I have an interesting issue. I have a string that's in html and I need to parse a table so that I can get the data I need out of that table and present it in a way that looks good on a mobile device. So I use regex and it works just fine but now I'm porting my code to using Kotlin and the solution I have is not porting over well. Here is what the solution looks currently:
var pointsParsing = Regex.Matches(htmlBody, "<td.*?>(.*?)</td>", RegexOptions.IgnoreCase | RegexOptions.Compiled);
var pointsSb = new StringBuilder();
for (var i = 0; i < pointsParsing.Count; i+= 2)
{
var pointsTitle = pointsParsing[i].Groups[1].Value.Replace("&", "&");
var pointsValue = pointsParsing[i+1].Groups[1].Value;
pointsSb.Append($"{pointsTitle} {pointsValue} {pointsVerbiage}\n");
}
return pointsSb.ToString();
as you see, each run in the loop I get two results from the regex search and as a result I tell the for loop to increment by two to avoid collision.
However I don't seem to have this ability within Kotlin, I know how to get the index in a for loop but no idea on how to tell it to skip by 2 so I don't accidentally get something I already parsed on the last loop lap.
how would I tell the for loop to work the way I need it to in Kotlin?
You might be looking for chunked which lets you split an iterable into chunks of e.g. 2 elements:
ptsListResults.chunked(2).forEach { data -> // data is a list of (up to) two elements
val pointsTitle = data[0].groups[1]!!.value
val pointsValue = data[1].groups[1]!!.value
// etc
}
so that's more explicit about breaking your list up into meaningful chunks, and operating within the structure of those chunks, rather that manipulating indices.
There's also windowed which is a bit more complex and gives you more options, one of which is disallowing partial windows (i.e. chunks at the end that don't have the required number of elements). Probably doesn't apply here but just so's you know!
I found a solution that looks to work and thought I'd share.
thanks to this SO answer I see how you can skip over the indexes.
val pointsListSearch = "<td.*?>(.*?)</td>".toRegex()
val pointsListSearchResults = pointsListSearch.findAll(htmlBody)
val pointsSb = StringBuilder()
val ptsListResults = pointsListSearchResults.toList()
for (i in ptsListResults.indices step 2)
{
val pointsTitle = ptsListResults[i].groups[1]!!.value
val pointsValue = ptsListResults[i+1].groups[1]!!.value
pointsSb.append("${pointsTitle}: ${pointsValue}")
}
Related
I start with a list of integers from 1 to 1000 in listOfRandoms.
I would like to left join on random from the createDatabase list.
I am currently using a find{} statement within a loop to do this but feel like this is too heavy. Is there not a better (quicker) way to achieve same result?
Psuedo Code
data class DatabaseRow(
val refKey: Int,
val random: Int
)
fun main() {
val createDatabase = (1..1000).map { i -> DatabaseRow(i, Random()) }
val listOfRandoms = (1..1000).map { j ->
val lookup = createDatabase.find { it.refKey == j }
lookup.random
}
}
As mentioned in comments, the question seems to be mixing up database and programming ideas, which isn't helping.
And it's not entirely clear which parts of the code are needed, and which can be replaced. I'm assuming that you already have the createDatabase list, but that listOfRandoms is open to improvement.
The ‘pseudo’ code compiles fine except that:
You don't give an import for Random(), but none of the likely ones return an Int. I'm going to assume that should be kotlin.random.Random.nextInt().
And because lookup is nullable, you can't simply call lookup.random; a quick fix is lookup!!.random, but it would be safer to handle the null case properly with e.g. lookup?.random ?: -1. (That's irrelevant, though, given the assumption above.)
I think the general solution is to create a Map. This can be done very easily from createDatabase, by calling associate():
val map = createDatabase.associate{ it.refKey to it.random }
That should take time roughly proportional to the size of the list. Looking up values in the map is then very efficient (approx. constant time):
map[someKey]
In this case, that takes rather more memory than needed, because both keys and values are integers and will be boxed (stored as separate objects on the heap). Also, most maps use a hash table, which takes some memory.
