Creating a dictionary with multiple values per key from a list of dictionaries

Question

I have the following list of dictionaries:

listofdics = [{'StrId': 11, 'ProjId': 1},{'StrId': 11,'ProjId': 2},
              {'StrId': 22, 'ProjId': 3},{'StrId': 22, 'ProjId': 4},
              {'StrId': 33, 'ProjId': 5},{'StrId': 33, 'ProjId': 6},
              {'StrId': 34, 'ProjId': 7}]

I need to get all ProjId values for StrId that are duplicate. So this is the output I'm looking for:

new_listofdics = [{11:[1,2]}, {22:[3,4]}, {33:[5,6]], {34:[7]}]

I wrote a function that creates a list of dictionaries with StrId values as keys, and a list with all ProjId that share the same key as values. Here it is:

def compare_projids(listofdics):
    proj_ids_dups = {} 

    for row in listofdics:       
        id_value = row['StrId']
        proj_id_value = row['ProjId']
        proj_ids_dups[id_value]=proj_id_value

        if row['StrId'] == id_value:
            sum_ids = []
            sum_ids.append(proj_id_value)  
        proj_ids_dups[id_value]=sum_ids
     return proj_ids_dups

This is the output I get now:

new_listofdics=  {33: [6], 34: [7], 11: [2], 22: [4]}

What I see is that append replaces each ProjId value with the last one iterated, instead of adding them at the end of the list.

How can I fix this?...

Answer 1

It's unclear why you need to have such output new_listofdics = [{11:[1,2]}, {22:[3,4]}, {33:[5,6]], {34:[7]}] , because it's better to have just dict object.

So program would look like this

>>> from collections import defaultdict
>>> listofdics = [{'StrId': 11, 'ProjId': 1},{'StrId': 11,'ProjId': 2},
              {'StrId': 22, 'ProjId': 3},{'StrId': 22, 'ProjId': 4},
              {'StrId': 33, 'ProjId': 5},{'StrId': 33, 'ProjId': 6},
              {'StrId': 34, 'ProjId': 7}]
>>> output = defaultdict(list)
>>> for item in listofdics:
...     output[item.get('StrId')].append(item.get('ProjId'))
>>> dict(output)
{11: [1, 2], 22: [3, 4], 33: [5, 6], 34: [7]}

It's much easier to go through that dict that your desired output.