Python：如何在匹配之间获取字符串？

Question

我有

FILE = open("file.txt", "r") #long text file
TEXT = FILE.read()

#long identification code with dots (.) and slashes (-)
regex = "process \d\d\d\d\d\d\d\-\d\d\.\d\d\d\d\.\d+\.\d\d\.\d\d\d\d"
SRC = re.findall(regex, TEXT, flags=re.IGNORECASE|re.MULTILINE)

如何在第一次出现SRC[i]的第一个字符和下一个出现的第一个字符SRC[i+1]之间获取文本，依此类推？无法找到任何直截了当的满意答案......

更多信息编辑：

pattern = 'process \d{7}\-\d{2}\.\d{4}\.\d+\.\d{2}\.\d{4}'

sample_input = "Process 1234567-89.1234.12431242.12.1234 -  text title and long text description with no assured pattern Process 2234567-89.1234.12431242.12.1234 : chars and more text Process 3234567-89.1234.12431242.12.1234 - more text process 3234567-89.1234.12431242.12.1234 (...)"

sample_output[0] = "Process 1234567-89.1234.12431242.12.1234 -  text title and long text description with no assured pattern "
sample_output[1] = "Process 2234567-89.1234.12431242.12.1234 : chars and more text "
sample_output[2] = "Process 3234567-89.1234.12431242.12.1234 - more text "
sample_output[3] = "process 3234567-89.1234.12431242.12.1234    "

Answer 1

假设您有一个字符串some_str = 'abcARelevant_SubstringAcba'，并且您希望字符串位于第一个A和第二个A之间;即期望的输出是'Relevant_Substring'。

您可以使用以下行找到A中some_str出现次数的索引：
inds = [a.start() for a in re.finditer('A', some_str)]

现在inds = [3, 22]。现在some_str[inds[0]+1:inds[1]将包含'Relevant_Substring'。

这应该可以扩展到您的问题。

编辑：这是一个具体的例子。

假设您有一个包含以下文本的文件“file.txt”：

Stuff I don't want.
0
Stuff I do want.
1
More stuff I don't want.

您想要使用所有数字（0-9）作为分隔符。因此，上面的0和1都将充当分隔符。请尝试以下代码：

import re
with open("file.txt", "r") as file:
    data = file.read()
patt = re.compile('[0-9]')
inds = [a.start() for a in re.finditer(patt, data)]
print data[inds[0]+1:inds[1]]

这应打印出Stuff I do want.

Answer 2

您可以使用此正则表达式：

(Process \d{7}\-\d{2}\.\d{4}\.\d+\.\d{2}\.\d{4}.*?)(?=Process)|(Process \d{7}\-\d{2}\.\d{4}\.\d+\.\d{2}\.\d{4}.*)

<强> Working demo

）

匹配信息

MATCH 1
1.  [0-105] `Process 1234567-89.1234.12431242.12.1234 -  text title and long text description with no assured pattern `
MATCH 2
1.  [105-168]   `Process 2234567-89.1234.12431242.12.1234 : chars and more text `
MATCH 3
1.  [168-221]   `Process 3234567-89.1234.12431242.12.1234 - more text `
MATCH 4
2.  [221-267]   `Process 3234567-89.1234.12431242.12.1234 (...)`

您可以使用此代码：

sample_input = "Process 1234567-89.1234.12431242.12.1234 -  text title and long text description with no assured pattern Process 2234567-89.1234.12431242.12.1234 : chars and more text Process 3234567-89.1234.12431242.12.1234 - more text process 3234567-89.1234.12431242.12.1234 (...)"
m = re.match(r"(Process \d{7}\-\d{2}\.\d{4}\.\d+\.\d{2}\.\d{4}.*?)(?=Process)|(Process \d{7}\-\d{2}\.\d{4}\.\d+\.\d{2}\.\d{4}.*)", sample_input)
m.group(1)       # The first parenthesized subgroup.
m.groups()       # Return a tuple containing all the subgroups of the match, from 1 up to however many groups are in the pattern

Answer 3

你不需要在两个字符之间找到一个字符串：

some_str = 'abcARelevant_SubstringAcba'
print some_str.split("A",2)[1]
Relevant_Substring