Python: Looping and nested dictionaries

Question

I have 3 tuples:

o = (0, 1)
n = ((1, 2), (1,))
c = ((30, 70), (20,))

that I want to put in a nested dictionary. The desired output is:

{'0':{'1':30, '2':70}, '1':{'1':20}}

I tried the following:

for x in enumerate(o):
    graph[str(o[x])] = {}
    for y in enumerate(n):
        for z in enumerate(y):
            graph[str(o[x])][n[x][z]] = c[x][z]

Which does not work, I am not sure how to proceed.

Answer 1

One problem is arising from the fact that (1) and (20) are not tuples (as pointed out by Ilja Everilä in the comments). Since they are actually of type int , you can not iterate over them or use [] to index.

First a note about enumerate() , which you were using incorrectly. This function returns a tuple of (index, value) . More info here .

If changing the source data is not an option, you can modify your code a little to get the desired result. Loop over the tuples as before, but use isinstance to check if the element is an int .

graph = {}
for i, x in enumerate(o):
    graph[x] = {}
    if isinstance(n[i], int):
        graph[x][n[i]] = c[i]
    else:
        for j in range(len(n[i])):
            graph[x][n[i][j]] = c[i][j]
print graph
#{0: {1: 30, 2: 70}, 1: {1: 20}}

This else block of this code can be simplified further, using zip() :

graph = {}
for i, x in enumerate(o):
    graph[x] = {}
    if isinstance(n[i], int):
        graph[x][n[i]] = c[i]
    else:
        for y, z in zip(n[i], c[i]):
            graph[x][y] = z
print graph
#{0: {1: 30, 2: 70}, 1: {1: 20}}

However, the better solution is to change the source data n and c to contain only tuples:

def make_tuple(x):
    return x if isinstance(x, tuple) else (x, )

new_n = map(make_tuple, n)
new_c = map(make_tuple, c)

Then the code above no longer needs the isinstance() check, and can be simplified to:

graph = {}
for i, x in enumerate(o):
    graph[x] = {}
    for y, z in zip(new_n[i], new_c[i]):
        graph[x][y] = z
print graph
#{0: {1: 30, 2: 70}, 1: {1: 20}}

We can simplify further into:

graph = {}
for ko, N, C in zip(o, new_n, new_c):
    graph[ko] = {}
    for kn, vc in zip(N, C):
        graph[ko][kn] = vc
print graph
#{0: {1: 30, 2: 70}, 1: {1: 20}}

Or if to the compact one liner (direct translation of the code block above to dict comprehension):

graph = {ko: {kn:vc for kn, vc in zip(N, C)} for ko, N, C in zip(o, new_n, new_c)}
print graph
#{0: {1: 30, 2: 70}, 1: {1: 20}}