Sort a dictionary alphabetically, and print it by frequency

Question

I am running python 2.7.2 on a mac.

I have a simple dictionary:

dictionary= {a,b,c,a,a,b,b,b,b,c,a,w,w,p,r}

I want it to be printed and have the output like this:

Dictionary in alphabetical order:
    a  4
    b  5
    c  2
    p  1
    r  1
    w  2

But what I'm getting is something like this...

a  1
a  1
a  1
a  1
b  1
.
.
.
w  1

This is the code I am using.

new_dict = []


    for word in dictionary.keys():
        value = dictionary[word]
        string_val = str(value)
        new_dict.append(word + ": " + string_val)

    sorted_dictionary = sorted(new_dict)

    for entry in sorted_dictionary:
        print entry

Can you please tell me where is the mistake? (By the way, I'm not a programmer but a linguist, so please go easy on me.)

Answer 1

What you're using is not a dictionary, it's a set! :)

And sets doesn't allow duplicates.

What you probably need is not dictionaries, but lists.

A little explanation

Dictionaries have keys, and each unique keys have their own values:

my_dict = {1:'a', 2:'b', 3:'c'} 

You retrieve values by using the keys:

>>> my_dict [1]
'a'

On the other hand, a list doesn't have keys.

my_list = ['a','b','c']

And you retrieve the values using their index:

>>> my_list[1]
'b'

Keep in mind that indices starts counting from zero, not 1.

Solving The Problem

Now, for your problem. First, store the characters as a list:

l = ['a', 'b', 'c', 'a', 'a', 'b', 'b', 'b', 'b', 'c', 'a', 'w', 'w', 'p', 'r']

Next, we'll need to know what items are in this list:

items = []
for item in l:
    if item not in items:
        items.append(item)

This is pretty much equal to items = set(l) (the only difference is that this is a list). But just to make things clear, hope you understand what the code does.

Here is the content of items:

>>> items
['a', 'b', 'c', 'w', 'p', 'r']

With that done, we will use lst.count() method to see the number of a char's occurence in your list, and the built-in function sorted() to sort the items:

for item in sorted(items): #iterates through the sorted items.
    print item, l.count(item)

Result:

a 4
b 5
c 2
w 2
p 1
r 1

Hope this helps!!