Coding Three Stop Codons - Part I

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Do curso por Duke University

Java Programming: Solving Problems with Software

2552 classificações

Duke University

Java Programming: Solving Problems with Software

2552 classificações

Curso 1 de 4 em Specialization Object Oriented Programming in Java

Building upon Duke’s “Programming and the Web for Beginners” course, you will grow in your ability to be a problem-solver and programmer by writing Java programs. We are excited that you've chosen to learn Java with us! You will soon be able to solve problems that you find interesting and useful by practicing a design process you will learn in this course. Java concepts presented in this course include basic control flow; object-oriented concepts including classes and methods; custom, open-source classes; and a design-implement-debug process that you can continue to leverage as you build your experience and knowledge of programming principles. The custom, open-source classes will provide a consistent application programming interface (API) so that you can write programs that access and transform images, websites, comma separated value (CSV) files and other data files using simple and accessible coding idioms built on the Java Iterable interface. You will learn to design, test, and debug programs and to find and read Java documentation that can serve as the basis for continued study of Java and programming. After completing this course you will be able to: 1. Edit, compile, and run a Java program; 2. Use conditionals and for loops appropriately in a Java program; 3. Use Java API documentation in writing programs. 4. Debug a Java program using the scientific method; 5. Develop a Java method to solve a specific problem; 6. Develop a set of test cases as part of developing a program; 7. Create a class with multiple methods that work together to solve a problem; and 8. Use divide-and-conquer design techniques for a program that uses more than one method.

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Strings in Java

This module begins with a short presentation from Raluca Gordân, an assistant professor in Duke University’s Center for Genomic and Computational Biology, about an important problem genomics scientists encounter regularly: how to identify genes in a strand of DNA. To tackle this problem, you will need to understand strings: series of characters such as letters, digits, punctuation, etc. After learning about Java methods that work with strings, you will be able to find genes within a DNA string as well as tackle other string related problems, such as finding all of the links in a web page. By the end of this module, you will be able to: (1) Use important methods for the Java String class; (2) Use conditionals, for loops, and while loops appropriately in a Java program; (3) Find patterns in the data represented by strings to help develop the algorithm for your program; (4) Understand the importance of designing programs that keep different data processing steps separate; (5) Use the StorageResource iterable for this course to store some data for further processing; and (6) Rely on Java documentation to better understand how to use different Java packages and classes.

Introduction0:48

Conceptual Understanding4:01

While Loops9:30

While Loop Syntax and Semantics3:06

Coding While Loops6:45

Three Stop Codons5:03

Coding Three Stop Codons - Part I7:45

Coding Three Stop Codons - Part II4:35

Logical And / Or8:01

Coding And / Or6:52

Finding Multiple Genes5:16

Translating to Code8:00

Conheça os instrutores

Owen Astrachan
Professor of the Practice
Computer Science
Robert Duvall
Lecturer
Computer Science
Andrew D. Hilton
Associate Professor of the Practice
Electrical and Computer Engineering
Susan H. Rodger
Professor of the Practice
Computer Science

Hi, welcome to another version of Gene-Finding.

In today's episode, we're going to look for three different stopCodons.

In the previous coding exercises, you've seen code that found one StopCodon,

making sure with a multiple three away from the StartCodon.

In this version of the code that you've already seen described previously,

we're going to look for three different StopCodons.

As you can see here we have TAA, TAG, and TGA, and

in order to find each of those we are going to write another method.

A method called findStopCodon that we saw outlined

before with our seven step method.

Up here I've got this method, findStopCodon, and

it has three parameters, dnaStr, the string of DNA in which we're searching for

a gene, startIndex, the location at which we're going to begin the search.

And stopCodon, the specific codon we're going to be looking for.

We've got the code from our seven step process.

And what we're going to do here is, write the code that goes along with this.

You can see here as we saw in the previous version of code,

I'm going to create a variable currIndex and I'm going to make it the location of

the first stopCodon, starting at startIndex + 3.

So as we've seen before the way we do that is to create an integer variable currIndex

and we right currIndex gets the location in dnaStr

of the stopCodon, starting at startIndex.

