In D language in operator allows to check if a value is present in a sorted random access range.
But if I want to check if a value is present in an unsorted and not random access range, how to do it?
While I agree with Lupus that countUntil does the job, there's a different function that may have less overhead and a more reasonable name: canFind:
import std.algorithm.searching : canFind;
if (haystack.canFind(needle)) {
// ...
}
Use std.algorithm.searching : countUntil
import std.algorithm.searching : countUntil
if (array.countUntil(lookingFor) != -1) {
// . . .
}
count Until is like indexOf in many other languages.
https://dlang.org/phobos/std_algorithm_searching.html#countUntil
Related
This question already has answers here:
Split a list into groups of consecutive elements based on a condition in Kotlin
(4 answers)
Closed 7 months ago.
I want to groupBy a list of items by its value, but only if subsequent, and ignore grouping otherwise:
input:
val values = listOf("Apple", "Apple", "Grape", "Grape", "Apple", "Cherry", "Cherry", "Grape")
output: {"Apple"=2, "Grape"=2, "Apple"=1, "Cherry"=2, "Grape"=1}
There's no built in option for this in Kotlin - it has to be custom, so there are many different options.
Because you need to keep track of the previous element, to compare the current one against, you need to have some sort of state. To achieve this you could use zipWithNext or windowed to group elements. Or use fold and accumulate the values into a list - removing and adding the last element depending on whether there's a break in the sequence.
To try and keep things a bit more clearer (even if it breaks the norms a bit) I recommend using vars and a single loop. I used the buildList { } DSL, which creates a clear scope for the operation.
val result: List<Pair<String, Int>> = buildList {
var previousElement: String? = null
var currentCount: Int = 0
// iterate over each incoming value
values.forEach { currentElement: String ->
// currentElement is new - so increment the count
currentCount++
// if we have a break in the sequence...
if (currentElement != previousElement) {
// then add the current element and count to our output
add(currentElement to currentCount)
// reset the count
currentCount = 0
}
// end this iteration - update 'previous'
previousElement = currentElement
}
}
Note that result will match the order of your initial list.
You cloud use MultiValueMap which can has duplicated keys. Since there is no native model you should implement yourself or use the open-source library.
Here is a reference.
Map implementation with duplicate keys
For comparison purposes, here's a short but inefficient solution written in the functional style using fold():
fun <E> List<E>.mergeConsecutive(): List<Pair<E, Int>>
= fold(listOf()) { acc, e ->
if (acc.isNotEmpty() && acc.last().first == e) {
val currentTotal = acc.last().second
acc.dropLast(1) + (e to currentTotal + 1)
} else
acc + (e to 1)
}
The accumulator builds up the list of pairs, incrementing its last entry when we get a duplicate, or appending a new entry when there's a different item. (You could make it slightly shorter by replacing the currentTotal with a call to let(), but that would be even harder to read.)
It uses immutable Lists and Pairs, and so has to create a load of temporary ones as it goes — which makes this pretty inefficient (𝒪(𝑛²)), and I wouldn't recommend it for production code. But hopefully it's instructive.
There is a data class as fruits.
data class Fruits(
val code: String, //Unique
val name: String
)
The base list indexed items with boolean variable is as below.
val indexList: MutableList<Boolean> = MutableList(baseFruitList.size) { false }
Now the Favourite Indexed list is as below
val favList: MutableList<Boolean> = MutableList(favFruitList.size) { true}
I want a combined full list which basically has the fav item indicated as true.
Ex:
baseFruitList = {[FT1,apple],[FT2,grapes],[FT3,banana],[FT4,mango],[FT5,pears]}
favList = {[FT2,grapes],[FT4,mango]}
The final index list should have
finalIndexed = {false,true,false,true,false}
How can we achieve in Kotlin, without iterating through each element.
You can do
val finalIndexed = baseFruitList.map { it in favList }
assuming, like #Tenfour04 is asking, that name is guaranteed to be a specific value (including matching case) for a specific code (since that combination is how a data class matches another, e.g. for checking if it's in another list)
If you can't guarantee that, this is safer:
val finalIndexed = baseFruitList.map { fruit ->
favList.any { fav.code == fruit.code }
}
but here you have to iterate over all the favs (at least until you find a match) looking to see if one has the code.