Since the key is (according to comments) “an ascending list starting from a random number, like 18123..19123”, in this particular case it can instead be stored in an IntArray without any boxing. As you say, array indexes start from 0, so using the key directly would need a huge array and use only the last few cells — but if you know the start key, you could simply subtract that from the array index each time.
Creating such an array would be a bit more complex, for example:
val minKey = createDatabase.minOf{ it.refKey }
val maxKey = createDatabase.maxOf{ it.refKey }
val array = IntArray(maxKey - minKey + 1)
for (row in createDatabase)
array[row.refKey - minKey] = row.random
You'd then access values with:
array[someKey - minKey]
…which is also constant-time.
Some caveats with this approach:
If createDatabase is empty, then minOf() will throw a NoSuchElementException.
If it has ‘holes’, omitting some keys inside that range, then the array will hold its default value of 0 — you can change that by using the alternative IntArray constructor which also takes a lambda giving the initial value.)
Trying to look up a value outside that range will give an ArrayIndexOutOfBoundsException.
Whether it's worth the extra complexity to save a bit of memory will depend on things like the size of the ‘database’, and how long it's in memory for; I wouldn't add that complexity unless you have good reason to think memory usage will be an issue.
I am solving leetcode solution. The question is
Given an integer array nums sorted in non-decreasing order, return an array of the squares of each number sorted in non-decreasing order.
Example 1:
Input: nums = [-4,-1,0,3,10]
Output: [0,1,9,16,100]
Explanation: After squaring, the array becomes [16,1,0,9,100].
After sorting, it becomes [0,1,9,16,100].
Example 2:
Input: nums = [-7,-3,2,3,11]
Output: [4,9,9,49,121]
I solved this through map and then use sorted() for sorting purpose and lastly converted in toIntArray().
My solution
class Solution {
fun sortedSquares(nums: IntArray): IntArray {
return nums.map { it * it }.sorted().toIntArray()
}
}
After all I am taking a look in the discuss success, I found this solution
class Solution {
fun sortedSquares(A: IntArray): IntArray {
// Create markers to use to navigate inward since we know that
// the polar ends are (possibly, but not always) the largest
var leftMarker = 0
var rightMarker = A.size - 1
// Create a marker to track insertions into the new array
var resultIndex = A.size - 1
val result = IntArray(A.size)
// Iterate over the items until the markers reach each other.
// Its likely a little faster to consider the case where the left
// marker is no longer producing elements that are less than zero.
while (leftMarker <= rightMarker) {
// Grab the absolute values of the elements at the respective
// markers so they can be compared and inserted into the right
// index.
val left = Math.abs(A[leftMarker])
val right = Math.abs(A[rightMarker])
// Do checks to decide which item to insert next.
result[resultIndex] = if (right > left) {
rightMarker--
right * right
} else {
leftMarker++
left * left
}
// Once the item is inserted we can update the index we want
// to insert at next.
resultIndex--
}
return result
}
}
The guy also mention in the title Kotlin -- O(n), 95% time, 100% space
So my solution is equal in time and space complexity with other solution with efficient time and space? Or Is there any better solution?
So my solution is equal in time and space complexity with other solution with efficient time and space?
No, your solution runs in O(n log n) time, as it relies on sorted(), which likely runs in O(n log n). Since the alternative solution does not sort the items, it indeed runs on O(n) time. Both solutions use O(n) space, although your solution uses three times as much space (each of map, sorted and toIntArray create a copy of the input).
