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Nyarts Gnaw
1,079 PointsWhy do I miss words in string when loop this way?
Just as I said in my title, why is that?
# 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.
def word_count(words):
dict = {}
test = words[:].split()
for key in words:
count = 0
while True:
try:
test.remove(key)
count += 1
dict.update({key:count})
except:
break
return dict
2 Answers
Steven Parker
231,269 PointsFor one thing, the argument that you named "words" is actually a string, so when you iterate over it (for key in words:) your key will be just one single letter of the string each time.
You have some other issues too. Why not try using the hints given in the challenge comments? They can help you create a more compact solution, as well as helping resolve some of the issues.
Jon Hockley
3,781 PointsFollow the advice given to you in the challenge. You are not using the .split() method on the string as it advises you.
I have run your code and when you use the for loop on the dictionary, each individual character is being assigned to key.
Nyarts Gnaw
1,079 PointsHey man I change my code some and I get the right answers. To be honest I didn't notice the green words and I didn't know about "X in Y " could be used as condition in if. I've changed that in my function
def word_count(words): dict = {} test = words[:].split() for key in words: count = 0 while True: try: test.remove(key) count += 1 dict.update({key:count}) except: break return dict