This program takes a text string and creates a list of words. Any attached punctuation marks are removed and the words are converted to lower case for sorting. Now you can generate a dictionary of the words and their frequencies. The result is displayed using a sorted list of dictionary keys. A fair amount of comments have been added to aid in the understanding of the program code.

Comments
i doubt you see this, but your code was extremely helpful even 10yrs later and on a newer version of python. thank you very much!!
# word frequency in a text
# tested with Python24    vegaseat    25aug2005

# Chinese wisdom ...
str1 = """Man who run in front of car, get tired.
Man who run behind car, get exhausted."""
print "Original string:"
print str1

print

# create a list of words separated at whitespaces
wordList1 = str1.split(None)

# strip any punctuation marks and build modified word list
# start with an empty list
wordList2 = []
for word1 in wordList1:
    # last character of each word
    lastchar = word1[-1:]
    # use a list of punctuation marks
    if lastchar in [",", ".", "!", "?", ";"]:
        word2 = word1.rstrip(lastchar)
    else:
        word2 = word1
    # build a wordList of lower case modified words
    wordList2.append(word2.lower())

print "Word list created from modified string:"
print wordList2

print

# create a wordfrequency dictionary
# start with an empty dictionary
freqD2 = {}
for word2 in wordList2:
    freqD2[word2] = freqD2.get(word2, 0) + 1
    
# create a list of keys and sort the list
# all words are lower case already
keyList = freqD2.keys()
keyList.sort()

print "Frequency of each word in the word list (sorted):"
for key2 in keyList:
  print "%-10s %d" % (key2, freqD2[key2])

The stripping of characters can be better handled this way:

<snippet>
chars = ",.!?;"
word2 = word.rstrip(chars);
</snippet>

This will strip more than one characters in the end if needed.

For building the wordlist, you could just use this:

wordlist = str.split(None)
wordlist2 = []
for word in wordlist:
    wordlist2.append((word.strip(string.punctuation)).lower())
text="Hier goes your text"

def countWords(list_item):
    d,words= {},list_item.split()
    for i in xrange(len(words)):
        item = words[i]
        count = words.count(item)
        d[item] = count
    return sorted(d.items(), key = lambda item: item[1], reverse=True)

print countWords(text)

Stefan stop post your code in 8 years old Code Snippets post.
Yes Python has moved on,and your soultion is not better at all.

>>> from collections import Counter
>>> Counter('Her goes your text text'.lower().split())
Counter({'text': 2, 'goes': 1, 'her': 1, 'your': 1})

Edited 2 Years Ago by snippsat

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