如何计算字符串中字符出现的次数?
如。“a”在“Mary had a little lamb”中出现了4次。
如何计算字符串中字符出现的次数?
如。“a”在“Mary had a little lamb”中出现了4次。
当前回答
“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()
其他回答
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))
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
要查找句子中字符的出现情况,您可以使用下面的代码
首先,我从句子中取出了唯一的字符,然后我计算了每个字符在句子中的出现次数,其中包括空格的出现次数。
ab = set("Mary had a little lamb")
test_str = "Mary had a little lamb"
for i in ab:
counter = test_str.count(i)
if i == ' ':
i = 'Space'
print(counter, i)
以上代码的输出如下所示。
1 : r ,
1 : h ,
1 : e ,
1 : M ,
4 : a ,
1 : b ,
1 : d ,
2 : t ,
3 : l ,
1 : i ,
4 : Space ,
1 : y ,
1 : m ,
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)
正则表达式?
import re
my_string = "Mary had a little lamb"
len(re.findall("a", my_string))