String to dictionary word count and display

Question

I have a homework question which asks:

Write a function print_word_counts(filename) that takes the name of a file as a parameter and prints an alphabetically ordered list of all words in the document converted to lower case plus their occurrence counts (this is how many times each word appears in the file).

I am able to get an out of order set of each word with it's occurrence; however when I sort it and make it so each word is on a new line the count disappears.

    import re

def print_word_counts(filename):
    input_file = open(filename, 'r')
    source_string = input_file.read().lower()
    input_file.close()
    words = re.findall('[a-zA-Z]+', source_string)    

    counts = {}
    for word in words:
        counts[word] = counts.get(word, 0) + 1

    sorted_count = sorted(counts)
    print("\n".join(sorted_count))

When I run this code I get:

a
aborigines
absence
absolutely
accept
after

and so on.

What I need is:

a: 4
aborigines: 1
absence: 1
absolutely: 1
accept: 1
after: 1

I'm not sure how to sort it and keep the values.

Answer 1

It's a homework question, so I can't give you the full answer, but here's enough to get you started. Your mistake is in this line

sorted_count = sorted(counts)

Firstly, you cant sort a dictionary by nature. Secondly, what this does is take the keys of the dictionary, sorts them, and returns a list.

You can just print the value of counts, or, if you really need them in sorted order, consider changing the dictionary items into a list, then sorting them.

lst = list(count.items())

#sort and return lst