How to use re.findAll(), re.sub(), and set() in Python

1. re.findall()

re.findall(): The function returns a list containing all matching items. Returns all strings matching pattern in string, in the form of list/array.

It can be seen from the function prototype code that the findall() function has three parameters:

1. pattern: the ‘ pattern string in the regular expression ’ ;

2. string: the original string that currently needs to be processed (search and replace);

3. flags: optional parameter, indicating the matching mode used during compilation (such as ignoring size writing, multi-line mode, etc.), in numeric form, the default is 0

# 示例代码
import re
text1 = '北京市海淀区不存在的38街区不想工作大厦99号'
res = re.findall(r'\d+', text1)
print(type(res))
print(res)
# output
# <class &#39;list&#39;>
# [&#39;38&#39;, &#39;99&#39;]

2. re.sub()

re.sub(): The function replaces all matching items with the selected text and returns the result.

It can be seen from the function prototype code that the re.sub() function has five parameters:

1. pattern: ‘ pattern in regular expressions String’ ;

2. repl: the string that needs to be replaced, that is, replacing the matched pattern with repl; it can be a function;

3. string: currently needs to be processed ( Find and replace) the original string; ;

5. flags: Optional parameter, indicating the matching mode used during compilation (such as ignoring case, multi-line mode, etc.), in numeric form, the default is 0

# 将所有匹配到的‘数字串' 替换为 '520‘
text1 = '北京市海淀区不存在的38街区不想工作大厦99号'
res = re.re(r'\d+', 520)
print(type(res))
print(res)
# output，返回值res结果是str
# <class &#39;str&#39;>
# 北京市海淀区不存在的520街区不想工作大厦520号

3. set ()

set(): One of python’s built-in functions, creates an unordered set of non-repeating elements. Supports calculation of intersection, difference, and union.

# 为list数组l1 去重
l1 = [1, 1, 2, 2, 2, 3, 4]
s1 = set(l1)
print(type(s1))
print(s1)
# output，返回类型是 set
# <class &#39;set&#39;>
# {1, 2, 3, 4}

# 计算l1 和 l2 的交集
l1 = [1, 1, 2, 2, 2, 3, 4]
l2 = [2, 3, 3, 4, 5, 6, 6]
s1 = set(l1)
s2 = set(l2)
u = s1 & s2
print(type(u))
print(u)
# output，返回结果类型set
# <class &#39;set&#39;>
# {2, 3, 4}

# 计算l1 和 l2 的并集, 并集符号 ‘｜'，intersection
l1 = [1, 1, 2, 2, 2, 3, 4]
l2 = [2, 3, 3, 4, 5, 6, 6]
s1 = set(l1) # {1, 2, 3, 4}
s2 = set(l2) # {2, 3, 4, 5, 6}
u = s1 | s2
print(type(u))
print(u)
# output，返回结果类型set, 计算 {1, 2, 3, 4} 和 {2, 3, 4, 5, 6} 的并集
# <class &#39;set&#39;>
# {1, 2, 3, 4, 5, 6}

# 计算差集，diff
l1 = [1, 1, 2, 2, 2, 3, 4]
l2 = [2, 3, 3, 4, 5, 6, 6]
s1 = set(l1) # {1, 2, 3, 4}
s2 = set(l2) # {2, 3, 4, 5, 6}
print(s2)
u = s1 - s2
print(type(u))
print(u)
# output，返回结果是set
# <class &#39;set&#39;>
# {1}

# set内也可以传入字符串，会自动转换成list类型
text1 = '北京市海淀区海淀区不想上班不想上班'
res = set(text1)
print(res) # 内部元素是一个个的字，去重 且 无序
# output
# <class &#39;set&#39;>
# {&#39;上&#39;, &#39;北&#39;, &#39;班&#39;, &#39;海&#39;, &#39;淀&#39;, &#39;京&#39;, &#39;不&#39;, &#39;想&#39;, &#39;区&#39;, &#39;市&#39;}

The above is the detailed content of How to use re.findAll(), re.sub(), and set() in Python. For more information, please follow other related articles on the PHP Chinese website!