Groupby Results to Dictionary of Lists

GroupBy results to dictionary of lists

You could groupby on Column1 and then take Column3 to apply(list) and call to_dict?

In [81]: df.groupby('Column1')['Column3'].apply(list).to_dict()
Out[81]: {0: [1], 1: [2, 3, 5], 2: [1, 2], 3: [4, 5], 4: [1], 5: [1, 2, 3]}

Or, do

In [433]: {k: list(v) for k, v in df.groupby('Column1')['Column3']}
Out[433]: {0: [1], 1: [2, 3, 5], 2: [1, 2], 3: [4, 5], 4: [1], 5: [1, 2, 3]}

Pandas groupby and get dict in list

You can use itertuples and defulatdict:

itertuples returns named tuples to iterate over dataframe:

for row in df.itertuples():
    print(row)
Pandas(Index=0, x=1, y=3, label=1.0)
Pandas(Index=1, x=4, y=2, label=1.0)
Pandas(Index=2, x=5, y=5, label=2.0)

So taking advantage of this:

from collections import defaultdict
dictionary = defaultdict(list)
for row in df.itertuples():
    dummy['x'] = row.x
    dummy['y'] = row.y
    dummy['index'] = row.Index
    dictionary[row.label].append(dummy)

dict(dictionary)
> {1.0: [{'x': 1, 'y': 3, 'index': 0}, {'x': 4, 'y': 2, 'index': 1}],
 2.0: [{'x': 5, 'y': 5, 'index': 2}]}

In Python Creating dictionary from the group-by result and making list for the unique elements

Group, apply list, turn into dict.

result = df.groupby(['pid', 'wips', 'be'])['rownum'].apply(list).reset_index().to_dict(orient='records')

Convert result from groupby on multiple columns to list of dictionaries of dictionaries

Let us try defining a function dictify that takes the input argument as the multiindex pandas series and return the nested dictionary in the required format

def dictify(s):
    if s.index.nlevels == 1: return s.to_dict()
    return {k: dictify(g.droplevel(0)) for k, g in s.groupby(level=0)}

records = [{k: v} for k, v in dictify(df).items()]

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>>> df

DE          SW  Tech  
2021-01-01  s1  t1          1
2021-04-02  s1  t1        699
                t2        268
2021-04-06  s1  t1          2
            s2  t1        466
                t2         81
                t3        954
Name: COUNT(*), Length: 474, dtype: int64

>>> records

[{'2021-01-01': {'s1': {'t1': 1}}},
 {'2021-04-02': {'s1': {'t1': 699, 't2': 268}}},
 {'2021-04-06': {'s1': {'t1': 2}, 's2': {'t1': 466, 't2': 81, 't3': 954}}}]

Convert pandas.groupby to dict

You can use dict with tuple / list applied on your groupby:

res = dict(tuple(d.groupby('a')))

A memory efficient alternative to dict is to create a groupby object and then use get_group:

res = d.groupby('a')
res.get_group(1)  # select dataframe where column 'a' = 1

In cases where the resulting table requires a minor manipulation, like resetting the index, or removing the groupby column, continue to use a dictionary comprehension.

res = {k: v.drop('a', axis=1).reset_index(drop=True) for k, v in d.groupby('a')}

How to convert pandas groupby result into a consolidated dictionary?

Try adding this code using reindex, repeat, reset_index, get_dummies, groupby and to_dict:

groupings = groupings.reset_index()
groupings.columns = [i if i != 0 else 'Total' for i in groupings.columns]
groupings = groupings.reindex(groupings.index.repeat(groupings['Total'])).reset_index(drop=True)
groupings['Total'] = 1
print(pd.get_dummies(groupings, columns=['Player', 'Status'], prefix='', prefix_sep='').groupby(['Country', 'State', 'Team'], as_index=False).sum().to_dict('records'))

Output:

[{'Country': 'Australia', 'State': 'Victoria', 'Team': 'United', 'Total': 4, 'Davis': 0, 'Dudley': 0, 'James': 0, 'Jones': 1, 'Kumar': 0, 'Singh': 0, 'Smith': 3, 'NBA': 1, 'NBA Legend': 0, 'Normal': 3}, {'Country': 'India', 'State': 'Karnataka', 'Team': 'Beasts', 'Total': 1, 'Davis': 0, 'Dudley': 0, 'James': 0, 'Jones': 0, 'Kumar': 1, 'Singh': 0, 'Smith': 0, 'NBA': 0, 'NBA Legend': 0, 'Normal': 1}, {'Country': 'India', 'State': 'Punjab', 'Team': 'Steelers', 'Total': 1, 'Davis': 0, 'Dudley': 0, 'James': 0, 'Jones': 0, 'Kumar': 0, 'Singh': 1, 'Smith': 0, 'NBA': 0, 'NBA Legend': 0, 'Normal': 1}, {'Country': 'United States', 'State': 'California', 'Team': 'Lakers', 'Total': 2, 'Davis': 1, 'Dudley': 0, 'James': 1, 'Jones': 0, 'Kumar': 0, 'Singh': 0, 'Smith': 0, 'NBA': 1, 'NBA Legend': 1, 'Normal': 0}, {'Country': 'United States', 'State': 'New York', 'Team': 'Globetrotters', 'Total': 1, 'Davis': 0, 'Dudley': 1, 'James': 0, 'Jones': 0, 'Kumar': 0, 'Singh': 0, 'Smith': 0, 'NBA': 0, 'NBA Legend': 0, 'Normal': 1}]

Pandas DataFrame.groupby() to dictionary with multiple columns for value

Customize the function you use in apply so it returns a list of lists for each group:

df.groupby('Column1')[['Column2', 'Column3']].apply(lambda g: g.values.tolist()).to_dict()
# {0: [[23, 1]], 
#  1: [[5, 2], [2, 3], [19, 5]], 
#  2: [[56, 1], [22, 2]], 
#  3: [[2, 4], [14, 5]], 
#  4: [[59, 1]], 
#  5: [[44, 1], [1, 2], [87, 3]]}

If you need a list of tuples explicitly, use list(map(tuple, ...)) to convert:

df.groupby('Column1')[['Column2', 'Column3']].apply(lambda g: list(map(tuple, g.values.tolist()))).to_dict()
# {0: [(23, 1)], 
#  1: [(5, 2), (2, 3), (19, 5)], 
#  2: [(56, 1), (22, 2)], 
#  3: [(2, 4), (14, 5)], 
#  4: [(59, 1)], 
#  5: [(44, 1), (1, 2), (87, 3)]}

python : group by multiple dictionary keys

You can use groupby() to group dicts in combination with itemgetter():

from itertools import groupby
from operator import itemgetter

list_pts = [
    {'city': 'Madrid', 'year': '2017', 'date': '05/07/2017', 'pts': 7},
    {'city': 'Madrid', 'year': '2017', 'date': '14/11/2017', 'pts': 5},
    {'city': 'Londres', 'year': '2018', 'date': '25/02/2018', 'pts': 5},
    {'city': 'Paris', 'year': '2019', 'date': '17/04/2019', 'pts' : 4},
    {'city': 'Londres', 'year': '2019', 'date': '15/06/2019', 'pts': 8},
    {'city': 'Paris', 'year': '2019', 'date': '21/08/2019', 'pts': 8},
    {'city': 'Londres', 'year': '2019', 'date': '04/12/2019', 'pts': 2}
]

city_year_getter = itemgetter('city', 'year')
date_pts_getter = itemgetter('date', 'pts')

result = []
for (city, year), objs in groupby(sorted(list_pts, key=city_year_getter), 
                                  city_year_getter):
    dates, ptss = zip(*map(date_pts_getter, objs))
    result.append({
        'city': city,
        'year': year,
        'date': list(dates),
        'pts': list(ptss)
    })

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