OpenEuphoria

slicing of sequences

• Posted by James Powell <Wizard at DJO.COM> Apr 26, 1997
• 1144 views

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!