I want to make a array in array and get one by index form in Kotlin.
for example, I make a this array [ (1, 12(real data is Bitmap)) , (2, 24(same)), (3, 36) ]
so I can get array(index) = 12
how can I make this form of array and get data by index like above?
Maybe Map is what you need:
val map = mapOf(1 to 12, 2 to 24, 3 to 36)
val twelve = map[1]
It is a collection that holds pairs of objects (keys and values) and supports efficiently retrieving the value corresponding to each key.
To add data to a map we can use mutableMapOf function:
val map = mutableMapOf<Int, Bitmap>()
val bitmap: Bitmap = ...
map[4] = bitmap
If you just want an array of bytes, use byteArrayOf:
val array = byteArrayOf(12, 24, 36)
println(array[0]) // 12
ByteArray is the equivalent of Java's byte[].
Note: There is also intArrayOf, floatArrayOf, doubleArrayOf etc.
Since you asked for an array in an array as well:
val arrayOfArrays = arrayOf(byteArrayOf(1, 2, 3), byteArrayOf(24), byteArrayOf(36))
println(arrayOfArrays[0][1]) // 2
In this case the type of arrayOfArrays will be Array<ByteArray> and you need arrayOf to construct that.
Related
var number: [Int] = [1,2,3,4]
var newArray: [Int] = []
for i in 0...number.count-1{
newArray = number[i] * number[i+1]
}
print(newArray)
I want output like this: [1 * 2, 2 * 3, 3 * 4].
I just don't get it where is the problem...
var number: [Int] = [1,2,3,4]
let things = zip(number, number.dropFirst()).map(*)
Whenever you need to turn something like [1, 2, 3, 4] into pairs (1, 2), (2, 3) etc, then the AdjacentPairs method is useful - in Swift Algorithms package - https://github.com/apple/swift-algorithms/blob/main/Sources/Algorithms/AdjacentPairs.swift
Or you can zip a collection with its dropFirst for the same result.
And whenever you need to turn an [A]s into an [B]s then map with a function that turns As into Bs. So in this example you want to turn an array of tuples of Int, like [(1,2), (2,3), (3,4)] into array of Int, like [2, 6, 12] by multiplying the 2 Ints together, so map with *
The benefit of writing it this way is you would avoid the issues with your array mutation, getting index values wrong, running off the ends of arrays etc, and it's often easier to read and think about if you express it without the indices and assignments.
The problem that the compiler flags you is that you assign a single Int value to an array of Int. The following line will resolve that immediate issue:
newArray.append(number[i] * number[i+1])
This should pass compilation but then create a runtime error at execution. The reason is that when you try to fetch number[i+1] when i == number.count-1, you actually fetch number[number.count]. This entry does not exist with 0-based indices. To get 3 sums out of 4 array entries, your loop should iterate 3 times:
for i in 0 ..< number.count-1 {
Or, if you prefer closed ranges:
for i in 0 ... number.count-2 {
A more Swifty way would be to use map, as #Dris suggested. The return type for map is implicitly given by the result of the multiplication, so you can write:
let newArray = number.indices.dropLast().map { i in
number[i] * number[i+1]
}
You can use map()
let numbers = [1,2,3,4]
let newArray = numbers.enumerated().map { $1 * numbers[($0 + 1) % numbers.count] }
May be you should not loop to count-1 but stop before and add result to array :
for i in 0..<number.count-1 {
newArray.append(number[i] * number[i+1])
}
I have an array of size 300x5. In this the column with index 3 consists if some index and column with index 4 consists of corresponding values.
I have created new array in which I am trying to assign the values in index 4 at index 3 locations in this new array. I tried this but it throws an error.
new_arr[old_arr[:,3]] = old_arr[:,4]
One of the example related to what I want to do
new_arr = np.ones((200,1))
new_arr[[2,3,4]] = [22,44,11]
It throws an error
ValueError: shape mismatch: value array of shape (3,) could not be broadcast to indexing result of shape (3,1)
With this code : new_arr[old_arr[:,3]] you try to access new_arr that index come from values are in old_arr[:,3] and you got IndexError.
Is this help you?
new_arr = np.zeros((300, 5))
new_arr[:,3] = old_arr[:,4]
For edited question you need reshape:
new_arr = np.ones((200,1))
new_arr[[2,3,4]] = np.array([2,4,6]).reshape(3,1)
# OR
# new_arr[2:5] = np.array([22,44,11]).reshape(3,1)
I've got an array that contains Integers as the one shown below:
val my_array = Array(10, 20, 6, 31, 0, 2, -2)
I need to get the maximum 3 elements of this array along with their corresponding indices (either using a single function or two separate funcs).
For example, the output might be something like:
// max values
Array(31, 20, 10)
// max indices
Array(3, 1, 0)
Although the operations look simple, I was not able to find any relevant functions around.
