Create dictionary with length of string as key and string as value without import statement?-CodePudding

Can you create a dictionary with length of string as key and string as value without an import statement?

With an import statement it would look like this:

strings =  ["zone", "abigail", "theta", "form", "libe", "zas"] 
from itertools import groupby
dict_of_len = {k: set(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']}

If possible I would like two options, where for one of the options the value is a list of values, and another option where the value is a set of values.

I tried this by myself but I keep getting difficulties when there are multiple values for the same key.

one of my failed attempts

for i in strings:
   dictio = dict()
   set1 = set()
   set1.add(i)
   dictio[len(i)] = set1
print(dictio)

output

{3: {'zas'}}

CodePudding user response：

You can use dict.setdefault:

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

CodePudding user response：

This would be easier with a defaultdict, but you can do it without any imports.

To use lists as values:

strings =  ["zone", "abigail", "theta", "form", "libe", "zas"] 
dict_of_len = {}
for string in strings:
    ls = len(string)
    if ls in dict_of_len:
        dict_of_len[ls].append(string)
    else:
        dict_of_len[ls] = [string]

To use sets as values, much the same but with a set instead of a list, and with add instead of append.

strings =  ["zone", "abigail", "theta", "form", "libe", "zas"] 
dict_of_len = {}
for string in strings:
    ls = len(string)
    if ls in dict_of_len:
        dict_of_len[ls].add(string)
    else:
        dict_of_len[ls] = {string}

CodePudding user response：

Use setdefault method of dict:

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

# List version
dict_of_len = {}
for word in strings:
    dict_of_len.setdefault(len(word), []).append(word)
print(dict_of_len)


# Sets version
dict_of_len = {}
for word in strings:
    s = dict_of_len.setdefault(len(word), set())
    dict_of_len[len(word)] = s.union([word])
print(dict_of_len)

Output:

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