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Welcome back. I'm going to work a problem

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that involves traversing a nested data structure,

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and accumulating a value.

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I've got here a variable called big list.

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As you can see, it has some nested lists some inner lists inside of the big list.

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In fact, it's got three levels of nesting that we can see right away,

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because we've got the three square brackets here.

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So, it's going to create a list that has some elements in it.

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That first element of that list is itself going to be

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a list that has some elements in it,

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and the first element of that has a list containing one and two.

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So, in order to traverse this and pull out a word like one,

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or seven, or nine,

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I'm going to have to go three levels into this.

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So, that's some combination of three square brackets and three four loops.

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In this case, I really want to go through every single word,

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because I'm supposed to create a dictionary

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word counts that contains all the words in big list as keys.

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So, I'm going to have as a key,

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the word one, and as a value,

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however many times it appears anywhere deep this nested list,

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and two, however many times it shows up.

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So, that's the dictionary I'm supposed to create.

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To do that I'm going to have to find all of the words anywhere in the big list.

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So, let's get started on this.

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As always, I'd like to make a little template.

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I'm supposed to create a dictionary word counts.

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I'm going to accumulate that dictionary,

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so I'm going to start with an empty one,

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and at the end, I'm going to print word counts just so I can check on my work.

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I'm going to clear out all of the markings that I have here,

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and I can try running this.

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Of course it's not going to give me the right answer,

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but at least it's going to get me started.

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You can see that I've failed the test because my actual value was not what was expected,

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but we have a good start.

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Just going to make my code window narrower,

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so that it doesn't overlap with the output window.

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Alright. So, we can see that what we've output is just the dictionary.

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That's empty we need to traverse through big list,

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and since we're going to have multiple levels here,

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I'm going to use a little mnemonic where I say,

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for inner list level one,

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in big list; IO one is going to be bound to one of the inner lists.

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Initially, its going to be bound to one that begins here,

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and keeps going off the screen.

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So, that's an inner list,

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I'm going to have to descend into that as well.

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So, I'm going to say for each second level list in the first level list,

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I'm going to do something,

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and because I need to descend a third level,

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I think I'm going to need three for loops here.

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I see that I needed a one instead of a two there.

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Now, I think that this is the right thing,

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but I always like to check my work.

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So, I'm just going to see if this gets me the thing that I'm looking for,

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and sure enough that prints out all the words.

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I'm realizing now that I really shouldn't have called my

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iterator variable at level three, IO three,

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I really should have called it word,

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because that's going to help me remember what kind of thing I have there.

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Okay, I change my variable name for my IO three to word,

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and I still got exactly the same output.

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Now, instead of printing out all these words,

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this is where I want to use them to update the word counts dictionary.

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Instead of printing the word,

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I'm going to check if the word is already in word counts,

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and this is a pattern we've seen quite a bit before.

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If we've already seen the word,

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then we want to update its count,

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we want to increment it by one.

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If we haven't seen this word before,

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then we have to put it into the dictionary for the very first time,

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and its initial value ought to be one because we've now seen that word one time.

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So, if we've seen it before,

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we increment its count,

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if we haven't seen before,

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we set its count to one.

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I think that should get us what we want.

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So, associated with the word one is the count of four,

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that means It's happening four times somewhere in here,

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and you can see it is showing up more than once somewhere there.

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So, that's how to work your way up to a solution for a problem.

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Remember that if the data you want to extract is nested three levels deep,

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you're going to need to descend three levels with your code.

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Either square brackets, or four loops.

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In this case we have three indented four loops one inside the other,

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and we have the first level of dissent,

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the second level, and notice that the iterator variable for

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the first time becomes the sequence for the second one,

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and the iterator variable at the second level

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becomes the sequence at the third level and so on.

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Hopefully, you'll be able to do some of the other exercises

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on this exercise's page at the end of the chapter,

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and you'll be able to do some of these on your own.

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Good luck. See you next time.