计算字符串中字符的出现次数

#1 楼

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

#2 楼

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

#3 楼

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

#4 楼

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

#5 楼

myString.count('a');

#6 楼

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

#7 楼

"aabc".count("a")

#8 楼

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}

#9 楼

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))

#10 楼

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

#11 楼

str = "count a character occurance"

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

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

#12 楼

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)

#13 楼

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

#14 楼

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()

#15 楼

>>> 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

#16 楼

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

#17 楼

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

print(count)

#18 楼

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))

#19 楼

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

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

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

print(count)

#20 楼

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