I have a data class that describes a chef by their name and their skill level and two lists of chefs with various skill levels.
data class Chef(val name: String, val level: Int)
val listOfChefsOne = listOf(
Chef("Remy", 9),
Chef("Linguini", 7))
val listOfChefsTwo = listOf(
Chef("Mark", 6),
Chef("Maria", 8))
I'm to write a function that takes these two lists and creates a list of pairs
so that the two chefs in a pair skill level's add up to 15. The challenge is to do this using only built in list functions and not for/while loops.
println(pairChefs(listOfChefsOne, listOfChefsTwo))
######################################
[(Chef(name=Remy, level=9), Chef(name=Mark, level=6)),
(Chef(name=Linguini, level=7), Chef(name=Maria, level=8))]
As I mentioned previously I'm not to use any for or while loops in my implementation for the function. I've tried using the forEach function to create a list containing all possible pairs between two lists, but from there I've gotten lost as to how I can filter out only the correct pairs.
I think the clue is in the question here!
I've tried using the forEach function to create a list containing all possible pairs between two lists, but from there I've gotten lost as to how I can filter out only the correct pairs.
There's a filter function that looks perfect for this…
To keep things clear, I'll split out a function for generating all possible pairs. (This is my own, but bears a reassuring resemblance to part of this answer! In any case, you said you'd already solved this bit.)
fun <A, B> Iterable<A>.product(other: Iterable<B>)
= flatMap{ a -> other.map{ b -> a to b }}
The result can then be:
val result = listOfChefsOne.product(listOfChefsTwo)
.filter{ (chef1, chef2) -> chef1.level + chef2.level == 15 }
Note that although this is probably the simplest and most readable way, it's not the most efficient for large lists. (It takes time and memory proportional to the product of the sizes of the two lists.) You could improve large-scale performance by using streams (which would take the same time but constant memory). But for this particular case, it might be even better to group one of the lists by level, then for each element of the other list, you could directly look up a Chef with 15 - its level. (That would time proportional to the sum of the sizes of the two lists, and space proportional to the size of the first list.)
Here is the pretty simple naive solution:
val result = listOfChefsOne.flatMap { chef1 ->
listOfChefsTwo.mapNotNull { chef2 ->
if (chef1.level + chef2.level == 15) {
chef1 to chef2
} else {
null
}
}
}
println(result) // prints [(Chef(name=Remy, level=9), Chef(name=Mark, level=6)), (Chef(name=Linguini, level=7), Chef(name=Maria, level=8))]
val matrix: FloatArray = emptyArray<Float>().toFloatArray()
This definitely works, it just looks pretty ugly. Is there no method to create XXXArray directly? Did I miss something?
You have three options for creating a FloatArray:
val arr1 = floatArrayOf(.1f)
val arr2 = FloatArray(12)
And, as you are doing already, emptyArray.
floatArrayOf works exactly like you'd expect; creates an array of the items with a corresponding size. It works just like arrayOf, just with a different return type.
The second one creates one defined by size. I just set the size to 12 as a demo, but you get the idea. The second one is roughly equivalent to float[] arr2 = new float[12];.
By default, it sets all the values to 0, but you can customize that with FloatArray(12) { 1f }, where 1f can be any number you want to initialize all the items in the array as. You don't need that if you just want to set it to 0 though.
You can use the floatArrayOf method to create such an array, as described in the documentation.
I wrote the following code:
val src = (0 until 1000000).toList()
val dest = ArrayList<Double>(src.size / 2 + 1)
for (i in src)
{
if (i % 2 == 0) dest.add(Math.sqrt(i.toDouble()))
}
IntellJ (in my case AndroidStudio) is asking me if I want to replace the for loop with operations from stdlib. This results in the following code:
val src = (0 until 1000000).toList()
val dest = ArrayList<Double>(src.size / 2 + 1)
src.filter { it % 2 == 0 }
.mapTo(dest) { Math.sqrt(it.toDouble()) }
Now I must say, I like the changed code. I find it easier to write than for loops when I come up with similar situations. However upon reading what filter function does, I realized that this is a lot slower code compared to the for loop. filter function creates a new list containing only the elements from src that match the predicate. So there is one more list created and one more loop in the stdlib version of the code. Ofc for small lists it might not be important, but in general this does not sound like a good alternative. Especially if one should chain more methods like this, you can get a lot of additional loops that could be avoided by writing a for loop.
