Use python programming to count all unique words for all files in a given directory. the output must be words and their corresponding frequency. note: • to get all files under a directory, use os. listdir(directory name) • to check if a character is an alpha character, you can use the following example expression (3 is not a word character): o > > > str = "hello3hello" o > > > str = str. lower() # to be all lowercase first o > > > str[5] > ‘a’ and str[5] < ‘z’ o > > > false • to debug your code, create a folder with at least 3 text files. you don’t need to submit your example folder and text files.

import re

word_Frequency = {}

#ITime to input

input_file = open('input.txt', 'r')

#The program considers UpperCase and LowerCase words as different

input_string = input_file.read()

#Searching all words containing [a-z]/[A-z] value as string

all_match_words = re.findall(r'\b[a-zA-Z]{1,15}\b', input_string)

for word in all_match_words:

count = word_Frequency.get(word,0)

word_Frequency[word] = count + 1  

frequency_list = word_Frequency.keys()  

for words in frequency_list:

print(words,word_Frequency[words])

answer:

put more,

answer:

d. attempt to find the owner, then leave your information if you cannot find them.

explanation:

d. represents both (b) and (c).

more logical as well.

b and c are to complicated.

glad i could , give me a follow, so if you need with any other question, im here.