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

拿起这个用户的评论:

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]))

检查“年代”。可以通过如下方式过滤两个数组的元组:

a[1][a[0]=='s']

旁注:它的工作原理类似于集合包的Counter()，只是在numpy中，无论如何都要导入numpy。你也可以在单词列表中计算唯一的单词。

你可以使用循环和字典。

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

“不使用计数查找字符串中需要的字符”方法。

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()

正则表达式?

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

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

print(count)