Pandas str.contains 用于部分字符串的精确匹配
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Pandas str.contains for exact matches of partial strings
提问by endangeredoxen
I have a DataFrame (I'll call it test) with a column containing file paths and I want to filter the data using a partial path.
我有一个 DataFrame（我称之为test），其中有一列包含文件路径，我想使用部分路径过滤数据。
full_path
0 C:\data\Data Files\BER\figure1.png
1 C:\data\Data Files\BER\figure2.png
2 C:\data\Previous\Error\summary.png
3 C:\data\Data Files\Valx2.png
4 C:\data\Data Files\Valx2.png
5 C:\data\Microscopy\defect.png
The partial path to find is:
找到的部分路径是：
ex = 'C:\data\Microscopy'
I've tried str.containsbut,
我试过了，str.contains但是
test.full_path.str.contains(ex)
0 False
1 False
2 False
3 False
4 False
5 False
I would have expected a value of Truefor index 5. At first I thought the problem might be with the path strings not actually matching due to differences with the escape character, but:
我本来希望True索引 5的值为。起初我认为问题可能是由于与转义字符不同，路径字符串实际上不匹配，但是：
ex in test.full_path.iloc[5]
equals True. After some digging, I'm thinking the argument to str.containsis supposed to be a regex expression so maybe the "\"s in the partial path are messing things up?
等于True. 经过一番挖掘，我str.contains认为 to的论点应该是一个正则表达式，所以也许部分路径中的“\”把事情搞砸了？
I also tried:
我也试过：
test.full_path.apply(lambda x: ex in x)
but this gives NameError: name 'ex' is not defined. These DataFrames can have a lot of rows in them so I'm also concerned that the applyfunction might not be very efficient.
但这给NameError: name 'ex' is not defined. 这些 DataFrame 中可以有很多行，所以我也担心该apply函数可能不是很有效。
Any suggestions on how to search a DataFrame column for exactpartial string matches?
关于如何在 DataFrame 列中搜索精确的部分字符串匹配的任何建议？
Thanks!
谢谢！
采纳答案by DSM
You can pass regex=Falseto avoid confusion in the interpretation of the argument to str.contains:
您可以传递regex=False以避免混淆参数的解释str.contains：
>>> df.full_path.str.contains(ex)
0 False
1 False
2 False
3 False
4 False
5 False
Name: full_path, dtype: bool
>>> df.full_path.str.contains(ex, regex=False)
0 False
1 False
2 False
3 False
4 False
5 True
Name: full_path, dtype: bool
(Aside: your lambda x: ex in xshould have worked. The NameError is a sign that you hadn't defined exfor some reason.)
（旁白：你lambda x: ex in x应该已经工作了。NameError 是你ex由于某种原因没有定义的标志。）
