CODEWITHSUNDEEP
pythonpython-3.xlist
Saad Hasan
Saad Hasan

Reputation: 152

Creating a dictionary from 2 lists by storing multiple values with repeated keys

There are 2 lists. What I am trying to do is to find the occurrence of first list elements and will hold the values of second list for a key in first list and in the end, it will become dictionary holding specific keys from list 1 and values from list 2

Input:

list1 = ['A', 'A', 'B', 'B', 'C', 'D']
list2 = [1, 2, 3, 4, 5, 6]

Expected Output:

{'A': [1, 2], 'B': [3, 4], 'C': [5], 'D':[6]}

Current Output:

{'A': [1, 2], 'B': [4], 'C': []}

What I tried so far,

values = [list2[0]]
key = list1[0]
dic = {}
for i in range(1, len(list1)):
    if list1[i] == list1[i - 1]:
        values.append(list2[i])
    elif list1[i] != list1[i - 1]:
        dic.update({key: values})
        values = []
        key = list1[i]
print(dic)

List 1 and List 2 are always equal in length and sorted

Upvotes: 2

Views: 54

Answers (3)

Talha Tayyab
Talha Tayyab

Reputation: 27385

You can also use defaultdict from collections

from collections import defaultdict

list1 = ['A', 'A', 'B', 'B', 'C', 'D']

list2 = [1, 2, 3, 4, 5, 6]

d = defaultdict(list)
for key, value in zip(list1, list2):
    d[key].append(value)

print(dict(d))
{'A': [1, 2], 'B': [3, 4], 'C': [5], 'D': [6]}

Upvotes: 1

blhsing
blhsing

Reputation: 106553

You can use dict.setdefault to initialize each distinct key with a new list to append values to:

output = {}
for key, value in zip(list1, list2):
    output.setdefault(key, []).append(value)

output would become:

{'A': [1, 2], 'B': [3, 4], 'C': [5], 'D': [6]}

Demo: https://replit.com/@blhsing/InstructiveNoxiousCollaborativesoftware

Upvotes: 1

Lorenz Hetterich
Lorenz Hetterich

Reputation: 306

There are multiple small issues with your code:

  1. you are missing the last entry in the dict, because the dict is only updated, if the value changes. However, for the last value (in your case 'D'), this is not the case.
  2. by initializing values with an empty list, you loose the first value

These issues could be fixed the following way:

  1. add dic.update({key: values}) below the loop
  2. change values = [] to values = [list2[i]]

Upvotes: 1

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