识别字典列表中特定键的相同值

Question

I have a list of dictionaries that look like this:

[
    {'ServiceID': 20, 'primary': '20', 'secondary': '12'},
    {'ServiceID': 20, 'primary': '20', 'secondary': '12'},
    {'ServiceID': 20, 'primary': '20', 'secondary': '12'},
    {'ServiceID': 16, 'primary': '16', 'secondary': '8'},
    {'ServiceID': 20, 'primary': '20', 'secondary': '12'},
    {'ServiceID': 8,  'primary': '8',  'secondary': '16'},
    {'ServiceID': 12, 'primary': '12', 'secondary': '20'},
    {'ServiceID': 8,  'primary': '8',  'secondary': '16'}
]

I would like create a new sorted dictionary where the we have the following:

key = value of 'ServiceID'
key = value of how many times that particular 'ServiceID' is listed as a 'primary'
key = value of how many times that particular 'ServiceID' is listed as a 'secondary'

For example:

[
    {'ServiceID': 8, 'primaryCount': 2, 'secondaryCount': 1},
    {'ServiceID': 12, 'primaryCount': 1, 'secondaryCount': 4},
    {'ServiceID': 16, 'primaryCount': 1, 'secondaryCount': 2},
    {'ServiceID': 120, 'primaryCount': 4, 'secondaryCount': 1}
]

Code that I have so far that doesn't quite seem to do what I desire, meaning that I am unsure as to how to appropriately increment the number of primaries and secondaries across the entire for loop as well as how to only ensure I am capturing the uniques for the 'ServiceID'

I believe there is something wrong with my logic:

temp_count_list = list()
temp_primary_counts = 0
temp_secondary_counts = 0

for sub_dict in new_list:
    temp_dict = dict()

    temp_dict['ServiceID'] = sub_dict['ServiceID']
    
    if sub_dict['ServiceID'] == int(sub_dict['primarySlice']):
        temp_dict['primaryCount'] = temp_primary_counts +=1

    if sub_dict['ServiceID'] == int(sub_dict['secondarySlice']):
        temp_dict['secondaryCount'] = temp_secondary_counts +=1

    temp_count_list.append(temp_dict)

Answer 1

Basic idea is, get all the ServiceID, primary, secondary in a dict (in code k), and then for each unique ServiceID count the frequency of that ServiceID in the primary and secondary.

l = [
    {'ServiceID': 20, 'primary': '20', 'secondary': '12'},
    {'ServiceID': 20, 'primary': '20', 'secondary': '12'},
    {'ServiceID': 20, 'primary': '20', 'secondary': '12'},
    {'ServiceID': 16, 'primary': '16', 'secondary': '8'},
    {'ServiceID': 20, 'primary': '20', 'secondary': '12'},
    {'ServiceID': 8,  'primary': '8',  'secondary': '16'},
    {'ServiceID': 12, 'primary': '12', 'secondary': '20'},
    {'ServiceID': 8,  'primary': '8',  'secondary': '16'}
]

k =     {'ServiceID': [], 'primaryCount': [], 'secondaryCount': []}

for i in l:
    k['ServiceID'].append(i['ServiceID'])
    k['primaryCount'].append(i['primary'])
    k['secondaryCount'].append(i['secondary'])

res = {'ServiceID': 0, 'primaryCount': [], 'secondaryCount': []}

result = []

for i in sorted(set(k['ServiceID'])):
    res['ServiceID']=i
    res['primaryCount'] = k['primaryCount' ].count(str(i))
    res['secondaryCount'] = k['secondaryCount' ].count(str(i))
    result.append(res)
    res = {'ServiceID': 0, 'primaryCount': [], 'secondaryCount': []}

print(result)

output

[
 {'ServiceID': 8, 'primaryCount': 2, 'secondaryCount': 1},
 {'ServiceID': 12, 'primaryCount': 1, 'secondaryCount': 4}, 
 {'ServiceID': 16, 'primaryCount': 1, 'secondaryCount': 2},
 {'ServiceID': 20, 'primaryCount': 4, 'secondaryCount': 1}
]

