How to loop through a list of dictionary and extract those with the same 'name' and 'school' into a new list while getting their other values in it

Question

I have this list of dictionary and I would like to get those with the same exact value of ‘name’ and ‘school’ into a new list and also getting their ‘age’ merged into a list as well and the rest of the dictionary that is not identical to just add into the list as per usual..

Here is an example of the list of dictionary

[{'name': 'Jane', 'age':12, 'school': 'SIT'}, {'name': 'John', 'age':13, 'school': 'SMU'},{'name': 'Jane', 'age':14, 'school': 'SIT'}, {'name': 'Jane', 'age':16, 'school': 'SIT'}, {'name': 'John', 'age':13, 'school': 'NUS'}]

and I would like it to make it into something like this..

[{'name': 'Jane', 'age': [12,14,16], 'school': 'SIT'}, {'name': 'John', 'age': 13, 'school': 'SMU'}, {'name': 'John', 'age':13, 'school': 'NUS'}]

using Python.. please help!

tried using counter, loops but still can’t get it to work..

Asked By: Qinisterr

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Source

Answer 1

x =  [{'name': 'Jane', 'age':12, 'school': 'SIT'}, {'name': 'John', 'age':13, 'school': 'SMU'},{'name': 'Jane', 'age':14, 'school': 'SIT'}, {'name': 'Jane', 'age':16, 'school': 'SIT'}, {'name': 'John', 'age':13, 'school': 'NUS'}]

new_x = {}
for r in x:
    if r['name'] in new_x.keys():
        if not isinstance(new_x[r['name']]['age'], list):
            new_x[r['name']]['age'] = [new_x[r['name']]['age']]
        if r['age'] not in new_x[r['name']]['age']:
            new_x[r['name']]['age'].append(r['age'])
    else:
        new_x[r['name']] = {'age': r['age'], 'school': r['school']}
z = [v.update(name=k) for k, v in new_x.items()]
z = [v for k, v in new_x.items()]

Answered By: Ashutosh Kulkarni

Answer 2

Here is a universal solution to your problem. Only name and school are considered "special". All other keys, like age are converted to list when a new value has to be added.

l = [
    {"name": "Jane", "age": 12, "school": "SIT"},
    {"name": "John", "age": 13, "school": "SMU"},
    {"name": "Jane", "age": 14, "school": "SIT"},
    {"name": "Jane", "age": 16, "school": "SIT"},
    {"name": "John", "age": 13, "school": "NUS"},
]
r = {}
for x in l:
    id = f"{x['name']}-{x['school']}"
    if id in r:
        for k,v in x.items():
            if k not in ["name", "school"]:
                if k in r[id]:
                    if isinstance(r[id][k], list):
                        r[id][k].append(v)
                    else:
                        r[id][k] = [r[id][k], v]
                else:
                    r[id][k] = v
    else:
        r[id] = x

result = [x for x in r.values()]

Answered By: MartinsM

Answer 3

You could use itertools.groupby().

Example:

import itertools
from pprint import pprint

data = [{'name': 'Jane', 'age':12, 'school': 'SIT'}, {'name': 'John', 'age':13, 'school': 'SMU'},{'name': 'Jane', 'age':14, 'school': 'SIT'}, {'name': 'Jane', 'age':16, 'school': 'SIT'}, {'name': 'John', 'age':13, 'school': 'NUS'}]

keyfunc = lambda x: (x["name"], x["school"])
# needs to be sorted to use groupby
data.sort(key=keyfunc)
output = []
for k,v in itertools.groupby(data, key=keyfunc):
    this_group = {
        "name": k[0],
        "school": k[1],
        "age": [i["age"] for i in v],
    }
    output.append(this_group)

pprint(output)

The output is:

[{'age': [12, 14, 16], 'name': 'Jane', 'school': 'SIT'},
 {'age': [13], 'name': 'John', 'school': 'NUS'},
 {'age': [13], 'name': 'John', 'school': 'SMU'}]

If you wish to go with the solution based on a buffer dictionary, please check out the dict.setdefault() method.

Example:

buffer = {}
for i in data:
    buffer.setdefault((i["name"], i["school"]), []).append(i["age"])

For reference:

Answered By: moooeeeep

How to loop through a list of dictionary and extract those with the same 'name' and 'school' into a new list while getting their other values in it

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