问题：从pandas DataFrame中删除名称包含特定字符串的列
我有一个带有以下列名称的pandas数据框：
Result1，Test1，Result2，Test2，Result3，Test3等…
我想删除名称包含单词“ Test”的所有列。这样的列数不是静态的，而是取决于先前的功能。
我怎样才能做到这一点？
I have a pandas dataframe with the following column names:
Result1, Test1, Result2, Test2, Result3, Test3, etc…
I want to drop all the columns whose name contains the word “Test”. The numbers of such columns is not static but depends on a previous function.
How can I do that?
回答 0
import pandas as pd
import numpy as np
array=np.random.random((2,4))
df=pd.DataFrame(array, columns=('Test1', 'toto', 'test2', 'riri'))
print df
Test1 toto test2 riri
0 0.923249 0.572528 0.845464 0.144891
1 0.020438 0.332540 0.144455 0.741412
cols = [c for c in df.columns if c.lower()[:4] != 'test']
df=df[cols]
print df
toto riri
0 0.572528 0.144891
1 0.332540 0.741412
import pandas as pd
import numpy as np
array=np.random.random((2,4))
df=pd.DataFrame(array, columns=('Test1', 'toto', 'test2', 'riri'))
print df
Test1 toto test2 riri
0 0.923249 0.572528 0.845464 0.144891
1 0.020438 0.332540 0.144455 0.741412
cols = [c for c in df.columns if c.lower()[:4] != 'test']
df=df[cols]
print df
toto riri
0 0.572528 0.144891
1 0.332540 0.741412
回答 1
这是一个很好的方法：
df = df[df.columns.drop(list(df.filter(regex='Test')))]
Here is one way to do this:
df = df[df.columns.drop(list(df.filter(regex='Test')))]
回答 2
便宜，快捷和惯用语： str.contains
在最新版本的熊猫中，可以在索引和列上使用字符串方法。在这里，str.startswith似乎很合适。
要删除以给定子字符串开头的所有列：
df.columns.str.startswith('Test')
# array([ True, False, False, False])
df.loc[:,~df.columns.str.startswith('Test')]
toto test2 riri
0 x x x
1 x x x
对于不区分大小写的匹配，可以将基于正则表达式的匹配与str.containsSOL锚一起使用：
df.columns.str.contains('^test', case=False)
# array([ True, False, True, False])
df.loc[:,~df.columns.str.contains('^test', case=False)]
toto riri
0 x x
1 x x
如果可能使用混合类型，则也要指定na=False。
Cheaper, Faster, and Idiomatic: str.contains
In recent versions of pandas, you can use string methods on the index and columns. Here, str.startswith seems like a good fit.
To remove all columns starting with a given substring:
df.columns.str.startswith('Test')
# array([ True, False, False, False])
df.loc[:,~df.columns.str.startswith('Test')]
toto test2 riri
0 x x x
1 x x x
For case-insensitive matching, you can use regex-based matching with str.contains with an SOL anchor:
df.columns.str.contains('^test', case=False)
# array([ True, False, True, False])
df.loc[:,~df.columns.str.contains('^test', case=False)]
toto riri
0 x x
1 x x
if mixed-types is a possibility, specify na=False as well.
回答 3
您可以使用“过滤器”过滤出您想要的列
import pandas as pd
import numpy as np
data2 = [{'test2': 1, 'result1': 2}, {'test': 5, 'result34': 10, 'c': 20}]
df = pd.DataFrame(data2)
df
c result1 result34 test test2
0 NaN 2.0 NaN NaN 1.0
1 20.0 NaN 10.0 5.0 NaN
现在过滤
df.filter(like='result',axis=1)
得到..
result1 result34
0 2.0 NaN
1 NaN 10.0
You can filter out the columns you DO want using ‘filter’
import pandas as pd
import numpy as np
data2 = [{'test2': 1, 'result1': 2}, {'test': 5, 'result34': 10, 'c': 20}]
df = pd.DataFrame(data2)
df
c result1 result34 test test2
0 NaN 2.0 NaN NaN 1.0
1 20.0 NaN 10.0 5.0 NaN
Now filter
df.filter(like='result',axis=1)
Get..
result1 result34
0 2.0 NaN
1 NaN 10.0
回答 4
可以整齐地用以下一行完成此操作：
df = df.drop(df.filter(regex='Test').columns, axis=1)
This can be done neatly in one line with:
df = df.drop(df.filter(regex='Test').columns, axis=1)
回答 5
使用DataFrame.select方法：
In [38]: df = DataFrame({'Test1': randn(10), 'Test2': randn(10), 'awesome': randn(10)})
In [39]: df.select(lambda x: not re.search('Test\d+', x), axis=1)
Out[39]:
awesome
0 1.215
1 1.247
2 0.142
3 0.169
4 0.137
5 -0.971
6 0.736
7 0.214
8 0.111
9 -0.214
Use the DataFrame.select method:
In [38]: df = DataFrame({'Test1': randn(10), 'Test2': randn(10), 'awesome': randn(10)})
In [39]: df.select(lambda x: not re.search('Test\d+', x), axis=1)
Out[39]:
awesome
0 1.215
1 1.247
2 0.142
3 0.169
4 0.137
5 -0.971
6 0.736
7 0.214
8 0.111
9 -0.214
回答 6
此方法可以完成所有事情。其他许多答案也会创建副本，但效率不高：
df.drop(df.columns[df.columns.str.contains('Test')], axis=1, inplace=True)
This method does everything in place. Many of the other answers create copies and are not as efficient:
df.drop(df.columns[df.columns.str.contains('Test')], axis=1, inplace=True)
回答 7
不要丢下 赶上你想要的相反。
df = df.filter(regex='^((?!badword).)*$').columns
Don’t drop. Catch the opposite of what you want.
df = df.filter(regex='^((?!badword).)*$').columns
回答 8
最简单的方法是：
resdf = df.filter(like='Test',axis=1)
the shortest way to do is is :
resdf = df.filter(like='Test',axis=1)
回答 9
删除包含正则表达式的列名称列表时的解决方案。我更喜欢这种方法，因为我经常编辑下拉列表。对下拉列表使用负过滤器正则表达式。
drop_column_names = ['A','B.+','C.*']
drop_columns_regex = '^(?!(?:'+'|'.join(drop_column_names)+')$)'
print('Dropping columns:',', '.join([c for c in df.columns if re.search(drop_columns_regex,c)]))
df = df.filter(regex=drop_columns_regex,axis=1)
Solution when dropping a list of column names containing regex. I prefer this approach because I’m frequently editing the drop list. Uses a negative filter regex for the drop list.
drop_column_names = ['A','B.+','C.*']
drop_columns_regex = '^(?!(?:'+'|'.join(drop_column_names)+')$)'
print('Dropping columns:',', '.join([c for c in df.columns if re.search(drop_columns_regex,c)]))
df = df.filter(regex=drop_columns_regex,axis=1)
