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Python Python Collections (2016, retired 2019) Dictionaries Word Count

Posted by ASHOK Gudivada
ASHOK Gudivada
11,701 Points

Challenge task 1 of 1 in word_count.py 

String = "I do not like it Sam I Am"
>>> string_lower = String.lower()
>>> string_dict = {}
>>> for word in string_lower.split(" "):
...     if word not in string_dict:
...             string_dict[word] = string_lower.count(word+" ")
...
>>> string_dict
{'do': 1, 'like': 1, 'sam': 1, 'i': 2, 'am': 1, 'it': 1, 'not': 1}
>>>

it work but not passing, can someone help me plz

wordcount.py 
# 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):
    string_lower = string.lower()
    string_dict = {}
    for word in string_lower.split(" "):
        if word not in string_dict:
            string_dict[word] = string_lower.count(word+" ")
    return string_dict

6 Answers

Chris Freeman
MOD
Chris Freeman
Treehouse Moderator 68,423 Points
Chris Freeman
Chris Freeman
Treehouse Moderator 68,423 Points

The task will not pass splitting on SPACE. It needs to split on Whitespace. Use split() without arguments for this.

ASHOK Gudivada
ASHOK Gudivada
11,701 Points

Bummer! Hmm, didn't get the expected output. Be sure you're lowercasing the string and splitting on all whitespace!

def word_count(string):
    string_lower = string.lower()
    string_dict = {}
    for word in string_lower.split():
        if word not in string_dict:
            string_dict[word] = string_lower.count(word+" ")
    return string_dict
Chris Freeman
Chris Freeman
Treehouse Moderator 68,423 Points
Chris Freeman
Chris Freeman
Treehouse Moderator 68,423 Points

Testing your code in the REPL reveals the issue:

>>> word_count("I see too many i's in this string")
{'in': 2, 'this': 1, 'string': 1, "i's": 1, 'many': 1, 'too': 1, 'see': 1, 'i': 5}

Using count() is not specific enough. Try assigning string_lower.split() to a variable, then see if word is in this list.

To see the difference between .split() and .split("") check the split docs

ASHOK Gudivada
ASHOK Gudivada
11,701 Points
ASHOK Gudivada
ASHOK Gudivada
11,701 Points 
>>> string_dict = {}
>>>
>>> my_str = "I see too many i's in this string"
>>> string_lower = my_str.lower()
>>> for word in string_lower.split(" "):
...     if word not in string_dict:
...             string_dict[word] = string_lower.count(word+" ")
...
>>> string_dict
{"i's": 1, 'string': 0, 'i': 1, 'many': 1, 'this': 1, 'see': 1, 'too': 1, 'in': 1}

please check this sir, it giving me right answer.

Chris Freeman
Chris Freeman
Treehouse Moderator 68,423 Points
Chris Freeman
Chris Freeman
Treehouse Moderator 68,423 Points

In your test string has a count of zero. That's not correct. Using a SPACE is not sufficient for redundant suffixes and will miss the last word of the string if there is not a trailing space.

Try this to see the error: "The cat sat at the mat":

>>> word_count("The cat sat at the mat")
{'cat': 1, 'sat': 1, 'mat': 1, 'the': 2, 'at': 4}
ASHOK Gudivada
ASHOK Gudivada
11,701 Points
ASHOK Gudivada
ASHOK Gudivada
11,701 Points

But can I use count in some other way to complete the Challenge sir?

Chris Freeman
Chris Freeman
Treehouse Moderator 68,423 Points
Chris Freeman
Chris Freeman
Treehouse Moderator 68,423 Points

Using str.count() will not work due to the false positives when a word is a subset of another word.

You are not far from the answer. Currently, you are:

  • check if word has not been count yet
  • scan string with count
  • store count in dict with word as key

What about trying:

  • check if word not in string_dict.keys() (not been counted)
  • if yes, then simply increment the count in the dict: string_dict[word] += 1
  • if no, then add the new word to the dict with an initial count of 1: string_dict[word] = 1
ASHOK Gudivada
ASHOK Gudivada
11,701 Points
ASHOK Gudivada
ASHOK Gudivada
11,701 Points

Thank you *Chris Freeman*

Sameera Sy
Sameera Sy
2,349 Points
Sameera Sy
Sameera Sy
2,349 Points

I have done the same thing as what Chris Freeman is suggesting, still no luck. Code runs in ideone with expected output but doesn't clear here! Not sure what this issue is about!

Ideone link : http://ideone.com/I6PpmI

def word_count(mystr):
    myl = mystr.lower().split()
    dic = {}
    for value in myl:
        if dic.has_key(value):
           dic[value]+=1
        else:
            dic[value]=1
    return dic
Chris Freeman
Chris Freeman
Treehouse Moderator 68,423 Points
Chris Freeman
Chris Freeman
Treehouse Moderator 68,423 Points

The dict.has_key() was removed in Python 3. Use in instead.

ASHOK Gudivada
ASHOK Gudivada
11,701 Points
ASHOK Gudivada
ASHOK Gudivada
11,701 Points 
def word_count(mystr):
    myl = mystr.lower().split()
    dic = {}
    for value in myl:
        if value in dic:
            dic[value]+=1
        else:
            dic[value]=1
    return dic

my solution to the challenge