Rather than explaining some big problem, I'll skip all that and list the small loop I am struggling with. Anyways, I have to print the key of a map, so I am using a special way to print the key by switching the value and the key around.
for (int i = 0; i < elementData.length; i++){
System.out.print("[" + i + "]");
for (Entry<HashEntry<E>, Integer> entry : foob.entrySet()){
if (entry.getValue().equals(i)){
System.out.print(entry.getKey().toString());
}
}
}
This is my goal: Print [0][1][2][3] like that all the way to 20. Along with that, 9 numbers will go in between those numbers in parens randomly, based on my program.
Here is my result:
[0][1]HashSet$HashEntry#7d4991ad[2][3][4]HashSet$HashEntry#4554617cHashSet$HashEntry#28d93b30[5][6][7][8][9]HashSet$HashEntry#232204a1[10][11]
So there's just some trick to make it not print all this machine language looking stuff. Anyways, what do I have to do? Looks like 1 thing was supposed to come after [1], 2 things after [4], something after [9], and so on.
Thanks!
So there's just some trick to make it not print all this machine language looking stuff?
Yea.
Don't try to print an instance of a class that doesn't override Object.toString(). That "machine language looking stuff" is simply the output of Object.toString().
However, I suspect that your real code is doing this:
if (entry.getValue().equals(i)){
System.out.print(entry.toString());
}
because "HashSet$HashEntry#7d4991ad" looks like the output you would get if you printed a HashSet.HashEntry object. (The other possibility is that you have used HashSet.HashEntry objects as keys in your Map.)
Related
A simple line of code:
String thing = "Something";
thing += " something else" + " and more.";
IntelliJ IDEA offers to change this line into 4 other ways to accomplish the same result:
String.format()
StringBuilder.append()
java.text.MessageFormat.format()
Replace += with =
Why? What is so wrong with +=? Can anyone explain this please? Thanks.
In general case, the result is not the same. + operator allocates intermediate string, which requires additional memory.
String thing = "Something";
thing += " something else" + " and more."; // two allocations: to compute string inside expression ```" something else" + " and more."```` and to execute ```+=```
By using StringBuilder you don't allocate intermediate result, e.g. less memory is needed.
In your case with short lines and without loops no actual performance boost is expected. However with loops you receive O(N^2) complexity. Please check this answer with explained example.
I am assuming you are referring to this:
This is not actually a fix "problem" with your code, as indicated by the pencil icon. Fixes to problems with your code are identified with a lightbulb, such as not using the value of thing:
The pencil just means "here's a shortcut to change your code, so you don't have to change it manually". Changing a a += b to a = a + b usually isn't that much work, but other things, like changing a for-each loop to a regular for loop, is.
This could be useful if you suddenly remembered that you need the index of the array for something.
I have an array list with some names inside it (first and last names). What I have to do is go through each "first name" and see how many times a character (which the user specifies) shows up at the end of every first name in the array list, and then print out the number of times that character showed up.
public int countFirstName(char c) {
int i = 0;
for (Name n : list) {
if (n.getFirstName().length() - 1 == c) {
i++;
}
}
return i;
}
That is the code I have. The problem is that the counter (i) doesn't add 1 even if there is a character that matches the end of the first name.
