CODEWITHSUNDEEP
pythonlistdictionaryzip
Windy71
Windy71

Reputation: 909

Pythonic method to create 2 lists, one of which is a repeating set of strings and turn into a dict

I have a list of 'widget types', ["C","A","B"] and need to make a dict for those types to correspond to their respective ID's, 1='C', 2='A', 3 = 'B', 4= 'C', 5='A', 6= 'B', 7 = 'C', etc. I already know how to do it, I just wondered if there was a more elegant pythonic way of achieving it than that below, also it would be good to have the length of the dict/lists expandable, i.e. in the example it is 10 but in reality it would be many hundreds of widgets long.

Existing working code

wid_letter = ['C', 'A', 'B']
widgets_per_row = 10
wid_letter = wid_letter * widgets_per_row 
#print("wid_letter : ", wid_letter, '\n')



ID = (list(range(1,widgets_per_row  + 1,1)))
wid_type = []

# Make list of widget_letters to match length of widget ID
for i in ID:
    wid_type.append(wid_letter[i-1])
print(len(wid_type))

# Turn the two lists into a dict
wid_ID_dict = dict(zip(ID, wid_type))
print('\n'*2, wid_ID_dict,'\n')

print("Widget_type: ", wid_ID_dict[1])

Upvotes: 1

Views: 43

Answers (4)

Jacek Błocki
Jacek Błocki

Reputation: 563

wid = {i:w for i,w in enumerate(wid_letter[:widgets_per_row],1)}

Dict comprehension above may be your quest item, wid is wid_ID_dict in your code.

Upvotes: 0

Rustam Garayev
Rustam Garayev

Reputation: 2692

Instead of overriding your list, you can just use module to get the index of letter you need. Then simple dictionary comprehension will do the job:

wid_letter = ['C', 'A', 'B']
widgets_per_row = 10

wid_ID_dict = {i + 1: wid_letter[i%len(wid_letter)] for i in range(widgets_per_row)}

Result would look like this:

{1: 'C', 2: 'A', 3: 'B', 4: 'C', 5: 'A', 6: 'B', 7: 'C', 8: 'A', 9: 'B', 10: 'C'}

Upvotes: 2

Loic RW
Loic RW

Reputation: 444

This leads to the same result, is this what you are looking for?

from itertools import cycle

wid_letter = ['C', 'A', 'B']
letter_cycle = cycle(wid_letter)
widgets_per_row = 10

wid_ID_dict = {i: next(letter_cycle) for i in range(1,widgets_per_row+1)}

print(wid_ID_dict)

Upvotes: 1

Sakurai
Sakurai

Reputation: 106

Try：

wid_ID_dict = dict(enumerate(wid_letter, start=1))

Upvotes: 0

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