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List of lists to dictionary of specific format


By : poineer911
Date : September 12 2020, 04:00 PM
seems to work fine The error you are getting is due to the fact you are trying to set a list as a dictionary key. The following example will reproducer the same results:
code :
>>> d = dict()
>>> d[['a', 'b']] = 123
Traceback (most recent call last):
  File "<pyshell#18>", line 1, in <module>
    d[['a', 'b']] = 123
TypeError: unhashable type: 'list'
{ ['a','b']: 2, 
  ['c','d']: 2, 
  ['a','b','c']: 3
}
import inflect
p = inflect.engine()

l=[['a','b'],['c','d'],['a','b','c']]
new_list = list()
count = 1
already_used_labels = set()

# The nested loop...
for inner_list in l:
    for label in inner_list:
        if label in already_used_labels:
            continue
        new_list.append({'data': {'id': p.number_to_words(count), 'label': label}})
        count += 1
        already_used_labels.add(label)
[
 {'data': {'id': 'one', 'label': 'a'}}, 
 {'data': {'id': 'two', 'label': 'b'}},
 {'data': {'id': 'three', 'label': 'c'}}, 
 {'data': {'id': 'four', 'label': 'd'}}
]


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Convert python list of lists into dictionary format


By : Vinayak Nanda
Date : March 29 2020, 07:55 AM
may help you . That's pretty simple. The dict constructor already accepts an iterable of two-element-iterables. The only extra difficulty here is that your keys and values are in reverse order.
code :
>>> mylist = [[14.76, u'2017-10-11T06:00:00Z'], [14.75, u'2017-10-11T07:00:00Z'], [14.21, u'2017-10-11T08:00:00Z']]
>>> dict(map(reversed, mylist))
{u'2017-10-11T07:00:00Z': 14.75, u'2017-10-11T06:00:00Z': 14.76, u'2017-10-11T08:00:00Z': 14.21}*

create nested dictionary from specific elements of 2 list of lists


By : M. Shif
Date : March 29 2020, 07:55 AM
help you fix your problem The trouble you're running into here is that dict() (and dictionary comprehensions as well) will favor the last key-value pair it encounters if it runs into duplicate keys, e.g.:
code :
dict([(1,2), (1,3)]) == {1: 3}
result = {}

for arr, fa in zip(fa_n, serv_n):   
    key = arr[3]

    if key in result:
        result[key][arr[4]] = fa[1]
    else:
        result[key] = {arr[4]: fa[1]}

print result

How to filter list lines of a dictionary of lists according to a specific number?


By : Sinan Thottan
Date : March 29 2020, 07:55 AM
wish help you to fix your issue If your lists are already sorted, we can use bisect to find the place between the "Experiencer" and "Status" entries:
code :
from bisect import bisect

l=[(2310, "Experiencer: 'like hours'", 212, 222), (2310, "Experiencer: 'two'", 1035,1038), (2310, "Experiencer: 'Anakin'", 1560, 1566), (2310, "Experiencer: ' '", 1619, 1620), (2310, "Experiencer: 'Tatooine'", 1726, 1734), (2310, "Experiencer: 'Anakin'", 1775, 1781), (2310, "Experiencer: 'Master Qui-Gon'", 1863, 1877), (2310, "Experiencer: 'half'", 1883, 1887), (2310, "Experiencer: 'One'", 2114, 2117), (2310, "Experiencer: 'Anakin'", 2180, 2186), (2310, "Stimulus: 'One'", 2484, 2487), (2310, "Stimulus: 'Anakin'", 2564, 2570), (2310, "Stimulus: 'Padme'", 2739, 2744)]

right_index = bisect(l, (2310, "F"))  # "F" comes between "Experiencer" and "Status" 
lower, higher = l[right_index-1], l[right_index]

print(lower, higher, sep="\n")

# (2310, "Experiencer: 'Anakin'", 2180, 2186)
# (2310, "Stimulus: 'One'", 2484, 2487)
from bisect import bisect

def get_boundary(l):  # This assumes all lists in your dict have at least 2 items
    if len(l) < 2:
        return l
    right_index = bisect(l, (l[0][0], "F"))  
    return [l[right_index-1], l[right_index]]

print({key: get_boundary(value) for key, value in d.items()})
{'hate': [(2310, "Experiencer: 'Anakin'", 2180, 2186), 
          (2310, "Stimulus: 'One'", 2484, 2487)], 
 'confirmation': [(4132, "Experiencer: 'Qui-Gon's'", 4100, 4109), 
                  (4132, "Stimulus: 'Anakin'", 4281, 4287)]
}

How to Have a dictionary that appends multiple values via list, from another list in specific format


By : Meena Vedam
Date : March 29 2020, 07:55 AM
will be helpful for those in need There are two reasonable ways to go about making a dictionary that contains lists that you add to, rather than single items. The first is to check for an existing value before adding a new one. The second is to use a more sophisticated data structure, that takes care of creating the lists whenever it is necessary.
Here's a quick example of the first approach:

How to return the list of dictionary in specific format in python


By : FreePvp
Date : September 22 2020, 06:00 AM
wish of those help Grouping of elements can be done by itertools.groupby, removing the duplicates by set.
For example:
code :
from pprint import pprint
from itertools import groupby

l = [{'pf': 'ABC', 'src': 'SCI'}, {'pf': 'ABC', 'src': 'MC'}, {'pf': 'ASR', 'src': 'CC'}, {'pf': 'ASR', 'src': 'CC'}, {'pf': 'ASR', 'src': 'ACS'}, {'pf': 'ABC', 'src': 'CC'}, {'pf': 'NS', 'src': 'ATT'}]

out = {'customers': [], 'platform': 'all'}
for v, g in groupby(sorted(l, key=lambda k: k['pf']), lambda k: k['pf']):
    out['customers'].append( {'pf': v, 'src': list(set(val['src'] for val in g))} )

pprint(out)
{'customers': [{'pf': 'ABC', 'src': ['MC', 'SCI', 'CC']},
               {'pf': 'ASR', 'src': ['ACS', 'CC']},
               {'pf': 'NS', 'src': ['ATT']}],
 'platform': 'all'}
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