Question

我有一个包含许多多行记录的结构化文本文件。每条记录都应该有一个关键的唯一字段。我需要阅读一系列这些文件，查找非唯一键字段并用唯一值替换键值。

我的脚本正在识别需要替换的所有字段。我将这些字段存储在字典中，其中键是非唯一字段，值是唯一值列表。

例如：

 {
 "1111111111" : ["1234566363", "5533356775", "6443458343"]
 }

我想做的是只阅读每个文件一次，找到＆＃34; 1111111111＆＃34; （dict键）并用第一个键值替换第一个匹配，用第二个键值替换第二个匹配值等。

我正在尝试使用正则表达式，但我不确定如何在不多次循环文件的情况下构造合适的RE

这是我目前的代码：

def multireplace(Text, Vars):
    dictSorted = sorted(Vars, key=len, reverse=True)
    regEx = re.compile('|'.join(map(re.escape, dictSorted)))
    return regEx.sub(lambda match: Vars[match.group(0)], Text)

text = multireplace(text, find_replace_dict)

它适用于单个键：值组合，但如果：value是列表，则无法编译：

return regEx.sub(lambda match: Vars[match.group(0)], Text , 1)
TypeError: sequence item 1: expected str instance, list found

可以在不通过文件循环多次的情况下更改功能吗？

Answer 1

看看并阅读评论。如果有什么事情没有意义，请告诉我：

import re

def replace(text, replacements):
    # Make a copy so we don't destroy the original.
    replacements = replacements.copy()

    # This is essentially what you had already.
    regex = re.compile("|".join(map(re.escape, replacements.keys())))

    # In our lambda, we pop the first element from the array. This way,
    # each time we're called with the same group, we'll get the next replacement.
    return regex.sub(lambda m: replacements[m.group(0)].pop(0), text)

print(replace("A A B B A B", {"A": ["A1", "A2", "A3"], "B": ["B1", "B2", "B3"]}))

# Output:
# A1 A2 B1 B2 A3 B3

<强>更新

要在下面的评论中帮助解决此问题，请尝试使用此版本，该版本将准确告诉您哪个字符串用完了替代品：

import re

def replace(text, replacements):

    # Let's make a method so we can do a little more than the lambda.
    def make_replacement(match):
        try:
            return replacements[match.group(0)].pop(0)
        except IndexError:
            # Print out debug info about what happened
            print("Ran out of replacements for {}".format(match.group(0)))
            # Re-raise so the process still exits.
            raise

    # Make a copy so we don't destroy the original.
    replacements = replacements.copy()

    # This is essentially what you had already.
    regex = re.compile("|".join(map(re.escape, replacements.keys())))

    # In our lambda, we pop the first element from the array. This way,
    # each time we're called with the same group, we'll get the next replacement.
    return regex.sub(make_replacement, text)

print(replace("A A B B A B A", {"A": ["A1", "A2", "A3"], "B": ["B1", "B2", "B3"]}))

# Output:
# A1 A2 B1 B2 A3 B3

使用正则表达式

1 个答案: