Function against pandas column not generating expected output - python-3.x

Question

I am trying to min-max scale a single column in a dataframe.
I am following this: Writing Min-Max scaler function
My code:
import numpy as np
import pandas as pd
df = pd.DataFrame(np.random.randint(0, 100, size=(100, 4)), columns=list('ABCD'))
print(df, '\n')
y = df['A'].values
def func(x):
return [round((i - min(x)) / (max(x) - min(x)), 2) for i in x]
df['E'] = func(y)
print(df)
df['E'] is just df['A'] / 100.
Not sure what I am missing, but my result is incorrect.

Answer 1

IIUC, are you trying to do something like this?
import numpy as np
import pandas as pd
df = pd.DataFrame(np.random.randint(0, 100, size=(100, 4)), columns=list('ABCD'))
print(df, '\n')
def func(x):
return [round((i - min(x)) / (max(x) - min(x)), 2) for i in x]
df_out = df.apply(func).add_prefix('Norm_')
print(df_out)
print(df.join(df_out))
Output:
A B C D
0 91 59 44 5
1 85 44 57 17
2 6 65 37 46
3 40 50 3 40
4 73 58 47 53
.. .. .. .. ..
95 94 76 22 66
96 70 99 40 59
97 96 84 85 24
98 43 51 59 60
99 31 5 55 89
[100 rows x 4 columns]
Norm_A Norm_B Norm_C Norm_D
0 0.93 0.60 0.44 0.05
1 0.87 0.44 0.58 0.17
2 0.06 0.66 0.37 0.47
3 0.41 0.51 0.03 0.41
4 0.74 0.59 0.47 0.54
.. ... ... ... ...
95 0.96 0.77 0.22 0.67
96 0.71 1.00 0.40 0.60
97 0.98 0.85 0.86 0.24
98 0.44 0.52 0.60 0.61
99 0.32 0.05 0.56 0.91
[100 rows x 4 columns]
A B C D Norm_A Norm_B Norm_C Norm_D
0 91 59 44 5 0.93 0.60 0.44 0.05
1 85 44 57 17 0.87 0.44 0.58 0.17
2 6 65 37 46 0.06 0.66 0.37 0.47
3 40 50 3 40 0.41 0.51 0.03 0.41
4 73 58 47 53 0.74 0.59 0.47 0.54
.. .. .. .. .. ... ... ... ...
95 94 76 22 66 0.96 0.77 0.22 0.67
96 70 99 40 59 0.71 1.00 0.40 0.60
97 96 84 85 24 0.98 0.85 0.86 0.24
98 43 51 59 60 0.44 0.52 0.60 0.61
99 31 5 55 89 0.32 0.05 0.56 0.91
[100 rows x 8 columns]

Answer 2

Also consider that using apply() with a function is typically quite inefficient. Try to use vectorized operations whenever you can...
This is a more efficient expression to normalize each column according to the minimum and maximum for that column:
min = df.min() # per column
max = df.max() # per column
df.join(np.round((df - min) / (max - min), 2).add_prefix('Norm_'))
That's much faster than using apply() on a function. For your sample DataFrame:
%timeit df.join(np.round((df - df.min()) / (df.max() - df.min()), 2).add_prefix('Norm_'))
9.89 ms ± 102 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
While the version with apply takes about 4x longer:
%timeit df.join(df.apply(func).add_prefix('Norm_'))
45.8 ms ± 1.16 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
But this difference grows quickly with the size of the DataFrame. For example, with a DataFrame with size 1,000 x 26, I get
37.2 ms ± 269 µs for the version using vectorized instructions, versus 19.5 s ± 1.82 s for the version using apply, around 500x faster!

Answer 3

Not sure what you are after. Your max and minimum are near known because of the number range.
df.loc[:,'A':'D'].apply(lambda x : x.agg({'min','max'}))
and if all you need is df['E'] is just df['A'] / 100.
why not;
df['E']=df['A']/100
y=df['E'].values
y
Please dont mark me down just trying to get some clarity