That's exactly what it says here.

Find stopCodon starting from (startIndex + 3), so I'd better make

sure I'm moving that correctly, and store that value in currIndex.

Now as long as current index is not equal to -1, and as you already learned

that's a while loop, so we're going to say while currIndex is not equal to -1.

And when I type the right curly brace as you see here,

I'm going to make sure I put in a left curly brace at the same time.

Indented the same way, and one thing I can do there to make sure that I've got it

right is to compile my program, so

I'm compiling it you can see I've got no syntax errors, so I know I'm on shape.

Now, as long as currIndex is not equal to -1, so I've just coded that,

check to see if the currIndex- startIndex is a multiple of 3.

I'm going to store that difference.

I'm going to say int diff gets currIndex- startIndex.

Now you don't need a new variable for that.

As you saw in the previous video, you can just say,

if currIndex- startIndex mod 3 is equal to 0.

But this sometimes easier conceptually to say if (diff % 3 == 0),

that means, and you can see I've already got my curly braces in there.

Return current index because it's my answer.

So I'm just going to say return currIndex.

That's what it says here.

Check if currIndex- startIndex is a multiple of three, and

if so, currIndex is the answer return it.

So I've got those two things taken care of.

And it says, if not, update currIndex.

So if not, that would be else.

And if you think carefully,

you might see that, although I don't need the else because I've already returned.

But it's probably easier to write it that way in terms of thinking about it.

It says, update currIndex, looking for

stopCodon again, starting from currIndex plus one.

So, you can see up there how I look for it.

So, I'm going to do the same thing.

CurrIndex gets dnastr.indexOf, I’m still looking for

stopCodon and remember, that's a parameter to this method.

And I'm starting, according to our comments at currIndex + 1.

So, currIndex + 1 and if we exit

the loop we didn't find the stopCodon,

so we return dnastr.length.

You can see here that the loop is going to continue

while currIndex is not negative 1.

If it is equal to negative 1 we've exhausted the string,

the dna string in which we are searching.

And so it says return dnaStr.length.

As we'll see later when we write this code, that's to ensure

that when we find the minimum value that our code works properly.

So I'm going to make sure that my method looks right, find the stopCodon

and I started at index+3, so that's the last comment I have.

I'm going to get rid of that comment now.

So I've stored in currIndex the location of stopCodon while that's not -1,

if it's a multiple of 3 I'm done, and if it's not I need to keep searching.

My class compiles, but just compiling doesn't indicate at all that I'm done.

I've written some test code down here, testFindStop.

And you can see as you saw in the previous coding video.

I've created a strand of DNA and I'm going to look for TAA.

Which is one of the stop codons, and I've got it in two different places.

When I run the method you'll see what it does, and

then I'll describe how that testing works.

So here's my workbench, I'm going to right click on this and make a new object.

It comes in the object work bench.

I'm going to right click again and I'm going to run test find stop.

And you can see it printed tests finished.

That means all the tests work correctly.

Because if there was an error, my print statements would have indicated that.

Lets look at the testing code so you can see how I did that.

In testing findStop, the first thing I did was look for

the stopCodon TAA, starting at index 0.

And you can see here, I've got the indices printed above the string.

So TAA here is at index 9.

9 minus 0 Is in fact a multiple of three, so

it should have returned 9, that it found TAA, and if it wasn't 9, I got an error.

To see how it might work if it returned to wrong thing.

Supposed I said of dex is not equal to ten, now that would be a mistake,

because we know it returns nine and it's supposed to return nine.

You'll see in this case when I run it, that it will print something.

Right-click to create the object.

Right-click to invoke the findStops.

And you can see error on 9.

Now, that wasn't an error because I knew I was supposed to get 9.

All these tests print something if there was an error.

And if there is no error, they don't print anything.

My last print statement prints when the tests are finished.

So what we have now is this method findStopCodon written and

tested and we are ready to move on to the next method which was

the same method that you saw before for finding just a single stopCodon.

And now we're going to find any occurrence of a stopCodon.

Any occurrence of three different stopCodons.

We're going to find TAA or TAG or TGA.

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