Count frequency of words in a list and sort by fre

Question

I am using Python 3.3

I need to create two lists, one for the unique words and the other for the frequencies of the word.

I have to sort the unique word list based on the frequencies list so that the word with the highest frequency is first in the list.

I have the design in text but am uncertain how to implement it in Python.

The methods I have found so far use either Counter or dictionaries which we have not learned. I have already created the list from the file containing all the words but do not know how to find the frequency of each word in the list. I know I will need a loop to do this but cannot figure it out.

Here's the basic design:

                 original list = ["the", "car",....]
                 newlst = []
                 frequency = []
                 for word in the original list
                       if word not in newlst
                           newlst.append(word)
                           set frequency = 1
                       else
                           increase the frequency
                 sort newlst based on frequency list 

Answer 1

The ideal way is to use a dictionary that maps a word to it's count. But if you can't use that, you might want to use 2 lists - 1 storing the words, and the other one storing counts of words. Note that order of words and counts matters here. Implementing this would be hard and not very efficient.

Answer 2

Using Counter would be the best way, but if you don't want to do that, you can implement it yourself this way.

# The list you already have
word_list = ['words', ..., 'other', 'words']
# Get a set of unique words from the list
word_set = set(word_list)
# create your frequency dictionary
freq = {}
# iterate through them, once per unique word.
for word in word_set:
    freq[word] = word_list.count(word) / float(len(word_list))

freq will end up with the frequency of each word in the list you already have.

You need float in there to convert one of the integers to a float, so the resulting value will be a float.

Edit:

If you can't use a dict or set, here is another less efficient way:

# The list you already have
word_list = ['words', ..., 'other', 'words']
unique_words = []
for word in word_list:
    if word not in unique_words:
        unique_words += [word]
word_frequencies = []
for word in unique_words:
    word_frequencies += [float(word_list.count(word)) / len(word_list)]
for i in range(len(unique_words)):
    print(unique_words[i] + ": " + word_frequencies[i])

The indicies of unique_words and word_frequencies will match.

Answer 3

words = file("test.txt", "r").read().split() #read the words into a list.
uniqWords = sorted(set(words)) #remove duplicate words and sort
for word in uniqWords:
    print words.count(word), word

Answer 4

use this

from collections import Counter
list1=['apple','egg','apple','banana','egg','apple']
counts = Counter(list1)
print(counts)
# Counter({'apple': 3, 'egg': 2, 'banana': 1})

Answer 5

Here is code support your question is_char() check for validate string count those strings alone, Hashmap is dictionary in python

def is_word(word):
   cnt =0
   for c in word:

      if 'a' <= c <='z' or 'A' <= c <= 'Z' or '0' <= c <= '9' or c == '$':
          cnt +=1
   if cnt==len(word):
      return True
  return False

def words_freq(s):
  d={}
  for i in s.split():
    if is_word(i):
        if i in d:
            d[i] +=1
        else:
            d[i] = 1
   return d

 print(words_freq('the the sky$ is blue not green'))

Answer 6

Try this:

words = []
freqs = []

for line in sorted(original list): #takes all the lines in a text and sorts them
    line = line.rstrip() #strips them of their spaces
    if line not in words: #checks to see if line is in words
        words.append(line) #if not it adds it to the end words
        freqs.append(1) #and adds 1 to the end of freqs
    else:
        index = words.index(line) #if it is it will find where in words
        freqs[index] += 1 #and use the to change add 1 to the matching index in freqs