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Start your free trialWilliam Sserubiri
2,677 PointsHey there, my attempted solution for this problem ```word_count("I do not like it Sam I Am")``` failed please advise
I have tried testing out my solution in many other apps and it works flawlessly. I am not sure what I am missing in my function
#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(string):
lower_string = string.lower()
string_list = lower_string.split(" ")
final = {}
count = 1
for word in string_list:
if word in final.keys():
final[word] = count + 1
else:
final[word] = count
return final
2 Answers
Steven Parker
231,275 PointsPerhaps you're testing with "lucky" data.
The "Bummer" message gave you a hint: "Be sure you're lowercasing the string and splitting on all whitespace!" To split on "all whitespace" the argument to "split" should be left empty or set to None
.
Also, it looks like "count" will continue to grow as the loop runs, but a new entry should always have the value set to 1, and an existing one should be incremented by one from its own current value.
William Sserubiri
2,677 PointsThanks, Steven I have solved it
def word_count(string):
lower_string = str(string)
lower_string = lower_string.lower()
string_list = lower_string.split()
final = {}
count = 1
for word in string_list:
if word in final.keys():
final[word] = final[word] + 1
else:
final[word] = 1
return final
print(word_count("I do not like it Sam I Am"))
Steven Parker
231,275 PointsGood job!
William Sserubiri
2,677 PointsWilliam Sserubiri
2,677 PointsThanks, Steven but I have tried to make some modifications but it still fails