On 15 Mar 2014, at 20:54 , Mike Miller <mbmille...@gmail.com> wrote:
> On Sat, 15 Mar 2014, peter dalgaard wrote:
>
>> I don't think so. I think some of your numbers differ sufficiently from
>> numbers with only a few digits to the right of the decimal that write.table
>> needs to write them with increased precision. You didn't read them like
>> that, didn't you? You did some calculations, and then it _looked like_ the
>> results have <= 6 digits after the decimal point?
>
> These have different representations:
>
> 1-0.995
> 2-1.995
>
> write(c(1-0.995, 2-1.995), file="data1.txt", sep="\n")
> write.table(c(1-0.995, 2-1.995), file="data2.txt", row.names=F, col.names=F)
>
> $ head -2 data[12].txt
> ==> data1.txt <==
> 0.005
> 0.005
>
> ==> data2.txt <==
> 0.005
> 0.00499999999999989
>
>
>> The digits= setting has nothing to do with this, write.table alway does its
>> damndest to avoid loss of precision. This _is_ in help(write.table):
>>
>> In almost all cases the conversion of numeric quantities is
>> governed by the option ‘"scipen"’ (see ‘options’), but with the
>> internal equivalent of ‘digits = 15’. For finer control, use
>> ‘format’ to make a character matrix/data frame, and call
>> ‘write.table’ on that.
>
> Yes! This was my mistake: write() not write.table() is controlled by
> options(digits=7). Repeating the write() command using a different number of
> digits:
>
>> options(digits=15)
>> getOption("digits")
> [1] 15
>> write(c(1-0.995, 2-1.995), file="data1.txt", sep="\n")
>
> $ cat data1.txt
> 0.005
> 0.00499999999999989
>
> I don't know why it shows 17 digits and doesn't round to 15, but it is
> showing that the numbers are different, for some reason.
>
Aiding my weakening eyesight a little:
0.004 999 999 999 999 89
Notice that that makes 15 _significant_ digits.
> Do you understand why there is a difference between 1-0.995 and 2-1.995 in
> their internal representations?
Let's see, that'll be like
1 - 2/3 vs. 10 - 29/3
on a decimal computer if someone is perverse enough to give input in base 3
(i.e., 1.0 - 0.2 ternary vs. 101.0 - 100.2 ternary). Assume that the computer
is floating point with 3 significant digits (and possibly taking some liberties
compared to what real computers really do), we have
1 = 1.000 * 10^0
10 = 1.000 * 10^1
2/3 = 0.667 * 10^0
29/3 = 0.967 * 10^1
1 - 2/3 = 0.333 * 10^0
10 - 29/3 = 0.033 * 10^1 = 0.330 * 10^0
So, yes, I think I do understand how these things can happen.
-pd
>
> Mike
--
Peter Dalgaard, Professor,
Center for Statistics, Copenhagen Business School
Solbjerg Plads 3, 2000 Frederiksberg, Denmark
Phone: (+45)38153501
Email: pd....@cbs.dk Priv: pda...@gmail.com
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