But really, if code is the unique identifier here, why not just store those in your favList?
favList = listOf("FT2", "FT4") // or a Set would be more efficient, and more correct!
val finalIndexed = baseFruitList.map { it.code in favList }
I don't know what you mean about "without iterating through each element" - if you mean without an explicit indexed for loop, then you can use these simple functions like I have here. But there's always some amount of iteration involved. Sets are always an option to help you minimise that
here is example of the list. I want to make dynamic where maybe the the value will become more.
val list = arrayListOf("A", "B", "C", "A", "A", "B") //Maybe they will be more
I want the output like:-
val result = list[i] + " size: " + list[i].size
So the output will display every String with the size.
A size: 3
B size: 2
C size: 1
If I add more value, so the result will increase also.
You can use groupBy in this way:
val result = list.groupBy { it }.map { it.key to it.value.size }.toMap()
Jeoffrey's way is better actually, since he is using .mapValues() directly, instead of an extra call to .toMap(). I'm just leaving this answer her since
I believe that the other info I put is relevant.
This will give a Map<String, Int>, where the Int is the count of the occurences.
This result will not change when you change the original list. That is not how the language works. If you want something like that, you'd need quite a bit of work, like overwriting the add function from your collection to refresh the result map.
Also, I see no reason for you to use an ArrayList, especially since you are expecting to increase the size of that collection, I'd stick with MutableList if I were you.
I think the terminology you're looking for is "frequency" here: the number of times an element appears in a list.
You can usually count elements in a list using the count method like this:
val numberOfAs = list.count { it == "A" }
This approach is pretty inefficient if you need to count all elements though, in which case you can create a map of frequencies the following way:
val freqs = list.groupBy { it }.mapValues { (_, g) -> g.size }
freqs here will be a Map where each key is a unique element from the original list, and the value is the corresponding frequency of that element in the list.
This works by first grouping elements that are equal to each other via groupBy, which returns a Map<String, List<String>> where each key is a unique element from the original list, and each value is the group of all elements in the list that were equal to the key.
Then mapValues will transform that map so that the values are the sizes of the groups instead of the groups themselves.
An improved approach, as suggested by #broot is to make use of Kotlin's Grouping class which has a built-in eachCount method:
val freqs = list.groupingBy { it }.eachCount()
My apologies if this is a simple basic info that I should be knowing. This is the first time I am trying to use Java 8 streams and other features.
I have two ArrayLists containing same type of objects. Let's say list1 and list2. Let's say the lists has Person objects with a property "employeeId".
The scenario is that I need to merge these lists. However, list2 may have some objects that are same as in list1. So I am trying to remove the objects from list2 that are same as in list1 and get a result list that then I can merge in list1.
I am trying to do this with Java 8 removeIf() and stream() features. Following is my code:
public List<PersonDto> removeDuplicates(List<PersonDto> list1, List<PersonDto> list2) {
List<PersonDto> filteredList = list2.removeIf(list2Obj -> {
list1.stream()
.anyMatch( list1Obj -> (list1Obj.getEmployeeId() == list2Obj.getEmployeeId()) );
} );
}
The above code is giving compile error as below:
The method removeIf(Predicate) in the type Collection is not applicable for the arguments (( list2Obj) -> {})
So I changed the list2Obj at the start of "removeIf()" to (<PersonDto> list2Obj) as below:
public List<PersonDto> removeDuplicates(List<PersonDto> list1, List<PersonDto> list2) {
List<PersonDto> filteredList = list2.removeIf((<PersonDto> list2Obj) -> {
list1.stream()
.anyMatch( list1Obj -> (list1Obj.getEmployeeId() == list2Obj.getEmployeeId()) );
} );
}
This gives me an error as below:
Syntax error on token "<", delete this token for the '<' in (<PersonDto> list2Obj) and Syntax error on token(s), misplaced construct(s) for the part from '-> {'
I am at loss on what I really need to do to make it work.
Would appreciate if somebody can please help me resolve this issue.
I've simplified your function just a little bit to make it more readable:
public static List<PersonDto> removeDuplicates(List<PersonDto> left, List<PersonDto> right) {
left.removeIf(p -> {
return right.stream().anyMatch(x -> (p.getEmployeeId() == x.getEmployeeId()));
});
return left;
}
Also notice that you are modifying the left parameter, you are not creating a new List.
You could also use: left.removeAll(right), but you need equals and hashcode for that and it seems you don't have them; or they are based on something else than employeeId.
Another option would be to collect those lists to a TreeSet and use removeAll:
TreeSet<PersonDto> leftTree = left.stream()
.collect(Collectors.toCollection(() -> new TreeSet<>(Comparator.comparing(PersonDto::getEmployeeId))));
TreeSet<PersonDto> rightTree = right.stream()
.collect(Collectors.toCollection(() -> new TreeSet<>(Comparator.comparing(PersonDto::getEmployeeId))));
leftTree.removeAll(rightTree);
I understand you are trying to merge both lists without duplicating the elements that belong to the intersection. There are many ways to do this. One is the way you've tried, i.e. remove elements from one list that belong to the other, then merge. And this, in turn, can be done in several ways.
