[R] which(df$name=="A") takes ~1 second! (df is very large), but can it be speeded up?

[Emmanuel Levy]

Dear Peter and Henrik,

Thanks for your replies - this helps speed up a bit, but I thought
there would be something much faster.

What I mean is that I thought that a particular value of a level
could be accessed instantly, similarly to a "hash" key.

Since I've got about 6000 levels in that data frame, it means that
making a list L of the form
L[[1]] = values of name "1"
L[[2]] = values of name "2"
L[[3]] = values of name "3"
...
would take ~1hour.

Best,

Emmanuel




2008/8/12 Henrik Bengtsson <hb at stat.berkeley.edu>:
> To simplify:
>
> n <- 2.7e6;
> x <- factor(c(rep("A", n/2), rep("B", n/2)));
>
> # Identify 'A':s
> t1 <- system.time(res <- which(x == "A"));
>
> # To compare a factor to a string, the factor is in practice
> # coerced to a character vector.
> t2 <- system.time(res <- which(as.character(x) == "A"));
>
> # Interestingly enough, this seems to be faster (repeated many times)
> # Don't know why.
> print(t2/t1);
>    user   system  elapsed
> 0.632653 1.600000 0.754717
>
> # Avoid coercing the factor, but instead coerce the level compared to
> t3 <- system.time(res <- which(x == match("A", levels(x))));
>
> # ...but gives no speed up
> print(t3/t1);
>    user   system  elapsed
> 1.041667 1.000000 1.018182
>
> # But coercing the factor to integers does
> t4 <- system.time(res <- which(as.integer(x) == match("A", levels(x))))
> print(t4/t1);
>     user    system   elapsed
> 0.4166667 0.0000000 0.3636364
>
> So, the latter seems to be the fastest way to identify those elements.
>
> My $.02
>
> /Henrik
>
>
> On Tue, Aug 12, 2008 at 7:31 PM, Peter Cowan <cowan.pd at gmail.com> wrote:
>> Emmanuel,
>>
>> On Tue, Aug 12, 2008 at 4:35 PM, Emmanuel Levy <emmanuel.levy at gmail.com> wrote:
>>> Dear All,
>>>
>>> I have a large data frame ( 2700000 lines and 14 columns), and I would like to
>>> extract the information in a particular way illustrated below:
>>>
>>>
>>> Given a data frame "df":
>>>
>>>> col1=sample(c(0,1),10, rep=T)
>>>> names = factor(c(rep("A",5),rep("B",5)))
>>>> df = data.frame(names,col1)
>>>> df
>>>   names col1
>>> 1      A    1
>>> 2      A    0
>>> 3      A    1
>>> 4      A    0
>>> 5      A    1
>>> 6      B    0
>>> 7      B    0
>>> 8      B    1
>>> 9      B    0
>>> 10     B    0
>>>
>>> I would like to tranform it in the form:
>>>
>>>> index = c("A","B")
>>>> col1[[1]]=df$col1[which(df$name=="A")]
>>>> col1[[2]]=df$col1[which(df$name=="B")]
>>
>> I'm not sure I fully understand your problem, you example would not run for me.
>>
>> You could get a small speedup by omitting which(), you can subset by a
>> logical vector also which give a small speedup.
>>
>>> n <- 2700000
>>> foo <- data.frame(
>> +       one = sample(c(0,1), n, rep = T),
>> +       two = factor(c(rep("A", n/2 ),rep("B", n/2 )))
>> +       )
>>> system.time(out <- which(foo$two=="A"))
>>   user  system elapsed
>>  0.566   0.146   0.761
>>> system.time(out <- foo$two=="A")
>>   user  system elapsed
>>  0.429   0.075   0.588
>>
>> You might also find use for unstack(), though I didn't see a speedup.
>>> system.time(out <- unstack(foo))
>>   user  system elapsed
>>  1.068   0.697   2.004
>>
>> HTH
>>
>> Peter
>>
>>> My problem is that the command:  *** which(df$name=="A") ***
>>> takes about 1 second because df is so big.
>>>
>>> I was thinking that a "level" could maybe be accessed instantly but I am not
>>> sure about how to do it.
>>>
>>> I would be very grateful for any advice that would allow me to speed this up.
>>>
>>> Best wishes,
>>>
>>> Emmanuel
>>
>> ______________________________________________
>> R-help at r-project.org mailing list
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>>
>