Create SEQUENCE based dictionary from list

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By : user3042697

Date : November 29 2020, 12:01 PM

this one helps. What you are trying to achieve is not clear to me. But, if I understand, you want to transform a list of dicts into a tree (or a forest if you have more than one root), following a predefined list of groups. The groups are ordered from the roots to the leaves:

code :

groups = ['benefit_categ_seq', 'benefit_category_name', 'comp_seq', 'insurance_category_name', 'sale_line_id']

def nest(L, path):
    *init_path, last_path = path # split before the last element
    root = {}
    for d in L:
        e = root # start at the root
        for g in init_path: # follow the path
            e = e.setdefault(d[g], {}) # get or create the subtree
        e[d[last_path]] = d # add the dict to the leaf

    return root

from pprint import pprint
pprint (nest(result, groups))
{1: {'Standard Benefits': {1: {'A': {34353: {'benefit_categ_seq': 1,
                                             'benefit_category_name': 'Standard '
                                                                      'Benefits',
                                             'benefit_name': 'TPA',
                                             'comp_seq': 1,
                                             'insurance_category_name': 'A',
                                             'sale_line_id': 34353},
                                     34355: {'benefit_categ_seq': 1,
                                             'benefit_category_name': 'Standard '
                                                                      'Benefits',
                                             'benefit_name': 'TPA',
                                             'comp_seq': 1,
                                             'insurance_category_name': 'A',
                                             'sale_line_id': 34355}},
                               'B': {34354: {'benefit_categ_seq': 1,
                                             'benefit_category_name': 'Standard '
                                                                      'Benefits',
                                             'benefit_name': 'TPA',
                                             'comp_seq': 1,
                                             'insurance_category_name': 'B',
                                             'sale_line_id': 34354},
                                     34356: {'benefit_categ_seq': 1,
                                             'benefit_category_name': 'Standard '
                                                                      'Benefits',
                                             'comp_seq': 1,
                                             'insurance_category_name': 'B',
                                             'sale_line_id': 34356}}}}}}

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create dictionary from list - in sequence

By : Nanosoft

Date : March 29 2020, 07:55 AM

it helps some times Dictionaries don't have any order, use collections.OrderedDict if you want the order to be preserved. And instead of using indices use an iterator.

code :

>>> from collections import OrderedDict
>>> lis = ['a', 1, 'b', 2, 'c', 3, 'd', 4]
>>> it = iter(lis)
>>> OrderedDict((k, next(it)) for k in it)
OrderedDict([('a', 1), ('b', 2), ('c', 3), ('d', 4)])

How to create list based on elements from dictionary?

By : James

Date : March 29 2020, 07:55 AM

With these it helps I have this list of dictionaries: , You can pass a generator to sorted:

code :

>>> codes = sorted( d['code'] for d in L )
>>> codes
['AF', 'AM', 'UE']

codes = sorted([ d['code'] for d in L ])

R -- Create Variable in List of Data Frames Based on a Sequence

By : user2850740

Date : March 29 2020, 07:55 AM

help you fix your problem If you want to iterate over two lists/vectors simultaneously, you can either use mapply() or Map. Here's code using the latter

code :

Map(function(x,z) cbind(x,z=z), ldf, 1:4)

How to create a list based on same value of a dictionary key

By : Adam Ward

Date : March 29 2020, 07:55 AM

I hope this helps . Iterate over group to sort out mins and maxs to separate keys of the dictionary:

code :

def get_temp(temp):
    return temp['date']

lst = []
for key, group in itertools.groupby(data, get_temp):
    groups = list(group)
    d = {}
    d['date'] = key
    d['temp_min'] = [x['temp_min'] for x in groups]
    d['temp_max'] = [x['temp_max'] for x in groups]
    lst.append(d)

print(lst)

Create a dictionary based on same key pairs from dictionary inside a list

By : Ashish Wadekar

Date : March 29 2020, 07:55 AM

like below fixes the issue defaultdict
You can use collections.defaultdict with iteration. Given an input list of dictionaries L:

code :

from collections import defaultdict

d = defaultdict(list)

for i in L:
    d[i['c1'].strip()+'#'+i['c2']].append(i['keywords'])

print(d)

defaultdict(list,
            {'Cars#Class': ['muv', 'hatchback', 'suv', 'sedan', 'coupe'],
             'Cars#FuelType': ['electric', 'diesel', 'cng', 'petrol']})