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Tyler Nordmann
1,083 PointsI need help with the word count in python collections. I run the script on my own and it works, but submission says no.
So I created a file in Workspaces to test my code and it works perfectly! Problem is that when I try to submit it it doesn't work. Anyone know why???
def word_count(phrase):
space_index = []
key_list = []
phrase = phrase.lower()
count = 1
for letter in phrase:
upto = phrase[0:count]
if letter == " ":
space_index.append(len(upto)-1)
count += 1
count = 0
key_list = [phrase[0:space_index[0]]]
for num in space_index:
try:
nums = space_index[count:count+2]
key_list.append(phrase[nums[0]+1:nums[1]])
count += 1
except IndexError:
continue
key_list.append(phrase[space_index[len(space_index)-1]+1:len(phrase)])
counter = []
count= 0
for num in key_list:
counter.append(count)
count = 0
for item in key_list:
for thing in key_list:
if thing == item:
counter[count] += 1
count += 1
my_dict = dict(zip(key_list, counter))
return my_dict
3 Answers
Chris Freeman
Treehouse Moderator 68,423 PointsYou solution is certainly complex. While it may work on the sample input, it raises error when run on trivial inputs: "I" and "".
Hint: try using the .split() method to divide a string on whitespace. It returns a list of the words. You can loop over this new wordlist to count words.
Post back if you need more help. Good luck!!!
Tyler Nordmann
1,083 PointsChris,
Thank you for the .split() advice! It greatly helped simplify my code, but it still is not working during submission. Any thoughts why? I have tried it with many different strings and it seems to work for all of them.
def word_count(phrase):
phrase = phrase.lower()
key_list = phrase.split()
counter = []
count= 0
for num in key_list:
counter.append(count)
count = 0
for item in key_list:
for thing in key_list:
if thing == item:
counter[count] += 1
count += 1
my_dict = dict(zip(key_list, counter))
return my_dict
Chris Freeman
Treehouse Moderator 68,423 PointsI couldn't figure out what the error was so I tried running your code as is and it passed the challenge.
Chris Freeman
Treehouse Moderator 68,423 PointsIt's very unusual to have two nested loops that operate on the same iterable. In this case, looping over key_list affectively is counting all the occurrences for every item. This is effectively an order n-squared performance that will quickly degrade as the word list grows.
A solution to consider is how to walk down the key_list one time while updating an existing dictionary each time a word is seen.
Tyler Nordmann
1,083 PointsChris,
I ran it a few more times in a row and after the third try it accepted it. Thank you for the help. And yes, I am quite new to Python so the nested for loop is most likely extremely degrading. I will be sure to try alternate routes in the future. Thanks again for the help!