How to make a dictionary of dictionaries with multiple lists

Question

There is one index list which will be the key of the parent dictionary:

index = [1,2,3]

and then multiple lists of which will be the child dicts:

triangles = [4,5,6]
circles = [7,8,9]
squares = [10,11,12]

the sequential elements being the data, resulting in:

{1:{'triangles':4, 'circles':7, 'squares': 10},
 2: {'triangles': 5, 'circles': 8, 'squares': 11},
 3: {'triangles': 6, 'circles': 9, 'squares': 12}}

how can I do this ?

Do you think easier to do in pandas ?

Answer 1

You can zip the lists, create the subdicts and then zip the subdicts with the indices. No restrictions on the indices; they can be non-sequencial/non-numerical:

dct =  dict(zip(index, ({'triangles': i, 'circles': j, 'squares': k} 
                          for i,j,k in zip(triangles, circles, squares))))
print(dct)

---

{1: {'circles': 7, 'squares': 10, 'triangles': 4},
 2: {'circles': 8, 'squares': 11, 'triangles': 5},
 3: {'circles': 9, 'squares': 12, 'triangles': 6}}

---

On another note, if you only need sequential counts, the index list can be replaced with enumerate:

dct =  dict(enumerate(({'triangles': i, 'circles': j, 'squares': k} 
                          for i,j,k in zip(triangles, circles, squares)), 1))

Answer 2

Dict comprehnesions to the rescue!
Note, BTW, that the indices stored in index seem to be one-based although python lists are zero-based:

result =  {i : {'triangles' : triangles[i-1], 'circles' : circles[i-1], 'squares' : squares[i-1]} for i in index}