在多个文件中搜索多个regex，然后输出每个匹配项及其各自的fi

Question

我正在尝试将输出格式化为一个表。例如，所有匹配的文件都应该是列，而matches实例应该是行

这是我的密码：

import glob
import re
folder_path = "/home/e136320"
file_pattern = "/*.txt"

match_list = []

folder_contents = glob.glob(folder_path + file_pattern)

#Search for Emails
regex1= re.compile(r'\S+@\S+')
#Search for Phone Numbers
regex2 = re.compile(r'\d\d\d[-]\d\d\d[-]\d\d\d\d')
#Search for Physician's Name
regex3=re.compile(r'\b\w\w\.\w+\b')


for file in folder_contents:
    read_file = open(file, 'rt').read()
    words=read_file.split()
    for line in words:
        email=regex1.findall(line)
        phone=regex2.findall(line)
        for word in email:
            print(file,email)
        for word in phone:
            print(file,phone)

以下是我的输出：

('/home/e136320/sample.txt', ['bcbs@aol.com'])
('/home/e136320/sample.txt', ['James@aol.com'])
('/home/e136320/sample.txt', ['248-981-3420'])
('/home/e136320/wow.txt', ['soccerfif@yahoo.com'])
('/home/e136320/wow.txt', ['313-806-6666'])
('/home/e136320/wow.txt', ['444-444-4444'])
('/home/e136320/wow.txt', ['248-805-6233'])
('/home/e136320/wow.txt', ['maliva@gmail.com'])

有什么想法吗

Answer 1

我会尝试将您找到的项附加到一个列表中，以便组织结果并在循环之间持久化它们。然后你可以试着打印出来。你可以这样做：

import glob
import re

folder_path = "/home/e136320"
file_pattern = "/*.txt"

match_list = []

folder_contents = glob.glob(folder_path + file_pattern)

# Search for Emails
regex1= re.compile(r'\S+@\S+')

# Search for Phone Numbers
regex2 = re.compile(r'\d\d\d[-]\d\d\d[-]\d\d\d\d')

# Search for Physician's Name
regex3=re.compile(r'\b\w\w\.\w+\b')

results = {}

for file in folder_contents:
    read_file = open(file, 'rt').read()
    words=read_file.split()
    current_results = []

    for line in words:
        email=regex1.findall(line)
        phone=regex2.findall(line)

        for word in email:
            # Append email Regex matches to a list
            current_results.append(word)

        for word in phone:
            # Append phone Regex matches to a list
            current_results.append(word)

     # Save results per file in a dictionary
     # The file name is the key.
     results[file] = current_results

for key in results.keys():
    print(key, [str(item) for item in results[key]]