Question

这是我的尝试，我想要一个3个字母的窗口并走过字符串，但没有得到2的预期答案。我正在进入一些无限循环。想知道为什么。

def find_bob(s1):
    check_list = 'bob'
    count = 0
    n=0
    cmp_chars=0
    while n<=len(s1):
        while cmp_chars == s1[n:3]:
            if cmp_chars == check_list:
                count += 1
                continue
    return count

s1= 'azcbobobegghakl'
#check_list='bob'
val1 = find_bob(s1)

Answer 1

您可以使用files: "/etc/httpd/conf.d/vhost.conf": mode: "000644" owner: root group: root encoding: plain content: | NameVirtualHost *:80 <VirtualHost *:80> ServerName staging.example.com DocumentRoot "/opt/python/current/app" <Directory "/opt/python/current/app"> Options Indexes FollowSymLinks MultiViews AllowOverride All Require all granted </Directory> </VirtualHost> <VirtualHost *:80> ServerName www.example.com DocumentRoot "/opt/python/current/app" <Directory "/opt/python/current/app"> Options Indexes FollowSymLinks MultiViews AllowOverride All Require all granted </Directory> </VirtualHost>使用str.find()参数来查找bob，例如：

start

Answer 2

您可以将regex与zero width lookahead一起使用，并只计算匹配项：

>>> import re
>>> s1= 'azcbobobegghakl'
>>> len([m for m in re.finditer(r'(?=bob)', s1)])
2

Answer 3

Wondering why...

n=0
cmp_chars=0
while n<=len(s1):
    while cmp_chars == s1[n:3]:
        if cmp_chars == check_list:
            count += 1
            continue

n从零开始，并且在循环内不会改变，因此while条件始终为True。

while cmp_chars == s1[n:3]:不确定该怎么做，cmp_chars从零开始，永远不会等于一个字符串，所以这个循环条件总是为假。

尝试

while n<=len(s1):
    if s1[n:n+3] == check_list:
        count += 1
    n = n +1

如何使用Python计算字符串中单词出现次数

3 个答案: