compare index and column in data frame with dictionary - python-3.x

Question

I have a dictionary:
d = {'A-A': 1, 'A-B':2, 'A-C':3, 'B-A':5, 'B-B':1, 'B-C':5, 'C-A':3,
'C-B':4, 'C-C': 9}
and a list:
L = [A,B,C]
I have a DataFrame:
df =pd.DataFrame(columns = L, index=L)
I would like to fill each row in df by values in dictionary based on dictionary keys.For example:
A B C
A 1 2 3
B 5 1 5
C 3 4 9
I tried doing that by:
df.loc[L[0]]=[1,2,3]
df.loc[L[1]]=[5,1,5]
df.loc[L[2]] =[3,4,9]
Is there another way to do that especially when there is a huge data?
Thank you for help

Answer 1

Here is another way that I can think of:
import numpy as np
import pandas as pd
# given
d = {'A-A': 1, 'A-B':2, 'A-C':3, 'B-A':5, 'B-B':1, 'B-C':5, 'C-A':3,
'C-B':4, 'C-C': 9}
L = ['A', 'B', 'C']
# copy the key values into a numpy array
z = np.asarray(list(d.values()))
# reshape the array according to your DataFrame
z_new = np.reshape(z, (3, 3))
# copy it into your DataFrame
df = pd.DataFrame(z_new, columns = L, index=L)

Answer 2

This should do the trick, though it's probably not the best way:
for index in L:
prefix = index + "-"
df.loc[index] = [d.get(prefix + column, 0) for column in L]
Calculating the prefix separately beforehand is probably slower for a small list and probably faster for a large list.
Explanation
for index in L:
This iterates through all of the row names.
prefix = index + "-"
All of the keys for each row start with index + "-", e.g. "A-", "B-"… etc..
df.loc[index] =
Set the contents of the entire row.
[ for column in L]
The same as your comma thing ([1, 2, 3]) just for an arbitrary number of items. This is called a "list comprehension".
d.get( , 0)
This is the same as d[ ] but returns 0 if it can't find anything.
prefix + column
Sticks the column on the end, e.g. "A-" gives "A-A", "A-B"…