Hi,
I was wondering whether following code can be rewritten using map/grep
construct.
class A {
has $.a;
has $.b;
}
my @array= (
A.new(a='a', b='11'),
A.new(a='a', b='22'),
A.new(a='v', b='33'),
A.new(a='w', b='44'),
A.new(a='v', b='55')
);
my
Hi, like that?
class A { has $.a; has $.b };
my @array = A.new(a='a', b='11'),
A.new(a='a', b='22'),
A.new(a='v', b='33'),
A.new(a='w', b='44'),
A.new(a='v', b='55');
say @array.map({ .a = .b = $_ })
OUTPUT«a = 11 = A.new(a = a, b = 11) a = 22 =
Hi Tobias,
Almost. At least at my rakudo creates list of hash of hash and loses
data while converting to hash:
(a = 11 = A.new(a = a, b = 11), a = 22 = A.new(a =
a, b = 22), v = 33 = A.new(a = v, b = 33), w =
44 = A.new(a = w, b = 44), v = 55 = A.new(a = v, b
= 55)).list.item
On Fri, Jun 13,
Ok got it. But solution is neither more readable nor faster (IMHO only
- I didn't benchmark it)
class A { has $.a; has $.b };
my @array = A.new(a='a', b='11'),
A.new(a='a', b='22'),
A.new(a='v', b='33'),
A.new(a='w', b='44'),
A.new(a='v', b='55');
my %h = @array.map({
my
On 06/13/2014 08:15 PM, Kamil Kułaga wrote:
Ok got it. But solution is neither more readable nor faster (IMHO only
- I didn't benchmark it)
class A { has $.a; has $.b };
my @array = A.new(a='a', b='11'),
A.new(a='a', b='22'),
A.new(a='v', b='33'),
A.new(a='w', b='44'),
On 13 Jun 2014, at 12:36, Kamil Kułaga teodoz...@gmail.com wrote:
I was wondering whether following code can be rewritten using map/grep
construct.
class A {
has $.a;
has $.b;
}
my @array= (
A.new(a='a', b='11'),
A.new(a='a', b='22'),
A.new(a='v',
