在文件行中搜索相同的字符串

Question

在一个文件中，我有几行具有这种结构：

> Present one time: "Instance: ...Edition: ..."
> Present two times: "Instance: ...Edition: ...Instance: ...Edition: ..."
> Present n times: "Instance: ...Edition: ... [n] Instance: ...Edition: ..."

这个结构可以每行出现一次，也可以在同一行出现多次。 思路是逐行读取文件，将[=18代表的值隔离开来=]... 并将它们写入 excel 文件中。我可以做到，但如果上面的结构在行中出现一次，我只能隔离这些值。如果结构不止一次出现在线上，我只能保存第一个结构的值。这是我的代码：

#READ FILE
for i in fin:
    if "Instance:" in i:
        instance = ((i.split('Instance:'))[1].split('Edition')[0])
        worksheet.write(row, col, instance)
    if "Edition:" in i:
        edition = ((i.split('Edition:'))[1].split('\n')[0])
        worksheet.write(row, col, edition)
    row += 1

知道如何解决这个问题吗？

Answer 1

请注意，这仅在您的输入以空行（称为换行符）结尾时有效。如果没有，您可以这样添加：s += '\n'

s = '''Instance: A Edition: Limited
Instance: B Edition: Common Instance: C Edition: 2020 Instance: D Edition: Bla
'''

result = []
start_in = start_ed = None
for i in range(len(s)):
    # Reaching the end of a data item
    if s[i:i+9] == 'Instance:' or s[i] == '\n':
        if start_in and start_ed:
            result.append(
                (s[start_in:start_ed-8].strip(), s[start_ed:i].strip())
            )
            start_in = start_ed = None

    if s[i:i+9] == 'Instance:':
        start_in = i+9
    if s[i:i+8] == 'Edition:':
        start_ed = i+8

print(result)

[('A', 'Limited'), ('B', 'Common'), ('C', '2020'), ('D', 'Bla')]

编辑： 根据要求使用 Version 字段

s = '''Instance: A Edition: Limited Version: 1
Instance: B Edition: Common Version: 2 Instance: C Edition: 2020 Version: 3 Instance: D Edition: Bla Version: 4
'''

result = []
start_in = start_ed = start_vs = None
for i in range(len(s)):
    # Reaching the end of a data item
    if s[i:i+9] == 'Instance:' or s[i] == '\n':
        if start_in and start_ed and start_vs:
            result.append((
                s[start_in:start_ed-8].strip(),
                s[start_ed:start_vs-8].strip(),
                s[start_vs:i].strip()
            ))
            start_in = start_ed = start_vs = None

    if s[i:i+9] == 'Instance:':
        start_in = i+9
    if s[i:i+8] == 'Edition:':
        start_ed = i+8
    if s[i:i+8] == 'Version:':
        start_vs = i+8

print(result)

Answer 2

使用 regular expression 的替代解决方案。这更短但可能更难阅读和维护：

import re
r = re.findall(r'Instance:([\w|\s]+?)Edition:([\w|\s]+?)(?=Instance|\n)', s)

[(' A ', ' Limited'), (' B ', ' Common '), (' C ', ' 2020 '), (' D ', ' Bla')]

如果您不希望匹配项周围有空格，您可以像我在其他解决方案中那样对所有元素应用 strip，或者您可以修改正则表达式以读取 Instance: ([\w...

在文件行中搜索相同的字符串

Search same string in line of file

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string

substring

file