CODEWITHSUNDEEP
pythonpandas
user308827
user308827

Reputation: 21961

Drop rows with all zeros in pandas data frame

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!

Upvotes: 193

Views: 484786

Answers (16)

Justin Tang
Justin Tang

Reputation: 1

from io import StringIO

import pandas as pd

s = '''
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
'''
df = pd.read_csv(StringIO(s), sep=r'\s+', engine='python',index_col=0)
print(df)
print()
print(
    df.where(df != 0).dropna(how='all')
)

Upvotes: 0

JALO - JusAnotherLivngOrganism
JALO - JusAnotherLivngOrganism

Reputation: 1288

Accessing only the indices that are TRUE for the row sum > 0 should suffice:

ndf=df[df.sum(axis=1)>0]

Upvotes: 0

Gideon Kogan
Gideon Kogan

Reputation: 763

Following the example in the accepted answer, a more elegant solution:

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

   a  b
1  0  1
2  1  0
3  1  1

Upvotes: 5

majdoul jihane
majdoul jihane

Reputation: 21

this works for me new_df = df[df.loc[:]!=0].dropna()

Upvotes: 2

Denisa
Denisa

Reputation: 21

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.

Upvotes: 1

stackpopped
stackpopped

Reputation: 307

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)

Upvotes: 29

Yapi
Yapi

Reputation: 314

To drop all columns with values 0 in any row:

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

Upvotes: -2

Ikbel
Ikbel

Reputation: 2203

I think this solution is the shortest : 

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

Upvotes: 62

Kumar Prasanna
Kumar Prasanna

Reputation: 46

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

Try this command its perfectly working.

Upvotes: 0

bmc
bmc

Reputation: 857

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

Upvotes: 3

clocker
clocker

Reputation: 1366

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

Upvotes: 12

The Unfun Cat
The Unfun Cat

Reputation: 31898

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!

Upvotes: 35

8one6
8one6

Reputation: 13768

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)]

Upvotes: 230

U2EF1
U2EF1

Reputation: 13251

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

Upvotes: 149

Akavall
Akavall

Reputation: 86128

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

Upvotes: 5

8one6
8one6

Reputation: 13768

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]

Upvotes: 7

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