Question

In python Count the frequency of each word in a text file. Let the user choose a filename to read.

1. The program will count the frequency with which each word appears in the text.

2. Words which are the spelled the same but differ by case will be combined.

3. Punctuation should be removed

4. If the file does not exist, use a ‘try-execption’ block to handle the error

5. Output will list the words alphabetically, with the word in the first column and the count in the second column … i.e. as : 1287

Psuedo Code

printWds(data)

## print each word in our data and the respective count

## but we need them sorted, dictionaries have a nice module, sorted

## and lists have sort. for x in sorted(data.keys()):

print() ## look at the formatting above

return

wordFreq(fptr)

## create empty data type, dictionary or list?

freq = {} or freq = []

## read in the first line

line = fptr.readline()

punctChars = (….create a tuple of punctuation characters to eliminate… )

while line:

for c in punctChars:

## use the string replace module to substitute an empty char

line = line.replace(c, “”)

## create a list of separate words that make up the string

words = line.split()

for each word in our split:

#create a temporary word converted to lower using the lower()

tmp = word.lower()

## add this to your database, this is where dictionaries have the

## advantage over dictionaries … the get() method

freq[tmp] = freq.get(tmp, 0) + 1

read another line

return freq

Answer 1

NOTE: USE PYTHON3

def txt_list(path):
file = open(path, 'r')
file_arr = []
file_str = ''
punc = ['!','.','/',',',':',';','?','\'','"']
for word in file.read():
if(word not in punc):
file_str = file_str + word
file_arr = list(map(str,file_str.lower().split()))
return sorted(file_arr)
def word_freq(arr):
_dict = {}
for x in arr:
if x not in _dict:
_dict[x] = 1
else:
_dict[x]+= 1
return _dict

try:
path = input('Enter the file name: ')
txt = txt_list(path)
print(word_freq(txt))

except:
print("No such file exists.")

EXECUTION: