Fequency of letters with dictionary

[Jen_Sophia]

Hello guys! So I have been trying to fix this but I havent made it yet, so if someone knows please give me a hand!

so the instructions are this:

" Enter a text and count the number of occurrences of letters and punctuation symbols (except spaces) to form a dictionary of "Number-Character List". That is, the key of the dictionary is the number of occurrences, and the value is a list of letters and punctuation symbols with the same occurrence times."

For example, the dictionary corresponding to "apple" is {1:['a','l','e'], and 2:['p']}.

right now I have done this:

1 2 3 4 5

test_str=input('Enter some text:' ) res={} for keys in test_str:     res[keys]=res.get(keys,0)+1 print('The "Freq-chars" dictionary is: \n' + str(res))

and this is my output:

Output:
Enter some text:saudi nuclear program accelerates, raising tensions in a volatile region
The "Freq-chars" dictionary is: 
{'s': 5, 'a': 8, 'u': 2, 'd': 1, 'i': 7, ' ': 9, 'n': 6, 'c': 3, 'l': 4, 'e': 7, 'r': 6, 'p': 1, 'o': 4, 'g': 3, 'm': 1, 't': 3, ',': 1, 'v': 1}

but what the output should look like is:

the thing here is that mine right now is (letter:times) but I should have (time,letter) for example: 1:['a','t','c'] instead of 'a':1 , 't':1, etc.

Thank you so much for your help and comments!

[Yoriz]

now you have a dictionary with a count of letters.
create another empty dictionary.
create a for loop for key(old), value(old) of the dictionary you created
if the value(old) is a key in the new dictionary
append the key(old) to the key(new)
else
set the key(new) to a list with the value(old) inside.

it makes sense to me but might not to you basically you are flipping the key, & values between dictionary's and storing the values in a list.

[perfringo]

Some random ideas.

Task is to flip keys and values with possibility that there are several equal values. As dictionary keys are unique one can't directly flip them - Python will overwrite key-value pairs and keep only last one:

1 2 3

>>> d = {'ham': 3, 'spam': 4, 'bacon': 5, 'eggs': 4} >>> {v:k for k, v in d.items()} {3: 'ham', 4: 'eggs', 5: 'bacon'}

So the type of the value should be list. Another thing which should be taken care is problem of missing keys. In most basic form it can be solved as:

1 2 3 4 5 6 7 8 9

>>> d = {'ham': 3, 'spam': 4, 'bacon': 5, 'eggs': 4} >>> new = {} >>> for k, v in d.items():         # iterate over original dict key-value pairs ...     if v not in new:           # if v is not among new dict keys ...         new[v] = []            # then create key in new dict with value of empty list ...     new[v].append(k)           # append key as value to list ... >>> new {3: ['ham'], 4: ['spam', 'eggs'], 5: ['bacon']}

There are also more 'advanced' ways achieving same result. Like using dictionary get method as in counting letters:

1 2 3 4 5 6 7

>>> d = {'ham': 3, 'spam': 4, 'bacon': 5, 'eggs': 4} >>> new = {} >>> for k, v in d.items(): ...     new[v] = new.get(v, []) + [k] ... >>> new {3: ['ham'], 4: ['spam', 'eggs'], 5: ['bacon']}

Flipping keys and values can be done also using defaultdict, if needed it can be converted to regular dict:

1 2 3 4 5 6 7 8 9 10

>>> from collections import defaultdict >>> d = {'ham': 3, 'spam': 4, 'bacon': 5, 'eggs': 4} >>> new = defaultdict(list) >>> for k in d: ...    new[d[k]].append(k) ... >>> new defaultdict(<class 'list'>, {3: ['ham'], 4: ['spam', 'eggs'], 5: ['bacon']}) >>> dict(new) {3: ['ham'], 4: ['spam', 'eggs'], 5: ['bacon']}

However, maybe instead of flipping keys and values create dictionary needed right away? One way of doing it:

1 2 3 4 5 6 7 8 9 10 11 12

>>> s = 'ham spam eggs bacon' >>> set(s) {'g', 'h', 'a', 'm', 'p', 'e', 'n', 'c', ' ', 'b', 's', 'o'} # unique symbols in s, incl space >>> chars = set(s) - set(' ')                                # get rid of space, if space must be counted, keep it >>> chars {'g', 'h', 'a', 'm', 'p', 'e', 'n', 'c', 'b', 's', 'o'} >>> new = dict() >>> for char in chars: ...     new[s.count(char)] = new.get(s.count(char), []) + [char] ... >>> new {2: ['g', 'm', 's'], 1: ['h', 'p', 'e', 'n', 'c', 'b', 'o'], 3: ['a']}