I have an array:
my #e = <60.922 20.946 8.721 7.292 4.306 2.821 2.765 2.752 2.741 2.725>
I would like to divide every element in the array by the minimum, however
#e /= #e.min
produced a single element, which isn't correct.
I've read https://docs.raku.org/type/Array but I don't understand the basic elements of Raku for this.
How can I divide every item in the array by the same number?
You can use the raku hyper & compound metaoperators like this:
#a >>/=>> #a.min
>>X>> means "apply operator X over all list items (with more items on the left)"
/= means "assign result of divide operator / to replace each original left hand item"
Use / instead of /= if you want to return the list of results but leave #a unchanged functional programming style.
[Edited according to #lizmat & #Sebastian comments]
my #a = 2, 3, 5, 7, 10;
my $div = #a.min;
$_ /= $div for #a;
say #a; # [1 1.5 2.5 3.5 5]
When you iterate over an array, you get mutable elements.
put #e.map( * / #e.min );
OR
put #e.map: * / #e.min;
Sample Input:
my #e = <60.922 20.946 8.721 7.292 4.306 2.821 2.765 2.752 2.741 2.725>;
Sample Output:
22.356697 7.686606 3.200367 2.675963 1.580183 1.035229 1.014679 1.009908 1.005872 1
If you want to continue working with the resultant values, assign the output to a new variable. Or overwrite the original #e array using the .= "back-assignment" operator [ short for #e = #e.map( โฆ ) ]. In the Raku REPL:
~$ raku
Welcome to ๐๐๐ค๐ฎ๐๐จโข v2021.06.
Implementing the ๐๐๐ค๐ฎโข programming language v6.d.
Built on MoarVM version 2021.06.
To exit type 'exit' or '^D'
> my #e = <60.922 20.946 8.721 7.292 4.306 2.821 2.765 2.752 2.741 2.725>;
[60.922 20.946 8.721 7.292 4.306 2.821 2.765 2.752 2.741 2.725]
> #e .= map( * / #e.min );
[22.356697 7.686606 3.200367 2.675963 1.580183 1.035229 1.014679 1.009908 1.005872 1]
> put #e;
22.356697 7.686606 3.200367 2.675963 1.580183 1.035229 1.014679 1.009908 1.005872 1
>
Related
Isn't order of execution generally from left to right in Raku?
my #a = my #b = [9 , 3];
say (#a[1] - #a[0]) == (#b[1] R- #b[0]); # False {as expected}
say (#a.pop() - #a.pop()) == (#b.pop() R- #b.pop()); # True {Huh?!?}
This is what I get in Rakudo(tm) v2020.12 and 2021.07.
The first 2 lines make sense, but the third I can not fathom.
It is.
But you should realize that the minus infix operator is just a subroutine under the hood, taking 2 parameters that are evaluated left to right. So when you're saying:
$a - $b
you are in fact calling the infix:<-> sub:
infix:<->($a,$b);
The R meta-operator basically creates a wrap around the infix:<-> sub that reverses the arguments:
my &infix:<R->($a,$b) = &infix:<->.wrap: -> $a, $b { nextwith $b, $a }
So, if you do a:
$a R- $b
you are in fact doing a:
infix:<R->($a,$b)
which is then basically a:
infix:<->($b,$a)
Note that in the call to infix:<R-> in your example, $a become 3, and $b becomes 9 because the order of the arguments is processed left to right. This then calls infix:<->(3,9), producing the -6 value that you would also get without the R.
It may be a little counter-intuitive, but I consider this behaviour as correct. Although the documentation could probably use some additional explanation on this behaviour.
Let me emulate what I assumed was happening in line 3 of my code prefaced with #a is the same as #b is 9, 3 (big number then little number)
(#a.pop() - #a.pop()) == (#b.pop() R- #b.pop())
(3 - 9) == (3 R- 9)
( -6 ) == ( 6 )
False
...That was my expectation. But what raku seems to be doing is
(#a.pop() - #a.pop()) == (#b.pop() R- #b.pop())
#R meta-op swaps 1st `#b.pop()` with 2nd `#b.pop()`
(#a.pop() - #a.pop()) == (#b.pop() - #b.pop())
(3 - 9) == (3 - 9)
( -6 ) == ( -6 )
True
The R in R- swaps functions first, then calls the for values. Since they are the same function, the R in R- has no practical effect.
Side Note: In fuctional programming a 'pure' function will return the same value every time you call it with the same parameters. But pop is not 'pure'. Every call can produce different results. It needs to be used with care.
The R meta op not only reverses the operator, it will also reverse the order in which the operands will be evaluated.
sub term:<a> { say 'a'; '3' }
sub term:<b> { say 'b'; '9' }
say a ~ b;
a
b
ab
Note that a happened first.
