最简单的方法是一行代码:

'Mary had a little lamb'.count("a")

但是如果你想用这个也可以:

sentence ='Mary had a little lamb'
   count=0;
    for letter in sentence :
        if letter=="a":
            count+=1
    print (count)

要获得所有字母的计数，请使用集合。计数器:

>>> from collections import Counter
>>> counter = Counter("Mary had a little lamb")
>>> counter['a']
4

不使用Counter()， count和regex获得所有字符计数的另一种方法

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)

我是pandas库的粉丝，尤其是value_counts()方法。你可以用它来计算字符串中每个字符的出现次数:

>>> 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

正则表达式非常有用，如果你想要区分大小写(当然还有regex的所有功能)。

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))

请注意，regex版本的运行时间大约是它的十倍，只有当my_string非常长或代码处于深度循环中时，这才可能成为问题。

我不知道“最简单的”，但简单的理解可以做到:

>>> my_string = "Mary had a little lamb"
>>> sum(char == 'a' for char in my_string)
4

利用内置的和，生成器理解和bool是整数的子类的事实:如何乘字符等于'a'。