问题:在熊猫数据框中删除全零的行

我可以使用pandas dropna()功能来删除将部分或全部列设置为NA的行。是否存在用于删除所有列的值为0的行的等效函数?

P   kt  b   tt  mky depth
1   0   0   0   0   0
2   0   0   0   0   0
3   0   0   0   0   0
4   0   0   0   0   0
5   1.1 3   4.5 2.3 9.0

在此示例中,我们要删除数据帧的前4行。

谢谢!

I can use pandas dropna() functionality to remove rows with some or all columns set as NA‘s. Is there an equivalent function for dropping rows with all columns having value 0?

P   kt  b   tt  mky depth
1   0   0   0   0   0
2   0   0   0   0   0
3   0   0   0   0   0
4   0   0   0   0   0
5   1.1 3   4.5 2.3 9.0

In this example, we would like to drop the first 4 rows from the data frame.

thanks!


回答 0

事实证明,这可以很好地以矢量化方式表达:

> df = pd.DataFrame({'a':[0,0,1,1], 'b':[0,1,0,1]})
> df = df[(df.T != 0).any()]
> df
   a  b
1  0  1
2  1  0
3  1  1

It turns out this can be nicely expressed in a vectorized fashion:

> df = pd.DataFrame({'a':[0,0,1,1], 'b':[0,1,0,1]})
> df = df[(df.T != 0).any()]
> df
   a  b
1  0  1
2  1  0
3  1  1

回答 1

一线。无需移调:

df.loc[~(df==0).all(axis=1)]

对于那些喜欢对称的人,这也适用…

df.loc[(df!=0).any(axis=1)]

One-liner. No transpose needed:

df.loc[~(df==0).all(axis=1)]

And for those who like symmetry, this also works…

df.loc[(df!=0).any(axis=1)]

回答 2

我大约每月一次查找此问题,并且总是必须从评论中找出最佳答案:

df.loc[(df!=0).any(1)]

谢谢丹·艾伦!

I look up this question about once a month and always have to dig out the best answer from the comments:

df.loc[(df!=0).any(1)]

Thanks Dan Allan!


回答 3

用替换零,nan然后将所有条目的行都删除为nan。之后,将其替换nan为零。

import numpy as np
df = df.replace(0, np.nan)
df = df.dropna(how='all', axis=0)
df = df.replace(np.nan, 0)

Replace the zeros with nan and then drop the rows with all entries as nan. After that replace nan with zeros.

import numpy as np
df = df.replace(0, np.nan)
df = df.dropna(how='all', axis=0)
df = df.replace(np.nan, 0)

回答 4

我认为这种解决方案是最短的:

df= df[df['ColName'] != 0]

I think this solution is the shortest :

df= df[df['ColName'] != 0]

回答 5

我发现一些解决方案在查找时很有用,尤其是对于较大的数据集:

df[(df.sum(axis=1) != 0)]       # 30% faster 
df[df.values.sum(axis=1) != 0]  # 3X faster 

继续@ U2EF1中的示例:

In [88]: df = pd.DataFrame({'a':[0,0,1,1], 'b':[0,1,0,1]})

In [91]: %timeit df[(df.T != 0).any()]
1000 loops, best of 3: 686 µs per loop

In [92]: df[(df.sum(axis=1) != 0)]
Out[92]: 
   a  b
1  0  1
2  1  0
3  1  1

In [95]: %timeit df[(df.sum(axis=1) != 0)]
1000 loops, best of 3: 495 µs per loop

In [96]: %timeit df[df.values.sum(axis=1) != 0]
1000 loops, best of 3: 217 µs per loop

在更大的数据集上:

In [119]: bdf = pd.DataFrame(np.random.randint(0,2,size=(10000,4)))

In [120]: %timeit bdf[(bdf.T != 0).any()]
1000 loops, best of 3: 1.63 ms per loop

In [121]: %timeit bdf[(bdf.sum(axis=1) != 0)]
1000 loops, best of 3: 1.09 ms per loop

In [122]: %timeit bdf[bdf.values.sum(axis=1) != 0]
1000 loops, best of 3: 517 µs per loop

Couple of solutions I found to be helpful while looking this up, especially for larger data sets:

df[(df.sum(axis=1) != 0)]       # 30% faster 
df[df.values.sum(axis=1) != 0]  # 3X faster 

Continuing with the example from @U2EF1:

In [88]: df = pd.DataFrame({'a':[0,0,1,1], 'b':[0,1,0,1]})

In [91]: %timeit df[(df.T != 0).any()]
1000 loops, best of 3: 686 µs per loop

In [92]: df[(df.sum(axis=1) != 0)]
Out[92]: 
   a  b
1  0  1
2  1  0
3  1  1

In [95]: %timeit df[(df.sum(axis=1) != 0)]
1000 loops, best of 3: 495 µs per loop

In [96]: %timeit df[df.values.sum(axis=1) != 0]
1000 loops, best of 3: 217 µs per loop

On a larger dataset:

In [119]: bdf = pd.DataFrame(np.random.randint(0,2,size=(10000,4)))

