'Python/Pandas Dataframe replace 0 with median value
I have a python pandas dataframe with several columns and one column has 0 values. I want to replace the 0 values with the median or mean of this column.
data is my dataframeartist_hotness is the column
mean_artist_hotness = data['artist_hotness'].dropna().mean()
if len(data.artist_hotness[ data.artist_hotness.isnull() ]) > 0:
data.artist_hotness.loc[ (data.artist_hotness.isnull()), 'artist_hotness'] = mean_artist_hotness
I tried this, but it is not working.
Solution 1:[1]
I think you can use mask and add parameter skipna=True to mean instead dropna. Also need change condition to data.artist_hotness == 0 if need replace 0 values or data.artist_hotness.isnull() if need replace NaN values:
import pandas as pd
import numpy as np
data = pd.DataFrame({'artist_hotness': [0,1,5,np.nan]})
print (data)
artist_hotness
0 0.0
1 1.0
2 5.0
3 NaN
mean_artist_hotness = data['artist_hotness'].mean(skipna=True)
print (mean_artist_hotness)
2.0
data['artist_hotness']=data.artist_hotness.mask(data.artist_hotness == 0,mean_artist_hotness)
print (data)
artist_hotness
0 2.0
1 1.0
2 5.0
3 NaN
Alternatively use loc, but omit column name:
data.loc[data.artist_hotness == 0, 'artist_hotness'] = mean_artist_hotness
print (data)
artist_hotness
0 2.0
1 1.0
2 5.0
3 NaN
data.artist_hotness.loc[data.artist_hotness == 0, 'artist_hotness'] = mean_artist_hotness
print (data)
IndexingError: (0 True 1 False 2 False 3 False Name: artist_hotness, dtype: bool, 'artist_hotness')
Another solution is DataFrame.replace with specifying columns:
data=data.replace({'artist_hotness': {0: mean_artist_hotness}})
print (data)
aa artist_hotness
0 0.0 2.0
1 1.0 1.0
2 5.0 5.0
3 NaN NaN
Or if need replace all 0 values in all columns:
import pandas as pd
import numpy as np
data = pd.DataFrame({'artist_hotness': [0,1,5,np.nan], 'aa': [0,1,5,np.nan]})
print (data)
aa artist_hotness
0 0.0 0.0
1 1.0 1.0
2 5.0 5.0
3 NaN NaN
mean_artist_hotness = data['artist_hotness'].mean(skipna=True)
print (mean_artist_hotness)
2.0
data=data.replace(0,mean_artist_hotness)
print (data)
aa artist_hotness
0 2.0 2.0
1 1.0 1.0
2 5.0 5.0
3 NaN NaN
If need replace NaN in all columns use DataFrame.fillna:
data=data.fillna(mean_artist_hotness)
print (data)
aa artist_hotness
0 0.0 0.0
1 1.0 1.0
2 5.0 5.0
3 2.0 2.0
But if only in some columns use Series.fillna:
data['artist_hotness'] = data.artist_hotness.fillna(mean_artist_hotness)
print (data)
aa artist_hotness
0 0.0 0.0
1 1.0 1.0
2 5.0 5.0
3 NaN 2.0
Solution 2:[2]
use pandas replace method:
df = pd.DataFrame({'a': [1,2,3,4,0,0,0,0], 'b': [2,3,4,6,0,5,3,8]})
df
a b
0 1 2
1 2 3
2 3 4
3 4 6
4 0 0
5 0 5
6 0 3
7 0 8
df['a']=df['a'].replace(0,df['a'].mean())
df
a b
0 1 2
1 2 3
2 3 4
3 4 6
4 1 0
5 1 5
6 1 3
7 1 8
Solution 3:[3]
data['artist_hotness'] = data['artist_hotness'].map( lambda x : data.artist_hotness.mean() if x == 0 else x)
Solution 4:[4]
Found these very useful, although mask is really slow (not sure why).
I did this:
df.loc[ df['artist_hotness'] == 0 | np.isnan(df['artist_hotness']), 'artist_hotness' ] = df['artist_hotness'].median()
Solution 5:[5]
I think below code will solve your problem in one line.
data['artist_hotness'] = data['artist_hotness'].replace(0,
data['artist_hotness'].mean())
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
| Solution | Source |
|---|---|
| Solution 1 | |
| Solution 2 | jezrael |
| Solution 3 | Sailendra Pinupolu |
| Solution 4 | |
| Solution 5 |
