如何用适当的字典值替换句子中的字符串？答案

【问题标题】：How to replace strings in sentence with appropriate dictionary values?如何用适当的字典值替换句子中的字符串？
【发布时间】：2021-02-01 04:05:02
【问题描述】：

我有一本字典如下：

dict_ = { 
        'the school in USA' : 'some_text_1',
        'school' : 'some_text_1',
        'the holy church in brisbane' : 'some_text_2',
        'holy church' : 'some_text_2'
}

还有一个句子列表如下：

text_sent = ["Ram is going to the holy church in brisbane",\
             "John is going to holy church", \
             "shena is going to the school in USA", \
             "Jennifer is going to the school"]

我想用 text_sent 中的相应值替换出现的 dict_ 字典键。我这样做如下：

for ind, text in enumerate(text_sent) :
    for iterator in dict_.keys() :
        if iterator in text : 
            text_sent[ind] = re.sub(iterator, dict_[iterator], text)

for i in text_sent:
    print(i)

我得到的输出如下：

Ram is going to the some_text_2 in brisbane
John is going to some_text_2
shena is going to the some_text_1 in USA
Jennifer is going to the some_text_1

预期输出是：

Ram is going to some_text_2
John is going to some_text_2
shena is going to some_text_1
Jennifer is going to some_text_1

我需要的是，较长的字符串（例如，“布里斯班的圣堂”）需要替换，以防万一，句子中没有完整的字符串, 只有这样才能使用较小的版本（例如，'holy church'）而不是较长的版本来替换 text_sent 句子中的相应值。

【问题讨论】：

标签： python string list dictionary

【解决方案1】：

您可以使用re.sub 进行替换，使用str.join 格式化子字符串字典中的正则表达式：

import re
d = {'the school in USA': 'some_text_1', 'school': 'some_text_1', 'the holy church in brisbane': 'some_text_2', 'holy church': 'some_text_2'}
text_sent = ["Ram is going to the holy church in brisbane",\
         "John is going to holy church", \
         "shena is going to the School in USA", \
         "Jennifer is going to the school"]

r = [re.sub('|'.join(d), lambda x:d[x.group()], i, re.I) for i in text_sent]

输出：

['Ram is going to some_text_2', 'John is going to some_text_2', 'shena is going to some_text_1', 'Jennifer is going to the some_text_1']

【讨论】：

我是初级程序员。您能否解释一下 d[x.group()] 究竟是如何工作的？
@rishikeshfulari 当正则表达式找到匹配项时，re.sub 函数通过lambda 函数的参数提供对匹配值的访问。找到匹配项时调用lambda x:d[x.group()]，d[x.group()] 在d 中查找匹配项并返回其值。
如果我在“shena is going to the school in USA”中将“school”的大小写改为“School”，则无法替换。您能否建议更改此代码以使比较不区分大小写？我无法将完整句子的大小写转换为 lower()，因为我们需要保留其他单词的大小写。但是，如果字典中存在不同大小写的对应键，程序应该能够改变句子中单词的大小写。
@rishikeshfulari 请看我最近的编辑。 re.I 标志可用于忽略匹配文本的大小写。

【解决方案2】：

您可以为字典创建一个辅助列表，并根据其元素长度对其进行排序。

dict_ = {'the school in USA' : 'some_text_1',
         'school' : 'some_text_1',
         'the holy church in brisbane' : 'some_text_2',
         'holy church' : 'some_text_2'}

text_sent = ["Ram is going to the holy church in brisbane",
             "John is going to holy church",
             "shena is going to the school in USA",
             "Jennifer is going to the school"]

dict_keys = list(dict_.keys())
dict_keys.sort(key=len)
dict_keys.reverse()

text_sent_replaced = []
for text in text_sent:
    modified_text = text
    for key in dict_:
        modified_text = modified_text.replace(key,dict_[key])
    text_sent_replaced.append(modified_text)

print(text_sent_replaced)

【讨论】：

【解决方案3】：

主要问题是您没有添加break 语句。如果稍后在 dict_ 字典中有多个匹配项，则您将覆盖值。试试这个：

for ind, text in enumerate(text_sent) :
    for iterator in dict_.keys() :
        if iterator in text :
            text_sent[ind] = re.sub(iterator, dict_[iterator], text)
            break

【讨论】：

【解决方案4】：

只要替换的元素位于每行的末尾，这将在不使用 re 的情况下完成任务，就像您的示例中的情况一样：

for ind, text in enumerate(text_sent) :
    for iterator in dict_.keys() :
        if iterator in text :
            text_sent[ind] = text.split(iterator)[0] + dict_[iterator]

for i in text_sent:
    print(i)

#Prints:
#Ram is going to the some_text_2
#John is going to some_text_2
#shena is going to the some_text_1
#Jennifer is going to the some_text_1

【讨论】：