문자열에서 문자의 발생 수를 계산합니다.

문자열에서 문자의 발생 수를 계산합니다.

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문자열에서 문자의 발생 횟수를 계산하는 가장 간단한 방법은 무엇입니까?

예 :에 'a'나타나는 횟수를 계산'Mary had a little lamb'

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

sub범위에서 부분 문자열이 겹치지 않는 횟수를 반환합니다 [start, end]. 선택적 인수 start이며 end슬라이스 표기법으로 해석됩니다.

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

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count () 사용할 수 있습니다 .

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

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다른 답변에서 말했듯이 문자열 메서드 count ()를 사용하는 것이 아마도 가장 간단하지만 자주 수행하는 경우 collections.Counter를 확인하십시오 .

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

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정규 표현식일까요?

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

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myString.count('a');

여기에 더 많은 정보

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"aabc".count("a")

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str.count(a)문자열에서 단일 문자를 계산하는 가장 좋은 솔루션입니다. 그러나 더 많은 문자를 계산해야하는 경우 계산하려는 문자 수만큼 전체 문자열을 읽어야합니다.

이 작업에 대한 더 나은 접근 방식은 다음과 같습니다.

from collections import defaultdict

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

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

따라서 문자열에있는 모든 문자의 발생 횟수와 0존재하지 않는 경우 를 반환하는 사전이 있습니다.

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

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대소 문자를 구분하지 않는 카운터의 경우 하위 클래스를 지정하여 mutator 및 접근 자 메서드를 재정의 할 수 있습니다 defaultdict(기본 클래스의 메서드 는 읽기 전용 임).

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

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정규식은 대소 문자를 구분하지 않으려는 경우 매우 유용합니다 (물론 정규식의 모든 기능).

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))

정규식 버전은 실행 시간이 10 배 정도 소요됩니다. 이는 my_string이 엄청나게 길거나 코드가 깊은 루프 내에있는 경우에만 문제가 될 수 있습니다.

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This easy and straight forward function might help:

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

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

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

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str = "count a character occurance"

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

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

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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)

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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.

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"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()

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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))

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

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

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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)

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Using Count:

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

x = 3.

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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")

참고URL : https://stackoverflow.com/questions/1155617/count-the-number-occurrences-of-a-character-in-a-string