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as.factor and floating point numbers

5 messages · Tobias Fellinger, Andrew Simmons, Ebert,Timothy Aaron +1 more

Tobias Fellinger
# 
Hello,

I'm encountering the following error: 

In a package for survival analysis I use a data.frame is created, one column is created by applying unique on the event times while others are created by running table on the event times and the treatment arm.

When there are event times very close together they are put in the same factor level when coerced to factor while unique outputs both values, leading to different lengths of the columns.

Try this to reproduce: 
x <- c(1, 1+.Machine$double.eps)
unique(x)
table(x)

Is there a general best practice to deal with such issues?

Should calling table on floats be avoided in general?

What can one use instead? 

One could easily iterate over the unique values and compare all values with the whole vector but this are N*N comparisons, compared to N*log(N) when sorting first and taking into account that the vector is sorted.

I think for my purposes I'll round to a hundredth of a day before calling the function, but any advice on avoiding this issue an writing more fault tolerant code is greatly appreciated.

all the best, Tobias
Andrew Simmons
# 
R converts floats to strings with ~15 digits of accuracy, specifically
to avoid differentiating between 1 and 1 + .Machine$double.eps, it is
assumed that small differences such as this are due to rounding errors
and are unimportant.

So, if when making your factor, you want all digits, you could write
this: `as.factor(format(x, digits = 17L))`
On Wed, Jan 25, 2023 at 4:03 AM Tobias Fellinger <tobby at htu.at> wrote:
Ebert,Timothy Aaron
# 
Another option is to convert all times to base units or the sample rate from the analog-to-digital converter. If this is 100 milliseconds then use milliseconds rather than fractions of an hour or day. This approach might not help if the range in values spans more than 16 digits: slightly finer resolution than one year sampled in microseconds.
Tim

-----Original Message-----
From: R-help <r-help-bounces at r-project.org> On Behalf Of Andrew Simmons
Sent: Wednesday, January 25, 2023 4:13 AM
To: Tobias Fellinger <tobby at htu.at>
Cc: r-help at r-project.org
Subject: Re: [R] as.factor and floating point numbers

[External Email]

R converts floats to strings with ~15 digits of accuracy, specifically to avoid differentiating between 1 and 1 + .Machine$double.eps, it is assumed that small differences such as this are due to rounding errors and are unimportant.

So, if when making your factor, you want all digits, you could write
this: `as.factor(format(x, digits = 17L))`
On Wed, Jan 25, 2023 at 4:03 AM Tobias Fellinger <tobby at htu.at> wrote:
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Valentin Petzel
# 
Hello Tobias,

A factor is basically a way to get a character to behave like an integer. It 
consists of an integer with values from 1 to nlev, and a character vector 
levels, specifying for each value a level name.

But this means that factors only really make sense with characters, and 
anything that is not a character will be forced to be a character. Thus two 
values that are represented by the same value in as.character will be treated 
as the same.

Now this is probably reasonable most of the time, as numeric values will 
usually represent metric data, which tends to make little sense as factor. But 
if we want to do this we can easily build or own factors from floats, and even 
write some convenience wrapper around table, as shown in the appended file.

Best regards,
Valentin


Am Mittwoch, 25. J?nner 2023, 10:03:01 CET schrieb Tobias Fellinger:
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Tobias Fellinger
# 
Hello, 

I'll reply in one mail to all. 

Thank you for your suggestions. I already tried Andrews solution with 
increasing the digits. In the most extreme case I encountered I had to take 
the maximum possible digits in format but it worked. 

Tims solution is also a good workaround but in this case I would have to know 
much about the user input.

Valentins solution works and is surely the safest of the options but somehow 
more than I need. The case I encountered does not really need to deal with the 
levels, but just with the counts of every unique value across another 
variable.

After thinking about it a little bit longer I came up with another solution 
that works alright for my purposes: I use table on the ranks. Since in the 
case I encountered the vector does not have duplicates and is already sorted, 
I can use table on the ranks of the vector and get the counts in the right 
order.

Thanks Everyone, Tobias
On Mittwoch, 25. J?nner 2023 20:59:16 CET Valentin Petzel wrote: