itemsize * x.shape)))  # convert what you search into one analyzable pattern
numpy.sum(temp==xt)  # count of the searched pattern in the list of patterns



2 people think this answer is useful

You can use dictionary comprehension to create a neat one-liner. More about dictionary comprehension can be found here

>>>counts = {int(value): list(y).count(value) for value in set(y)}
>>>print(counts)
{0: 8, 1: 4}



This will create a dictionary with the values in your ndarray as keys, and the counts of the values as the values for the keys respectively.

This will work whenever you want to count occurences of a value in arrays of this format.

2 people think this answer is useful

You have a special array with only 1 and 0 here. So a trick is to use

np.mean(x)



which gives you the percentage of 1s in your array. Alternatively, use

np.sum(x)
np.sum(1-x)



will give you the absolute number of 1 and 0 in your array.

2 people think this answer is useful

Try this:

a = np.array([0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 1])
list(a).count(1)



2 people think this answer is useful

If you are interested in the fastest execution, you know in advance which value(s) to look for, and your array is 1D, or you are otherwise interested in the result on the flattened array (in which case the input of the function should be np.flatten(arr) rather than just arr), then Numba is your friend:

import numba as nb

@nb.jit
def count_nb(arr, value):
result = 0
for x in arr:
if x == value:
result += 1
return result



or, for very large arrays where parallelization may be beneficial:

@nb.jit(parallel=True)
def count_nbp(arr, value):
result = 0
for i in nb.prange(arr.size):
if arr[i] == value:
result += 1
return result



Benchmarking these against np.count_nonzero() (which also has a problem of creating a temporary array which may be avoided) and np.unique()-based solution

import numpy as np

def count_np(arr, value):
return np.count_nonzero(arr == value)


import numpy as np

def count_np2(arr, value):
uniques, counts = np.unique(a, return_counts=True)
counter = dict(zip(uniques, counts))
return counter[value] if value in counter else 0



for input generated with:

def gen_input(n, a=0, b=100):
return np.random.randint(a, b, n)



the following plots are obtained (the second row of plots is a zoom on the faster approach):

Showing that Numba-based solution are noticeably faster than the NumPy counterparts, and, for very large inputs, the parallel approach is faster than the naive one.

Full code available here.

1 people think this answer is useful

It involves one more step, but a more flexible solution which would also work for 2d arrays and more complicated filters is to create a boolean mask and then use .sum() on the mask.

>>>>y = np.array([0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 1])
8



1 people think this answer is useful

This can be done easily in the following method

y = np.array([0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 1])
y.tolist().count(1)



1 people think this answer is useful

Since your ndarray contains only 0 and 1, you can use sum() to get the occurrence of 1s and len()-sum() to get the occurrence of 0s.

num_of_ones = sum(array)
num_of_zeros = len(array)-sum(array)



1 people think this answer is useful
dict(zip(*numpy.unique(y, return_counts=True)))



Just copied Seppo Enarvi’s comment here which deserves to be a proper answer

0 people think this answer is useful

If you don’t want to use numpy or a collections module you can use a dictionary:

d = dict()
a = [0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 1]
for item in a:
try:
d[item]+=1
except KeyError:
d[item]=1



result:

>>>d
{0: 8, 1: 4}



Of course you can also use an if/else statement. I think the Counter function does almost the same thing but this is more transparant.

0 people think this answer is useful

For generic entries:

x = np.array([11, 2, 3, 5, 3, 2, 16, 10, 10, 3, 11, 4, 5, 16, 3, 11, 4])
n = {i:len([j for j in np.where(x==i)]) for i in set(x)}
ix = {i:[j for j in np.where(x==i)] for i in set(x)}



Will output a count:

{2: 2, 3: 4, 4: 2, 5: 2, 10: 2, 11: 3, 16: 2}



And indices:

{2: [1, 5],
3: [2, 4, 9, 14],
4: [11, 16],
5: [3, 12],
10: [7, 8],
11: [0, 10, 15],
16: [6, 13]}



0 people think this answer is useful

here I have something, through which you can count the number of occurrence of a particular number: according to your code

count_of_zero=list(y[y==0]).count(0)

print(count_of_zero)

// according to the match there will be boolean values and according to True value the number 0 will be return

0 people think this answer is useful

if you are dealing with very large arrays using generators could be an option. The nice thing here it that this approach works fine for both arrays and lists and you dont need any additional package. Additionally, you are not using that much memory.

my_array = np.array([0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 1])
sum(1 for val in my_array if val==0)
Out: 8



0 people think this answer is useful

This funktion returns the number of occurences of a variable in an array:

def count(array,variable):
number = 0
for i in range(array.shape):
for j in range(array.shape):
if array[i,j] == variable:
number += 1
return number



numpy.histogram(y, bins=y)