如何用re.search从Python字典中提取指定短语及对应坐标

问题背景 #

给定如下Python字典，键是完整语句，值是元组列表，列表中每个元素对应键里每个单词的坐标（比如Policy对应第一个坐标元组，general对应第二个，以此类推）：

ip_dict = {
    "Policy general commercial general liability for the individual": [
        (("x1", "y1"), ("w1", "h1")),
        (("x2", "y2"), ("w2", "h2")),
        (("x3", "y3"), ("w3", "h3")),
        (("x4", "y4"), ("w4", "h4")),
        (("x5", "y5"), ("w5", "h5")),
        (("x6", "y6"), ("w6", "h6")),
        (("x7", "y7"), ("w7", "h7")),
        (("x8", "y8"), ("w8", "h8"))
    ]
}

需求是用re.search搜索特定短语（比如commercial general liability），匹配成功后提取该短语及其对应的坐标，存入新字典。期望输出：

op_dict = {'commercial general liability': [(('x3', 'y3'), ('w3', 'h3')), (('x4', 'y4'), ('w4', 'h4')), (('x5', 'y5'), ('w5', 'h5'))]}

用户编写的代码会取出所有坐标，而非目标短语对应的部分：

import re

ip_dict = {
"Policy general commercial general liability for the individual": [
    (("x1", "y1"), ("w1", "h1")),
    (("x2", "y2"), ("w2", "h2")),
    (("x3", "y3"), ("w3", "h3")),
    (("x4", "y4"), ("w4", "h4")),
    (("x5", "y5"), ("w5", "h5")),
    (("x6", "y6"), ("w6", "h6")),
    (("x7", "y7"), ("w7", "h7")),
    (("x8", "y8"), ("w8", "h8"))
]
}

op_dict = {}

# Search for the key containing "commercial general liability"
for key in ip_dict.keys():
    if re.search(r"commercial.?\s.?general.?\s.?liability", key, flags = re.DOTALL):
        op_dict["commercial general liability"] = ip_dict[key]
        break

print(op_dict)
# 输出：
# {'commercial general liability': [(('x1', 'y1'), ('w1', 'h1')), (('x2', 'y2'), ('w2', 'h2')), (('x3', 'y3'), ('w3', 'h3')), (('x4', 'y4'), ('w4', 'h4')), (('x5', 'y5'), ('w5', 'h5')), (('x6', 'y6'), ('w6', 'h6')), (('x7', 'y7'), ('w7', 'h7')), (('x8', 'y8'), ('w8', 'h8'))]}

问题分析 #

这段代码只判断了目标短语是否存在，直接把整个坐标列表赋值给新字典，没有定位短语对应的单词在原语句中的位置，自然无法截取对应的坐标片段。

解决方案 #

核心思路是：先通过正则确认短语存在，再定位短语在原语句单词列表中的位置，最后根据位置截取对应坐标。实现代码如下：

import re

ip_dict = {
    "Policy general commercial general liability for the individual": [
        (("x1", "y1"), ("w1", "h1")),
        (("x2", "y2"), ("w2", "h2")),
        (("x3", "y3"), ("w3", "h3")),
        (("x4", "y4"), ("w4", "h4")),
        (("x5", "y5"), ("w5", "h5")),
        (("x6", "y6"), ("w6", "h6")),
        (("x7", "y7"), ("w7", "h7")),
        (("x8", "y8"), ("w8", "h8"))
    ]
}

target_phrase = "commercial general liability"
op_dict = {}

# 构建正则：转义短语避免语法冲突，允许单词间有任意空白（含换行）
pattern = re.compile(re.escape(target_phrase).replace(r'\ ', r'\s+'), flags=re.DOTALL)

for key, coords in ip_dict.items():
    match = pattern.search(key)
    if match:
        # 拆分原语句和目标短语为单词列表
        key_words = key.split()
        phrase_words = target_phrase.split()
        phrase_len = len(phrase_words)
        
        # 遍历找到短语在原单词列表中的起始索引
        for i in range(len(key_words) - phrase_len + 1):
            if key_words[i:i+phrase_len] == phrase_words:
                # 截取对应坐标并存入新字典
                op_dict[target_phrase] = coords[i:i+phrase_len]
                break
        break

print(op_dict)
# 输出：
# {'commercial general liability': [(('x3', 'y3'), ('w3', 'h3')), (('x4', 'y4'), ('w4', 'h4')), (('x5', 'y5'), ('w5', 'h5'))]}

代码说明 #

• re.escape(target_phrase)：转义短语中的特殊字符，避免和正则语法冲突；后续把转义后的空格替换为\s+，允许单词间有多个空白（包括换行）。
• 拆分原语句和目标短语为单词列表，通过遍历对比找到短语在原列表中的起始位置。
• 根据起始索引和短语长度，精准截取对应的坐标片段。

内容的提问来源于stack exchange，提问作者user21785694