If we use R, then b happens first instead.
say a R~ b;
b
a
ba
The problem is that in your code all of the pop calls are getting their data from the same source.
my #data = < a b a b >;
sub term:<l> { my $v = #data.shift; say "l=$v"; return $v }
sub term:<r> { my $v = #data.shift; say "r=$v"; return $v }
say l ~ r;
l=a
r=b
ab
say l R~ r;
r=a
l=b
ab
A way to get around that is to use the reduce meta operator with a list
[-](#a.pop, #a.pop) == [R-](#a.pop, #a.pop)
Or in some other way make sure the pop operations happen in the order you expect.
You could also just use the values directly from the array without using pop.
[-]( #a[0,1] ) == [R-]( #a[2,3] )
Let me emulate what happens by writing the logic one way for #a then manually reversing the operands for #b instead of using R:
my #a = my #b = [9 , 3];
sub apop { #a.pop }
sub bpop { #b.pop }
say apop - apop; # -6
say bpop - bpop; # -6 (operands *manually* reversed)
This not only appeals to my sense of intuition about what's going on, I'm thus far confused why you were confused and why Liz has said "It may be a little counter-intuitive" and you've said it is plain unintuitive!
I am trying to do this simple while loop but it's not working. The compiler is not giving any type of error or warning but when i try to run my function it ends up in an infinite loop:
let example x =
let k = ref x in
while (!k > 42) do
if ( (!k mod 5) == 0) then (
k:= !k/2
);
done;
if(!k<42) then (
Printf.printf "k is less than 42"
);
if(!k == 42) then (
Printf.printf "k is equal to 42"
)
;;
Well, your loop only modifies !k when !k mod 5 = 0. If !k isn't divisible by 5, it will never change in value. This suggests that the loop will either run 0 times or will run an infinite number of times.
You don't show any calls to example, but any call where you pass a value > 42 and not a multiple of 5 should loop infinitely it seems to me.
By the way, this is what #user2864740 was trying to point out. 43 is a value > 42 that's not divisible by 5.
(As a side comment, you should use = to compare values for equality. The == operator in OCaml is useful in limited cases. This often causes problems for people coming from other languages.)
I'm working on this weeks PerlWChallenge.
You are given an array of integers #A. Write a script to create an
array that represents the smaller element to the left of each
corresponding index. If none found then use 0.
Here's my approach:
my #A = (7, 8, 3, 12, 10);
my $L = #A.elems - 1;
say gather for 1 .. $L -> $i { take #A[ 0..$i-1 ].grep( * < #A[$i] ).min };
Which kinda works and outputs:
(7 Inf 3 3)
The Infinity obviously comes from the empty grep. Checking:
> raku -e "().min.say"
Inf
But why is the minimum of an empty Seq Infinity? If anything it should be -Infinity. Or zero?
It's probably a good idea to test for the empty sequence anyway.
I ended up using
take .min with #A[ 0..$i-1 ].grep( * < #A[$i] ) or 0
or
take ( #A[ 0..$i-1 ].grep( * < #A[$i] ) or 0 ).min
Generally, Inf works out quite well in the face of further operations. For example, consider a case where we have a list of lists, and we want to find the minimum across all of them. We can do this:
my #a = [3,1,3], [], [-5,10];
say #a>>.min.min
And it will just work, since (1, Inf, -5).min comes out as -5. Were min to instead have -Inf as its value, then it'd get this wrong. It will also behave reasonably in comparisons, e.g. if #a.min > #b.min { }; by contrast, an undefined value will warn.
TL;DR say min displays Inf.
min is, or at least behaves like, a reduction.
Per the doc for reduction of a List:
When the list contains no elements, an exception is thrown, unless &with is an operator with a known identity value (e.g., the identity value of infix:<+> is 0).
Per the doc for min:
a comparison Callable can be specified with the named argument :by
by is min's spelling of with.
To easily see the "identity value" of an operator/function, call it without any arguments:
say min # Inf
Imo the underlying issue here is one of many unsolved wide challenges of documenting Raku. Perhaps comments here in this SO about doc would best focus on the narrow topic of solving the problem just for min (and maybe max and minmax).
I think, there is inspiration from
infimum
(the greatest lower bound). Let we have the set of integers (or real
numbers) and add there the greatest element Inf and the lowest -Inf.
Then infimum of the empty set (as the subset of the previous set) is the
greatest element Inf. (Every element satisfies that is smaller than
any element of the empty set and Inf is the greatest element that
satisfies this.) Minimum and infimum of any nonempty finite set of real
numbers are equal.
Similarly, min in Raku works as infimum for some Range.
1 ^.. 10
andthen .min; #1
but 1 is not from 1 ^.. 10, so 1 is not minimum, but it is infimum
of the range.
