I'm trying to explore kotlin but when i came across list of strings like below.
I tried min and max functions with it.
And i initially thought it will give compile time error but i didn't get that.
And when i print min i got a555585887996669 as output which is the longest word in array.
val list = listOf<String>("a555585887996669","abtfcr6cr","abcde","abcd")
println(list.min()) //a555585887996669
I need to know on what basis it is returning this value
why min and max is supported to list of strings
The min() and max() extension functions operate on anything that can be compared. That includes numeric types, but also on anything that implements the Comparable interface, which is the standard way for objects to implement a natural ordering.
In this case, String implements Comparable; it uses lexicographic order (which is roughly the order of words in a dictionary), comparing characters pairwise until it finds a difference, or until one String ends. So for example "a" < "abc" < "b".
Collection ordering in Kotlin is explained here.
Have a look at this docu:
https://kotlinlang.org/api/latest/jvm/stdlib/kotlin.collections/min.html
Since String implements comparable, min will return the smallest value based on alpha-numeric sorting.
For eval. array by string length:
list.minBy { it -> it.length };
Related
I have string output:
1 4 2 1 4
I want to get each character in string to compare.
I did it to want to know whether the list is sorted yet.
It's not exactly clear to me what you are trying to achieve. Going by "to know whether the list is sorted", and assuming a list of integers, you can use tcl::mathop::< or tcl::mathop::<=, depending on whether you want to allow duplicate values:
if {[tcl::mathop::<= {*}$list]} {
puts "List is sorted"
} else {
puts "List is mixed up"
}
This will also work for ASCII comparison of strings. For more complex comparisons, like using dictionary rules or case insensitive, it's probably easiest to combine that with lsort along with the -indices option:
tcl::mathop::< {*}[lsort -indices -dictionary $list]
The -indices option returns the original index of each list element in sorted order. By checking if those indices are in incremental order, you know if the original list was already sorted.
Of course, if the point of the exercise was to avoid unnecessary sorting, then this is no use. But then again, bubble sort of an already sorted list is very fast and will basically do exactly the comparisons you described. So just sorting will probably be faster than first checking for a sorted list via a scripted loop.
To get each character in the string, do split $the_string "" (yes, on the empty string). That gives you a list of all the characters in the string; you can use foreach to iterate over them. Remember, you can iterate over two (or more) lists at once:
foreach c1 [split $the_string ""] c2 $target_comparison_list {
if {$c1 ne $c2} {
puts "The first not equal character is “$c1” when “$c2” was expected"
break
}
}
Note that it's rarely useful to continue comparison after a difference is found as the most common differences are (relative to the target string) insertions and deletions; almost everything after either of those will differ.
Seems like some use to knowing a good pattern to make an n-step composition or pipeline from a binary function. Maybe it's obvious or common knowledge.
What I was trying to do was R.either(predicate1, predicate2, predicate3, ...) but R.either is one of these binary functions. I thought R.composeWith might be part of a good solution but didn't get it to work right. Then I think R.o is at the heart of it, or perhaps R.chain somehow.
Maybe there's a totally different way to make an n-ary either that could be better than a "compose-with"(R.either)... interested if so but trying to ask a more general question than that.
One common way for converting a binary function into one that takes many arguments is by using R.reduce. This requires at least the arguments of the binary function and its return type to be the same type.
For your example with R.either, it would look like:
const eithers = R.reduce(R.either, R.F)
const fooOr42 = eithers([ R.equals("foo"), R.equals(42) ])
This accepts a list of predicate functions that will each be given as arguments to R.either.
The fooOr42 example above is equivalent to:
const fooOr42 = R.either(R.either(R.F, R.equals("foo")), R.equals(42))
You can also make use of R.unapply if you want to convert the function from accepting a list of arguments, to a variable number of arguments.
const eithers = R.unapply(R.reduce(R.either, R.F))
const fooOr42 = eithers(R.equals("foo"), R.equals(42))
The approach above can be used for any type that can be combined to produce a value of the same type, where the type has some "monoid" instance. This just means that we have a binary function that combines the two types together and some "empty" value, which satisfy some simple laws:
Associativity: combine(a, combine(b, c)) == combine(combine(a, b), c)
Left identity: combine(empty, a) == a
Right identity: combine(a, empty) == a
Some examples of common types with a monoid instance include:
arrays, where the empty list is the empty value and concat is the binary function.
numbers, where 1 is the empty value and multiply is the binary function
numbers, where 0 is the empty value and add is the binary function
In the case of your example, we have predicates (a function returning a boolean value), where the empty value is R.F (a.k.a (_) => false) and the binary function is R.either. You can also combine predicates using R.both with an empty value of R.T (a.k.a (_) => true), which will ensure the resulting predicate satisfies all of the combined predicates.
