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Lakshit Kerai
3,816 PointsBummer! IndentationError: return mydict
What's wrong with my code? I don't see any error!. when i run it on my local machine?!
# E.g. word_count("I am that I am") gets back a dictionary like:
# {'i': 2, 'am': 2, 'that': 1}
# Lowercase the string to make it easier.
# Using .split() on the sentence will give you a list of words.
# In a for loop of that list, you'll have a word that you can
# check for inclusion in the dict (with "if word in dict"-style syntax).
# Or add it to the dict with something like word_dict[word] = 1.
mydict = {}
def word_count(mystring):
listofwords = mystring.lower().split()
for word in listofwords:
if word in mydict:
mydict[word] += mydict[word] # update existing entry
else:
mydict[word] = 1 # add new entry
return mydict
2 Answers
Chris Freeman
Treehouse Moderator 68,457 PointsThere are two errors:
- mixing TABs and SPACEs in indentation. The standard practice is to uses 4 spaces
- the update existing entry should increment by 1. By adding the value to itself, you are doubling the value each loop.
Nick Osborne
4,612 PointsHi, looks like the error is just in the update existing entry. The problem being that if mydict[word] is bigger than 1 already then it will add too many. So just add 1. Hope that makes sense.
mydict = {}
def word_count(mystring):
listofwords = mystring.lower().split()
for word in listofwords:
if word in mydict:
mydict[word] += 1 # update existing entry
else:
mydict[word] = 1 # add new entry
return mydict
Lakshit Kerai
3,816 PointsLakshit Kerai
3,816 PointsYes, I wrote those lines in Sublime and mixed used tabs, spaces. My bad.