pandas dataframe columns scaling with sklearn
PythonPandasScikit LearnDataframePython Problem Overview
I have a pandas dataframe with mixed type columns, and I'd like to apply sklearn's min_max_scaler to some of the columns. Ideally, I'd like to do these transformations in place, but haven't figured out a way to do that yet. I've written the following code that works:
import pandas as pd
import numpy as np
from sklearn import preprocessing
scaler = preprocessing.MinMaxScaler()
dfTest = pd.DataFrame({'A':[14.00,90.20,90.95,96.27,91.21],'B':[103.02,107.26,110.35,114.23,114.68], 'C':['big','small','big','small','small']})
min_max_scaler = preprocessing.MinMaxScaler()
def scaleColumns(df, cols_to_scale):
for col in cols_to_scale:
df[col] = pd.DataFrame(min_max_scaler.fit_transform(pd.DataFrame(dfTest[col])),columns=[col])
return df
dfTest
A B C
0 14.00 103.02 big
1 90.20 107.26 small
2 90.95 110.35 big
3 96.27 114.23 small
4 91.21 114.68 small
scaled_df = scaleColumns(dfTest,['A','B'])
scaled_df
A B C
0 0.000000 0.000000 big
1 0.926219 0.363636 small
2 0.935335 0.628645 big
3 1.000000 0.961407 small
4 0.938495 1.000000 small
I'm curious if this is the preferred/most efficient way to do this transformation. Is there a way I could use df.apply that would be better?
I'm also surprised I can't get the following code to work:
bad_output = min_max_scaler.fit_transform(dfTest['A'])
If I pass an entire dataframe to the scaler it works:
dfTest2 = dfTest.drop('C', axis = 1)
good_output = min_max_scaler.fit_transform(dfTest2)
good_output
I'm confused why passing a series to the scaler fails. In my full working code above I had hoped to just pass a series to the scaler then set the dataframe column = to the scaled series.
Python Solutions
Solution 1 - Python
I am not sure if previous versions of pandas prevented this but now the following snippet works perfectly for me and produces exactly what you want without having to use apply
>>> import pandas as pd
>>> from sklearn.preprocessing import MinMaxScaler
>>> scaler = MinMaxScaler()
>>> dfTest = pd.DataFrame({'A':[14.00,90.20,90.95,96.27,91.21],
'B':[103.02,107.26,110.35,114.23,114.68],
'C':['big','small','big','small','small']})
>>> dfTest[['A', 'B']] = scaler.fit_transform(dfTest[['A', 'B']])
>>> dfTest
A B C
0 0.000000 0.000000 big
1 0.926219 0.363636 small
2 0.935335 0.628645 big
3 1.000000 0.961407 small
4 0.938495 1.000000 small
Solution 2 - Python
Like this?
dfTest = pd.DataFrame({
'A':[14.00,90.20,90.95,96.27,91.21],
'B':[103.02,107.26,110.35,114.23,114.68],
'C':['big','small','big','small','small']
})
dfTest[['A','B']] = dfTest[['A','B']].apply(
lambda x: MinMaxScaler().fit_transform(x))
dfTest
A B C
0 0.000000 0.000000 big
1 0.926219 0.363636 small
2 0.935335 0.628645 big
3 1.000000 0.961407 small
4 0.938495 1.000000 small
Solution 3 - Python
df = pd.DataFrame(scale.fit_transform(df.values), columns=df.columns, index=df.index)
This should work without depreciation warnings.
Solution 4 - Python
As it is being mentioned in pir's comment - the .apply(lambda el: scale.fit_transform(el)) method will produce the following warning:
> DeprecationWarning: Passing 1d arrays as data is deprecated in 0.17 > and will raise ValueError in 0.19. Reshape your data either using > X.reshape(-1, 1) if your data has a single feature or X.reshape(1, -1) > if it contains a single sample.
Converting your columns to numpy arrays should do the job (I prefer StandardScaler):
from sklearn.preprocessing import StandardScaler
scale = StandardScaler()
dfTest[['A','B','C']] = scale.fit_transform(dfTest[['A','B','C']].as_matrix())
-- Edit Nov 2018 (Tested for pandas 0.23.4)--
As Rob Murray mentions in the comments, in the current (v0.23.4) version of pandas .as_matrix() returns FutureWarning. Therefore, it should be replaced by .values:
from sklearn.preprocessing import StandardScaler
scaler = StandardScaler()
scaler.fit_transform(dfTest[['A','B']].values)
-- Edit May 2019 (Tested for pandas 0.24.2)--
As joelostblom mentions in the comments, "Since 0.24.0, it is recommended to use .to_numpy() instead of .values."
Updated example:
import pandas as pd
from sklearn.preprocessing import StandardScaler
scaler = StandardScaler()
dfTest = pd.DataFrame({
'A':[14.00,90.20,90.95,96.27,91.21],
'B':[103.02,107.26,110.35,114.23,114.68],
'C':['big','small','big','small','small']
})
dfTest[['A', 'B']] = scaler.fit_transform(dfTest[['A','B']].to_numpy())
dfTest
A B C
0 -1.995290 -1.571117 big
1 0.436356 -0.603995 small
2 0.460289 0.100818 big
3 0.630058 0.985826 small
4 0.468586 1.088469 small
Solution 5 - Python
You can do it using pandas only:
In [235]:
dfTest = pd.DataFrame({'A':[14.00,90.20,90.95,96.27,91.21],'B':[103.02,107.26,110.35,114.23,114.68], 'C':['big','small','big','small','small']})
df = dfTest[['A', 'B']]
df_norm = (df - df.min()) / (df.max() - df.min())
print df_norm
print pd.concat((df_norm, dfTest.C),1)
A B
0 0.000000 0.000000
1 0.926219 0.363636
2 0.935335 0.628645
3 1.000000 0.961407
4 0.938495 1.000000
A B C
0 0.000000 0.000000 big
1 0.926219 0.363636 small
2 0.935335 0.628645 big
3 1.000000 0.961407 small
4 0.938495 1.000000 small
Solution 6 - Python
I know it's a very old comment, but still:
Instead of using single bracket (dfTest['A']), use double brackets (dfTest[['A']]).
i.e: min_max_scaler.fit_transform(dfTest[['A']]).
I believe this will give the desired result.
Solution 7 - Python
(Tested for pandas 1.0.5)
Based on @athlonshi answer (it had ValueError: could not convert string to float: 'big', on C column), full working example without warning:
import pandas as pd
from sklearn.preprocessing import MinMaxScaler
scale = preprocessing.MinMaxScaler()
df = pd.DataFrame({
'A':[14.00,90.20,90.95,96.27,91.21],
'B':[103.02,107.26,110.35,114.23,114.68],
'C':['big','small','big','small','small']
})
print(df)
df[["A","B"]] = pd.DataFrame(scale.fit_transform(df[["A","B"]].values), columns=["A","B"], index=df.index)
print(df)
A B C
0 14.00 103.02 big
1 90.20 107.26 small
2 90.95 110.35 big
3 96.27 114.23 small
4 91.21 114.68 small
A B C
0 0.000000 0.000000 big
1 0.926219 0.363636 small
2 0.935335 0.628645 big
3 1.000000 0.961407 small
4 0.938495 1.000000 small