seq?

ThU — Tue, 15 Nov 2011 17:40:58 Z

One way is to use seq instead:
seq(i^2,i=1..10);

Another approach

Markiyan Hirnyk — Tue, 15 Nov 2011 23:03:40 Z

It is useful to have a few solutions.
> L := [];
> for j to 4 do L := [op(L), j^2] end do:
> L;
[1, 4, 9, 16]
S := convert(L, set);
{1, 4, 9, 16}

It is Ok

Fei Wang — Wed, 16 Nov 2011 06:13:19 Z

Thanks for both of you !

timing

pagan — Wed, 09 May 2012 22:59:48 Z

This is one of the archetypal (avoidable) inefficient programming techniques in Maple -- to build a list by concatenation which is a worse complexity class than using `seq`.

The significant consequence is that the timing disadvantage is a function of the number of elements created. The performance of list concatenation gets even worse relative to that of using `seq` as the problem size increases.

restart:
st:=time():
  K:=[seq(i^2,i=1..10^4)]:
time()-st;
                               0.

st :=time():
L := []:
for j to 10^4 do L := [op(L), j^2] end do:
time()-st;
                             0.960

evalb(K=L);
                              true

restart:
st:=time():
  K:=[seq(i^2,i=1..10^5)]:
time()-st;
                             0.020

restart:
st :=time():
L := []:
for j to 10^5 do L := [op(L), j^2] end do:
time()-st;
                             93.780

Above, the timing of list concatenation got worse by a factor of approximately 100 when the problem size was increased by a factor of 10.

But `seq` slows down by not too much more than the factor by which the problem size increases (up until much larger sizes, at least).

restart:
st:=time():
  K:=[seq(i^2,i=1..10^6)]:
time()-st;
                             0.170

restart:
st:=time():
  K:=[seq(i^2,i=1..10^7)]:
time()-st;
                             1.850

The inefficiency is not only for timing. It is also for the total memory allocation.

MaplePrimes - answers and comments on Question, A little problem

seq?

Another approach

It is Ok

timing