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

print(count)

我知道要求是数一个特定的字母。我在这里没有使用任何方法编写泛型代码。

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)

输出

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

现在如果你想要任何特定的字母频率，你可以像下面这样打印。

print(count['m'])
2

myString.count('a');

更多信息请点击这里

正则表达式?

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

Python 3

有两种方法可以做到这一点:

1)内置函数count()

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

2)不使用函数

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

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

print(count)

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

print(count)