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Start your free trialBradley Demerich
11,688 PointsOn one of the final code challenges in python; my code creates the desired output but comes up wrong.
the code challenge is as follows: Alright, this one might be a bit challenging but you've been doing great so far, so I'm sure you can manage it.
I need you to make a function named word_count. It should accept a single argument which will be a string. The function needs to return a dictionary. The keys in the dictionary will be each of the words in the string, lowercased. The values will be how many times that particular word appears in the string.
Check the comments below for an example.
my code looks like this:
# 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.
def word_count(x):
x = x.lower()
lx =x.split(" ")
sx = set(lx)
dx = {}
for s in sx:
c= 0
for l in lx:
if d == l:
c+=1
dx[s] = c
return dx
It generates the correct output when I test it in the console, so I'm not sure what i'm missing.
5 Answers
Bradley Demerich
11,688 PointsNevermind. I figured it out.
Prashant Nayak
2,300 Pointsdef word_count(val): val=val.lower() dist={} list1=val.split(" ")
for x in list1:
if x in dist:
dist[x] +=1
else:
dist[x]=1
return dist
Prashant Nayak
2,300 PointsI am getting not sure of the issue
Kevin Lassar
22,440 PointsHey Treehouse,
Any idea why my code submission wouldn't be passing? I've tested all inputs both lower, uppercase, duplicates etc..
def word_count(string):
split_words = []
dictionary = dict()
for word in string.split(" "):
split_words.append(word.lower())
for word in split_words:
dictionary[word.lower()] = split_words.count(word)
return dictionary
Code Challenege:
Alright, this one might be a bit challenging but you've been doing great so far, so I'm sure you can manage it.
I need you to make a function named word_count. It should accept a single argument which will be a string.
The function needs to return a dictionary. The keys in the dictionary will be each of the words in the string,
lowercased. The values will be how many times that particular word appears in the string.
Check the comments below for an example.
serafinsanchez13
1,540 PointsHere's what I got:
def word_count(string):
text_list = string.lower().split()
key_list = text_list
value_list = []
for words in key_list:
value_list.append(text_list.count(words))
zip_obj = zip(key_list, value_list)
word_dict = dict(zip_obj)
return(word_dict)