You're comparing the index of last character in the string to the required character, instead of the last character itself, which you can access with charAt:
String firstName = n.getFirstName()
if (firstName.charAt(firstName.length() - 1) == c) {
i++;
}
When you're setting out learning to code, there is a great value in using pencil and paper, or describing your algorithm ahead of time, in the language you think in. Most people that learn a foreign language start out by assembling a sentence in their native language, translating it to foreign, then speaking the foreign. Few, if any, learners of a foreign language are able to think in it natively
Coding is no different; all your life you've been speaking English and thinking in it. Now you're aiming to learn a different pattern of thinking, syntax, key words. This task will go a lot easier if you:
work out in high level natural language what you want to do first
write down the steps in clear and simple language, like a recipe
don't try to do too much at once
Had I been a tutor marking your program, id have been looking for something like this:
//method to count the number of list entries ending with a particular character
public int countFirstNamesEndingWith(char lookFor) {
//declare a variable to hold the count
int cnt = 0;
//iterate the list
for (Name n : list) {
//get the first name
String fn = n.getFirstName();
//get the last char of it
char lc = fn.charAt(fn.length() - 1);
//compare
if (lc == lookFor) {
cnt++;
}
}
return cnt;
}
Taking the bullet points in turn:
The comments serve as a high level description of what must be done. We write them aLL first, before even writing a single line of code. My course penalised uncommented code, and writing them first was a handy way of getting the requirement out of the way (they're a chore, right? Not always, but..) but also it is really easy to write a logic algorithm in high level language, then translate the steps into the language learning. I definitely think if you'd taken this approach you wouldn't have made the error you did, as it would have been clear that the code you wrote didn't implement the algorithm you'd have described earlier
Don't try to do too much in one line. Yes, I'm sure plenty of coders think it looks cool, or trick, or shows off what impressive coding smarts they have to pack a good 10 line algorithm into a single line of code that uses some obscure language features but one day it's highly likely that someone else is going to have to come along to maintain that code, improve it or change part of what it does - at that moment it's no longer cool, and it was never really a smart thing to do
Aominee, in their comment, actually gives us something like an example of this:
return (int)list.stream().filter(e -> e.charAt.length()-1)==c).count();
It's a one line implementation of a solution to your problem. Cool huh? Well, it has a bug* (for a start) but it's not the main thrust of my argument. At a more basic level: have you got any idea what it's doing? can you look at it and in 2 seconds tell me how it works?
It's quite an advanced language feature, it's trick for sure, but it might be a very poor solution because it's hard to understand, hard to maintain as a result, and does a lot while looking like a little- it only really makes sense if you're well versed in the language. This one line bundles up a facility that loops over your list, a feature that effectively has a tiny sub method that is called for every item in the list, and whose job is to calculate if the name ends with the sought char
It p's a brilliant feature, a cute example and it surely has its place in production java, but it's place is probably not here, in your learning exercise
Similarly, I'd go as far to say that this line of yours:
if (n.getFirstName().length() - 1 == c) {
Is approaching "doing too much" - I say this because it's where your logic broke down; you didn't write enough code to effectively implement the algorithm. You'd actually have to write even more code to implement this way:
if (n.getFirstName().charAt(n.getFirstName().length() - 1) == c) {
This is a right eyeful to load into your brain and understand. The accepted answer broke it down a bit by first getting the name into a temporary variable. That's a sensible optimisation. I broke it out another step by getting the last char into a temp variable. In a production system I probably wouldn't go that far, but this is your learning phase - try to minimise the number of operations each of your lines does. It will aid your understanding of your own code a great deal
If you do ever get a penchant for writing as much code as possible in as few chars, look at some code golf games here on the stack exchange network; the game is to abuse as many language features as possible to make really short, trick code.. pretty much every winner stands as a testament to condense that should never, ever be put into a production system maintained by normal coders who value their sanity
*the bug is it doesn't get the first name out of the Name object
Hey guys I am developing a project where I have 4 questions where someone can evaluate as (great, good, regular, and poor), and after that I would need to check how many people voted as great, how many voted as good, regular, and poor, for each of the 4 questions. So I would like to make a count to check the .txt and count how many times the word (great, good, regular, and poor) apears on it. I was trying to do it like in Python, where you only need a dictionary (or a counter) and simply do something like:
dict["great"] += 1
However, it isn't possible to do so in Java. Does anyone know any method that would be similar to this one in Java, or another way to do it simply (without having to create a lot of variables to save each question's answer).
Thank you very much for your help.
In java 8 the compute method was added to the Map interface. It may be a bit more complicated than in python, but it's probably the closest it gets to the python code:
Map<String, Integer> map = new HashMap<>();
String rating = ...
map.compute(rating, (key, oldValue) -> ((oldValue == null) ? 1 : oldValue+1));
The lambda expression passed as second parameter to compute receives the old value the key was mapped to as second parameter or null, if there was no mapping.
This is 100% possible in Java.
Use a HashMap to store the values.