It is useful for some algorithm, see the answer by Jonathan
Worthington or
q{3 1 3
-2
--
-5 10
}.lines
andthen .map: *.comb( /'-'?\d+/ )ยป.Int # (3, 1, 3), (-2,), (), (-5, 10)
andthen .map: *.min # 1,-2,Inf,-5
andthen .produce: &[min]
andthen .fmt: '%2d',',' # 1,-2,-2,-5
this (from the docs) makes sense to me
method min(Range:D:)
Returns the start point of the range.
say (1..5).min; # OUTPUT: ยซ1โคยป
say (1^..^5).min; # OUTPUT: ยซ1โคยป
and I think the infinimum idea is quite a good mnemonic for the excludes case which also could be 5.1^.. , 5.0001^.. etc.
I've come across this code at RosettaCode
constant #primes = 2, 3, { first * %% none(#_), (#_[* - 1], * + 2 ... Inf) } ... Inf;
say #primes[^10];
Inside the explicit generator block:
1- What or which sequence do the #_ s refer to?
2- What does the first * refer to?
3- What do the * in #_[* - 1] and the next * refer to?
4- How does the sequence (#_[* - 1], * + 2 ... Inf) serve the purpose of finding prime numbers?
Thank you.
The outer sequence operator can be understood as: start the sequence with 2 and 3, then run the code in the block to work out each of the following values, and keep going until infinity.
The sequence operator will pass that block as many arguments as it asks for. For example, the Fibonacci sequence is expressed as 1, 1, * + * ... Inf, where * + * is shorthand for a lambda -> $a, $b { $a + $b }; since this wishes for two parameters, it will be given the previous two values in the sequence.
When we use #_ in a block, it's as if we write a lambda like -> *#_ { }, which is a slurpy. When used with ..., it means that we wish to be passed all the previous values in the sequence.
The sub first takes a predicate (something we evaluate that returns true or false) and a list of values to search, and returns the first value matching the predicate. (Tip for reading things like this: whenever we do a call like function-name arg1, arg2 then we are always parsing a term for the argument, meaning that we know the * cannot be a multiplication operator here.)
The predicate we give to first is * %% none(#_). This is a closure that takes one argument and checks that it is divisible by none of the previous values in the sequence - for if it were, it could not be a prime number!
What follows, #_[* - 1], * + 2 ... Inf, is the sequence of values to search through until we find the next prime. This takes the form: first value, how to get the next value, and to keep going until infinity.
The first value is the last prime that we found. Once again, * - 1 is a closure that takes an argument and subtracts 1 from it. When we pass code to an array indexer, it is invoked with the number of elements. Thus #arr[* - 1] is the Raku idiom for "the last thing in the array", #arr[* - 2] would be "the second to last thing in the array", etc.
The * + 2 calculates the next value in the sequence, and is a closure that takes an argument and adds 2 to it. While we could in fact just do a simple range #_[* - 1] .. Inf and get a correct result, it's wasteful to check all the even numbers, thus the * + 2 is there to produce a sequence of odd numbers.
So, intuitively, this all means: the next prime is the first (odd) value that none of the previous primes divide into.
I have an array of 3 elements:
my #a = <x y z>;
Then I make a hash with Array values and make #a's contents one of its values:
my Array %h;
%h<a> = #a;
Later I retrieve this value and assign it to another array:
my #A = %h<a>;
But what I get in #A is not an array of 3 elements, but an array of one element, which is itself an array of 3 elements:
say #A; # [[x y z]]
say #A[0].elems; # 3
So, %h<a> was pushed into #A.
Where is the error in my code?
UPD: This seems to fix the problem, but doesn't improve my understanding. :)
my #A = #(%h<a>);
say #A; [x y z]
The reason this happens, is that the array #a must be put into a container to allow to be stored as a value in a Hash. And something inside a container will remain in the container when being stored into array #A. So, what you need is to get rid of the container when you assign to #A. Your solution is one way of getting rid of the container, but that creates an intermediate List, which could become expensive for large arrays.
As I said, you need to get rid of the container. Fortunately, we have a syntax for that: postfix <>:
my #a = <a b c>;
my %h = a => #a;
my #b = %h<a><>; # <-- note the <> here
dd #b; # Array #b = ["a", "b", "c"]
This would be the most efficient way if you really want #b to be mutable. Alternately, if you want #b to be just an alias of the original #a, you could also bind:
my #a = <a b c>;
my %h = a => #a;
my #b := %h<a>; # <-- note the := here
dd #b; # Array #a = ["a", "b", "c"]
Note that the dd output now shows #a as the name, as it now really is the same as #a: any changes to #a will now also show in #b and vice-versa. Which would be ok if you're not changing them anyway. If that's true in your situation, then this would be the most efficient way of doing this, both CPU-wise as well as memory-wise.