Create a function named word_count() that takes a string.

Question

I've been looking at code for so long I'm blocking out on what's wrong with my code.
Create a function named word_count() that takes a string.
Return a dictionary with each word in the string
as the key and the number of times it appears as the value.
```python
def wordcount(string):
    mydict ={}
    x = string.lower().split()
    for item in x:
        mydict[item] = 1
        if item in mydict:
            my_dict[item] +=1
return my_dict
```

William Li · Accepted Answer

python
my_dict[item] = 1
if item in my_dict:
  my_dict[item] +=1

because of these lines, my_dict[item] =1 comes first will always set the value lookup to 1, EVEN if the key already in the dictionary.
you should do
python
if item in my_dict:
  my_dict[item] +=1
else:
  my_dict[item] = 1

colingallagher · Answer

I was able to pass the challenge with this code:
python
def word_count(a_string):
  string_dict = {}
  for word in a_string.split():
    if word in string_dict:
        string_dict[word] += 1
    else:
        string_dict[word] = 1
  return string_dict

Hope that helps.
Cheers!
Source (https://teamtreehouse.com/forum/dont-even-know-where-to-start-with-this-one-challenge-task-1-of-1-in-wordcountpy)

Raymond James Cabrera · Answer

Here's my attempt.  The output I get for the string "blah blah blah yeah" is {blah: 3, yeah: 4}. It seems that the variable 'count' is continuing its iteration for the next word in my list. Is there another way to have separate iterators for each word in the list, or is Colin's solution the only way to do this?
```python
def wordcount(string):
  liststring = string.split()
  my_dict = {}
  count = 0
  for words in liststring:   
    mydict.update({words: count})
    if words in my_dict:
      count += 1
    else:
      count = 1
my_dict.update({words: count})
  return print(my_dict)
            ```

Tony McCabe · Answer

def wordcount(astring):
  stringdict = {}
  for word in astring.split():
    if word in stringdict:
        stringdict[word] += 1
    else:
        stringdict[word] = 1
  return stringdict 
Here is your answer again, just to make sure!

Adam Oliver · Answer

I knew it was in there somewhere. Thank you.

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Adam Oliver

Adam Oliver

Create a function named word_count() that takes a string.

Create a function named word_count() that takes a string.

Return a dictionary with each word in the string

as the key and the number of times it appears as the value.

5 Answers

William Li

William Li

colingallagher

colingallagher

Adam Oliver

Adam Oliver

Tony McCabe

Tony McCabe

Raymond James Cabrera

Raymond James Cabrera