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.
Mateen Ulhaq
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asked Jul 20, 2009 at 20:00
2
str.count[sub[, start[, end]]]
Return the number of non-overlapping occurrences of substring
sub
in the range[start, end]
. Optional argumentsstart
andend
are interpreted as in slice notation.
>>> sentence = 'Mary had a little lamb'
>>> sentence.count['a']
4
Mateen Ulhaq
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answered Jul 20, 2009 at 20:04
Ogre CodesOgre Codes
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2
You can use .count[]
:
>>> 'Mary had a little lamb'.count['a']
4
Mateen Ulhaq
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answered Jul 20, 2009 at 20:04
eduffyeduffy
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To get the counts of all letters, use collections.Counter
:
>>> from collections import Counter
>>> counter = Counter["Mary had a little lamb"]
>>> counter['a']
4
Mateen Ulhaq
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answered Aug 1, 2012 at 22:12
Brenden BrownBrenden Brown
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4
Regular expressions maybe?
import re
my_string = "Mary had a little lamb"
len[re.findall["a", my_string]]
answered Jul 20, 2009 at 20:03
Sinan TaifourSinan Taifour
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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.
tremendows
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answered Jul 20, 2009 at 20:05
Aaron FiAaron Fi
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myString.count['a'];
more info here
answered Jul 20, 2009 at 20:04
Finer ReclinerFiner Recliner
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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
answered Oct 2, 2016 at 2:33
Nuno AndréNuno André
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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]}
rayryeng
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answered Mar 20, 2018 at 13:19
0
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.
answered Dec 5, 2014 at 17:04
jafeldsjafelds
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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'.
answered Sep 20, 2020 at 8:55
Aivar PaalbergAivar Paalberg
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a = 'have a nice day'
symbol = 'abcdefghijklmnopqrstuvwxyz'
for key in symbol:
print[key, a.count[key]]
answered Oct 24, 2015 at 22:36
rookierookie
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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
answered Jan 19, 2020 at 9:23
1
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.
answered Mar 28, 2017 at 17:13
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]
answered Mar 7, 2019 at 9:04
a = "I walked today,"
c=['d','e','f']
count=0
for i in a:
if str[i] in c:
count+=1
print[count]
answered Sep 29, 2020 at 3:27
KquekKquek
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2
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
answered Feb 3, 2021 at 11:42
Rishi BansalRishi Bansal
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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 ,
answered Sep 23, 2021 at 5:06
5
"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[]
answered Jan 11, 2015 at 17:24
B-YB-Y
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1
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]
answered Sep 7, 2019 at 20:16
Dipen GajjarDipen Gajjar
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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]
answered Dec 21, 2021 at 13:34
3
str = "count a character occurence"
List = list[str]
print [List]
Uniq = set[List]
print [Uniq]
for key in Uniq:
print [key, str.count[key]]
answered Apr 1, 2016 at 1:29
1
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='