[R] customize the step value

Martin Maechler m@ech|er @end|ng |rom @t@t@m@th@ethz@ch
Fri Oct 29 17:04:20 CEST 2021


>>>>> Duncan Murdoch 
>>>>>     on Fri, 29 Oct 2021 09:07:31 -0400 writes:

    > On 29/10/2021 4:34 a.m., PIKAL Petr wrote:
    >> Hi
    >> 
    >> One has to be careful when using fractions in seq step.
    >> 
    >> Although it works for 0.5
    >>> (seq(0,10, .5) - round(seq(0,10,.5),2))==0
    >> [1] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE
    >> TRUE
    >> [16] TRUE TRUE TRUE TRUE TRUE TRUE
    >> 
    >> in case of 0.3 (or others) it does not always result in expected values (see
    >> FAQ 7.31 for explanation)
    >> 
    >>> (seq(0,10, .3) - round(seq(0,10,.3),2))==0
    >> [1]  TRUE  TRUE  TRUE FALSE  TRUE  TRUE FALSE  TRUE  TRUE FALSE  TRUE  TRUE
    >> [13] FALSE  TRUE  TRUE  TRUE  TRUE  TRUE FALSE  TRUE  TRUE  TRUE  TRUE FALSE
    >> [25] FALSE  TRUE  TRUE  TRUE  TRUE  TRUE  TRUE FALSE  TRUE  TRUE


    > Petr is right, it's unsafe to use fractional values for the step.  0.5 
    > works because it has a power of 2 in the denominator and so does the 
    > start value, but it's easy to make mistakes when you rely on that (e.g. 
    > changing the step size from 0.5 to 0.3 would break things).

    > A better idea is to modify a sequence of integers.  For example, to get 
    > 1.5 to 3.5 by 0.5, you can do (3:7)*0.5, and for 0 to 3 by 0.3, use 
    > (0:10)*0.3.


Well, but you will not get truly equidistant (to the last bit)
sequences also by that and people who are not aware of
FAQ 7.31  and its consequences do wrongly assume that

length(unique(diff(seqVec))) == 1

for any      seqVec <-  k * seq(....)   # k a "scalar" (of length 1)

but the reality of floating point arithmetic can be quite
different than pure math :

In this case and on my platform the two ways to construct the
sequence are even identical:

> identical((0:10)*0.3, seq(0, 3, by=.3))
[1] TRUE
> sv <- (0:10)*0.3
> length(unique(diff(sv))) # you'd like |-->  1 (the number 0.3 !)
[1] 5
>

Martin



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