Hi Sarah,
On 05/29/2015 12:04 PM, Sarah Goslee wrote:
On Fri, May 29, 2015 at 2:16 PM, Hervé Pagès <hpa...@fredhutch.org> wrote:
Hi Kate,
I found that matching the character vector to itself is a very
effective way to do this:
x <- c("a", "bunch", "of", "strings", "whose", "exact", "content",
"is", "of", "little", "interest")
ids <- match(x, x)
ids
# [1] 1 2 3 4 5 6 7 8 3 10 11
By using this trick, many manipulations on character vectors can
be replaced by manipulations on integer vectors, which are sometimes
way more efficient.
Hm. I hadn't thought of that approach - I use the
as.numeric(factor(...)) approach.
So I was curious, and compared the two:
set.seed(43)
x <- sample(letters, 10000, replace=TRUE)
system.time({
for(i in seq_len(20000)) {
ids1 <- match(x, x)
}})
# user system elapsed
# 9.657 0.000 9.657
system.time({
for(i in seq_len(20000)) {
ids2 <- as.numeric(factor(x, levels=letters))
}})
# user system elapsed
# 6.16 0.00 6.16
Using factor() is faster.
That's an unfair comparison, because you already know what the levels
are so you can supply them to your call to factor(). Most of the time
you don't know what the levels are so either you just do factor(x) and
let the factor() constructor compute the levels for you, or you compute
them yourself upfront with something like factor(x, levels=unique(x)).
library(microbenchmark)
microbenchmark(
{ids1 <- match(x, x)},
{ids2 <- as.integer(factor(x, levels=letters))},
{ids3 <- as.integer(factor(x))},
{ids4 <- as.integer(factor(x, levels=unique(x)))}
)
Unit: microseconds
expr min lq
{ ids1 <- match(x, x) } 245.979 262.2390
{ ids2 <- as.integer(factor(x, levels = letters)) } 214.115 219.2320
{ ids3 <- as.integer(factor(x)) } 380.782 388.7295
{ ids4 <- as.integer(factor(x, levels = unique(x))) } 332.250 342.6630
mean median uq max neval
267.3210 264.4845 268.348 293.894 100
226.9913 220.9870 226.147 314.875 100
402.2242 394.7165 412.075 481.410 100
349.7405 345.3090 353.162 383.002 100
More importantly, using factor() lets you
set the order of the indices in an expected fashion, where match()
assigns them in the order of occurrence.
head(data.frame(x, ids1, ids2))
x ids1 ids2
1 m 1 13
2 x 2 24
3 b 3 2
4 s 4 19
5 i 5 9
6 o 6 15
In a problem like Kate's where there are several columns for which the
same ordering of indices is desired, that becomes really important.
I'm not sure why which particular ID gets assigned to each string would
matter but maybe I'm missing something. What really matters is that each
string receives a unique ID. match(x, x) does that.
In Kate's problem, where the strings are in more than one column,
and you want the ID to be unique across the columns, you need to do
match(x, x) where 'x' contains the strings from all the columns
that you want to replace:
m <- matrix(c(
"X0001", "BYX859", 0, 0, 2, 1, "BYX859",
"X0001", "BYX894", 0, 0, 1, 1, "BYX894",
"X0001", "BYX862", "BYX894", "BYX859", 2, 2, "BYX862",
"X0001", "BYX863", "BYX894", "BYX859", 2, 2, "BYX863",
"X0001", "BYX864", "BYX894", "BYX859", 2, 2, "BYX864",
"X0001", "BYX865", "BYX894", "BYX859", 2, 2, "BYX865"
), ncol=7, byrow=TRUE)
x <- m[ , 2:4]
id <- match(x, x, nomatch=0, incomparables="0")
m[ , 2:4] <- id
No factor needed. No loop needed. ;-)
Cheers,
H.
If you take Bill Dunlap's modification of the match() approach, it
resolves both problems: matching against the pooled unique values is
both faster than the factor() version and gives the same result:
On Fri, May 29, 2015 at 1:31 PM, William Dunlap <wdun...@tibco.com> wrote:
match() will do what you want. E.g., run your data through
the following function.
f <- function (data)
{
uniqStrings <- unique(c(data[, 2], data[, 3], data[, 4]))
uniqStrings <- setdiff(uniqStrings, "0")
for (j in 2:4) {
data[[j]] <- match(data[[j]], uniqStrings, nomatch = 0L)
}
data
}
##
y <- data.frame(id = 1:5000, v1 = sample(letters, 5000, replace=TRUE),
v2 = sample(letters, 5000, replace=TRUE), v3 = sample(letters, 5000,
replace=TRUE), stringsAsFactors=FALSE)
system.time({
for(i in seq_len(20000)) {
ids3 <- f(data.frame(y))
}})
# user system elapsed
# 22.515 0.000 22.518
ff <- function(data)
{
uniqStrings <- unique(c(data[, 2], data[, 3], data[, 4]))
uniqStrings <- setdiff(uniqStrings, "0")
for (j in 2:4) {
data[[j]] <- as.numeric(factor(data[[j]], levels=uniqStrings))
}
data
}
system.time({
for(i in seq_len(20000)) {
ids4 <- ff(data.frame(y))
}})
# user system elapsed
# 26.083 0.002 26.090
head(ids3)
id v1 v2 v3
1 1 1 2 8
2 2 2 19 22
3 3 3 21 16
4 4 4 10 17
5 5 1 8 18
6 6 1 12 26
head(ids4)
id v1 v2 v3
1 1 1 2 8
2 2 2 19 22
3 3 3 21 16
4 4 4 10 17
5 5 1 8 18
6 6 1 12 26
Kate, if you're getting all zeros, check str(yourdataframe) - it's
likely that when you imported your data into R the strings were
already converted to factors, which is not what you want (ask me how I
know this!).
Sarah
On 05/29/2015 09:58 AM, Kate Ignatius wrote:
I have a pedigree file as so:
X0001 BYX859 0 0 2 1 BYX859
X0001 BYX894 0 0 1 1 BYX894
X0001 BYX862 BYX894 BYX859 2 2 BYX862
X0001 BYX863 BYX894 BYX859 2 2 BYX863
X0001 BYX864 BYX894 BYX859 2 2 BYX864
X0001 BYX865 BYX894 BYX859 2 2 BYX865
And I was hoping to change all unique string values to numbers.
That is:
BYX859 = 1
BYX894 = 2
BYX862 = 3
BYX863 = 4
BYX864 = 5
BYX865 = 6
But only in columns 2 - 4. Essentially I would like the data to look like
this:
X0001 1 0 0 2 1 BYX859
X0001 2 0 0 1 1 BYX894
X0001 3 2 1 2 2 BYX862
X0001 4 2 1 2 2 BYX863
X0001 5 2 1 2 2 BYX864
X0001 6 2 1 2 2 BYX865
Is this possible with factors?
Thanks!
K.
--
Hervé Pagès
Program in Computational Biology
Division of Public Health Sciences
Fred Hutchinson Cancer Research Center
1100 Fairview Ave. N, M1-B514
P.O. Box 19024
Seattle, WA 98109-1024
E-mail: hpa...@fredhutch.org
Phone: (206) 667-5791
Fax: (206) 667-1319
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