1. slicing of sequences
Im at a loss as how to proceed here. I know I can build a routine that will
do the following code, but I would rather use a built in Euphoria statement
than build my own, since it will be called often. ( I think that a built in
routine dealing with slicing sequences would be faster than any loop I
create to do the same thing.)
Anyway, here's a piece of sample code demonstrating what I would like to do,
Euphoria's responce, and an explanation of the result I think I should get.
-- Program Test.ex --
sequence a, b
a = {{1, 2, 3},
{4, 5, 6},
{7, 8, 9}}
b = a[1..3][3]
-- End Test.ex --
When run, Euphoria responds with:
b = a[1..3][3]
^Unknown Command
What I am trying to do is get the 3rd element of all the sequences in a. So
theoretically, b should contain {3, 6, 9}, ie. b = {a[1][3], a[2][3], a[3][3]}.
The reason I did this short test program is because I would like to write a
routine that contains the following code:
-- Code --
function Numof3in(sequence a)
atom numfound, done
numfound = 0
done = 0
while not done do
if find(3, a) then
a = a[(find(3, a) + 1)..length(a)]
numfound = numfound + 1
else
done = 1
end if
end while
return numfound
end function
for Da_Loop = 1 to Numof3in(Da_Seq[1..(length(Da_Seq))][4])
end for
-- End Code --
NOTE: No attempt has yet been made to see if the function Numof3in()
actually works. Hopefully, you get the idea of what I am trying to do with
this code.
Any ideas as how to procede using the built in slicing functions?
Could this be added to a future version of Euphoria, or am I asking to much?
Just a little question/idea I had.
James Powell
PS. Here's a little tip for other programmers. ; )
There are 2 ways to write bug free code, and only the 3rd way works!
2. slicing of sequences
- Posted by Mike Fowler <stoner at NELSUN.GEN.NZ>
Apr 26, 1997
-
Last edited Apr 27, 1997
<SNIP>
-> Anyway, here's a piece of sample code demonstrating what I would like
-> to do, Euphoria's responce, and an explanation of the result I think
-> I should get.
-> -- Program Test.ex --
-> sequence a, b
-> a = {{1, 2, 3},
-> {4, 5, 6},
-> {7, 8, 9}}
-> b = a[1..3][3]
-> -- End Test.ex --
well heres the fix :)
-- Start Test.ex --
sequence a, b
b = {} -- added
a = {{1, 2, 3},
{4, 5, 6},
{7, 8, 9}}
--b = a[1..3][3] -- removed
for c = 1 to length(a) do -- added
b = append(b, (a[c][3])) -- added
end for -- added
print(1, b) -- added
-- End Test.ex --
-> When run, Euphoria responds with:
-> b = a[1..3][3]
-> ^Unknown Command
i dont know why that is, but someone else will :)
Hope thats a help
Mike Fowler - mike.fowler at nelsun.gen.nz
o__ ---
_,>/'_ ---
(_) \(_) ---
Tips for the month: Bill your dentist for your time in surgery.
3. slicing of sequences
- Posted by Robert Craig <robert_craig at COMPUSERVE.COM>
Apr 26, 1997
-
Last edited Apr 27, 1997
James Powell writes:
> Could this be added to a future version of Euphoria, or am I asking to much?
I would recommend that you write your own Euphoria function
to do this "column" slice operation.
I do not expect that this feature will be added to Euphoria
for the following reasons:
1. I can't do the operation *that* much faster than you can.
2. I don't think the operation would be used in that many
programs, and when it was, speed would not necessarily be
critical.
Regards,
Rob Craig
Rapid Deployment Software
4. Re: slicing of sequences
James Powell wrote:
>
> -- Program Test.ex --
>
> sequence a, b
>
> a = {{1, 2, 3},
> {4, 5, 6},
> {7, 8, 9}}
> b = a[1..3][3]
>
> -- End Test.ex --
>
> When run, Euphoria responds with:
>
> b = a[1..3][3]
> ^Unknown Command
>
b = a[1..3] is the same as a, a = {{1,2,3},{4,5,6},{7,8,9}} and
the last index [3] doesn't exist so Euphoria complains.
So a[1] = {{1,2,3}}, a[2] = {{4,5,6}} etc.
You need to write b[1] = a[1][3]. Put it in a loop.
for i = 1 to length(a) do
b[i] = a[i][3]
end for
b = {3,6,9}
Marcel Kollenaar
