# 计算字符串中字符的出现次数

## 问题：计算字符串中字符的出现次数

What's the simplest way to count the number of occurrences of a character in a string?

e.g. count the number of times `'a'` appears in `'Mary had a little lamb'`

## 回答 0

str.count（sub [，start [，end]]）

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

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.

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

## 回答 1

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

You can use count() :

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

## 回答 2

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

As other answers said, using the string method count() is probably the simplest, but if you're doing this frequently, check out collections.Counter:

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

## 回答 3

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

Regular expressions maybe?

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

## 回答 4

``myString.count('a');``

``````myString.count('a');
``````

## 回答 5

Python-3.x：

``"aabc".count("a")``

str.count（sub [，start [，end]]）

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.

## 回答 6

`str.count(a)`是计算字符串中单个字符的最佳解决方案。但是，如果您需要计算更多的字符，则必须读取整个字符串与要计算的字符一样多的次数。

``````from collections import defaultdict

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

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

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

``````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``````

`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
``````

## 回答 7

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

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

This easy and straight forward function might help:

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

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

## 回答 8

``````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))``````

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.

## 回答 9

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

## 回答 10

``````str = "count a character occurance"

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

for key in Uniq:
print (key, str.count(key))``````
``````str = "count a character occurance"

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

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

## 回答 11

``````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)``````

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)
``````

## 回答 12

`count`绝对是计算字符串中字符出现次数的最简洁，最有效的方法，但是我尝试使用解决方案`lambda`，例如：

``````sentence = 'Mary had a little lamb'
sum(map(lambda x : 1 if 'a' in x else 0, sentence))``````

``4``

``````sentence = ['M', 'ar', 'y', 'had', 'a', 'little', 'l', 'am', 'b']
sum(map(lambda x : 1 if 'a' in x else 0, sentence))``````

``4``

`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.

## 回答 13

“不使用count来查找想要的字符串中的字符”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()``````

"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()
``````

## 回答 14

``````>>> 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``````

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
``````

## 回答 15

``````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))``````
``````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))
``````

## 回答 16

Python 3

1）内置函数count（）

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

2）不使用功能

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

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

print(count)``````

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)
``````

## 回答 17

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

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

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