I am trying to count an element in an array of objects.
long number = Stream.of(jobTitle).count();
System.out.println("There are " + number + " employees.");
What happens is that it will print out the message as many times as many employees have the same job title. Yet "number" stays always 1.
Any guiding would be much appreciated.
long number = Stream.of(jobTitle).count();
Counts the elements in a stream that contains one element.
It is not surprising that this operation always ends up with the exact same result.
Your code is equivalent to:
List<Whatever> titels = new ArrayList<>();
titels.put(oneEntry);
... print titels.size()
Long story short: that statement is nonsensical. What you probably meant was:
if (arbetstitel.equalsIgnoreCase(jobCount)){
g++;
or something alikw. Of course g is a rather bad name for a counter.
But the real answer here is: step back. Think what the problem is you intend to solve, and what the elements are you need to look at. The code you are showing here is simply not making (much) sense. I can't tell you how to fix it, because, as said: it is not clear what you try to achieve here.
A streamish way of counting:
long usersWithMatchingTitle = Arrays.stream(employees).filter(e -> e.getJobTitle().equalsIgnoreCase(jobTitleFromUser)).count();
Meaning: instead of manually iterating your array, you can turn the whole array into a stream, and then filter/count whatever you want to.
Please note: your code seems to only care about the first 30 elements in that array. If that is really what you want, you will need ...stream(employees).limit(30)...
You need to change the stream of command to define a proper Predicate for filter option.
Stream.of(employees).filter(e -> e.getJobTitle().equals(jobTitle)).count();
Related
This question already has answers here:
How do I determine whether an array contains a particular value in Java?
(30 answers)
Closed 2 years ago.
I'm working on a little project in java, and I want to make my algorithm more efficient.
What I'm trying to do is check if a given string is present in an array of strings.
The thing is, I know a few ways to check if a string is present in an array of strings, but the array I am working with is pretty big (around 90,000 strings) and I am looking for a way to make the search more efficient, and the only ways I know are linear search based, which is not good for an array of this magnitude.
Edit: So I tried implementing the advices that were given to me, but the code i wrote accordingly is not working properly, would love to hear your thoughts.`
public static int binaryStringSearch(String[] strArr, String str) {
int low = 0;
int high = strArr.length -1;
int result = -1;
while (low <= high) {
int mid = (low + high) / 2;
if (strArr[mid].equals(str)) {
result = mid;
return result;
}else if (strArr[mid].compareTo(str) < 0) {
low = mid + 1;
}else {
high = mid - 1;
}
}
return result;
}
Basically what it's supposed to do is return the index at which the string is present in the array, and if it is not in the array then return -1.
So you have a more or less fixed array of strings and then you throw a string at the code and it should tell you if the string you gave it is in the array, do I get that right?
So if your array pretty much never changes, it should be possible to just sort them by alphabet and then use binary search. Tom Scott did a good video on that (if you don't want to read a long, messy text written by someone who isn't a native english speaker, just watch this, that's all you need). You just look right in the middle and then check - is the string you have before or after the string in the middle you just read? If it is already precisely the right one, you can just stop. But in case it isn't, you can eliminate every string after that string in case it's after the string you want to find, otherwise every string that's before the just checked string. Of course, you also eliminate the string itself if it's not equal because - logic. And then you just do it all over again, check the string in the middle of the ones which are left (btw you don't have to actually delete the array items, it's enough just to set a variable for the lower and upper boundary because you don't randomly delete elements in the middle) and eliminate based on the result. And you do that until you don't have a single string in the list left. Then you can be sure that your input isn't in the array. So this basically means that by checking and comparing one string, you can't just eliminate 1 item like you could with checking one after the other, you can remove more then half of the array, so with a list of 256, it should only take 8 compares (or 9, not quite sure but I think it takes one more if you don't want to find the item but know if it exists) and for 65k (which almost matches your number) it takes 16. That's a lot more optimised.
If it's not already sorted and you can't because that would take way too long or for some reason I don't get, then I don't quite know and I think there would be no way to make it faster if it's not ordered, then you have to check them one by one.
Hope that helped!
Edit: If you don't want to really sort all the items and just want to make it a bit (26 times (if language would be random)) faster, just make 26 arrays for all letters (in case you only use normal letters, otherwise make more and the speed boost will increase too) and then loop through all strings and put them into the right array matching their first letter. That way it is much faster then sorting them normally, but it's a trade-off, since it's not so neat then binary search. You pretty much still use linear search (= looping through all of them and checking if they match) but you already kinda ordered the items. You can imagine that like two ways you can sort a buncha cards on a table if you want to find them quicker, the lazy one and the not so lazy one. One way would be to sort all the cards by number, let's just say the cards are from 1-100, but not continuously, there are missing cards. But nicely sorting them so you can find any card really quickly takes some time, so what you can do instead is making 10 rows of cards. In each one you just put your cards in some random order, so when someone wants card 38, you just go to the third row and then linearly search through all of them, that way it is much faster to find items then just having them randomly on your table because you only have to search through a tenth of the cards, but you can't take shortcuts once you're in that row of cards.
