Carmina
0
I have a correlation dataframe, and I'm trying to turn it into different lists:
A B C
A 1.000000 0.932159 -0.976221
B 0.932159 1.000000 -0.831509
C -0.976221 -0.831509 1.000000
The output I need is:
[A, B, 0.932159]
[A, C, -0.976221]
[B, A, 0.932159]
[B, C, -0.831509]
[C, A, -0.976221]
[C, B, -0.831509]
I have tried converting the dataframe into list, but I don't get what I need. Thanks
Stack the dataframe, reset the index, then exclude the rows which have identical 1st and 2nd column values, then create the list out of it:
out=df.stack().reset_index()
out=out[out.iloc[:,0].ne(out.iloc[:,1])].values.tolist()
OUTPUT
[['A', 'B', 0.932159],
['A', 'C', -0.976221],
['B', 'A', 0.932159],
['B', 'C', -0.831509],
['C', 'A', -0.976221],
['C', 'B', -0.831509]]
The simplest way I can think of for a tiny DataFrame like this is with a list comprehension:
corr
A B C
A 1.000000 0.932159 -0.976221
B 0.932159 1.000000 -0.831509
C -0.976221 -0.831509 1.000000
[[row, col, corr[col][row]] for row in corr.index for col in corr if row != col]
[['A', 'B', 0.932159],
['A', 'C', -0.976221],
['B', 'A', 0.932159],
['B', 'C', -0.831509],
['C', 'A', -0.976221],
['C', 'B', -0.831509]]
The longer-form way may be easier to read for a general audience:
result = []
for row in corr.index:
for col in corr:
if row != col:
result.append([row, col, corr[col][row]])
result
[['A', 'B', 0.932159],
['A', 'C', -0.976221],
['B', 'A', 0.932159],
['B', 'C', -0.831509],
['C', 'A', -0.976221],
['C', 'B', -0.831509]]
A simple list comprehension will be enough
from itertools import permutations
elements = df.index
out = [[val[0], val[1], df.loc[val[0], val[1]]]
for val in permutations(elements, 2)]
Output
[['A', 'B', 0.932159],
['A', 'C', -0.976221],
['B', 'A', 0.932159],
['B', 'C', -0.831509],
['C', 'A', -0.976221],
['C', 'B', -0.831509]]