Print frequency of the words along with word like (W1 : N1 ) (W2 : N2) with N1>N2 from a dictionary containing tuple in form of {(w1,n1),(w2,n2)}

Question

I am trying to create a function that prints the word frequency after taking out the unnecessary words from the given texts and prints out: firs the word, then number of times it occurred in the text. So far I have been able to clean the text, count the words, and create a dictionary of the words with number of times its occurring in the text. Now I want to print it in a sorted format where word with highest count gets printed first, along with number of times its occurring. Here is the code I have been able to come up with. What should I do next?

Here is a sample output of the dictionary:

{'welcome': 3, 'probe': 5, 'pm': 2, 'security': 6, 'breach': 2, 'propaganda': 4, 'sc': 2}

I want the output as

security: 6

probe: 5

Propaganda: 4

welcome: 3

enter code here

file = "/Users/learninguser/Documents/GitHub/Projects/News_analysis/news.txt"
f = open(file)
text = f.read()



def RemovePunchuations(text):
     # removes all the special characters supplied in
     # the variable punchuations from the supplied
     # text string
       newtext = text

    punctuations = '''!()-[]{};:'"\,<>./?@—#$%^&*_~'''

    for punchuation in punctuations:
        newtext = newtext.replace(punchuation, " ")

    return newtext




def TokenfromText(text):
    # take a string as input
    # return a list of words in the text

    newtext = text
    newnewtext = newtext.split()
    return_list = []
    for word in newnewtext:
        newword = word.lower()
        return_list.append(newword)
    return return_list




def FrequencyWithUninterestingRemoved(listofwords):
    # takes the tokens from text
    # calculates frequency of words supplied in the list
    # removes uninteresting words based on corpus supplied
    # returns a dictionary of frequency

    frequency = {}
    supplied_list = listofwords

    for word in supplied_list:
        if word not in frequency:
            frequency[word] = 1
        frequency[word] += 1

    uninteresting_words = ["the", "a", "to", "if", "is", "it", "of",
                       "and", "or", "an", "as", "i", "me", "my",
                       "we", "our", "ours", "you", "your", "yours",
                       "he", "she", "him", "his", "her", "hers",
                       "its", "they", "them",     "their", "what",
                       "which", "who", "whom", "this", "that",
                       "am", "are", "was", "were", "be", "been",
                       "being",     "have", "has", "had", "do",
                       "does", "did", "but", "at", "by", "with",
                       "from", "here", "when", "where", "how",
                       "all", "any", "both", "each", "few", "more",
                       "some", "such", "no", "nor", "too", "very",
                       "can", "will", "just", 'a', 'about', 'above',
                       'after', 'again', 'against', 'all', 'am',
                       'an', 'and', 'any', 'are', "aren't", 'as',
                       'at', 'be', 'because', 'been', 'before',
                       'didst', 'us', 'one',
                       'being', 'below', 'between', 'both',
                       'but', 'by', "can't", 'cannot', 'could',
                       "couldn't", 'did', "didn't", 'do', 'does',
                       "doesn't", 'doing', "don't", 'down', 'during',
                       'each', 'few', 'for', 'from', 'further', 'had',
                       "hadn't", 'has', "hasn't", 'have', "haven't",
                       'having',
                       'he', "he'd", "he'll", "he's", 'her', 'here', "here's",
                       'hers', 'herself', 'him', 'himself', 'his', 'how', "how's",
                       'i', "i'd", "i'll", "i'm", "i've", 'if', 'in', 'into', 'is',
                       "isn't", 'it', "it's", 'its', 'itself', "let's", 'me', 'more',
                       'most', "mustn't", 'my', 'myself', 'no', 'nor', 'not', 'of',
                       'off', 'on', 'once', 'only', 'or', 'other', 'ought', 'our',
                       'ours', 'ourselves', 'out', 'over', 'own', 'same', "shan't",
                       'she', "she'd", "she'll", "she's", 'should', "shouldn't", 'so',
                       'some', 'such', 'than', 'that', "that's", 'the', 'their', 'theirs',
                       'them', 'themselves', 'then', 'there', "there's", 'these', 'they',
                       "they'd", "they'll", "they're", "they've", 'this', 'those', 'through',
                       'to', 'too', 'under', 'until', 'up', 'very', 'was', "wasn't", 'we',
                       "we'd", "we'll", "we're", "we've", 'were', "weren't", 'what', "what's",
                       'when', "when's", 'where', "where's", 'which', 'while', 'who', "who's",
                       'whom', 'why', "why's", 'with', "won't", 'would', "wouldn't", 'you',
                       "you'd", "you'll", "you're", "you've", 'your', "thou", "thee", "thy",
                       "but", "man", 'yours', 'yourself', 'yourselves']

    for uword in uninteresting_words:
        if uword in frequency:
            del frequency[uword]
    return frequency



def Frequency_dict(text):
    text1 = RemovePunchuations(text)

    listofwords = TokenfromText(text1)

    frequencyofwords = FrequencyWithUninterestingRemoved(listofwords)

    return frequencyofwords


f_dict = Frequency_dict(text)

print(f_dict)

Answer 1

You can use pythons built in sorted function and then turn it back into a dict or just return the list of sorted tuples like so:

def Sorted_dict(text):
    tempDict = Frequency_dict(text)
    return dict(sorted(tempDict.items(), key=lambda kv: kv[-1], reverse=True))

f_dict = Sorted_dict(text)

Its turning the dict into a list of tuples with.items() then sorting using the value (kv[-1]) on each tuple. It uses reverse=True to make it go in decreasing order. Lastly, it converts that back into a dict and returns it.