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Start your free trialNicolas Tautiva
3,159 Pointswordcount.py runs as expected on Workspaces, but doesnยดt pass on the challenge
I tried several times... and after checking other posts I know my error is not the split("")...
What am I missing here??
# E.g. word_count("I do not like it Sam I Am") gets back a dictionary like:
# {'i': 2, 'do': 1, 'it': 1, 'sam': 1, 'like': 1, 'not': 1, 'am': 1}
# Lowercase the string to make it easier.
count = 1
wordcount = {}
def word_count(x):
word = x.lower().split() # hago lista. Cada palabra es un item en la lista
for z in word:
if z in wordcount.keys():
wordcount[z] = count + 1
else:
wordcount[z] = count
return wordcount
4 Answers
ASHOK Gudivada
11,701 Pointswordcount = {}
def word_count(x):
word = x.lower() # hago lista. Cada palabra es un item en la lista
for z in word.split():
if z in wordcount.keys():
wordcount[z] += 1
else:
wordcount[z] = 1
return wordcount
this is your code
>>> x = "I do not like it Sam I Am"
>>> count = 1
>>> wordcount = {}
>>> def word_count(x):
... word = x.lower().split()
... for z in word:
... if z in wordcount.keys():
... wordcount[z] = count + 1
... else:
... wordcount[z] = count
... return wordcount
...
>>> word_count(x)
{'do': 1, 'like': 1, 'sam': 1, 'i': 2, 'am': 1, 'it': 1, 'not': 1}
>>> x = "I do not like it Sam I Am am am sam sam sam"
>>> word_count(x)
{'do': 2, 'like': 2, 'sam': 2, 'i': 2, 'am': 2, 'it': 2, 'not': 2}
ASHOK Gudivada
11,701 Pointshttps://teamtreehouse.com/community/challenge-task-1-of-1-in-wordcountpy
Check this discussion out, may help you.
Nicolas Tautiva
3,159 PointsThanks ASHOK. I checked your code and we solved things in a different order... but all steps are there.
I tried the other suggested string and the count of words was correct.
I still can see what I am doing wrong.
Nicolas Tautiva
3,159 PointsOk... got it. It's working now. I saw my mistake.
Much appreciated ASHOK