Pandas Dataframe: Pairwise division of columns without replacement [duplicate] - python-2.7

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Fastest way to calculate difference in all columns
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I am trying to divide all columns by each column but only once (A/B but not B/A)
From Dividing each column by every other column and creating a new dataframe from the results
and thanks to #COLDSPEED, the following code performs the division of all columns by every column (and adds the corresponding new columns).
I cannot figure out how to avoid the pair duplication.
import pandas as pd
import numpy as np
np.random.seed(42)
df = pd.DataFrame(np.random.randint(0,9,size=(5, 3)), columns=list('ABC'))
ratio_df = pd.concat([df[df.columns.difference([col])].div(df[col], axis=0) \
for col in df.columns], axis=1)
print ratio_df
Which outputs:
Original dataframe
A B C
0 6 3 7
1 4 6 2
2 6 7 4
3 3 7 7
4 2 5 4
Resulting dataframe
B C A C A B
0 0.500000 1.166667 2.000000 2.333333 0.857143 0.428571
1 1.500000 0.500000 0.666667 0.333333 2.000000 3.000000
2 1.166667 0.666667 0.857143 0.571429 1.500000 1.750000
3 2.333333 2.333333 0.428571 1.000000 0.428571 1.000000
4 2.500000 2.000000 0.400000 0.800000 0.500000 1.250000
In row 0, the value for the first column B is B/A or 3/6 = 0.5 and the first column A is A/B or 6/3 = 2
I would like to keep only one result for the pair operation (eg only left column / right column).
A/B A/C B/C
0 2.000000 0.857143 0.428571
1 0.666667 2.000000 3.000000
2 0.857143 1.500000 1.750000
3 0.428571 0.428571 1.000000
4 0.400000 0.500000 1.250000
I was not able to find clues on this matter.
How could I resolve it?
Thanks!

Answer 1

Here's one approach -
idx0,idx1 = np.triu_indices(df.shape[1],1)
df_out = pd.DataFrame(df.iloc[:,idx0].values/df.iloc[:,idx1])
c = df.columns.values
df_out.columns = c[idx0]+'/'+c[idx1]
Sample run -
In [58]: df
Out[58]:
A B C
0 6 3 7
1 4 6 2
2 6 7 4
3 3 7 7
4 2 5 4
In [59]: df_out
Out[59]:
A/B A/C B/C
0 2.000000 0.857143 0.428571
1 0.666667 2.000000 3.000000
2 0.857143 1.500000 1.750000
3 0.428571 0.428571 1.000000
4 0.400000 0.500000 1.250000
Alternative way to get idx0 and idx1 -
from itertools import combinations
idx0,idx1 = np.array(list(combinations(range(df.shape[1]),2))).T