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

Henrik Bengtsson hb at stat.berkeley.edu
Wed Aug 13 04:56:05 CEST 2008

```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
>
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