Answer 2

You can do the following (l is your list):

d={i['ServiceID']:{'primaryCount':0, 'secondaryCount':0} for i in l}

for i in l:
    d[int(i['primary'])]['primaryCount']+=1
    d[int(i['secondary'])]['secondaryCount']+=1

res=[{'ServiceID':i, 'primaryCount': k['primaryCount'], 'secondaryCount': k['secondaryCount']} for i, k in d.items()]

Output:

>>> print(res)

[{'ServiceID': 20, 'primaryCount': 4, 'secondaryCount': 1}, {'ServiceID': 16, 'primaryCount': 1, 'secondaryCount': 2}, {'ServiceID': 8, 'primaryCount': 2, 'secondaryCount': 1}, {'ServiceID': 12, 'primaryCount': 1, 'secondaryCount': 4}]

Answer 3

It seems like the correct solution here would involve using collections.Counter s (or largely equivalently in this case, collections.defaultdict(int) s) to allow you to cheaply and easily increment counts without relying on them being adjacent in the input, and without using intermediate data structures that add pointless overhead; why build the result all at once when you can count the parts you care about with simpler code, then build the result with equally simple code from those simple counts? You don't actually use the 'ServiceID' field in the input, so you may as well just count efficiently, and convert back to the preferred format at the end:

import pprint  # For pretty-printing in the example
from collections import Counter

inp = [
    {'ServiceID': 20, 'primary': '20', 'secondary': '12'},
    {'ServiceID': 20, 'primary': '20', 'secondary': '12'},
    {'ServiceID': 20, 'primary': '20', 'secondary': '12'},
    {'ServiceID': 16, 'primary': '16', 'secondary': '8'},
    {'ServiceID': 20, 'primary': '20', 'secondary': '12'},
    {'ServiceID': 8,  'primary': '8',  'secondary': '16'},
    {'ServiceID': 12, 'primary': '12', 'secondary': '20'},
    {'ServiceID': 8,  'primary': '8',  'secondary': '16'}
]

primarycount = Counter()
secondarycount = Counter()
for d in inp:
   primarycount[int(d['primary'])] += 1      # Counts times seen as primary
   secondarycount[int(d['secondary'])] += 1  # Counts times seen as secondary

# Just to see intermediate results
print(primarycount)
print(secondarycount)

# Make new list mapping each thing seen to its counts
# The union of keys ensures anything with even one count in input appears in the output
# Sorting the union before iterating gets desired output order
result = [{'ServiceID': sid, 'primaryCount': primarycount[sid], 'secondaryCount': secondarycount[sid]}
          for sid in sorted(primarycount.keys() | secondarycount.keys())]
pprint.pprint(result)

Try it online!

which produces output:

Counter({20: 4, 8: 2, 16: 1, 12: 1})
Counter({12: 4, 16: 2, 8: 1, 20: 1})
[{'ServiceID': 8, 'primaryCount': 2, 'secondaryCount': 1},
 {'ServiceID': 12, 'primaryCount': 1, 'secondaryCount': 4},
 {'ServiceID': 16, 'primaryCount': 1, 'secondaryCount': 2},
 {'ServiceID': 20, 'primaryCount': 4, 'secondaryCount': 1}]

This might be slightly wrong if some ServiceID s might be seen in the input, but never as a primary or secondary (they won't appear in the output at all, rather than appearing with zero counts; unclear which is correct), or if primary or secondary values sometimes appear where the ServiceID corresponding never appears in the input (they'll show up in the output with counts, rather than being omitted; again, unclear on which is correct), but it's relatively trivial to fix. Flipping both behaviors would just involve changing primarycount.keys() | secondarycount.keys() primarycount.keys() | secondarycount.keys() to {d['ServiceID'] for d in inp} to ensure values come from the input ServiceID fields, not a combination of all values seen for primary and secondary . For the provided input, both approaches are equivalent (with the former being slightly faster in most cases, where there are many duplicate ServiceID s in the input).