Here's a straightforward way - zipWithIndex followed by sorting:
val (values, indices) = my_array
.zipWithIndex // add indices
.sortBy(t => -t._1) // sort by values (descending)
.take(3) // take first 3
.unzip // "unzip" the array-of-tuples into tuple-of-arrays
Here's another way to do it:
(my_array zip Stream.from(0)).
sortWith(_._1 > _._1).
take(3)
res1: Array[(Int, Int)] = Array((31,3), (20,1), (10,0))
I have a 1x10 structure array with plenty of fields and I would like to remove from the struct array the element with a specific value on one of the field variables.
I know the value im looking for and the field I should be looking for and I also know how to delete the element from the struct array once I find it. Question is how(if possible) to elegantly identify it without going through a brute force solution ie a for-loop that goes through elements of the struct array to compare with the value I m looking for.
Sample code: buyers as 1x10 struct array with fields:
id,n,Budget
and the variable to find in the id values like id_test = 12
You can use the fact that if you have an array of structs, and you use the dot referencing, this creates a comma-separated list. If you enclose this in [] it will attempt to create an array and if you enclose it in {} it will be coerced into a cell array.
a(1).value = 1;
a(2).value = 2;
a(3).value = 3;
% Into an array
[a.value]
% 1 2 3
% Into a cell array
{a.value}
% [1] [2] [3]
So to do your comparison, you can convert the field you care about into either an array of cell array to do the comparison. This comparison will then yield a logical array which you can use to index into the original structure.
For example
% Some example data
s = struct('id', {1, 2, 3}, 'n', {'a', 'b', 'c'}, 'Budget', {100, 200, 300});
% Remove all entries with id == 2
s = s([s.id] ~= 2);
% Remove entries that have an id of 2 or 3
s = s(~ismember([s.id], [2 3]));
% Find ones with an `n` of 'a' (uses a cell array since it's strings)
s = s(ismember({s.id}, 'a'));
I am trying to wrote a program to manage a Database through a Scala Gui, and have been running into alot of trouble formatting my data in such a way as to input it into a Table and have the Column Headers populate. To do this, I have been told I would need to use an Array[Array[Any]] instead of an ArrayBuffer[ArrayBuffer[String]] as I have been using.
My problem is that the way I am trying to fill these arrays is modular: I am trying to use the same function to draw from different tables in a MySQL database, each of which has a different number of columns and entries.
I have been able to (I think) define a 2-D array with
val Data = new Array[Array[String]](numColumns)(numRows)
but I haven't found any ways of editing individual cells in this new array.
Data(i)(j)=Value //or
Data(i,j)=Value
do not work, and give me errors about "Update" functionality
I am sure this can't possibly be as complicated as I have been making it, so what is the easy way of managing these things in this language?
You don't need to read your data into an Array of Arrays - you just need to convert it to that format when you feed it to the Table constuctor - which is easy, as demonstrated my answer to your other question: How do I configure the Column names in a Scala Table?
If you're creating a 2D array, the idiom you want is
val data = Array.ofDim[String](numColumms, numRows)
(There is also new Array[String](numColumns, numRows), but that's deprecated.)
You access element (i, j) of an Array data with data(i)(j) (remember they start from 0).
But in general you should avoid mutable collections (like Array, ArrayBuffer) unless there's a good reason. Try Vector instead.
Without knowing the format in which you're retrieving data from the database it's not possible to say how to put it into a collection.
Update:
You can alternatively put the type information on the left hand side, so the following are equivalent (decide for yourself which you prefer):
val a: Array[Array[String]] = Array.ofDim(2,2)
val a = Array.ofDim[String](2,2)
To explain the syntax for accessing / updating elements: as in Java, a multi-dimensional array is just an array of arrays. So here, a(i) is element i of a, which an Array[String], and so a(i)(j) is element j of that array, which is a String.
Luigi's answer is great, but I'd like to shed some light on why your code isn't working.
val Data = new Array[Array[String]](numColumns)(numRows)
does not do what you expect it to do. The new Array[Array[String]](numColumns) part does create an array of array of strings with numColumns entries, with all entries (arrys of strings) being null, and returns it. The following (numRows) then just calls the apply function on that returned object, which returns the numRowsth entry in that list, which is null.
You can try that out in the scala REPL: When you input
new Array[Array[String]](10)(9)
you get this as output:
res0: Array[String] = null
Luigi's solution, instead
Array.ofDim[String](2,2)
does the right thing:
res1: Array[Array[String]] = Array(Array(null, null), Array(null, null))
It's rather ugly, but you can update a multidimensional array with update
> val data = Array.ofDim[String](2,2)
data: Array[Array[String]] = Array(Array(null, null), Array(null, null))
> data(0).update(0, "foo")
> data
data: Array[Array[String]] = Array(Array(foo, null), Array(null, null))
Not sure about the efficiency of this technique.
Luigi's answer is great, but I just wanted to point out another way of initialising an Array that is more idiomatic/functional – using tabulate. This takes a function that takes the array cell coordinates as input and produces the cell value:
scala> Array.tabulate[String](4, 4) _
res0: (Int, Int) => String => Array[Array[String]] = <function1>
scala> val data = Array.tabulate(4, 4) {case (x, y) => x * y }
data: Array[Array[Int]] = Array(Array(0, 0, 0, 0), Array(0, 1, 2, 3), Array(0, 2, 4, 6), Array(0, 3, 6, 9))