[Tutor] Counting words

[Nicole Seitz]

Am Donnerstag,  4. April 2002 21:42 schrieben Sie:

> what is happening is the 'get' method of the dictionary is being called. 
> It is looking for the key specified by word and if it is not found it
> returns 0.
>
> This is equivalent to:
>
> if occurences.has_key(word):
>   occurences[word] = occurences[word] + 1
> else:
>   occurences[word] = 1


Now it's very clear, thanks!


>
> > Second question:
> > "Some" and "some" should be recognized as one word, the same is with
> > "BORING" and "boring". I  thought of string.lowercase as a possible
> > solution, but as it doesn't  work , I might be wrong. Any idea what to
> > do?
>
> word = string.lower(word)

I am so stupid! I do know this function from string module. I mixed up 
lower() and lowercase. I guess one should not program when he or she is 
totally overtired.It's much better when your brain's engaged:-)

> ['A', 'typical', 'line', '']
>
> notice the empty last entry.  Have you loop skip empty words.

Like this?

for word in reg.split(text):
    if word == "": continue
    word = string.lower(word)
    occurences[word] = occurences.get(word,0)+1




Many thanks.

Nicole