[R] rounding down with as.integer

[Richard M. Heiberger]

Interesting.  Following someone on this list today the goal is input
the data correctly.
My inclination would be to read the file as text, pad each number to
the right, drop the decimal point,
and then read it as an integer.
0 1 2 0.325 1.12 1.9
0.000 1.000 2.000 0.325 1.120 1.900
0000 1000 2000 0325 1120 1900

The pad step is the interesting step.

## 0 1 2 0.325 1.12 1.9
## 0.000 1.000 2.000 0.325 1.120 1.900
## 0000 1000 2000 0325 1120 1900

x.in <- scan(text="
0 1 2 0.325 1.12 1.9 1.
", what="")

padding <- c(".000", "000", "00", "0", "")

x.pad <- paste(x.in, padding[nchar(x.in)], sep="")

x.nodot <- sub(".", "", x.pad, fixed=TRUE)

x <- as.integer(x.nodot)


Rich


On Thu, Jan 1, 2015 at 1:21 PM, Mike Miller <mbmiller+l at gmail.com> wrote:
> On Thu, 1 Jan 2015, Duncan Murdoch wrote:
>
>> On 31/12/2014 8:44 PM, David Winsemius wrote:
>>>
>>>
>>> On Dec 31, 2014, at 3:24 PM, Mike Miller wrote:
>>>
>>>> This is probably a FAQ, and I don't really have a question about it, but
>>>> I just ran across this in something I was working on:
>>>>
>>>>> as.integer(1000*1.003)
>>>>
>>>> [1] 1002
>>>>
>>>> I didn't expect it, but maybe I should have.  I guess it's about the
>>>> machine precision added to the fact that as.integer always rounds down:
>>>>
>>>>
>>>>> as.integer(1000*1.003 + 255 * .Machine$double.eps)
>>>>
>>>> [1] 1002
>>>>
>>>>> as.integer(1000*1.003 + 256 * .Machine$double.eps)
>>>>
>>>> [1] 1003
>>>>
>>>>
>>>> This does it right...
>>>>
>>>>> as.integer( round( 1000*1.003 ) )
>>>>
>>>> [1] 1003
>>>>
>>>> ...but this seems to always give the same answer and it is a little
>>>> faster in my application:
>>>>
>>>>> as.integer( 1000*1.003 + .1 )
>>>>
>>>> [1] 1003
>>>>
>>>>
>>>> FYI - I'm reading in a long vector of numbers from a text file with no
>>>> more than three digits to the right of the decimal.  I'm converting them to
>>>> integers and saving them in binary format.
>>>>
>>>
>>> So just add 0.0001 or even .0000001 to all of them and coerce to integer.
>>
>>
>> I don't think the original problem was stated clearly, so I'm not sure
>> whether this is a solution, but it looks wrong to me.  If you want to round
>> to the nearest integer, why not use round() (without the as.integer
>> afterwards)?  Or if you really do want an integer, why add 0.1 or 0.0001,
>> why not add 0.5 before calling as.integer()?  This is the classical way to
>> implement round().
>>
>> To state the problem clearly, I'd like to know what result is expected for
>> any real number x.  Since R's numeric type only approximates the real
>> numbers we might not be able to get a perfect match, but at least we could
>> quantify how close we get.  Or is the input really character data?  The
>> original post mentioned reading numbers from a text file.
>
>
>
> Maybe you'd like to know what I'm really doing.  I have 1600 text files each
> with up to 16,000 lines with 3100 numbers per line, delimited by a single
> space.  The numbers are between 0 and 2, inclusive, and they have up to
> three digits to the right of the decimal.  Every possible value in that
> range will occur in the data.  Some examples numbers: 0 1 2 0.325 1.12 1.9.
> I want to multiply by 1000 and store them as 16-bit integers (uint16).
>
> I've been reading in the data like so:
>
>> data <- scan( file=FILE, what=double(), nmax=3100*16000)
>
>
> At first I tried making the integers like so:
>
>> ptm <- proc.time() ; ints <- as.integer( 1000 * data ) ; proc.time()-ptm
>
>    user  system elapsed
>   0.187   0.387   0.574
>
> I decided I should compare with the result I got using round():
>
>> ptm <- proc.time() ; ints2 <- as.integer( round( 1000 * data ) ) ;
>> proc.time()-ptm
>
>    user  system elapsed
>   1.595   0.757   2.352
>
> It is a curious fact that only a few of the values from 0 to 2000 disagree
> between the two methods:
>
>> table( ints2[ ints2 != ints ] )
>
>
>  1001  1003  1005  1007  1009  1011  1013  1015  1017  1019  1021  1023
> 35651 27020 15993 11505  8967  7549  6885  6064  5512  4828  4533  4112
>
> I understand that it's all about the problem of representing digital numbers
> in binary, but I still find some of the results a little surprising, like
> that list of numbers from the table() output.  For another example:
>
>> 1000+3 - 1000*(1+3/1000)
>
> [1] 1.136868e-13
>
>> 3 - 1000*(0+3/1000)
>
> [1] 0
>
>> 2000+3 - 1000*(2+3/1000)
>
> [1] 0
>
> See what I mean?  So there is something special about the numbers around
> 1000.
>
> Back to the quesion at hand:  I can avoid use of round() and speed things up
> a little bit by just adding a small number after multiplying by 1000:
>
>> ptm <- proc.time() ; R3 <- as.integer( 1000 * data + .1 ) ;
>> proc.time()-ptm
>
>    user  system elapsed
>   0.224   0.594   0.818
>
> You point out that adding .5 makes sense.  That is probably a better idea
> and I should take that approach under most conditions, but in this case we
> can add anything between 2e-13 and about 0.99999999999 and always get the
> same answer.  We also have to remember that if a number might be negative
> (not a problem for me in this application), we need to subtract 0.5 instead
> of adding it.
>
> Anyway, right now this is what I'm actually doing:
>
>> con <- file( paste0(FILE, ".uint16"), "wb" )
>> ptm <- proc.time() ; writeBin( as.integer( 1000 * scan( file=FILE,
>> what=double(), nmax=3100*16000 ) + .1 ), con, size=2 ) ; proc.time()-ptm
>
> Read 48013406 items
>    user  system elapsed
>  10.263   0.733  10.991
>>
>> close(con)
>
>
> By the way, writeBin() is something that I learned about here, from you,
> Duncan.  Thanks for that, too.
>
> Mike
>
> --
> Michael B. Miller, Ph.D.
> University of Minnesota
> http://scholar.google.com/citations?user=EV_phq4AAAAJ
>
>
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