Hi, AFAIK in Ruby the only (or the “coolest”) way to do something as:
if num in [1,2,3,4]
is by doing:
if [1,2,3,4].index(num)
Is it? any other “cooler” way? Thanks.
On Aug 20, 2008, at 5:32 PM, Iñaki Baz C. wrote:
–
Iñaki Baz C.
if [1,2,3,4].include?(num)
if (1…4) === num
Robert might chime in with an inject version, so I won’t do one of
those.
-Rob
On Wednesday 20 August 2008, Iñaki Baz C. wrote:
Hi, AFAIK in Ruby the only (or the “coolest”) way to do something as:
if num in [1,2,3,4]
is by doing:
if [1,2,3,4].index(num)
Is it? any other “cooler” way? Thanks.
if [1,2,3,4].include? num
It’s not that different, but it looks more like English.
Stefano
El Miércoles, 20 de Agosto de 2008, Rob B.
escribió:> if [1,2,3,4].include?(num)
if (1…4) === num
Thanks.
On Wed, Aug 20, 2008 at 2:32 PM, Iñaki Baz C. [email protected] wrote:
Hi, AFAIK in Ruby the only (or the “coolest”) way to do something as:
if num in [1,2,3,4]
is by doing:
if [1,2,3,4].index(num)
Is it? any other “cooler” way? Thanks.
class Object
def in? (other)
other.respond_to?(“include?”) && other.include?(self)
end
end
num.in? [1,2,3,4]
martin
On 20 Aug 2008, at 22:32, Iñaki Baz C. wrote:
Hi, AFAIK in Ruby the only (or the “coolest”) way to do something as:
if num in [1,2,3,4]
is by doing:
if [1,2,3,4].index(num)
Is it? any other “cooler” way? Thanks.
Well not saying this is actually a good idea, but…
class Object
def in(collection)
collection.include? self
end
end
if num.in [1,2,3,4]
…
end
On Wed, Aug 20, 2008 at 7:05 PM, Michael F.
[email protected] wrote:
num = 5
[*1…4].inject(num){|m,v| m == true ? m : (m == v || m) } == true
ary.inject([num, false]) {|(n,a), i| [n, a | (n == i)]}.last
martin
On Thu, Aug 21, 2008 at 6:41 AM, Rob B.
[email protected] wrote:
Is it? any other “cooler” way? Thanks.
РI̱aki Baz C.
if [1,2,3,4].include?(num)
if (1…4) === num
Robert might chime in with an inject version, so I won’t do one of those.
num = 5
[*1…4].inject(num){|m,v| m == true ? m : (m == v || m) } == true
But it’s far from cool, and my name isn’t Robert, but i hope it wakes
up the competitive spirit of all injectionalisits.
^ manveru
On Wed, Aug 20, 2008 at 8:03 PM, Mikael Høilund [email protected] wrote:
num = 5
ary.inject(false) { |m, v| true if m or v == num }
or even
ary.inject(false) { |m, v| m or v == num }
On Aug 21, 2008, at 4:58, Martin DeMello wrote:
On Wed, Aug 20, 2008 at 7:05 PM, Michael F.
[email protected] wrote:num = 5
[*1…4].inject(num){|m,v| m == true ? m : (m == v || m) } == trueary.inject([num, false]) {|(n,a), i| [n, a | (n == i)]}.last
num = 5
ary.inject(false) { |m, v| true if m or v == num }
2008/8/21 Martin DeMello [email protected]:
On Wed, Aug 20, 2008 at 8:03 PM, Mikael Høilund [email protected] wrote:
num = 5
ary.inject(false) { |m, v| true if m or v == num }or even
ary.inject(false) { |m, v| m or v == num }
Disclaimer: I would not use #inject in this case.
But since you asked…
We can even short circuit with #inject:
irb(main):003:0> i = 3
=> 3
irb(main):004:0> (1…4).inject(nil) {|b,a| break a if i == a}
=> 3
irb(main):005:0> i = 10
=> 10
irb(main):006:0> (1…4).inject(nil) {|b,a| break a if i == a}
=> nil
But then again, you could directly use #find or even better #include?.
