Count the number of occurrences of a character in a string

Question

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

eg 'a' appears in 'Mary had a little lamb' 4 times.

Answer 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

Answer 2

You can use count() :

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

Answer 3

As other answers said, using the string method count() is probably the simplest, but if you're doing this frequently, check out collections.Counter :

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

Answer 4

Regular expressions maybe?

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

Answer 5

myString.count('a');

更多信息在这里

Answer 6

Python-3.x:

"aabc".count("a")

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.

Answer 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

Answer 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)}

Answer 9

如果您想要不区分大小写（当然还有正则表达式的所有功能），正则表达式非常有用。

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))

Answer 10

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

Answer 11

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

Answer 12

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'.

Answer 13

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)

Answer 14

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

print(count)

Answer 15

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

Answer 16

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 ,

Answer 17

str = "count a character occurence"

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

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

Answer 18

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.

Answer 19

"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()

Answer 20

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)

Answer 21

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)

Answer 22

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

Answer 23

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[1][a[0]=='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.

Answer 24

Use count:

sentence = 'A man walked up to a door'
print(sentence.count('a'))
# 4

Answer 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))

Answer 26

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

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

Answer 27

Using Count:

string = "count the number of counts in string to count from."
x = string.count("count")

x = 3.

Answer 28

This will give you the occurrence of each characters in a string. O/P is also in string format:

def count_char(string1):
string2=""
lst=[]
lst1=[]
for i in string1:
    count=0
    if i not in lst:
        for j in string1:
            if i==j:
                count+=1
        lst1.append(i)
        lst1.append(count)
    lst.append(i)

string2=''.join(str(x) for x in lst1)
return string2 

print count_char("aabbacddaabbdsrchhdsdg")