I've CCd the list. Please use reply all when responding to the list.
Also please use plain text as HTML/RTF doesn't work on all
systems and code in particular often gets mangled.
On 01/06/15 23:59, Stephanie Quiles wrote:
Hello again,
here is the final code… I think :) please see below. Is this is the
easiest way to go about it? I appreciate your assistance!
defmain():
words = {}
count =0
Do you need count? What is its purpose?
withopen('words.txt')asdata:
forlineindata:
text = line.split()
forwordintext:
ifwordnot inwords:
words[word] =1
else:
words[word] +=1
Look into the setdefault() method of dictionaries.
It can replace the if/else above.
count +=1
print(words)
Aside from the two comments above, good job!
--
Alan G
Author of the Learn to Program web site
http://www.alan-g.me.uk/
http://www.amazon.com/author/alan_gauld
Follow my photo-blog on Flickr at:
http://www.flickr.com/photos/alangauldphotos
_______________________________________________
Tutor maillist - Tutor@python.org
To unsubscribe or change subscription options:
https://mail.python.org/mailman/listinfo/tutor
