Better way to solve this Python Code Challenge?

Question

It took me awhile to solve this Code Challenge (https://teamtreehouse.com/library/word-count). Finally did (yay!) but I have a feeling my solution isn't very elegant. Suggestions welcome!
python
def word_count(my_string):
  my_string = my_string.lower()
  word_list = my_string.split()
  word_dict = {}
  count = 0
  for word in word_list:
    if word in word_dict:
      count = word_dict[word]
      count += 1
      word_dict.update({word: count})
    else:
      word_dict[word] = 1
  return word_dict

Dino Paškvan · Accepted Answer

Hello, Ryan!
I wouldn't say there's a necessarily better way, but you can make the whole function less verbose. I'll add comments to your code first:
python
def word_count(my_string):
  my_string = my_string.lower()
  word_list = my_string.split()
  word_dict = {}
  count = 0 # You don't really need this outside of the for loop
  for word in word_list:
    if word in word_dict:
      count = word_dict[word] # Instead of doing all of this
      count += 1 # you can increment the value for that key
      word_dict.update({word: count}) # directly - word_dict[word] += 1
    else:
      word_dict[word] = 1
  return word_dict

So that would turn into:
python
def word_count(my_string):
  my_string = my_string.lower()
  word_list = my_string.split()
  word_dict = {}
  for word in word_list:
    if word in word_dict:
      word_dict[word] += 1
    else:
      word_dict[word] = 1
  return word_dict

If you want to make it even shorter, you can get rid of the word_list variable and iterate over the return value of the chained .lower().split() method call:
python
def word_count(my_string):
  word_dict = {}
  for word in my_string.lower().split():
    if word in word_dict:
      word_dict[word] += 1
    else:
      word_dict[word] = 1
  return word_dict

Kenneth Love · Answer

Since we talked about this in work chat, I'll post my two alternate solutions. Let's assume this string for all examples:
python
sentence = "If the facts don't fit the theory change the facts"

Using Counter
So Python has this library named collections that stores a lot of alternate collection types. One of those is Counter which, well, counts things.
```python
from collections import Count
def wordcount(mystring):
    words = my_string.lower().split()
    counts = Counter(words)
    return counts
```
```python
output = word_count(sentence)
Counter({'if': 1, 'the': 3, ...})
```
What's really awesome about Counter, though is being able to do this:
```python
output.most_common(1)
[('the', 3)]
```
Using setdefault
Python's dict object has a weird (to me) method called setdefault() that let's you set a default value if the dict doesn't have the key you want. Otherwise it returns the value of the key. Say:
```python
mydict = {'a': 5}
mydict.setdefault('a', 10)
5
mydict.setdefault('b', 10)
10
mydict
{'a': 5, 'b', 10)
```
So you can use that for word_count()
python
def word_count(my_string):
    words = my_string.lower().split()
    counter = {}
    for word in words:
        counter[word] = counter.setdefault(word, 0) + 1
    return counter

The counter[word] = counter.setdefault(word, 0) + 1 line trips some people up. We're setting the default to 0 and then incrementing it by 1. Basically we've negating the try/except that's needed for the usual approach.
So why didn't I teach either of these methods? They're way more special-case than using try/exceptand.get()` and the like. I'd rather you have common scenario tools than wizzbang special effect tools. But, I guess you can now consider yourself taught these nifty little tools. Be sure to check out collections (https://docs.python.org/3/library/collections.html?highlight=collections#module-collections) if you want more neat tools.

Chris Shaw · Answer

Hi Ryan,
Going off Jason's answer which is  the same thought process I had you can shorten your code all the way down to...
```python
def wordcount(mystring):
  word_dict = {}
  for word in mystring.split():
    if word in worddict:
      worddict[word] += 1
    else:
      worddict[word] = 1
  return word_dict
```

Jason Anello · Answer

Hi Ryan,
You can increment the count in your if block like this: word_dict[word] += 1
This would grab the current count for that key, add one to it and store it back in that same key.
This can replace the 3 lines you have in your if block.
You can then remove the count initialization before the for loop.

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Ryan Carson

Ryan Carson

Better way to solve this Python Code Challenge?

4 Answers

Dino Paškvan

Dino Paškvan

Jason Anello

Jason Anello

Chris Shaw

Chris Shaw

Kenneth Love

Kenneth Love

Using `Counter`

Using `setdefault`

Ryan Carson

Ryan Carson

Ken Alger

Ken Alger