In [120]: %timeit bdf[(bdf.T != 0).any()]
1000 loops, best of 3: 1.63 ms per loop

In [121]: %timeit bdf[(bdf.sum(axis=1) != 0)]
1000 loops, best of 3: 1.09 ms per loop

In [122]: %timeit bdf[bdf.values.sum(axis=1) != 0]
1000 loops, best of 3: 517 µs per loop

回答 6

import pandas as pd

df = pd.DataFrame({'a' : [0,0,1], 'b' : [0,0,-1]})

temp = df.abs().sum(axis=1) == 0      
df = df.drop(temp)

结果:

>>> df
   a  b
2  1 -1
import pandas as pd

df = pd.DataFrame({'a' : [0,0,1], 'b' : [0,0,-1]})

temp = df.abs().sum(axis=1) == 0      
df = df.drop(temp)

Result:

>>> df
   a  b
2  1 -1

回答 7

您可以使用快速lambda功能来检查给定行中的所有值是否均为0。然后,您可以将应用该结果的结果lambda用作仅选择与该条件匹配或不匹配的行的一种方式:

import pandas as pd
import numpy as np

np.random.seed(0)

df = pd.DataFrame(np.random.randn(5,3), 
                  index=['one', 'two', 'three', 'four', 'five'],
                  columns=list('abc'))

df.loc[['one', 'three']] = 0

print df
print df.loc[~df.apply(lambda row: (row==0).all(), axis=1)]

Yield:

              a         b         c
one    0.000000  0.000000  0.000000
two    2.240893  1.867558 -0.977278
three  0.000000  0.000000  0.000000
four   0.410599  0.144044  1.454274
five   0.761038  0.121675  0.443863

[5 rows x 3 columns]
             a         b         c
two   2.240893  1.867558 -0.977278
four  0.410599  0.144044  1.454274
five  0.761038  0.121675  0.443863

[3 rows x 3 columns]

You can use a quick lambda function to check if all the values in a given row are 0. Then you can use the result of applying that lambda as a way to choose only the rows that match or don’t match that condition:

import pandas as pd
import numpy as np

np.random.seed(0)

df = pd.DataFrame(np.random.randn(5,3), 
                  index=['one', 'two', 'three', 'four', 'five'],
                  columns=list('abc'))

df.loc[['one', 'three']] = 0

print df
print df.loc[~df.apply(lambda row: (row==0).all(), axis=1)]

Yields:

              a         b         c
one    0.000000  0.000000  0.000000
two    2.240893  1.867558 -0.977278
three  0.000000  0.000000  0.000000
four   0.410599  0.144044  1.454274
five   0.761038  0.121675  0.443863

[5 rows x 3 columns]
             a         b         c
two   2.240893  1.867558 -0.977278
four  0.410599  0.144044  1.454274
five  0.761038  0.121675  0.443863

[3 rows x 3 columns]

回答 8

另一种选择:

# Is there anything in this row non-zero?
# df != 0 --> which entries are non-zero? T/F
# (df != 0).any(axis=1) --> are there 'any' entries non-zero row-wise? T/F of rows that return true to this statement.
# df.loc[all_zero_mask,:] --> mask your rows to only show the rows which contained a non-zero entry.
# df.shape to confirm a subset.

all_zero_mask=(df != 0).any(axis=1) # Is there anything in this row non-zero?
df.loc[all_zero_mask,:].shape

Another alternative:

# Is there anything in this row non-zero?
# df != 0 --> which entries are non-zero? T/F
# (df != 0).any(axis=1) --> are there 'any' entries non-zero row-wise? T/F of rows that return true to this statement.
# df.loc[all_zero_mask,:] --> mask your rows to only show the rows which contained a non-zero entry.
# df.shape to confirm a subset.

all_zero_mask=(df != 0).any(axis=1) # Is there anything in this row non-zero?
df.loc[all_zero_mask,:].shape

回答 9

对我来说这段代码: df.loc[(df!=0).any(axis=0)] 没有用。它返回了确切的数据集。

相反,我用 df.loc[:, (df!=0).any(axis=0)]并删除了数据集中所有具有0值的列

该函数.all()删除了我的数据集中所有零值的所有列。

For me this code: df.loc[(df!=0).any(axis=0)] did not work. It returned the exact dataset.

Instead, I used df.loc[:, (df!=0).any(axis=0)] and dropped all the columns with 0 values in the dataset

The function .all() droped all the columns in which are any zero values in my dataset.


回答 10

df = df [~( df [ ['kt'  'b'   'tt'  'mky' 'depth', ] ] == 0).all(axis=1) ]

尝试使用此命令,即可正常运行。

df = df [~( df [ ['kt'  'b'   'tt'  'mky' 'depth', ] ] == 0).all(axis=1) ]

Try this command its perfectly working.


回答 11

要在任何行中删除所有值为0的列:

new_df = df[df.loc[:]!=0].dropna()

To drop all columns with values 0 in any row:

new_df = df[df.loc[:]!=0].dropna()

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