How do I count the number of occurrences of a character in a string?

e.g. 'a' appears in 'Mary had a little lamb' 4 times.

## Solution 1

str.count(sub[, start[, end]])

Return the number of non-overlapping occurrences of substring sub in the range [start, end]. Optional arguments start and end are interpreted as in slice notation.

>>> sentence = 'Mary had a little lamb'
>>> sentence.count('a')
4


## Solution 2

You can use .count() :

>>> 'Mary had a little lamb'.count('a')
4


## Solution 3

To get the counts of all letters, use collections.Counter:

>>> from collections import Counter
>>> counter = Counter("Mary had a little lamb")
>>> counter['a']
4


## Solution 4

Regular expressions maybe?

import re
my_string = "Mary had a little lamb"
len(re.findall("a", my_string))


## Solution 5

Python-3.x:

"aabc".count("a")


Return the number of non-overlapping occurrences of substring sub in the range [start, end]. Optional arguments start and end are interpreted as in slice notation.

## Solution 6

myString.count('a');


## Solution 7

str.count(a) is the best solution to count a single character in a string. But if you need to count more characters you would have to read the whole string as many times as characters you want to count.

A better approach for this job would be:

from collections import defaultdict

text = 'Mary had a little lamb'
chars = defaultdict(int)

for char in text:
chars[char] += 1


So you'll have a dict that returns the number of occurrences of every letter in the string and 0 if it isn't present.

>>>chars['a']
4
>>>chars['x']
0


For a case insensitive counter you could override the mutator and accessor methods by subclassing defaultdict (base class' ones are read-only):

class CICounter(defaultdict):
def __getitem__(self, k):
return super().__getitem__(k.lower())

def __setitem__(self, k, v):
super().__setitem__(k.lower(), v)

chars = CICounter(int)

for char in text:
chars[char] += 1

>>>chars['a']
4
>>>chars['M']
2
>>>chars['x']
0


## Solution 8

This easy and straight forward function might help:

def check_freq(x):
freq = {}
for c in set(x):
freq[c] = x.count(c)
return freq

check_freq("abbabcbdbabdbdbabababcbcbab")
{'a': 7, 'b': 14, 'c': 3, 'd': 3}


If a comprehension is desired:

def check_freq(x):
return {c: x.count(c) for c in set(x)}


## Solution 9

Regular expressions are very useful if you want case-insensitivity (and of course all the power of regex).

my_string = "Mary had a little lamb"
# simplest solution, using count, is case-sensitive
my_string.count("m")   # yields 1
import re
# case-sensitive with regex
len(re.findall("m", my_string))
# three ways to get case insensitivity - all yield 2
len(re.findall("(?i)m", my_string))
len(re.findall("m|M", my_string))
len(re.findall(re.compile("m",re.IGNORECASE), my_string))


Be aware that the regex version takes on the order of ten times as long to run, which will likely be an issue only if my_string is tremendously long, or the code is inside a deep loop.

## Solution 10

I don't know about 'simplest' but simple comprehension could do:

>>> my_string = "Mary had a little lamb"
>>> sum(char == 'a' for char in my_string)
4


Taking advantage of built-in sum, generator comprehension and fact that bool is subclass of integer: how may times character is equal to 'a'.

## Solution 11

a = 'have a nice day'
symbol = 'abcdefghijklmnopqrstuvwxyz'
for key in symbol:
print(key, a.count(key))


## Solution 12

I am a fan of the pandas library, in particular the value_counts() method. You could use it to count the occurrence of each character in your string:

>>> import pandas as pd
>>> phrase = "I love the pandas library and its value_counts() method"
>>> pd.Series(list(phrase)).value_counts()
8
a    5
e    4
t    4
o    3
n    3
s    3
d    3
l    3
u    2
i    2
r    2
v    2
    2
h    2
p    1
b    1
I    1
m    1
(    1
y    1
_    1
)    1
c    1
dtype: int64


## Solution 13

count is definitely the most concise and efficient way of counting the occurrence of a character in a string but I tried to come up with a solution using lambda, something like this :

sentence = 'Mary had a little lamb'
sum(map(lambda x : 1 if 'a' in x else 0, sentence))


This will result in :

4


Also, there is one more advantage to this is if the sentence is a list of sub-strings containing same characters as above, then also this gives the correct result because of the use of in. Have a look :

sentence = ['M', 'ar', 'y', 'had', 'a', 'little', 'l', 'am', 'b']
sum(map(lambda x : 1 if 'a' in x else 0, sentence))


This also results in :

4


But Of-course this will work only when checking occurrence of single character such as 'a' in this particular case.

