I find Lambda in Kotlin to be very confusing and on the top of it is "it".
There are two things I know about "it" and i.e.
If your Lambda has their own argument, you can replace its name with "it".
"It" is an automatically generated name for your Lambda, if it has
only one argument, and you don't specify a different argument name.
Still I don't understand what actually passes as "it".
For E.g. I wanted to apply modulo function on each element of a 3x3 matrix.
fun main(){
var result = Array(3) {
IntArray(3) { 3;2;4;6;7;9;12;11;23 }
}
result = Array(3){ IntArray(3) {it%2} }
println(result.joinToString("\n") { it.joinToString(" ") })
}
Here I assumed that "it" takes each element of the matrix which is clearly not the case as my output was:
0 1 0
0 1 0
0 1 0
So can you please explain me how "it" works, what is happening here? and what would be the correct way to implement this program?
Your line
result = Array(3){ IntArray(3) {it%2} }
isn't doing anything to the original Array that result is pointing at. You are creating a brand new group of array objects by calling the Array and IntArray constructors.
The lambda that you pass to the IntArray constructor has an input parameter that represents the array index, and the return value of your lambda is what will be put into the array at that index. So in this case it is the array index, and your lambda is returning 0 and 1 for even and odd indices respectively.
You are also instantiating your array incorrectly to begin with. Your lambda that you pass to that IntArray constructor is throwing away a bunch of pointless Int values and then returning 23 for each item. So you've created a 3x3 matrix that is completely filled with the number 23.
The correct syntax for creating an array with explicit values is to use arrayOf or intArrayOf.
val result = arrayOf(
intArrayOf(3, 2, 4),
intArrayOf(6, 7, 9),
intArrayOf(12, 11, 23)
)
To modify all the values of an array, you typically iterate the traditional way, not with a lambda:
for (innerArray in result) {
for (i in innerArray.indices)
innerArray[i] = innerArray[i] % 2
}
You were probably thinking of the map function, which lets you pass a lambda and returns a new List with the lambda function applied to every element of the input collection. Or when working with collections other than arrays, you can use forEach or onEach to iterate them without modifying them.
Related
The count is 4 at the end of the code below. I expected 0. Why is it 4? How can I get 0?
var count = 0;
"hello".forEach {
if(it == 'h')
{
println("Exiting the forEach loop. Count is $count");
return#forEach;
}
count++;
}
println("count is $count");
Output:
Exiting the forEach loop. Count is 0
count is 4
return#forEach does not exit forEach() itself, but the lambda passed to it ("body" of forEach()). Note that this lambda is executed several times - once per each item. By returning from it you actually skip only a single item, so this is similar to continue, not to break.
To workaround this you can create a label in the outer scope and return to it:
var count = 0;
run loop# {
"hello".forEach {
if(it == 'h')
{
println("Exiting the forEach loop. Count is $count");
return#loop;
}
count++;
}
}
This is described here: https://kotlinlang.org/docs/returns.html#return-at-labels
Note that the use of local returns in previous three examples is similar to the use of continue in regular loops. There is no direct equivalent for break, but it can be simulated by adding another nesting lambda and non-locally returning from it
It is 4 because the forEach call the lambda passed to it for each character in the string, so the return#forEach in your code return for the first element. You can use a for loop and use break to obtain 0.
return#forEach returns from the lambda function. But the forEach function is a higher-order function that calls the lambda repeatedly for each item in the iterator. So when you return from the lambda, you are only returning for that single item in the iterator. It is analogous to using continue in a traditional for loop.
If you want to exit iteration in a higher-order function completely, you have to use labels. And as I type this, I see another answer already shows how to do that, but I'll leave this in case the different explanation helps.
If your objective is to count the number of characters before 'h', you could do something like this:
val numCharsBeforeH = "hello".takeWhile { it != 'h' }.length
From your comment to Tenfour04's answer:
This is not very convenient. Why didn't the makers of Kotlin create a "break" equivalent?
Here is a quote of the "Loops" section of the Coding conventions:
Prefer using higher-order functions (filter, map etc.) to loops.
Exception: forEach (prefer using a regular for loop instead, unless
the receiver of forEach is nullable or forEach is used as part of a
longer call chain).
When making a choice between a complex expression using multiple
higher-order functions and a loop, understand the cost of the
operations being performed in each case and keep performance
considerations in mind.
Indeed, using a regular for loop with break does what you expect:
var count = 0;
for (char in "hello") {
if (char == 'h') {
println("Breaking the loop. Count is $count")
break
}
count++
}
println("count is $count")
Output:
Breaking the loop. Count is 0
count is 0
Except for very simple operations, there are probably better ways to do what you need than using forEach.
I'm looping through the depends array from a META6.json. I've loaded into a Hash using JSON::Fast. When I'm looping through it using a for loop, however, it only goes through the loop once, and the item is the same array:
use JSON::Fast;
my %meta = from-json(slurp("META6.json"));
for %meta<depends> -> $dependency {
dd $dependency;
}
This piece of code returns
Array $dependency = $["Config::Parser::toml:ver<1.0.1+>", "Config:api<1>:ver<1.3.5+>", "Dist::Helper:ver<0.21.0+>", "Hash::Merge", "Terminal::Getpass:ver<0.0.5+>", "zef"]
I'm expecting it to loop through the %meta<depends> 6 times, which each iteration holding a different element from that array.
