pandas 反转string.contains在python、pandas中
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Reversal of string.contains In python, pandas
提问by Xodarap777
I have something like this in my code:
我的代码中有这样的东西：
df2 = df[df['A'].str.contains("Hello|World")]
df2 = df[df['A'].str.contains("Hello|World")]
However, I want all the rows that don'tcontain either of Hello or World. How do I most efficiently reverse this?
但是，我想要所有不包含 Hello 或 World 的行。我如何最有效地扭转这种情况？
回答by DSM
You can use the tilde ~to flip the bool values:
您可以使用波浪号~翻转布尔值：
>>> df = pd.DataFrame({"A": ["Hello", "this", "World", "apple"]})
>>> df.A.str.contains("Hello|World")
0 True
1 False
2 True
3 False
Name: A, dtype: bool
>>> ~df.A.str.contains("Hello|World")
0 False
1 True
2 False
3 True
Name: A, dtype: bool
>>> df[~df.A.str.contains("Hello|World")]
A
1 this
3 apple
[2 rows x 1 columns]
Whether this is the most efficient way, I don't know; you'd have to time it against your other options. Sometimes using a regular expression is slower than things like df[~(df.A.str.contains("Hello") | (df.A.str.contains("World")))], but I'm bad at guessing where the crossovers are.
这是否是最有效的方式，我不知道；你必须根据你的其他选择来计时。有时使用正则表达式比诸如 之类的要慢df[~(df.A.str.contains("Hello") | (df.A.str.contains("World")))]，但我不擅长猜测交叉点在哪里。
回答by Martijn Pieters
The .contains()method uses regular expressions, so you can use a negative lookahead testto determine that a word is notcontained:
该.contains()方法使用正则表达式，因此您可以使用否定前瞻测试来确定不包含某个单词：
df['A'].str.contains(r'^(?:(?!Hello|World).)*$')
This expression matches any string where the words Helloand Worldare notfound anywhere in the string.
这种表达的哪里话任何字符串相匹配Hello，并World在未找到该字符串中的任何位置。
Demo:
演示：
>>> df = pd.DataFrame({"A": ["Hello", "this", "World", "apple"]})
>>> df['A'].str.contains(r'^(?:(?!Hello|World).)*$')
0 False
1 True
2 False
3 True
Name: A, dtype: bool
>>> df[df['A'].str.contains(r'^(?:(?!Hello|World).)*$')]
A
1 this
3 apple
