How do I count the occurrence of a certain item in an ndarray?

Question

How do I count the number of 0s and 1s in the following array?

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

---

y.count(0) gives:

numpy.ndarray object has no attribute count

Answer 1

The simplest,do comment if not necessary

import numpy as np
y = np.array([0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 1])
count_0, count_1 = 0, 0
for i in y_train:
    if i == 0:
        count_0 += 1
    if i == 1:
        count_1 += 1
count_0, count_1

Answer 2

For your case you could also look into numpy.bincount

In [56]: a = np.array([0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 1])

In [57]: np.bincount(a)
Out[57]: array([8, 4])  #count of zeros is at index 0, i.e. 8
                        #count of ones is at index 1, i.e. 4

Answer 3

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.ravel(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

These can be benchmarked against np.count_nonzero() (which also has a problem of creating a temporary array -- something that is avoided in the Numba solutions) and a np.unique()-based solution (which is actually counting all unique value values contrarily to the other solutions).

import numpy as np


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

import numpy as np


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

---

Since the support for "typed" dicts in Numba, it is also possible to have a function counting all occurrences of all elements. This competes more directly with np.unique() because it is capable of counting all values in a single run. Here is proposed a version which eventually only returns the number of elements for a single value (for comparison purposes, similarly to what is done in count_np_uniq()):

@nb.jit
def count_nb_dict(arr, value):
    counter = {arr[0]: 1}
    for x in arr:
        if x not in counter:
            counter[x] = 1
        else:
            counter[x] += 1
    return counter[value] if value in counter else 0

---

The input is generated with:

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

The timings are reported in the following plots (the second row of plots is a zoom on the faster approaches):

Showing that the simple Numba-based solution is fastest for smaller inputs and the parallelized version is fastest for larger inputs. They NumPy version is reasonably fast at all scales.

When one wants to count all values in an array, np.unique() is more performant than a solution implemented manually with Numba for sufficiently large arrays.

EDIT: It seems that the NumPy solution has become faster in recent versions. In a previous iteration, the simple Numba solution was outperforming NumPy's approach for any input size.

---

Full code available here.

Answer 4

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.

Answer 5

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.

Answer 6

Using numpy.unique:

import numpy
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 method using collections.Counter;

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

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

Answer 7

Filter and use len

Using len could be another option.

A = np.array([1,0,1,0,1,0,1])

Say we want the number of occurrences of 0.

A[A==0]  # Return the array where item is 0, array([0, 0, 0])

Now, wrap it around with len.

len(A[A==0])  # 3
len(A[A==1])  # 4
len(A[A==7])  # 0, because there isn't such item.

Answer 8

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

In [75]: boo = np.array([0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 1])
 
# use bool value `True` or equivalently `1`
In [77]: uniq, cnts = np.unique(boo, return_counts=1)
In [81]: uniq
Out[81]: array([0, 1])   #unique elements in input array are: 0, 1

In [82]: cnts
Out[82]: array([8, 4])   # 0 occurs 8 times, 1 occurs 4 times

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

Try this:

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

Answer 13

dict(zip(*numpy.unique(y, return_counts=True)))

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

Answer 14

using numpy.count

$ a = [0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 1]

$ np.count(a, 1)

Answer 15

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.

Answer 16

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)[0]]) for i in set(x)}
ix = {i:[j for j in np.where(x==i)[0]] 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]}

Answer 17

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 [1]: choices = np.random.randint(0, 100, 10000)

In [2]: %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 [3]: %timeit np.unique(choices, return_counts=True)
1000 loops, best of 3: 388 µs per loop

In [4]: %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)) !

Answer 18

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.

Answer 19

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

Answer 20

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()

Answer 21

Numpy has a module for this. Just a small hack. Put your input array as bins.

numpy.histogram(y, bins=y)

The output are 2 arrays. One with the values itself, other with the corresponding frequencies.

Answer 22

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)

Answer 23

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[1])))  # convert the 2d-array into an array of analyzable patterns
xt=numpy.ascontiguousarray(x).view(numpy.dtype((numpy.void, x.dtype.itemsize * x.shape[0])))  # convert what you search into one analyzable pattern
numpy.sum(temp==xt)  # count of the searched pattern in the list of patterns

Answer 24

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)

Answer 25

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.

Answer 26

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.

Answer 27

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.

Answer 28

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()

Answer 29

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])
>>>>mask = y == 0
>>>>mask.sum()
8

Answer 30

I'd use np.where:

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