It is probably also worth mentioning that you could alternatively just use R.anyPass :)
Thanks to #RedBassett for this Ressource (Kotlin problem solving): https://kotlinlang.org/docs/tutorials/koans.html
I'm aware this question exists here:
Creating a 4 digit Random Number using java with no repetition in digits
but I'm new to Kotlin and would like to explore the direct Kotlin features.
So as the title suggests, I'm trying to find a Kotlin specific way to nicely solve generate a 4 digit number (after that it's easy to make it adaptable for length x) without repeating digits.
This is my current working solution and would like to make it more Kotlin. Would be very grateful for some input.
fun createFourDigitNumber(): Int {
var fourDigitNumber = ""
val rangeList = {(0..9).random()}
while(fourDigitNumber.length < 4)
{
val num = rangeList().toString()
if (!fourDigitNumber.contains(num)) fourDigitNumber +=num
}
return fourDigitNumber.toInt()
}
So the range you define (0..9) is actually already a sequence of numbers. Instead of iterating and repeatedly generating a new random, you can just use a subset of that sequence. In fact, this is the accepted answer's solution to the question you linked. Here are some pointers if you want to implement it yourself to get the practice:
The first for loop in that solution is unnecessary in Kotlin because of the range. 0..9 does the same thing, you're on the right track there.
In Kotlin you can call .shuffled() directly on the range without needing to call Collections.shuffle() with an argument like they do.
You can avoid another loop if you create a string from the whole range and then return a substring.
If you want to look at my solution (with input from others in the comments), it is in a spoiler here:
fun getUniqueNumber(length: Int) = (0..9).shuffled().take(length).joinToString('')
(Note that this doesn't gracefully handle a length above 10, but that's up to you to figure out how to implement. It is up to you to use subList() and then toString(), or toString() and then substring(), the output should be the same.)
When calling List.indexOf(...), what are the advantages of returning -1 rather than null if the value isn't present?
For example:
val list = listOf("a", "b", "c")
val index = list.indexOf("d")
print(index) // Prints -1
Wouldn't it be a cleaner result if index was null instead? If it had an optional return type, then it would be compatible with the elvis operator :? as well as doing things such as index?.let { ... }.
What are the advantages of returning -1 instead of null when there are no matches?
Just speculations but i could think of two reasons:
The first reason is to be compatible with Java and its List.indexOf
As the documentation states:
Returns:
the index of the first occurrence of the specified element in this list, or -1 if this list does not contain the element
The second reason is to have the same datatype as kotlins binarySearch.
Return the index of the element, if it is contained in the list within the specified range; otherwise, the inverted insertion point (-insertion point - 1). The insertion point is defined as the index at which the element should be inserted, so that the list (or the specified subrange of list) still remains sorted.
Where the negative values actually hold additional information where to insert the element if absent. But since the normal indexOf method works on unsorted collections you can not infer the insertion position.
To add to the definitive answer of #Burdui, another reason of such behavior is that -1 return value can be expressed with the same primitive Int type as the other possible results of indexOf function.
If indexOf returned null, it would require making its return type nullable, Int?, and that would cause a primitive return value being boxed into an object. indexOf is often used in a tight loop, for example, when searching for all occurrences of a substring in a string, and having boxing on that hot path could make the cost of using indexOf prohibitive.
On the other hand, there definitely can be situations where performance does not so matter, and returning null from indexOf would make code more expressive. There's a request KT-8133 to introduce indexOfOrNull extension for such situations.
Meanwhile a workaround with calling .takeIf { it >= 0 } on the result of indexOf allows to achieve the same.
I need to store an alphanumeric string in an integer column on one of my models.
I have tried:
#result.each do |i|
hex_id = []
i["id"].split(//).each{|c| hex_id.push(c.hex)}
hex_id = hex_id.join
...
Model.create(:origin_id => hex_id)
...
end
When I run this in the console using puts hex_id in place of the create line, it returns the correct values, however the above code results in the origin_id being set to "2147483647" for every instance. An example string input is "t6gnk3pp86gg4sboh5oin5vr40" so that doesn't make any sense to me.
Can anyone tell me what is going wrong here or suggest a better way to store a string like the aforementioned example as a unique integer?
Thanks.
Answering by request form OP
It seems that the hex_id.join operation does not concatenate strings in this case but instead sums or performs binary complement of the hex values. The issue could also be that hex_id is an array of hex-es rather than a string, or char array. Nevertheless, what seems to happen is reaching the maximum positive value for the integer type 2147483647. Still, I was unable to find any documented effects on array.join applied on a hex array, it appears it is not concatenation of the elements.
On the other hand, the desired result 060003008600401100500050040 is too large to be recorded as an integer either. A better approach would be to keep it as a string, or use different algorithm for producing a number form the original string. Perhaps aggregating the hex values by an arithmetic operation will do better than join ?