This is the given problem statement(attached in the image). It uses the concepts of lists and dictio-CodePudding

I've attached the problem statement in the image. I somehow managed to get the "similar" output in the form of a list because dictionary needs to have a "unique" key all the time. Here's my code:

l=[]

words=[a for a in 'aaabccdd']
for i in words:
    l.append((i,words.count(i)))
h = list(set(l))   
print(h)

output of my code is:

[('a', 3), ('d', 2), ('b', 1), ('c', 2)]

but I want it in the form of a dictionary like this :

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

where count(frequency) acts as a key to the dictionary and elements with common frequency comes in the form of a list as shown above.

Click here for image

CodePudding user response：

Since you want to go from

[('a', 3), ('d', 2), ('b', 1), ('c', 2)]

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

Here is the piece of code to achieve this:

from collections import defaultdict

d = defaultdict(list)
for word, count in h:
    d[count].append(word)

print(d)

Or the full program:

from collections import defaultdict


l=[]

words=[a for a in 'aaabccdd']
for i in words:
    l.append((i,words.count(i)))
h = list(set(l))   


d = defaultdict(list)
for word, count in h:
    d[count].append(word)

print(d)

if you are not allowed to use defaultdict, you can use the following:

l=[]

words=[a for a in 'aaabccdd']
for i in words:
    l.append((i,words.count(i)))
h = list(set(l))   


d = {}
for word, count in h:
    if count in d:
        d[count].append(word)
    else:
        d[count] = [word]
print(d)

CodePudding user response：

from collections import Counter
new_dict = {}
for  key,val in dict(Counter('aaabccdd')).items():
    if val not in new_dict:
        new_dict[val] = key
    elif type(new_dict[val]) == list:
        new_dict[val].append(key)
    else:
        new_dict[val] = [new_dict[val]] [key]
print (new_dict)

output:

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