将函数应用于 Pandas Dataframe 的单列

Question

I am trying to apply a function to a single column of my dataframe (specifically, normalization).

The dataframe looks like this:

     Euclidian        H         N       Volume
222   0.012288  0.00518  0.011143   85203000.0
99    1.296833 -0.80266  1.018583   17519400.0
98    1.618482 -0.60979  1.499213   16263900.0
211   2.237388  0.38073 -2.204757   38375400.0
175   2.313548  0.35656 -2.285907   66974200.0
102   3.319342  3.01295 -1.392897   33201000.0
7     3.424589 -0.31313  3.410243   97924700.0
64    3.720370 -0.03526  3.720203  116514000.0
125   3.995138  0.27396  3.985733   80526200.0
210   4.999969  0.46453  4.978343   70612100.0

The dataframe is named 'discrepancies', and my code is as such:

max = discrepancies['Volume'].max()
discrepancies['Volume'].apply(lambda x: x/max)
return discrepancies

But the column values do not change. I cannot find anywhere in the documentation to apply to single columns, they only talk about applying to all columns or all rows:

https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.DataFrame.apply.html

Thank you

Answer 1

If it is just a single column, you don't need to use apply . Directly divide the column using its max will do.

discrepancies['Volume'] = discrepancies['Volume'] / discrepancies['Volume'].max()

Answer 2

Since single columns do not need apply also we need assign it back

max = discrepancies['Volume'].max()
discrepancies['some col']=discrepancies['Volume']/max

Also series you can use map

max = discrepancies['Volume'].max()
discrepancies['Volume'].map(lambda x: x/max)

Answer 3

the problem with your code is that pandas.apply returns the result as new data frame. (there is inplace attribute for lots of pandas functions but not apply )

to correct you code you should do:

max = discrepancies['Volume'].max()
discrepancies['Volume'] = discrepancies['Volume'].apply(lambda x: x/max)
return discrepancies

or you can use @YOBEN_S answer.