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Peter Lynch
946 PointsWord_count.py
Am I close to the solution or way off base?
# 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(string):
string = string.lower()
string = list(string.split())
for word in string:
count = 0
if word in string:
count = count + 1
return word_dict[word] = count
2 Answers
Carlos Federico Puebla Larregle
21,073 PointsYou are close, you can do the lower() and split() in one line. Remember that the split() function returns a list so it is a little bit redundant to put the "list()" too. You could use a dictionary variable to check "if" the current value of the list already exists. You could do it like this:
def word_count(string_value):
list_value = string_value.lower().split()
dict_value = {}
for word in list_value:
if word in dict_value.keys():
dict_value[word] += 1
else:
dict_value[word] = 1
return dict_value
I hope that helps
Peter Lynch
946 PointsThanks Carlos helped me understand much better.