For good measure, this is the output of dd %meta<depends> from the example:
Array %meta = $["Config::Parser::toml:ver<1.0.1+>", "Config:api<1>:ver<1.3.5+>", "Dist::Helper:ver<0.21.0+>", "Hash::Merge", "Terminal::Getpass:ver<0.0.5+>", "zef"]
Why is the loop not looping the way I expected?
EDIT: I'm using the latest Rakudo Star:
This is Rakudo Star version 2018.04.1 built on MoarVM version 2018.04.1
implementing Perl 6.c.
Even though %meta<depends> contains an Array, it is contained inside an item (container). The for statement looks at that and decides there's only 1 thing to iterate over (the container).
This is easily remedied: by suffixing .list you convert the item to something Iterable, and thus it will iterate over the Array in the container:
for %meta<depends>.list -> $dependency {
A slightly shorter syntax for this is #():
for #(%meta<depends>) -> $dependency {
EDIT: or use the syntax suggested by jjmerelo, which decontainerizes the element, and thus exposes the underlying Array to for:
for %meta<depends><> -> $dependency {
This is a pitfall. Essentially this is like:
my $var = ['a', 'b', 'c'];
for $var -> $v {
dd $v;
}
Which gives you: $["a", "b", "c"]
If you iterate an array with # sigil it already acts as an Array, but when you have a list inside a scalar it will return the Array and not iterate inside it.
The solution is to use .list to make it act as a list instead of as a scalar.
I am pretty confused with both functions fold() and reduce() in Kotlin, can anyone give me a concrete example that distinguishes both of them?
fold takes an initial value, and the first invocation of the lambda you pass to it will receive that initial value and the first element of the collection as parameters.
For example, take the following code that calculates the sum of a list of integers:
listOf(1, 2, 3).fold(0) { sum, element -> sum + element }
The first call to the lambda will be with parameters 0 and 1.
Having the ability to pass in an initial value is useful if you have to provide some sort of default value or parameter for your operation. For example, if you were looking for the maximum value inside a list, but for some reason want to return at least 10, you could do the following:
listOf(1, 6, 4).fold(10) { max, element ->
if (element > max) element else max
}
reduce doesn't take an initial value, but instead starts with the first element of the collection as the accumulator (called sum in the following example).
For example, let's do a sum of integers again:
listOf(1, 2, 3).reduce { sum, element -> sum + element }
The first call to the lambda here will be with parameters 1 and 2.
You can use reduce when your operation does not depend on any values other than those in the collection you're applying it to.
The major functional difference I would call out (which is mentioned in the comments on the other answer, but may be hard to understand) is that reduce will throw an exception if performed on an empty collection.
listOf<Int>().reduce { x, y -> x + y }
// java.lang.UnsupportedOperationException: Empty collection can't be reduced.
This is because .reduce doesn't know what value to return in the event of "no data".
Contrast this with .fold, which requires you to provide a "starting value", which will be the default value in the event of an empty collection:
val result = listOf<Int>().fold(0) { x, y -> x + y }
assertEquals(0, result)
So, even if you don't want to aggregate your collection down to a single element of a different (non-related) type (which only .fold will let you do), if your starting collection may be empty then you must either check your collection size first and then .reduce, or just use .fold
val collection: List<Int> = // collection of unknown size
val result1 = if (collection.isEmpty()) 0
else collection.reduce { x, y -> x + y }
val result2 = collection.fold(0) { x, y -> x + y }
assertEquals(result1, result2)
Another difference that none of the other answers mentioned is the following:
The result of a reduce operation will always be of the same type (or a super type) as the data that is being reduced.
We can see that from the definition of the reduce method:
public inline fun <S, T : S> Iterable<T>.reduce(operation: (acc: S, T) -> S): S {
val iterator = this.iterator()
if (!iterator.hasNext()) throw UnsupportedOperationException("Empty collection can't be reduced.")
var accumulator: S = iterator.next()
while (iterator.hasNext()) {
accumulator = operation(accumulator, iterator.next())
}
return accumulator
}
On the other hand, the result of a fold operation can be anything, because there are no restrictions when it comes to setting up the initial value.
So, for example, let us say that we have a string that contains letters and digits. We want to calculate the sum of all the digits.
We can easily do that with fold:
val string = "1a2b3"
val result: Int = string.fold(0, { currentSum: Int, char: Char ->
if (char.isDigit())
currentSum + Character.getNumericValue(char)
else currentSum
})
//result is equal to 6
reduce - The reduce() method transforms a given collection into a single result.
val numbers: List<Int> = listOf(1, 2, 3)
val sum: Int = numbers.reduce { acc, next -> acc + next }
//sum is 6 now.
fold - What would happen in the previous case of an empty list? Actually, there’s no right value to return, so reduce() throws a RuntimeException
In this case, fold is a handy tool. You can put an initial value by it -
val sum: Int = numbers.fold(0, { acc, next -> acc + next })
Here, we’ve provided initial value. In contrast, to reduce(), if the collection is empty, the initial value will be returned which will prevent you from the RuntimeException.
