Find all the patterns of “1(0+)1” in a given string using Python Regex
本教程中,我们将编写一个程序,该程序使用regexes查找字符串中所有出现 1(0+)1 的位置。我们在 Python 中有一个 re 模块来帮助我们处理正则表达式。
我们来看一个示例案例。
Input: string = "Sample 1(0+)1 string with 1(0+)1 unnecessary patterns 1(0+)1" Output: Total number of pattern maches are 3 ['1(0+)1', '1(0+)1', '1(0+)1']
请按照以下步骤编写程序代码。
算法
1. Import the re module. 2. Initialise a string. 3. Create a regex object using regular expression which matches the pattern using the re.compile(). Remember to pass a raw string to the function instead of the usual string. 4. Now, match all the occurrence of the pattern using regex object from the above step and regex_object.findall() method. 5. The above steps return a match object and print the matched patterns using match_object.group() method.
我们来看看代码。
示例
# importing the re module import re # initializing the string string = "Sample 1(0+)1 string with 1(0+)1 unnecessary patterns 1(0+)1" # creating a regex object for our patter regex = re.compile(r"\d\(\d\+\)\d") # this regex object will find all the patter ns which are 1(0+)1 # storing all the matches patterns in a variable using regex.findall() method result = regex.findall(string) # result is a match object # printing the frequency of patterns print(f"Total number of pattern maches are {len(result)}") print() # printing the matches from the string using result.group() method print(result)
输出
如果您运行以上代码,您将得到以下输出。
Total number of pattern maches are 3 ['1(0+)1', '1(0+)1', '1(0+)1']
结论
如果您对本教程有任何疑问,可以在评论部分中提出来。
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