List to Dictionary Conversion with Multiple Values Per Key

list to dictionary conversion with multiple values per key?

from collections import defaultdict

d1 = defaultdict(list)

for k, v in l:
    d1[k].append(v)

d = dict((k, tuple(v)) for k, v in d1.items())

d contains now {1: ('A', 'B'), 2: ('C',)}

d1 is a temporary defaultdict with lists as values, which will be converted to tuples in the last line. This way you are appending to lists and not recreating tuples in the main loop.

Handling a dictionary with multiple values per key

You almost had it, you need to create a list for each key and then just append the timestamps.

Try this:

team2goals = {}
<loop through list>
    timestamp = xValue
    if name not in team2goals:
        team2goals[name] = []
    team2goals[name].append(timestamp)

Converting list of lists to a dictionary with multiple values for a key

You can iterate through the key-value pairs in the list of lists, but unpack the value as a list to accommodate the possible lack of a value:

s = [['A','B'],['A','C'],['A','D'],['B','A'],['C','B'],['C','D'],['D','B'],['E']]
output = {}
for k, *v in s:
    if v:
        output.setdefault(k, []).extend(v)
    else:
        output[k] = None

output becomes:

{'A': ['B', 'C', 'D'], 'B': ['A'], 'C': ['B', 'D'], 'D': ['B'], 'E': None}

Or if you don't mind that keys without a value get an empty list instead of None, you can simply do:

output = {}
for k, *v in s:
    output.setdefault(k, []).extend(v)

output would then become:

{'A': ['B', 'C', 'D'], 'B': ['A'], 'C': ['B', 'D'], 'D': ['B'], 'E': []}

List of dictionaries with multiple values per key as Dataframe

pd.DataFrame(d).stack().apply(
    pd.Series).add_prefix(
    'Value').reset_index().rename(columns={'level_0':'index', 'level_1':'Date'})

Output

    index Date               Value0 Value1  Value2  Value3
0   0     2020-12-31 00:00:00   123  456    789     321
1   1     2021-05-06 00:00:00   999  888    777     666

How convert multiple keys into single of dictionary and create Dictionary with multiple values per key in python

I think this is what you're looking for:

d = {
    "Parameter_1": {
        "2021-11-16 14:29:00": 319.56,
        "2021-11-16 15:16:00": 319.56,
        "2021-11-16 15:17:00": 319.56,
        "2021-11-17 00:00:00": 335.48,
        "2021-11-17 00:01:00": 335.48,
        "2021-11-17 00:02:00": 335.48,
        "2021-11-18 00:00:00": 355.45,
        "2021-11-18 00:01:00": 355.45,
        "2021-11-18 00:03:00": 355.45,
},
    "Parameter_2": {
        "2021-11-16 14:29:00": 319.56,
        "2021-11-16 15:16:00": 319.56,
        "2021-11-16 15:17:00": 319.56,
        "2021-11-17 00:00:00": 335.48,
        "2021-11-17 00:01:00": 335.48,
        "2021-11-17 00:02:00": 335.48,
         "2021-11-18 00:00:00": 355.45,
        "2021-11-18 00:01:00": 355.45,
        "2021-11-18 00:03:00": 355.45,
}}

r = dict()

for k, v in d.items():
    r[k] = dict()
    for k_, v_ in v.items():
        r[k].setdefault(k_[:10], []).append(v_)

print(r)

Output:

{'Parameter_1': {'2021-11-16': [319.56, 319.56, 319.56], '2021-11-17': [335.48, 335.48, 335.48], '2021-11-18': [355.45, 355.45, 355.45]}, 'Parameter_2': {'2021-11-16': [319.56, 319.56, 319.56], '2021-11-17': [335.48, 335.48, 335.48], '2021-11-18': [355.45, 355.45, 355.45]}}

Merge two lists into one dict that may contain several values for each key

Python does not have multidicts, so you have two options:

1. use multidicts from an existing library e.g. werkzeug's MultiDict, when initialised with lists of pairs, will associate multiple values to keys which are present multiple times (unlike dict which will only keep the last value)

   system_data_dict = werkzeug.datastructures.MultiDict(zip(system, instrument))
   system_data_dict.getlist('System A') # => ['Instrument 1', 'Instrument 2']
   

2. alternatively, do that by hand by using a regular loop, even if it's feasible using a comprehension (which I'm not sure about) it's going to look dreadful: use a defaultdict or the dict.setdefault method to define a dict mapping keys to lists and append values every time

   system_data_dict = {}
   for k, v in zip(system, instrument):
      system_data_dict.setdefault(k, []).append(v)
   system_data_dict['System A'] # => ['Instrument 1', 'Instrument 2']
   

Python convert list to dict with multiple key value

Several ways to do this, this is one:
EDIT: my first solution gave a list for every value, but you only require a list when there is more than one value for a key.

my_list = ['key1=value1', 'key2=value2', 'key3=value3-1', 'value3-2', 'value3-3', 'key4=value4', 'key5=value5', 'value5-1', 'value5-2', 'key6=value6']

my_dict = {}
current_key = None
for item in my_list:
    if '=' in item:
        current_key, value = item.split('=')
        # This puts a string as the value
        my_dict[current_key] = value  
    else:
        # Check if the value is already a list
        if not isinstance(my_dict[current_key], list):
            # If value is not a list, create one
            my_dict[current_key] = [my_dict[current_key]]

        my_dict[current_key].append(item)

import pprint
pprint.pprint(my_dict)

Gives:

{'key1': 'value1',
 'key2': 'value2',
 'key3': ['value3-1', 'value3-2', 'value3-3'],
 'key4': 'value4',  
 'key5': ['value5', 'value5-1', 'value5-2'],
 'key6': 'value6'}

You might wish to make it more robust by checking if current_key is None. I'll leave that to you.

Convert list of tuples of several values into dictionary in Python 3.8

My approach is bit different, probably you can checkout this answer too:

user_list = [("a", 1), ("b", 2), ("a", 3), ("b", 1), ("a", 2), ("c", 1)]

final = {}
for item in user_list:
    final.update(
        {
            item[0]: [item[1]] + final[item[0]] if final.get(item[0]) else [item[1]]
        }
    )
print(final)

output: {'a': [2, 3, 1], 'b': [1, 2], 'c': [1]}

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