Iterate over list of dict and group item by key

问题

I have a list of dicts:

my_list = [
    {'name': 'AAA', 'date': '2018-05-14', 'price': 20.0},
    {'name': 'AAA', 'date': '2018-05-15', 'price': 22.0},
    {'name': 'AAA', 'date': '2018-05-16', 'price': 30.0},
    {'name': 'BBB', 'date': '2018-05-14', 'price': 15.0},
    {'name': 'BBB', 'date': '2018-05-15', 'price': 32.0}
    ]

My question is how can I iterate over that list to produce a list in this format?

parsed_list = [
    {'name': 'AAA', 'data': [['2018-05-14', 20.0], ['2018-05-15', 22.0], ['2018-05-16', 30.0]]},
    {'name': 'BBB', 'data': [['2018-05-14', 15.0], ['2018-05-15', 32.0]]}
    ]

I tried approach described in this question: Python: group list items in a dict but I need a different output format and I can't figure out what I need to change.

回答1:

One way may be to use defaultdict:

from collections import defaultdict

my_list = [
    {'name': 'AAA', 'date': '2018-05-14', 'price': 20.0},
    {'name': 'AAA', 'date': '2018-05-15', 'price': 22.0},
    {'name': 'AAA', 'date': '2018-05-16', 'price': 30.0},
    {'name': 'BBB', 'date': '2018-05-14', 'price': 15.0},
    {'name': 'BBB', 'date': '2018-05-15', 'price': 32.0}
    ]

tmp = defaultdict(list)

for item in my_list:
    tmp[item['name']].append([item['date'],item['price']])

parsed_list = [{'name':k, 'data':v} for k,v in parsed_list.items()]
print(parsed_list)

Result:

[{'name': 'AAA', 'data': [['2018-05-14', 20.0], 
                          ['2018-05-15', 22.0], ['2018-05-16', 30.0]]}, 
 {'name': 'BBB', 'data': [['2018-05-14', 15.0], ['2018-05-15', 32.0]]}]

回答2:

One way is to use itertools.groupby():

from itertools import groupby
print([{"name": key, "data": [(g['date'], g['price']) for g in group]} 
       for key, group in groupby(my_list, lambda x: x['name'])])
#[{'name': 'AAA', 'data': [('2018-05-14', 20.0), ('2018-05-15', 22.0), ('2018-05-16', 30.0)]},
# {'name': 'BBB', 'data': [('2018-05-14', 15.0), ('2018-05-15', 32.0)]}]

The first argument to groupby is an iterable, in this case my_list.

The second argument is a function that defines how to create the groups, in this case you extract the key name.

Note: this will group consecutive items with the same name, so it assumes that my_list is already sorted with respect to name. If not you can sort first using:

my_list = sorted(my_list, key=lambda x: x['name'])

Then we can iterate over all the (key, group) pairs from the output of groupby() and do a list comprehension.

Inside the list comprehension, we do a dict comprehension {"name": key, "data": [(g['date'], g['price']) for g in group]} to build a dictionary of the form {'name': key, 'data': [[date,price]]} for each name.

来源：https://stackoverflow.com/questions/50491984/iterate-over-list-of-dict-and-group-item-by-key

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python

dictionary

grouping