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This lesson is going to be a very, very quick exercise.
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So I'll insert a markdown cell here,
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so the exercise will consist of checking if a word is part of the vocabulary.
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So we've got 2500 words in the vocabulary and what I'd like to do is I'd like you to write the code
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to check if a particular word is in this vocabulary. As a challenge,
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can you write a line of code that checks if a particular word is part of the vocabulary? Your code should
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return True if the word is among the 2500 words that comprise the vocabulary and
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False otherwise.
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And I'd like you to check the following words individually:
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'machine', 'learning', 'fun', 'learn', 'data', 'science', 'app' and 'brewery'.
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I'll give you a few seconds to pause the video and give this a shot.
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Did you have a go?
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Well first, let me show you the more inefficient way. You can use our dataframe of vocabulary words and
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select the "VOCAB_WORD" column and then check if any of these words is equal to the word 'machine'.
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So if I hit Shift+Enter on this I'll get a huge, huge result
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and because I get a whole series of booleans I can't check these all individually.
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So what I would do then is wrap this in the "any" function.
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So in this case I can find out,
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yes, the word 'machine' is amongst the vocabulary words that are most frequent. Now there's actually a better
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way because we've already learned about sets.
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So I'll just add a little comment here and I'll write "inefficient". The better way of doing this, of checking
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membership in a collection is to use the Python sets. I can convert our vocabulary words to a set by
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using the "set" keyword and then feeding in our column "vocab.VOCAB_WORD".
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This converts our vocabulary to a set
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and this is very, very efficient at checking membership. And the way you would do this is using this "in" keyword.
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So if I have the word 'machine' in single quotes followed by the 'in' keyword and then our set here, I get
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a much, much better way of checking for membership.
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And this quite frankly is the better way.
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Now if you're just executing one line of code, you might be like "Well, why is this inefficient?", right.
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Why is using this "any" and this logical condition here inefficient?
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And the thing is - yes, you wouldn't actually notice in this case, right.
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It's
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a very, very quick thing to execute.
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But if you're running a loop, if you're running something that runs thousands and thousands of times,
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then you will actually see a massive difference in the computer time using sets versus another type
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of data structure. Going back up to our list, let's see which words were actually among the 2500.
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So 'machine' is True.
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What about 'learning'? 'learning' is False,
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right.
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So 'learning' is not amongst the words, but that could be because of the stemming, right.
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So if I write 'learn' instead, then it is indeed included. The word 'fun' is as well. The word 'data' is as well.
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The word 'science' is as well. The word 'app' also among the 2500 most common words in our data set. And the
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word 'brewery' is not.
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Yeah.
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So this one is not included.
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'brewer' also not,
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and 'brew' also not. Nobody uses this word.
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All right.
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So that completes the challenge. And the rationale for putting these challenges always into the lessons
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is because programming is really like a sport.
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Right.
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You can't really read about it to get good and you can't just copy code to get good at it.
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You actually have to do it and it's kind of similar how nobody really reads the book on how to surf and
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then jumps on a surfboard and, you know, surfs around and knows how to surf, doesn't work.
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Now it seems obvious with the surfing example, but with programming you really have to think about it
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in the same way.
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And that's why I've got another exercise for you.
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This one will fall into the realm of data exploration. I'll see in the next lesson.