Create dictionary with length of string as key and string as value

Question

I have a list with random strings, as the one below:

strings =  ["zone", "abigail", "theta", "form", "libe", "zas"]

What I want is a dictionary with len of those strings (non unique, linear as they are in the list) as keys, and strings as values. Seems banal, but I am trying to do this in one line, find a nicer way to write it.

I tried this:

dict_of_len = {len(element):element for element in strings}

but somehow I get this as an output:

{4: 'libe', 7: 'abigail', 5: 'theta', 3: 'zas'}

len(dict_of_len) equals 4. What am I doing wrong?

What I would like to have is this:

{4: 'zone', 7: 'abigail', 5: 'theta', 4: 'form', 4: 'libe', 3: 'zas'}

dict key should be unique

Olvin Roght
– Olvin Roght

2019-09-03 22:15:48 +00:00
Commented Sep 3, 2019 at 22:15 — Olvin Roght
– Olvin Roght, Commented Sep 3, 2019 at 22:15
That actually explains everything, thank you very much

Luigi
– Luigi

2019-09-03 22:19:54 +00:00
Commented Sep 3, 2019 at 22:19 — Luigi
– Luigi, Commented Sep 3, 2019 at 22:19

Djaouad · Accepted Answer · 2019-09-03 23:18:00Z

4

dict keys must be unique, a better alternative solution to your problem is storing lengths as keys (as you did before) and lists of strings with that length as values, you can do this using itertools.groupby (which requires that the data be sorted by the criteria it will be grouped by):

strings =  ["zone", "abigail", "theta", "form", "libe", "zas"] 

from itertools import groupby

dict_of_len = {k: list(g) for k, g in groupby(sorted(strings, key = len), len)}

print(dict_of_len)

Output:

{3: ['zas'], 4: ['zone', 'form', 'libe'], 5: ['theta'], 7:['abigail']}

edited Sep 3, 2019 at 23:18

answered Sep 3, 2019 at 22:20

Djaouad

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Comments

Peter DeGlopper · Accepted Answer · 2019-09-03 22:30:01Z

0

As an alternative to the groupby solution, which requires pre-sorting, you can use a defaultdict or use setdefault to concisely accumulate a list of values:

from collections import defaultdict
dict_of_len = defaultdict(list)
for str in strings:
    dict_of_len[len(str)].append(str)

Or, using the older idiom:

dict_of_len = {}
for str in strings:
    dict_of_len.setdefault(len(str), []).append(str)

But there's nothing wrong with the groupby solution as long as you don't mind sorting.

answered Sep 3, 2019 at 22:30

Peter DeGlopper

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Comments

Ace · Accepted Answer · 2022-09-08 19:49:19Z

0

i use this:

dict_of_len = {element:len(element) for element in strings}

answered Sep 8, 2022 at 19:49

Ace

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1 Comment

Andrew Over a year ago

That would make a dictionary where the keys are strings and the values are lengths, which isn't what the OP asked for

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