1. Euphorize THIS!
-- Here's code that creates a list of coordinates,
-- and then deletes each n-th element of that list
-- I would like some feedback on a more Euphorish
-- way to accomplish the task. Any takers?
-- Thanks, Irv
include get.e
constant n = 4, -- each n-th element
scrn = 1
sequence dyn_list, new_list
dyn_list = {} -- create an empty dynamic list
new_list = {}
function GetCoords()
integer x,y
x = prompt_number("Coord X:",{1,640})
y = prompt_number("Coord Y:",{1,480})
return({x,y})
end function
function delete(sequence list, integer item)
return list[1..item-1] & list[item+1..length(list)]
end function
dyn_list = {GetCoords()} -- get the first set of coords
while 1 do
dyn_list = append(dyn_list,GetCoords()) -- get the next coords
? dyn_list
if match(dyn_list[1], dyn_list[length(dyn_list)]) then -- compare first and
last list items
exit -- quit when equal
end if
end while
printf(scrn,"You have entered %d items\n",length(dyn_list))
printf(scrn,"Now removing each %dth item\n",n)
puts(scrn,"Before: ") ? dyn_list
-----------------------------------
-- Here's where I need some help.--
-- Isn't there a more Euphorish --
-- way to do this? Maybe without --
-- duplicating sequences and so --
-- tying up more memory? Tnx --
-----------------------------------
for i = 1 to length(dyn_list) do
if remainder(i,n) then
new_list = append(new_list,dyn_list[i])
end if
end for
puts(scrn,"After: ") ? new_list
2. Re: Euphorize THIS!
Erm, will this suffice?
-- Code starts here --
integer i, max
i = 0 max = length(dyn_list)
for count = n to max by n do
dyn_list = dyn_list[1..(count - i) - 1] & dyn_list[(count - i) +
1..length(dyn_list)]
i += 1
end for
-- Code ends here --
I tested it, and it works.. I hope this is what you want.
The AfterBeat
Irv Mullins wrote:
> -- Here's code that creates a list of coordinates,
> -- and then deletes each n-th element of that list
> -- I would like some feedback on a more Euphorish
> -- way to accomplish the task. Any takers?
> -- Thanks, Irv
>
> include get.e
> constant n = 4, -- each n-th element
> scrn = 1
>
> sequence dyn_list, new_list
> dyn_list = {} -- create an empty dynamic list
> new_list = {}
>
> function GetCoords()
> integer x,y
> x = prompt_number("Coord X:",{1,640})
> y = prompt_number("Coord Y:",{1,480})
> return({x,y})
> end function
>
> function delete(sequence list, integer item)
> return list[1..item-1] & list[item+1..length(list)]
> end function
>
> dyn_list = {GetCoords()} -- get the first set of coords
> while 1 do
> dyn_list = append(dyn_list,GetCoords()) -- get the next coords
> ? dyn_list
> if match(dyn_list[1], dyn_list[length(dyn_list)]) then -- compare first
> and last list items
> exit -- quit when equal
> end if
> end while
>
> printf(scrn,"You have entered %d items\n",length(dyn_list))
> printf(scrn,"Now removing each %dth item\n",n)
> puts(scrn,"Before: ") ? dyn_list
> -----------------------------------
> -- Here's where I need some help.--
> -- Isn't there a more Euphorish --
> -- way to do this? Maybe without --
> -- duplicating sequences and so --
> -- tying up more memory? Tnx --
> -----------------------------------
> for i = 1 to length(dyn_list) do
> if remainder(i,n) then
> new_list = append(new_list,dyn_list[i])
> end if
> end for
> puts(scrn,"After: ") ? new_list
3. Re: Euphorize THIS!
On Sat, 23 Jan 1999 11:40:20 -0800, The AfterBeat <afterbeat at GEOCITIES.COM>
wrote:
>Erm, will this suffice?
>
>-- Code starts here --
>integer i, max
>
>i = 0 max = length(dyn_list)
>for count = n to max by n do
> dyn_list = dyn_list[1..(count - i) - 1] & dyn_list[(count - i) +
> 1..length(dyn_list)]
> i += 1
>end for
>-- Code ends here --
>
>I tested it, and it works.. I hope this is what you want.
>
> The AfterBeat
>
Yup. That frees up memory that the second sequence was taking up.
I wonder if there is a way to do it with Euphoria's sequence
functions - so as to avoid the for/loop?
Irv
4. Re: Euphorize THIS!
- Posted by Mike <michael at IGRIN.CO.NZ>
Jan 25, 1999
-
Last edited Jan 26, 1999
>>Erm, will this suffice?
>>
>>-- Code starts here --
>>integer i, max
>>
>>i = 0 max = length(dyn_list)
>>for count = n to max by n do
>> dyn_list = dyn_list[1..(count - i) - 1] & dyn_list[(count - i) +
1..length(dyn_list)]
>> i += 1
>>end for
>>-- Code ends here --
>>
>>I tested it, and it works.. I hope this is what you want.
Excuse me, but isn't this code somewhat inefficient? The line within the
for-loop causes the *entire* list to contract at each iteration.Try the
following code. I tested it against the above and i figured that its quicker
(well at least for a 1-d sequence of integers 4000 elements long) by a
factor of about 100....
(I hope it gives the right answer!)
function nth(sequence s, integer n) -- remove each nth item from the list
integer rem,len,c,d,p
len=length(s)
rem=remainder(len,n)
c=n-1
d=n-2
p=n
for i=n+1 to len - rem by n do
s[p..p + d]=s[i..i + d]
p+=c
end for
if rem then s[p..p + rem - 1] = s[len - rem + 1..len] end if
s=s[1..p + rem - 1]
return s
end function
