Python wordfreq.py 40511.7. Non-sequential CollectionsThis says to set the count associated with word w to be one more than the currentcount for w.There is one small complication with using a dictionary here. The first timewe encounter a word, it will not yet be in counts. Attempting to access a non-existent key produces a run-time KeyError. To guard against this, we need adecision in our algorithm:if w is already in counts:add one to the count for welse:set count for w to 1This decision ensures that the first time a word is encountered, it will be enteredinto the dictionary with a count of 1.One way to implement this decision is to use the in operator:if w in counts:counts [w] = counts [w] + 1else:counts [w] = 1A more elegant approach is to use the get method:counts [w] counts.get (w,0) + 1If w is not already in the dictionary, this get will return 0, and the result is thatthe entry for w is set to 1.The dictionary updating code will form the heart of our program. We justneed to fill in the parts around it. We will need to split our text document into asequence of words. Before splitting, though, it's useful to convert all the text tolowercase (so occurrences of "Foo" match "foo") and eliminate punctuation (so"foo," matches "foo"). Here's the code to do those three tasks:fname = input ("File to analyze: ")#read file as one long stringtext= open(fname, "r").read()#convert all letters to lower casetext= text.lower()JA list (passwd.keys())['turing', 'bill', 'newuser', 'guido'] list (passwd.values())['genius', 'bluescreen', 'ImANewbie', 'superprogrammer'] list (passwd.items())[('turing', 'genius'), ('bill', 'bluescreen'),\('newuser', 'ImANewbie'), ('guido', 'superprogrammer')] "bill" in passwdTrue 'fred' in passwdFalse passwd.get('bill', 'unknown')'bluescreen' passwd.get('john', 'unknown')unknown' passwd.clear() passwd11.7.3 Example Program: Word FrequencyLet's write a program that analyzes text documents and counts how many timeseach word appears in the document. This kind of analysis is sometimes usedas a crude measure of the style similarity between two documents and is alsoused by automatic indexing and archiving programs (such as Internet searchengines).At the highest level, this is just a multi-accumulator problem. We need acount for each word that appears in the document. We can use a loop that iter-ates through each word in the document and adds one to the appropriate count.The only catch is that we will need hundreds or thousands of accumulators, onefor each unique word in the document. This is where a Python dictionary comesin handy.We will use a dictionary where the keys are strings representing words in thedocument and the values are ints that count how many times the word appears.Let's call our dictionary counts. To update the count for a particular word, w,we just need a line of code something like this:counts [w]= counts [w] + 1WQ11

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Python wordfreq.py 40511.7. Non-sequential CollectionsThis says to set the count associated with word w to be one more than the currentcount for w.There is one small complication with using a dictionary here. The first timewe encounter a word, it will not yet be in counts. Attempting to access a non-existent key produces a run-time KeyError. To guard against this, we need adecision in our algorithm:if w is already in counts:add one to the count for welse:set count for w to 1This decision ensures that the first time a word is encountered, it will be enteredinto the dictionary with a count of 1.One way to implement this decision is to use the in operator:if w in counts:counts [w] = counts [w] + 1else:counts [w] = 1A more elegant approach is to use the get method:counts [w] counts.get (w,0) + 1If w is not already in the dictionary, this get will return 0, and the result is thatthe entry for w is set to 1.The dictionary updating code will form the heart of our program. We justneed to fill in the parts around it. We will need to split our text document into asequence of words. Before splitting, though, it's useful to convert all the text tolowercase (so occurrences of &#34;Foo&#34; match &#34;foo&#34;) and eliminate punctuation (so&#34;foo,&#34; matches &#34;foo&#34;). Here's the code to do those three tasks:fname = input (&#34;File to analyze: &#34;)#read file as one long stringtext= open(fname, &#34;r&#34;).read()#convert all letters to lower casetext= text.lower()JA >>> list (passwd.keys())['turing', 'bill', 'newuser', 'guido']>>> list (passwd.values())['genius', 'bluescreen', 'ImANewbie', 'superprogrammer']>>> list (passwd.items())[('turing', 'genius'), ('bill', 'bluescreen'),\('newuser', 'ImANewbie'), ('guido', 'superprogrammer')]>>> &#34;bill&#34; in passwdTrue>>> 'fred' in passwdFalse>>> passwd.get('bill', 'unknown')'bluescreen'>>> passwd.get('john', 'unknown')unknown'>>> passwd.clear()>>> passwd11.7.3 Example Program: Word FrequencyLet's write a program that analyzes text documents and counts how many timeseach word appears in the document. This kind of analysis is sometimes usedas a crude measure of the style similarity between two documents and is alsoused by automatic indexing and archiving programs (such as Internet searchengines).At the highest level, this is just a multi-accumulator problem. We need acount for each word that appears in the document. We can use a loop that iter-ates through each word in the document and adds one to the appropriate count.The only catch is that we will need hundreds or thousands of accumulators, onefor each unique word in the document. This is where a Python dictionary comesin handy.We will use a dictionary where the keys are strings representing words in thedocument and the values are ints that count how many times the word appears.Let's call our dictionary counts. To update the count for a particular word, w,we just need a line of code something like this:counts [w]= counts [w] + 1WQ11

Accepted Answer

code for find frequency of words in file using dictionary -
 
file = open("bii.txt", "r")d = {}for…