If it’s for integers I’d probably just do
irb(main):010:0> i = 3
=> 3
irb(main):011:0> i >= 1 && i <= 4
=> true
irb(main):012:0> (1…4) === i
=> true
Here’s a different approach:
irb(main):013:0> pat = (1…4).inject(0) {|x,a| x | 1 << a}
=> 30
irb(main):014:0> pat.to_s 2
=> “11110”
irb(main):015:0> pat[i] == 1
=> true
Kind regards
robert
On Wed, Aug 20, 2008 at 5:32 PM, Iñaki Baz C. [email protected] wrote:
Hi, AFAIK in Ruby the only (or the “coolest”) way to do something as:
if num in [1,2,3,4]
is by doing:
if [1,2,3,4].index(num)
Is it? any other “cooler” way? Thanks.
num.between?(1,4)
El Jueves, 21 de Agosto de 2008, Gregory B.
escribió:>
num.between?(1,4)
Great, I didn’t know! 
But in my case I’m using custom objects instead of Fixnum or String.
Thanks a lot.
On Thu, Aug 21, 2008 at 1:51 PM, Iñaki Baz C. [email protected] wrote:
num.between?(1,4)
Great, I didn’t know!
But in my case I’m using custom objects instead of Fixnum or String.
Sure, this is implemented by Comparable, so you just need to include
that module and implement <=>
On Thu, Aug 21, 2008 at 1:38 PM, Robert K.
[email protected] wrote:
But then again, you could directly use #find or even better #include?.
In case of ranges #include? really should be your choice, it seems to
be optimized as one
might have expected…
511/12 > cat find-include.rb && ruby find-include.rb
vim: sw=2 ts=2 ft=ruby expandtab tw=0 nu:
require ‘benchmark’
R = (1…1_000)
N = 5
Benchmark::bmbm do |bm|
bm.report(“find”) do
N.times do
R.each do |ele|
R.find{ |e| e == ele}
R.find{ |e| e.zero? }
end
end
end
bm.report(“include?”) do
N.times do
R.each do |ele|
R.include? ele
R.include? 0
end
end
end
end
Rehearsal --------------------------------------------
find 6.797000 0.000000 6.797000 ( 6.828000)
include? 0.000000 0.000000 0.000000 ( 0.016000)
----------------------------------- total: 6.797000sec
user system total real
find 6.438000 0.000000 6.438000 ( 6.828000)
include? 0.016000 0.000000 0.016000 ( 0.016000)
HTH
Robert
–
use.inject do |as, often| as.you_can - without end
endless loops might run for quite some time though
–
http://ruby-smalltalk.blogspot.com/
There’s no one thing that’s true. It’s all true.
From: Iñaki Baz C. [mailto:[email protected]]
> num.between?(1,4)
Great, I didn’t know!
careful. it just test the boundaries.
2.5.between?(1,4)
=> true
[1,2,3,4].include? 2.5
=> false
system “qri between”
---------------------------------------------------- Comparable#between?
obj.between?(min, max) => true or false
Returns false if obj <=> min is less than zero or if anObject <=>
max is greater than zero, true otherwise.
3.between?(1, 5) #=> true
6.between?(1, 5) #=> false
'cat'.between?('ant', 'dog') #=> true
'gnu'.between?('ant', 'dog') #=> false
kind regards -botp
On Fri, Aug 22, 2008 at 4:51 AM, Peña, Botp [email protected] wrote:
From: Iñaki Baz C. [mailto:[email protected]]
> num.between?(1,4)
Great, I didn’t know!
careful. it just test the boundaries.
The same warning holds for Range#include? of course2.5.between?(1,4)
=> true
(1…4).include? 2.5
=> true
R.
http://ruby-smalltalk.blogspot.com/
There’s no one thing that’s true. It’s all true.