## Solution 14

An alternative way to get all the character counts without using Counter(), count and regex

counts_dict = {}
for c in list(sentence):
if c not in counts_dict:
counts_dict[c] = 0
counts_dict[c] += 1

for key, value in counts_dict.items():
print(key, value)


## Solution 15

a = "I walked today,"
c=['d','e','f']
count=0
for i in a:
if str(i) in c:
count+=1

print(count)


## Solution 16

I know the ask is to count a particular letter. I am writing here generic code without using any method.

sentence1 =" Mary had a little lamb"
count = {}
for i in sentence1:
if i in count:
count[i.lower()] = count[i.lower()] + 1
else:
count[i.lower()] = 1
print(count)


output

{' ': 5, 'm': 2, 'a': 4, 'r': 1, 'y': 1, 'h': 1, 'd': 1, 'l': 3, 'i': 1, 't': 2, 'e': 1, 'b': 1}


Now if you want any particular letter frequency, you can print like below.

print(count['m'])
2


## Solution 17

To find the occurrence of characters in a sentence you may use the below code

Firstly, I have taken out the unique characters from the sentence and then I counted the occurrence of each character in the sentence these includes the occurrence of blank space too.

ab = set("Mary had a little lamb")

test_str = "Mary had a little lamb"

for i in ab:
counter = test_str.count(i)
if i == ' ':
i = 'Space'
print(counter, i)


Output of the above code is below.

1 : r ,
1 : h ,
1 : e ,
1 : M ,
4 : a ,
1 : b ,
1 : d ,
2 : t ,
3 : l ,
1 : i ,
4 : Space ,
1 : y ,
1 : m ,


## Solution 18

"Without using count to find you want character in string" method.

import re

def count(s, ch):

pass

def main():

s = raw_input ("Enter strings what you like, for example, 'welcome': ")

ch = raw_input ("Enter you want count characters, but best result to find one character: " )

print ( len (re.findall ( ch, s ) ) )

main()


## Solution 19

Python 3

Ther are two ways to achieve this:

1) With built-in function count()

sentence = 'Mary had a little lamb'
print(sentence.count('a'))


2) Without using a function

sentence = 'Mary had a little lamb'
count = 0

for i in sentence:
if i == "a":
count = count + 1

print(count)


## Solution 20

the easiest way is to code in one line:

'Mary had a little lamb'.count("a")


but if you want can use this too:

sentence ='Mary had a little lamb'
count=0;
for letter in sentence :
if letter=="a":
count+=1
print (count)


## Solution 21

str = "count a character occurence"

List = list(str)
print (List)
Uniq = set(List)
print (Uniq)

for key in Uniq:
print (key, str.count(key))


## Solution 22

Taking up a comment of this user:

import numpy as np
sample = 'samplestring'
np.unique(list(sample), return_counts=True)


Out:

(array(['a', 'e', 'g', 'i', 'l', 'm', 'n', 'p', 'r', 's', 't'], dtype='<U1'),
array([1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1]))


Check 's'. You can filter this tuple of two arrays as follows:

a[a=='s']


Side-note: It works like Counter() of the collections package, just in numpy, which you often import anyway. You could as well count the unique words in a list of words instead.

## Solution 23

No more than this IMHO - you can add the upper or lower methods

def count_letter_in_str(string,letter):
return string.count(letter)


## Solution 24

You can use loop and dictionary.

def count_letter(text):
result = {}
for letter in text:
if letter not in result:
result[letter] = 0
result[letter] += 1
return result


## Solution 25

spam = 'have a nice day'
var = 'd'

def count(spam, var):
found = 0
for key in spam:
if key == var:
found += 1
return found
count(spam, var)
print 'count %s is: %s ' %(var, count(spam, var))