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[英]Search string in dataframe column that contains lists of string and return complete dataframe
[英]Return sentence that contains search string
目标: print()
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)
Output:
['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)
追溯:
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
所需的 Output:
My sentence contains PHRASE here.
如果还有什么我可以添加到帖子中,请告诉我。
将 lambda 替换为list-comprehension 。
使用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)
追溯:
My sentence contains PHRASE here
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