CODEWITHSUNDEEP
python
CaptainDriftwood
CaptainDriftwood

Reputation: 5708

python group records together by field

I have a group of records that I am wanting to group together on two separate fields. Each record is a python dictionary. One of the fields is a date value and the other is a number field. I.e.,:

h = [{'date': 20170728, 'group': 121, ...},
     {'date': 20170729, 'group': 131, ...},
     ...]

Now, if I wanted to group certain groups together, say any group that is in [123, 134, 145] but has the same date, group them together, but every other group gets grouped together by themselves, how would I accomplish this?

I'm using the following code:

grouped_list = []
for date, items in groupby(h, key=itemgetter('date'):
    g = list(items)
    grouped_list.append(g)

The output that I'm looking for is the following:

grouped_list = [
                [records that have a distinct date value and group],
                [records that have a distinct date but are in the group [123, 134, 145],
               etc.]

The records in groups 123, 134, and 145 should not have their own respective lists in the grouped_list. They should be grouped together in a list.

Upvotes: 1

Views: 661

Answers (1)

pchiquet
pchiquet

Reputation: 3187

You can write a custom function to compute the keys to group your records, something like this:

from itertools import groupby

records = [
      {'date': 20170728, 'group': 121},
      {'date': 20170729, 'group': 131}, 
      {'date': 20170729, 'group': 134},
      {'date': 20170729, 'group': 145}, 
]
grouped_groups = [123, 134, 145]

def compute_groupby_key(entry): 
      return "%d-%d" % (
            entry['date'],
            grouped_groups[0] if entry['group'] in grouped_groups else entry['group']
      )

grouped_records = [list(entries) for key, entries in groupby(records, compute_groupby_key)]

here grouped_records contains:

[
    [{'date': 20170728, 'group': 121}],
    [{'date': 20170729, 'group': 131}],
    [{'date': 20170729, 'group': 134}, {'date': 20170729, 'group': 145}]]
]

Upvotes: 1

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