find the first value greater than d, in a list (special one : each value is the double of the previous...)
then give me the index of this value (with special case ...)
i = ([0.5,1.0,2.0,4.0,8.0,16.0,32.0,64.0,128.0,256.0,512.0,1024.0,2048.0].select {|v| v if v >= d }).first
zl = d <0.5 ? 0 : [0.5,1.0,2.0,4.0,8.0,16.0,32.0,64.0,128.0,256.0,512.0,1024.0,2048.0].index(i)
find the first value greater than d, in a list (special one : each
value is the double of the previous...)
then give me the index of this value (with special case ...)
i =
([0.5,1.0,2.0,4.0,8.0,16.0,32.0,64.0,128.0,256.0,512.0,1024.0,2048.0].selec
t {|v| v if v >= d }).first
zl = d <0.5 ? 0 :
[0.5,1.0,2.0,4.0,8.0,16.0,32.0,64.0,128.0,256.0,512.0,1024.0,2048.0].index(
i)
joss
What about this?
a.each_index{|i| break i if a[i] > d}
The only problem is that it returns a if no element greater than d is found.
You could turn it into a method, though:
def find_greater array, value
array.each_index{|i| return i if array[i] > value}
nil
end
find the first value greater than d, in a list (special one : each
value is the double of the previous...)
then give me the index of this value (with special case ...)
i =
([0.5,1.0,2.0,4.0,8.0,16.0,32.0,64.0,128.0,256.0,512.0,1024.0,2048.0].select
{|v| v if v >= d }).first
zl = d <0.5 ? 0 :
[0.5,1.0,2.0,4.0,8.0,16.0,32.0,64.0,128.0,256.0,512.0,1024.0,2048.0].index(i)
Hi Joss,
In this specific case, you could simply do:
(Math.log(d)/Math.log(2) + 1).ceil
Which is a different kind of solution, and doesn't need the array.
a = [0.5,1.0,2.0,4.0,8.0,16.0,32.0,64.0,128.0,256.0,512.0,1024.0,2048.0]
d = 7.0
# version 1: matches your verbal specification
zl = a.each_with_index { |v,i| break i if v > d }
# version 2: matches your sample code's behaviour
zl = a.each_with_index { |v,i| break i if v >= d }
However this gives different behaviour to your code in the event that d is
greater than the last element in the array (yours returns nil, mine returns
the whole array). Maybe this is OK; you can always stick a 1.0/0.0
(infinity) at the end of the array. Or:
zl = nil; a.each_with_index { |v,i| break zl = i if v >= d }
But in this particular case you also could use your power-of-two property
and not construct or search an array at all:
zl = d < 0.5 ? 0 : (d*2).to_i.to_s(2).length
B.
···
On Fri, Mar 09, 2007 at 04:50:08PM +0900, Josselin wrote:
Could I write it in just 1 line ?
find the first value greater than d, in a list (special one : each
value is the double of the previous...)
then give me the index of this value (with special case ...)
i =
([0.5,1.0,2.0,4.0,8.0,16.0,32.0,64.0,128.0,256.0,512.0,1024.0,2048.0].select
{|v| v if v >= d }).first
zl = d <0.5 ? 0 :
[0.5,1.0,2.0,4.0,8.0,16.0,32.0,64.0,128.0,256.0,512.0,1024.0,2048.0].index(i)
irb(main):015:0>
arr=[0.5,1.0,2.0,4.0,8.0,16.0,32.0,64.0,128.0,256.0,512.0,1024.0,2048.0]
=> [0.5, 1.0, 2.0, 4.0, 8.0, 16.0, 32.0, 64.0, 128.0, 256.0, 512.0,
1024.0, 2048.0]
irb(main):016:0> idx = arr.to_enum(:each_with_index).inject(nil)
{|x,(e,i)| break i if e > 100}
=> 8
irb(main):017:0> idx = arr.to_enum(:each_with_index).inject(nil)
{|x,(e,i)| break i if e > 1000000}
=> nil
Kind regards
robert
···
2007/3/9, Josselin <josselin@wanadoo.fr>:
Could I write it in just 1 line ?
find the first value greater than d, in a list (special one : each
value is the double of the previous...)
then give me the index of this value (with special case ...)
i =
([0.5,1.0,2.0,4.0,8.0,16.0,32.0,64.0,128.0,256.0,512.0,1024.0,2048.0].select
{|v| v if v >= d }).first
zl = d <0.5 ? 0 :
[0.5,1.0,2.0,4.0,8.0,16.0,32.0,64.0,128.0,256.0,512.0,1024.0,2048.0].index(i)