Depending on the requirements, there can be so many ways to deal with it. It's better to use a collection class for the rich API available OOTB.
Are the strings supposed to be unique i.e. the duplicate strings need to be discarded automatically and the insertion order does not matter: Use Set<String> set = new HashSet<>() and then you can use Set#contains to check the presence of a particular string.
Are the strings supposed to be unique i.e. the duplicate strings need to be discarded automatically and also the insertion order needs to be preserved: Use Set<String> set = new LinkedHashSet<>() and then you can use Set#contains to check the presence of a particular string.
Can the list contain duplicate strings. If yes, you can use a List<String> list = new ArrayList<>() to benefit from its rich API as well as get rid of the limitation of fixed size (Note: the maximum number of elements can be Integer.MAX_VALUE) beforehand. However, a List is navigated always in a sequential way. Despite this limitation (or feature), the can gain some efficiency by sorting the list (again, it's subject to your requirement). Check Why is processing a sorted array faster than processing an unsorted array? to learn more about it.
You could use a HashMap which stores all the strings if
Contains query is very frequent and lookup strings do not change frequently.
Memory is not a problem (:D) .
im working on a problem where i have to obtain all permutations of an arraylist of numbers. The only restriction is that any number cant start with 0, so if we have [0,1,2] we would obtain
[1,2,0]
[1,0,2]
[2,0,1]
[2,1,0]
i know how to do this with 3 loops but the thing is that i have to repeat this to different sets of numbers with differentes sizes, so i need one method that i can apply to different sets of numbers but i have no clue on how to do this. I imagine i have to used some kind of recursive function but i dont know how to implement it so the numbers cant start with a 0. Any ideas? please dont just post the code i want to understand the problem, thank you in advantage!!
Curious question! Interesting code kata.
I naively think I would have a recursive method that takes:
a list of the items currently chosen by the caller
a set of the items available for the callee
The method would iterate over the set to chose 1 more item and call itself with the list extended by this item, and the set reduced by this item. Upon return, remove from list, add back to set and go on with next item (take a defensive copy of the set of course).
If the current list is empty, the selected first item cannot be 0, as per your rules. If you must collect the permutations somewhere (not just print), a 3rd argument would be required for a collection or an observer.
The recursion obvioulsy stops when the available set is empty, at which point the permutation is sent to the collection or observer.
If items can repeat, you may have benefit from sorting them first in order to skip electing the same item again at a given position.
Beware this quires a recursion depth of N, for N items. But the danger is minimal because even with N=10000, it may not stackoverflow, but the CPU time to complete would be order(N!) (probably end of universe...)
You could solve this recursively as described here: Permutation of an ArrayList of numbers using recursion.
The only thing that is missing there is your restriction with the zeros, which could be solved somehow like this (the loop is taken from the example above):
for (List<Integer> al : myLists) {
// The part you need to add:
if (al.get(0) == 0) {
continue;
}
String appender = "";
for (Integer i : al) {
System.out.print(appender + i);
appender = " ";
}
System.out.println();
}
You basically check the first element of each permutation and skip the ones with a leading zero. The continue jumps to the next iteration of the loop and therefore to the next permutation.
I want to declare integers, while the program is running.
I run the program, and then I give it via System.in.println an integer and repeat this as long as I want.
I want the program to give those integers a name of a certain type for, for example a(i) or a[i], dunno, (it should be handy) and then a(i) represents the the i'th integer I gave the program.
My idea is then that I can use those elements by their name just like, if I had declared them in the first place.
For example add two integers together.
For example I defined a method add+, which waits for 2 integer and then adds them. For example I write:
add
a(2)
a(47)
(then I would get here the result.)
I don't think implementing the add function is difficult. However I don't know, how to let the program count the number of inputs or how to let it declare and use variables.
First: Welcome to programming java; it will be a long road.
Here are some hints:
Use a List<Integer> to hold the sequence of numbers entered by the user.
Actually instanciate a concreate List class, for example LinkedList<Integer>'. If you need to access the elements by index, use anArrayList`.\
Each time the user enters a number, create a new Integer and userList.add(newInteger);
Simple sample
List<Integer> userList = new LinkedList<Integer>();
for (index = 0; index < 9; ++index)
{
Integer newInteger = new Integer(index);
userList.add(newInteger);
}
for (Integer current : userList)
{
System.out.println(current);
}
Yeah, I am following the conversation.