One of these ways would be to keep the employee ids of one list in a HashSet and then use removeIf on the other list, with a predicate that checks whether each element has an employee id that is contained in the set. This is better than using anyMatch on the second list for each element of the first list, because HashSet.contains runs in O(1) amortized time. Here's a sketch of the solution:
// Determine larger and smaller lists
boolean list1Smaller = list1.size() < list2.size();
List<PersonDto> smallerList = list1Smaller ? list1 : list2;
List<PersonDto> largerList = list1Smaller ? list2 : list1;
// Create a Set with the employee ids of the larger list
// Assuming employee ids are long
Set<Long> largerSet = largerList.stream()
.map(PersonDto::getEmployeeId)
.collect(Collectors.toSet());
// Now remove elements from the smaller list
smallerList.removeIf(dto -> largerSet.contains(dto.getEmployeeId()));
The logic behind this is that HashSet.contains will take the same time for both a large and a small set, because it runs in O(1) amortized time. However, traversing a list and removing elements from it will be faster on smaller lists.
Then, you are ready to merge both lists:
largerList.addAll(smallerList);
While exploring the documented example raised in this perl6 question that was asked here recently, I found that the final implementation option - (my interpretation of the example is that it provides three different ways to do something) - doesn't work. Running this;
class HTTP::Header does Associative {
has %!fields handles <iterator list kv keys values>;
sub normalize-key ($key) { $key.subst(/\w+/, *.tc, :g) }
method EXISTS-KEY ($key) { %!fields{normalize-key $key}:exists }
method DELETE-KEY ($key) { %!fields{normalize-key $key}:delete }
method push (*#_) { %!fields.push: #_ }
multi method AT-KEY (::?CLASS:D: $key) is rw {
my $element := %!fields{normalize-key $key};
Proxy.new(
FETCH => method () { $element },
STORE => method ($value) {
$element = do given $value».split(/',' \s+/).flat {
when 1 { .[0] } # a single value is stored as a string
default { .Array } # multiple values are stored as an array
}
}
);
}
}
my $header = HTTP::Header.new;
say $header.WHAT; #-> (Header)
$header<Accept> = "text/plain";
$header{'Accept-' X~ <Charset Encoding Language>} = <utf-8 gzip en>;
$header.push('Accept-Language' => "fr"); # like .push on a Hash
say $header<Accept-Language>.perl; #-> $["en", "fr"]
... produces the expected output. Note that the third last line with the X meta-operator assigns a literal list (built with angle brackets) to a hash slice (given a flexible definition of "hash"). My understanding is this results in three seperate calls to method AT-KEY each with a single string argument (apart from self) and therefore does not exersise the default clause of the given statement. Is that correct?
When I invent a use case that excersises that part of the code, it appears to fail;
... as above ...
$header<Accept> = "text/plain";
$header{'Accept-' X~ <Charset Encoding Language>} = <utf-8 gzip en>;
$header{'Accept-Language'} = "en, fr, cz";
say $header<Accept-Language>.perl; #-> ["en", "fr", "cz"] ??
# outputs
(Header)
This Seq has already been iterated, and its values consumed
(you might solve this by adding .cache on usages of the Seq, or
by assigning the Seq into an array)
in block at ./hhorig.pl line 20
in method <anon> at ./hhorig.pl line 18
in block <unit> at ./hhorig.pl line 32
The error message provides an awesome explanation - the topic is a sequence produced by the split and is now spent and hence can't be referenced in the when and/or default clauses.
Have I correctly "lifted" and implemented the example? Is my invented use case of several language codes in the one string wrong or is the example code wrong/out-of-date? I say out-of-date as my recollection is that Seq's came along pretty late in the perl6 development process - so perhaps, this code used to work but doesn't now. Can anyone clarify/confirm?
Finally, taking the error message into account, the following code appears to solve the problem;
... as above ...
STORE => method ($value) {
my #values = $value».split(/',' \s+/) ;
$element = do given #values.flat {
when 1 { $value } # a single value is stored as a string
default { #values } # multiple values are stored as an array
}
}
... but is it an exact equivalent?
That code works now (Rakudo 2018.04) and prints
$["en", "fr", "cz"]
as intended. It was probably a bug which was eventually solved.