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Date: Sat, 18 Jun 2011 19:09:18 -0500
From: Tim Chase <python.list@tim.thechases.com>
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To: Cathy James <nambo4jb@gmail.com>
Subject: Re: NEED HELP-process words in a text file
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On 06/18/2011 06:21 PM, Cathy James wrote:

>      freq = [] #empty dict to accumulate words and word length

While you say you create an empty dict, using "[]" creates an 
empty *list*, not a dict.  Either your comment is wrong or your 
code is wrong. :)  Given your usage, I presume you want a dict, 
not a list.

>      for line in filename:
>          punc = string.punctuation + string.whitespace#use Python's
> built-in punctuation and whiitespace

Since you don't change "punc" in your loop, you'd get better 
performance by hoisting this outside of the loop so it's only 
evaluated once.  Not that it should matter *that* greatly, but 
it's just a bad-code-smell.

>          for i, word in enumerate (line.replace (punc, "").lower().split()):

.replace() doesn't operate on sets of characters, but rather 
strings.  So unless your line contains the exact text in "punc" 
(unlikely), that replacement is a NOP.  There are a couple ways 
to go about removing unwanted characters:

- make a set of those characters and produce a resulting string 
from things not in that set:

  punc_set = set(punc)
  line = ''.join(c for c in line if c not in punc_set)

- use a regexp to strip them out...something like

   punc_re = re.compile("[" + re.escape(punc) + "]")
   ...
   line = punc_re.sub('', line)

- use string translations.  I'm not as familiar with these, but 
the following seemed to work for me, abusing the 2nd 
"deletechars" parameter for your particular use-case:

   line = line.translate(None, punc)

I don't see .translate(None) documented anywhere.  My random 
effort seemed to work in 2.6, but fails in 2.5 and prior.  YMMV.

>              if word in freq:
>                  freq[word] +=1 #increment current count if word already in dict
>
>              else:
>                  freq[word] = 0 #if punctuation encountered,
> frequency=0 word length = 0

Again, your 2nd comment disagrees with your code.  As an aside, 
if you're using 2.5 or greater, I'd use 
collections.defaultdict(int) as the accumulator:

   freq = collections.defaultdict(int)
   ...
   freq[word] += 1
   # no need to check presence

>          for word in freq.items():
>              print("Length /t"+"Count/n"+ freq[word],+'/t' +
> len(word))#print word count and length of word separated by a tab

Where to begin:

- Your escapes are using "/" instead of "\" for <tab> and 
<newline> which I expect will mess up the formatting.

- You're also labeling them "Length/Count" but printing 
"count/length".

- you're iterating over freq.items() but that should be written as

   for word, count in freq.items():

or

   for word in freq:

-  Additionally, adding the bits together makes it somewhat hard 
to understand.

I'd use something like

   for word, count in freq.items():
     print("Word \tLength \tCount\n%s \t%i \t%i" % (
       word, len(word), count))

-tkc