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!
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)]
I think this solution is the shortest :
df= df[df['ColName'] != 0]
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
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
this works for me
new_df = df[df.loc[:]!=0].dropna()
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.
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
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]
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)
To drop all columns with values 0 in any row:
new_df = df[df.loc[:]!=0].dropna()
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!
df = df [~( df [ ['kt' 'b' 'tt' 'mky' 'depth', ] ] == 0).all(axis=1) ]
Try this command its perfectly working.
Source: Stackoverflow.com