如何在Python字符串中提取字典？

Question

example_string = "Bla bla {"value": "1"} bla bla"

有没有一种方法可以搜索字符串，目标是仅提取字典而不使用正则表达式？

我正在尝试在网站的脚本标签中提取字典。

编辑 ：我在下面添加了确切的字符串。 不想仅仅因为我不熟悉正则表达式而一直在寻找一种更简单的方法（性能不是问题）。 我用正则表达式尝试了建议的解决方案，但没有成功。

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"""

Answer 1

假设字符串中的字典不超过一个，并且没有其他花括号，并且该字典采用与JSON兼容的格式（所有这些对于您的示例都是正确的，但我不知道它们是否对所有示例都是正确的您的真实数据），您可以使用简单的字符串操作将其提取出来：

prefix, openbrace, rest = s.partition('{')
j, closebrace, suffix = rest.rpartition('}')
if openbrace and closebrace:
    j = openbrace + j + closebrace
    d = json.loads(j)

如果这些假设不成立，那就会困难得多。

首先，您需要提取匹配的括号对。 像这样：

braces = 0
start = None
for i, c in enumerate(s):
    if c == '{':
        if not braces: start = I
        braces += 1
    elif c == '}':
        braces -= 1
        if not braces:
            yield s[start:I+1]

除非字符串中可以有括号，否则您需要跳过它们，并且需要以适用于JavaScript字符串或源语言实际上是任何形式的方式处理它，包括适当的反斜杠转义引用规则等。 。 而且，如果您的字符串可以包含HTML实体转义符（和引号），那么您也需要处理获胜。

然后，您必须解析每个字典。 例如，如果它们是任意JavaScript代码的字符串，则不能将其解析为JSON或Python dict，因为它们都不能够处理完美有效的JS对象，例如{abc: 1} （请注意，引号引起来），因此您需要编写自己的解析器。

而且，如果字典甚至不一定是文字，例如{abc: spam} ，其中spam是变量的名称，非平凡的表达式等，那么您将需要使用JS解释器，feed在适当的环境中，然后解释代码。

Answer 2

如果不使用正则表达式，则必须在看到第一个{同时对字符串进行迭代，然后将其后的所有内容保存到结束} （或str.partition使用str.partition ）。

如果字典本身有一个字典，则还必须计算开括号，并且只有在找到相应的右括号时才停止保存。

如果存在无与伦比的寄生虫，那还是一个问题，要解决起来并不容易。 使用正则表达式更好地坚持解决方案： https : //stackoverflow.com/a/51861961/2648551

Answer 3

您可以使用堆栈查找匹配的花括号而无需使用正则表达式

import ast
def find_dicts(s):
    stack = []
    buffer=""
    for ch in s:
        if ch == "{":
            buffer += "{"
            stack.append(ch)
        elif ch == "}":
            stack.pop(-1)
            buffer += "}"
            if not stack:
              yield ast.literal_eval(buffer)
              buffer = ""   
        elif stack:
            buffer += ch

print(list(find_dicts("""Bla bla {"value": "1"} bla bla""")            ))

我认为这也应该处理嵌套的字典https://repl.it/repls/JuicyWoodenTranslation

以及字符串中的多个dict