I am just a bit frustrated, because I can't really write any interesting or practical java programs (yet), because my knowledge isn't that big yet.
First I tried to find out, if there was a way to add elements to array, because arrays seemed to me very useful, because each element of an array already has an address. I googled, and it seems that is not possible.
I might be able to use the idea with the list, but it seems to be that the length of the list has to have a limit and actually I wanted to avoid that.
Am working on some programming homework and am a bit lost. The project is to select the even/odd elements of a listarray and store in another array. It is not the even numbers in each element, but the elements themselves so if an array had values "1,2,5,7,9" and returned the even elements it would give "1, 5, 9". Also have to use recursion. Would anyone be able to give me a starting point or some advice. Though about starting with 2 elements and taking 2nd element and then building up from that, but don't know how it would add on the 2nd pass
public static ArrayList<Integer> even(ArrayList<Integer> list)
ArrayList<Integer> evenlist = ListMethods.deepClone(tList);//make copy of list
if (evenlist.size()<=1) // The list is empty or has one element
{
// return null;// Return the list as is
}
if
(evenlist.size()==2)
{
//return right element
//call method again
//add to list
}
Psuedocode
int[] evens,odds;
function categorize(List<Integer> in,int idx)
if(idx>=in.length)
return
int cur = in[idx]
if(even), add to evens
else add to odds
categorize(in,idx+1)
This sounds similar to the homework I just completed, so if it is (And you're in my class!), I'll not tell you to use any terminology we haven't covered as I know it can be daunting trying to discover something new for practicals (beyond what we have to do).
First, set your exit condition. As you've already said, you have to create a new ArrayList out of the existing one. You are going to remove items from the existing ArrayList, storing the integers that are at even (or odd) indices, until the list is empty.
So your exit condition is:
if (evenList is Empty)
return evenList;
Then, work your way through the steps. I would advise determining if the Array you start with has an even of odd number of steps, something like this:
if (evenList has Even Elements)
int holderForIntsAtEvenElements = last evenList EVEN element
Note we start at the last element, so when you are coming OUT of the recursive method, this will be the last one added to your new ArrayList, and thus it'll be in numerical order. You might find this post interesting to do this: What does this boolean return mean?
We then want to remove the last element from the list and recursively call the method again.
Finally, when we hit our exit condition and start to come out, we want to add the ints we've been storing to them, e.g.:
evenList.add(holderForIntsAtEvenElements);
return evenList;
That doesn't solve one problem, which is what to do with the very first element if the list does NOT have an even number of elements - however, I'll let you try and solve that!
That's a good mix of code and pseudo code and will hopefully help to get you on the right track.
You could use a simple for loop like this:
for (int i = 0; i < list.size(); i += 2) {
System.out.println(list.get(i));
}
If you have to use recursion, here's an outline of the steps you might take. (I won't tell you exactly what to do because you haven't tried anything and it is like homework.)
Take first element and store it
Remove (new) first element from list
Call self
:)
I have
int[] code = new int[10];
It has the following values:
code[0] = 1234;
code[1] = 2222;
code[2] = 2121;
code[3] = 4321;
code[4] = 3333;
code[5] = 2356;
The code in this case refers to the serial number of the files.
The user is suppose to enter the code of the file to remove that specific file.
Let's say user enter 3333 as the code to remove.
code[4] = 3333 would be removed and code[5] = 2356 will move up to take its place. See below...
code[0] = 1234;
code[1] = 2222;
code[2] = 2121;
code[3] = 4321;
code[4] = 2356;
How would I tackle this problem?
I read up that using an Array List would make my life much easier.
However, I was told to just use an array.
Any help please? :)
How would I tackle this problem?
Allocate a new array and copy all of the values that you want to keep from the existing array to the new one.
You could also update the array in place, filling the "hole" at the end of the array with some special value that can't be a legal code.
Since this is a "sounds like homework" question, I'll leave you to figure out how to code it. (It is pretty simple. Just a loop, a test, and some careful manipulation of a second index.)
That's simply not possible. Arrays have a fixed size, so you'll have to make a second array with its size reduced by 1 and copy all values except the one you want to keep. Or keep track of the "working size" of the array separately, at which point you're begun to reimplement ArrayList.
You can simply set the value to something which is improbable (like -1 to save yourself the trouble of moving around and adjust arrays using System.arrayCopy or the likes. This of course assumes that the aim is to get the functionality working. If "moving" elements is an absolute requirement, you'd have to create a new array as mentioned by another comment here.