Simple Answer
Result of both reduce and fold is "a list of items will be transformed into a single item".
In case of fold,we provide 1 extra parameter apart from list but in case of reduce,only items in list will be considered.
Fold
listOf("AC","Fridge").fold("stabilizer") { freeGift, itemBought -> freeGift + itemBought }
//output: stabilizerACFridge
In above case,think as AC,fridge bought from store & they give stabilizer as gift(this will be the parameter passed in the fold).so,you get all 3 items together.Please note that freeGift will be available only once i.e for the first iteration.
Reduce
In case of reduce,we get items in list as parameters and can perform required transformations on it.
listOf("AC","Fridge").reduce { itemBought1, itemBought2 -> itemBought1 + itemBought2 }
//output: ACFridge
The difference between the two functions is that fold() takes an initial value and uses it as the accumulated value on the first step, whereas the first step of reduce() uses the first and the second elements as operation arguments on the first step.
val seq1 = sequenceOf(1, 2, 3)
val seq2 = sequenceOf(5, 6, 7)
sequenceOf(seq1, seq2).flatten().forEach { ... }
That's how I'm doing sequence concatenation but I'm worrying that it's actually copying elements, whereas all I need is an iterator that uses elements from the iterables (seq1, seq2) I gave it.
Is there such a function?
Your code doesn't copy the sequence elements, and sequenceOf(seq1, seq2).flatten() actually does what you want: it generates a sequence that takes items first from seq1 and then, when seq1 finishes, from seq2.
Also, operator + is implemented in exactly this way, so you can just use it:
(seq1 + seq2).forEach { ... }
The source of the operator is as expected:
public operator fun <T> Sequence<T>.plus(elements: Sequence<T>): Sequence<T> {
return sequenceOf(this, elements).flatten()
}
You can take a look at the implementation of .flatten() in stdlib that uses FlatteningSequence, which actually switches over the original sequences' iterators. The implementation can change over time, but Sequence is intended to be as lazy as possible, so you can expect it to behave in a similar way.
Example:
val a = generateSequence(0) { it + 1 }
val b = sequenceOf(1, 2, 3)
(a + b).take(3).forEach { println(it) }
Here, copying the first sequence can never succeed since it's infinite, and iterating over (a + b) takes items one by one from a.
Note, however, that .flatten() is implemented in a different way for Iterable, and it does copy the elements. Find more about the differences between Iterable and Sequence here.
Something else you might need to do is create a sequence of sequences:
val xs = sequence {
yield(1)
yield(2)
}
val twoXs = sequence {
yieldAll(xs)
// ... interesting things here ...
yieldAll(xs)
}
This doesn't do anything that xs + xs doesn't do, but it gives you a place to do more complex things.
In the code below:
var verticesCount: Int // to read a vertices count for graph
// Reading until we get a valid vertices count.
while (!Assertions.checkEnoughVertices(
verticesCount = consoleReader.readInt(null, Localization.getLocStr("type_int_vertices_count"))))
// The case when we don't have enough vertices.
println(String.format(Localization.getLocStr("no_enough_vertices_in_graph"),
Assertions.CONFIG_MIN_VERTICES_COUNT))
val resultGraph = Graph(verticesCount)
we are getting next error on the last line:
Error:(31, 33) Kotlin: Variable 'verticesCount' must be initialized
Assertions.checkEnoughVertices accepts a safe type variable as an argument (verticesCount: Int), so it's impossible for verticesCount to be uninitialized or null here (and we're getting no corresponding errors on those lines).
What's going on on the last line when already initialized variable becomes uninitialized again?
The syntax you've used denotes a function call with named arguments, not the assignment of a local variable. So verticesCount = is just an explanation to the reader that the value which is being passed here to checkEnoughVertices corresponds to the parameter of that function named verticesCount. It has nothing to do with the local variable named verticesCount declared just above, so the compiler thinks you've still to initialize that variable.
In Kotlin, the assignment to a variable (a = b) is not an expression, so it cannot be used as a value in other expressions. You have to split the assignment and the while-loop condition to achieve what you want. I'd do this with an infinite loop + a condition inside:
var verticesCount: Int
while (true) {
verticesCount = consoleReader.readInt(...)
if (Assertions.checkEnoughVertices(verticesCount)) break
...
}
val resultGraph = Graph(verticesCount)
Well, technically it is possible to assign values to variables in the while condition - and anything else you might want to do there, too.
The magic comes from the also function:
Try this: (excuse the completely useless thing this is doing...)
var i = 10
var doubleI: Int
while ((i * 2).also { doubleI = it } > 0) {
i--
println(doubleI)
}
Any expression can be "extended" with "something to do" by calling also which takes the expression it is called upon as the it parameter and executes the given block. The value also returns is identical to its caller value.
Here's a very good article to explain this and much more: https://medium.com/#elye.project/mastering-kotlin-standard-functions-run-with-let-also-and-apply-9cd334b0ef84