My question is what is considered good practice in Kotlin. Should I stick to for loops or am I missing something and it does not work as I think it works.
If you are concerned about performance, what you need is Sequence. For example, your above code will be
val src = (0 until 1000000).toList()
val dest = ArrayList<Double>(src.size / 2 + 1)
src.asSequence()
.filter { it % 2 == 0 }
.mapTo(dest) { Math.sqrt(it.toDouble()) }
In the above code, filter returns another Sequence, which represents an intermediate step. Nothing is really created yet, no object or array creation (except a new Sequence wrapper). Only when mapTo, a terminal operator, is called does the resulting collection is created.
If you have learned java 8 stream, you may found the above explaination somewhat familiar. Actually, Sequence is roughly the kotlin equivalent of java 8 Stream. They share similiar purpose and performance characteristic. The only difference is Sequence isn't designed to work with ForkJoinPool, thus a lot easier to implement.
When there is multiple steps involved or the collection may be large, it's suggested to use Sequence instead of plain .filter {...}.mapTo{...}. I also suggest you to use the Sequence form instead of your imperative form because it's easier to understand. Imperative form may become complex, thus hard to understand, when there are 5 or more steps involved in the data processing. If there is just one step, you don't need a Sequence, because it just creates garbage and gives you nothing useful.
You're missing something. :-)
In this particular case, you can use an IntProgression:
val progression = 0 until 1_000_000 step 2
You can then create your desired list of squares in various ways:
// may make the list larger than necessary
// its internal array is copied each time the list grows beyond its capacity
// code is very straight forward
progression.map { Math.sqrt(it.toDouble()) }
// will make the list the exact size needed
// no copies are made
// code is more complicated
progression.mapTo(ArrayList(progression.last / 2 + 1)) { Math.sqrt(it.toDouble()) }
// will make the list the exact size needed
// a single intermediate list is made
// code is minimal and makes sense
progression.toList().map { Math.sqrt(it.toDouble()) }
My advice would be to choose whichever coding style you prefer. Kotlin is both object-oriented and functional language, meaning both of your propositions are correct.
Usually, functional constructs favor readability over performance; however, in some cases, procedural code will also be more readable. You should try to stick with one style as much as possible, but don't be afraid to switch some code if you feel like it's better suited to your constraints, either readability, performance, or both.
The converted code does not need the manual creation of the destination list, and can be simplified to:
val src = (0 until 1000000).toList()
val dest = src.filter { it % 2 == 0 }
.map { Math.sqrt(it.toDouble()) }
And as mentioned in the excellent answer by #glee8e you can use a sequence to do a lazy evaluation. The simplified code for using a sequence:
val src = (0 until 1000000).toList()
val dest = src.asSequence() // change to lazy
.filter { it % 2 == 0 }
.map { Math.sqrt(it.toDouble()) }
.toList() // create the final list
Note the addition of the toList() at the end is to change from a sequence back to a final list which is the one copy made during the processing. You can omit that step to remain as a sequence.
It is important to highlight the comments by #hotkey saying that you should not always assume that another iteration or a copy of a list causes worse performance than lazy evaluation. #hotkey says:
Sometimes several loops. even if they copy the whole collection, show good performance because of good locality of reference. See: Kotlin's Iterable and Sequence look exactly same. Why are two types required?
And excerpted from that link:
... in most cases it has good locality of reference thus taking advantage of CPU cache, prediction, prefetching etc. so that even multiple copying of a collection still works good enough and performs better in simple cases with small collections.
#glee8e says that there are similarities between Kotlin sequences and Java 8 streams, for detailed comparisons see: What Java 8 Stream.collect equivalents are available in the standard Kotlin library?