[Rd] Inaccuracy in seq() function (PR#7503)
(Ted Harding)
Ted.Harding at nessie.mcc.ac.uk
Wed Jan 12 21:52:26 CET 2005
On 12-Jan-05 vstolin at lordabbett.com wrote:
> [...]
> When generating the sequence using seq() function with
> non-integer numbers result is somewhat unpredictable.
> Example:
>> v1<-seq(1.60,1.90,.05)
>> v2<-c(1.60,1.65,1.70,1.75,1.80,1.85,1.90)
>> v1-v2
> [1] 0.000000e+00 2.220446e-16 2.220446e-16 0.000000e+00 0.000000e+00
> 0.000000e+00 2.220446e-16
>> v1==v2
> [1] TRUE FALSE FALSE TRUE TRUE TRUE FALSE
>> v1>1.70 & v2<1.80
> [1] FALSE FALSE TRUE TRUE FALSE FALSE FALSE
This sort of thing is inevitable as a consequence of the
fact that most decimal fractions are not stored accurately
in a finite binary representation. It's not a bug in seq().
Example:
1.65-(1.60+0.05)
[1] -2.220446e-16
(compare your second term in v1-v2).
Personally I always prefer to use scaled integer sequences
when creating fractional sequences:
v3 <- seq(160,190,5)/100
v1-v3
[1] 0.000000e+00 2.220446e-16 2.220446e-16 0.000000e+00
[5] 0.000000e+00 0.000000e+00 2.220446e-16
v3-v2
[1] 0 0 0 0 0 0 0
so my method agrees with your hand-forced v2.
But be watchful:
v4 <- seq(160,190,5)*0.01
v3-v4
[1] 0.000000e+00 -2.220446e-16 0.000000e+00 0.000000e+00
[5] 0.000000e+00 0.000000e+00 -2.220446e-16
so "/100" is not the same as "*0.01"! And you can similarly
check that not only is v4 != v3, but neither is v4 != v1 nor
is v4 != v2. So now we have three different results (v1,
v2 == v3, v4).
Since (1.60+0.05) corresponds to the method used by seq(),
and is not equal to 1.65, you might argue that seq() is wrong,
but since the method used by seq() consists of adding the "by"
the result of seq() is correct -- in its own terms!
If you know that your intended sequence should only have a
fixed number of decimal places (in your case 2), you can force
all methods to give the same result by using round():
v1a <- round(seq(1.60,1.90,0.05),2)
v2a <- round(c(1.60,1.65,1.70,1.75,1.80,1.85,1.90),2)
v3a <- round(seq(160,190,5)/100,2)
v4a <- round(seq(160,190,5)*0.01,2)
v1a-v2a
[1] 0 0 0 0 0 0 0
v1a-v3a
[1] 0 0 0 0 0 0 0
v1a-v4a
[1] 0 0 0 0 0 0 0
so now they are indeed all equal. Also note that
v2-v2a
[1] 0 0 0 0 0 0 0
v3-v3a
[1] 0 0 0 0 0 0 0
so my "seq(160.190.5)/100" gives the same result as the rounded
result, and as your hand-forced version. This is why I prefer it!
Best wishes,
Ted.
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Date: 12-Jan-05 Time: 20:52:26
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