Initialize List to a variable in a Dictionary inside a loop

Question

I have been working for a while in Python and I have solved this issue using \"try\" and \"except\", but I was wondering if there is another method to solve it.

Bas

Answer 1

Your code is not appending elements to the lists; you are instead replacing the list with single elements. To access values in your existing dictionaries, you must use indexing, not attribute lookups (item['name'], not item.name).

Use collections.defaultdict():

from collections import defaultdict

example_dictionary = defaultdict(list)
for item in root_values:
    example_dictionary[item['name']].append(item['value'])

defaultdict is a dict subclass that uses the __missing__ hook on dict to auto-materialize values if the key doesn't yet exist in the mapping.

or use dict.setdefault():

example_dictionary = {}
for item in root_values:
    example_dictionary.setdefault(item['name'], []).append(item['value'])

Answer 2

List and dictionary comprehensions can help here ...

Given

In [72]: root_values
Out[72]:
[{'name': 'red', 'value': 2},
 {'name': 'red', 'value': 3},
 {'name': 'red', 'value': 2},
 {'name': 'green', 'value': 7},
 {'name': 'green', 'value': 8},
 {'name': 'green', 'value': 9},
 {'name': 'blue', 'value': 4},
 {'name': 'blue', 'value': 4},
 {'name': 'blue', 'value': 4}]

A function like item() shown below can extract values with specific names:

In [75]: def item(x): return [m['value'] for m in root_values if m['name']==x]
In [76]: item('red')
Out[76]: [2, 3, 2]

Then, its just a matter of dictionary comprehension ...

In [77]: {x:item(x) for x in ['red', 'green', 'blue']  }
Out[77]: {'blue': [4, 4, 4], 'green': [7, 8, 9], 'red': [2, 3, 2]}