I have the following Pandas DataFrame:
In [31]:
import pandas as pd
sample = pd.DataFrame({'Sym1': ['a','a','a','d'],'Sym2':['a','c','b','b'],'Sym3':['a','c','b','d'],'Sym4':['b','b','b','a']},index=['Item1','Item2','Item3','Item4'])
In [32]: print(sample)
Out [32]:
Sym1 Sym2 Sym3 Sym4
Item1 a a a b
Item2 a c c b
Item3 a b b b
Item4 d b d a
and I want to find the elegant way to get the distance between each Item
according to this distance matrix:
In [34]:
DistMatrix = pd.DataFrame({'a': [0,0,0.67,1.34],'b':[0,0,0,0.67],'c':[0.67,0,0,0],'d':[1.34,0.67,0,0]},index=['a','b','c','d'])
print(DistMatrix)
Out[34]:
a b c d
a 0.00 0.00 0.67 1.34
b 0.00 0.00 0.00 0.67
c 0.67 0.00 0.00 0.00
d 1.34 0.67 0.00 0.00
For example comparing Item1
to Item2
would compare aaab
-> accb
-- using the distance matrix this would be 0+0.67+0.67+0=1.34
Ideal output:
Item1 Item2 Item3 Item4
Item1 0 1.34 0 2.68
Item2 1.34 0 0 1.34
Item3 0 0 0 2.01
Item4 2.68 1.34 2.01 0
this is doing twice as much work as needed, but technically works for non-symmetric distance matrices as well ( whatever that is supposed to mean )
pd.DataFrame ( { idx1: { idx2:sum( DistMatrix[ x ][ y ]
for (x, y) in zip( row1, row2 ) )
for (idx2, row2) in sample.iterrows( ) }
for (idx1, row1 ) in sample.iterrows( ) } )
you can make it more readable by writing it in pieces:
# a helper function to compute distance of two items
dist = lambda xs, ys: sum( DistMatrix[ x ][ y ] for ( x, y ) in zip( xs, ys ) )
# a second helper function to compute distances from a given item
xdist = lambda x: { idx: dist( x, y ) for (idx, y) in sample.iterrows( ) }
# the pairwise distance matrix
pd.DataFrame( { idx: xdist( x ) for ( idx, x ) in sample.iterrows( ) } )
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