For example:
HashMap counts = new HashMap<String, Integer>();
counts.put("great", 0);
counts.put("good", 0);
counts.put("regular", 0);
counts.put("poor", 0);
Now, suppose you read in a string input.
To increase the counter, do :
counts.put(input, counts.get(input) + 1);
This will increase the counter in that position by 1.
Use counts.get(input) to get the counter of input string.
I'm using Blue J and would love to be able to test out small fragments of code without having to create a new program, create a class, and write a method. Does Blue J have any way to run simple segments of code (like trying out a new method) without creating a whole new program?? Similar to operating in the Python shell rather than a new file.
The code pad allows you to run some code and instantiate objects. It is helpful for testing small bits of code, but not the type of tool that allows for complete methods as far as I understand. The code pad can be accessed under View -> Show code pad.
Sorry, I'm not able to comment on answers yet - you can write as much code in the code pad as you want, really, probably including full methods if you really wanted to - the only differences between the code pad and regular coding are 1. You don't need to write any 'boilerplate', ie code does not have to be in a method or class structure, and 2., BlueJ will evaluate each line as you enter it, rather than all at once.
Don't forget that SHIFT + ENTER goes to the next line without 'sending' what you've typed to the compiler, too - though in most cases you can write line-by-line, this can help you practice formatting and keep multiple statements (with semicolons of course) from getting unreadable and confusing to debug.
Just as a proof-of-concept, try the below out in the BlueJ codepad. It should work either as one block (using shift+enter) OR as a series of individual lines:
int apples = 5;
int pears = 4;
int fruit = apples + pears;
System.out.println("we have " + apples + " apples and " + pears + " pears.");
System.out.println("that's " + fruit + " pieces of fruit");
It seems, however, that while coming up with examples I've found a limitation. While if (apples > pears) System.out.println("We have more apples than pears"); works fine, it doesn't seem to like me adding an else statement. But if/else statements are probably reaching the threshold where you are better off writing your code out properly, anyway. The code pad is more useful for "how does this work" type of queries that beginners like us tend to have, or to test an individual line before putting it into your method. I often use it to experiment with formatting/structure problems like "is it possible to do evaluations within a println statement" (It is - in the above example we could have just had System.out.println("that's " + (apples + pears) + " pieces of fruit");
This is about my assignment so I would appreciate generic answers with explanation.
I have a for-loop which I have to make for an online ordering system. The for-loop and ordering all works fine but if the same item is put in twice, then it forgets the previous order for the same item and only remembers the latest one. I am required to use simple arrays and for-loops so I would appreciate if the solution/help was also of this basic level.
My code looks something like this (NOTE: The following is just an example of what part of the loop looks like--this is not a complete program):
for (int i = 0; i < 3; i++) { //I changed the loop so there's no confusion about
//what I am actually asking about.
if (order.equalsIgnoreCase(computer) {
int price = quantity[i] + itemcode;
}
}
To explain further, this loop and the if statement work perfectly if a certain item is only ordered once. But if the user enters, say, an order for a computer once and then after 3 more orders, orders another computer, then the output does not add the previous order in the new order but only remembers the latest one.
I would appreciate any work around suggested for this but again, since this is for my studies, I would appreciate explanations rather than direct solutions.
Please ask me questions in case this is not clear.
"Forgets" suggests that you are overwriting something in your code rather than, say, just incrementing. Go through your code, see what parts of it gets reset when you place a new order. For instance, if you are doing
quantity[1] = getNumberFromUser("How many apples?");
then this would obviously erase the old value each time. If you want to merely increment the number of apples, do something like
quantity[1] += getNumberFromUser("How many apples?");
Another general advice is to use print statements to debug your code. That way you can see for yourself what really happens. Learning to use a real debugger would also be of great benefit.
if you have two or more typres of products and want to calculate the price for all the orders together then you can try the following code,, i think thats very simple,,
int price=0;
for (int i = 0; i < 3; i++) { //I changed the loop so there's no confusion about
//what I am actually asking about.
if (order.equalsIgnoreCase(computer) {
price += quantity[i] + itemcode; //use the previous value of price
}
}
or else if you want to have history for each product separately then you have to try the same with a array of price for each product type..
If you cant get the answer then comment here,,