Return 个包含搜索字符串的句子

Question

目标：print() 个句子，来自长文本，基于短语。

我可以拆分句号，sentence = sentence.split('.')。

代码 1:

phrase = 'PHRASE'

sentence = "Foo. My sentence contains PHRASE here. Bar."
sentence = sentence.split('.')

print(sentence)

输出：

['Foo', ' My sentence contains PHRASE here', ' Bar', '']

现在，我需要能够对任何和所有句子类型进行这项工作： . ! ?。然后从包含 phrase.

的列表中提取元素

代码 2：

phrase = 'PHRASE'

sentence = "Foo. My sentence contains PHRASE here. Bar."
sentence = sentence.split('.')
sentence = [s for s in sentence if phrase in s]
sentence = sentence[0]
print(sentence)

回溯：

Traceback (most recent call last):
  File "/usr/lib/python3.8/py_compile.py", line 144, in compile
    code = loader.source_to_code(source_bytes, dfile or file,
  File "<frozen importlib._bootstrap_external>", line 846, in source_to_code
  File "<frozen importlib._bootstrap>", line 219, in _call_with_frames_removed
  File "./prog.py", line 16
    sentence = lambda sentence : for s in enumerate(sentence): if phrase in s: return s
                                 ^
SyntaxError: invalid syntax

During handling of the above exception, another exception occurred:

Traceback (most recent call last):
  File "<string>", line 1, in <module>
  File "/usr/lib/python3.8/py_compile.py", line 150, in compile
    raise py_exc
py_compile.PyCompileError:   File "./prog.py", line 16
    sentence = lambda sentence : for s in enumerate(sentence): if phrase in s: return s
                                 ^
SyntaxError: invalid syntax

期望输出：

My sentence contains PHRASE here.

如果我还有什么要补充的，请告诉我post。

Answer 1

用 list-comprehension 替换了 lambda。

二手 str.replace('?', '.').

代码：

phrase = 'PHRASE'

sentence = "Foo! My sentence contains PHRASE here! Bar?"
sentence = sentence.replace('!', '.')
sentence = sentence.replace('?', '.')
sentence = sentence.split('.')
sentence = [s for s in sentence if phrase in s]
sentence = sentence[0]
print(sentence)

回溯：

My sentence contains PHRASE here

Return 个包含搜索字符串的句子

Return sentence that contains search string

python

string

lambda

split

syntax-error