Article Outline
search for “does-not-contain” on a dataframe in pandas
问题来源:做项目时,想拿到不符合条件的所有数据,比如:通话类型有好多种(主叫、被叫、呼转……),现在想分析所有非主叫数据,那么问题就来了。
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方法一:df[~df.col.str.contains(word)]
>>> 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]
注意:
- 似乎
df[~(df.A.str.contains("Hello") | (df.A.str.contains("World")))]
比上面使用正则,速度会快点- 获取“非”数据的条数:
(~df.col3.str.contains('u|z')).sum()
方法二:
df[df["col"].str.contains('this'|'that')==False]
>>> df = pd.DataFrame({"A": ["Hello", "this", "World", "apple"]})
>>> df[df['A'].str.contains("Hello|World")==False]
A
1 this
3 apple
# 多个条件情况下:
# df[df["col1"].str.contains('this|that')==False and df["col2"].str.contains('foo|bar')==True]
方法三:
>>> 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