Hi,
We've had several solutions, and I was curious about their relative efficiency.
Here's a test with a moderately large data vector:
> library("microbenchmark")
> set.seed(123) # for reproducibility
> x <- sample(xc, 1e4, replace=TRUE) # "data"
> microbenchmark(John = John <- xn[x],
+ Rich = Rich <- xn[match(x, xc)],
+ Jeff = Jeff <- {
+ n <- as.integer( sub( "[a-i]$", "", x ) )
+ d <- match( sub( "^\\d+", "", x ), letters[1:9] )
+ d[ is.na( d ) ] <- 0
+ n + d / 10
+ },
+ David = David <- as.numeric(gsub("a", ".3",
+ gsub("b", ".5",
+ gsub("c", ".7", x)))),
+ times=1000L
+ )
Unit: microseconds
expr min lq mean median uq max neval cld
John 228.816 345.371 513.5614 503.5965 533.0635 10829.08 1000 a
Rich 217.395 343.035 534.2074 489.0075 518.3260 15388.96 1000 a
Jeff 10325.471 13070.737 15387.2545 15397.9790 17204.0115 153486.94 1000 b
David 14256.673 18148.492 20185.7156 20170.3635 22067.6690 34998.95 1000 c
> all.equal(John, Rich)
[1] TRUE
> all.equal(John, David)
[1] "names for target but not for current"
> all.equal(John, Jeff)
[1] "names for target but not for current" "Mean relative difference:
0.1498243"
Of course, efficiency isn't the only consideration, and aesthetically (and no
doubt subjectively) I prefer Rich Heiberger's solution. OTOH, Jeff's solution
is more general in that it generates the correspondence between letters and
numbers. The argument for Jeff's solution would, however, be stronger if it
gave the desired answer.
Best,
John
> On Jul 10, 2020, at 3:28 PM, David Carlson <dcarl...@tamu.edu> wrote:
>
> Here is a different approach:
>
> xc <- c("1", "1a", "1b", "1c", "2", "2a", "2b", "2c")
> xn <- as.numeric(gsub("a", ".3", gsub("b", ".5", gsub("c", ".7", xc))))
> xn
> # [1] 1.0 1.3 1.5 1.7 2.0 2.3 2.5 2.7
>
> David L Carlson
> Professor Emeritus of Anthropology
> Texas A&M University
>
> On Fri, Jul 10, 2020 at 1:10 PM Fox, John <j...@mcmaster.ca> wrote:
> Dear Jean-Louis,
>
> There must be many ways to do this. Here's one simple way (with no claim of
> optimality!):
>
> > xc <- c("1", "1a", "1b", "1c", "2", "2a", "2b", "2c")
> > xn <- c(1, 1.3, 1.5, 1.7, 2, 2.3, 2.5, 2.7)
> >
> > set.seed(123) # for reproducibility
> > x <- sample(xc, 20, replace=TRUE) # "data"
> >
> > names(xn) <- xc
> > z <- xn[x]
> >
> > data.frame(z, x)
> z x
> 1 2.5 2b
> 2 2.5 2b
> 3 1.5 1b
> 4 2.3 2a
> 5 1.5 1b
> 6 1.3 1a
> 7 1.3 1a
> 8 2.3 2a
> 9 1.5 1b
> 10 2.0 2
> 11 1.7 1c
> 12 2.3 2a
> 13 2.3 2a
> 14 1.0 1
> 15 1.3 1a
> 16 1.5 1b
> 17 2.7 2c
> 18 2.0 2
> 19 1.5 1b
> 20 1.5 1b
>
> I hope this helps,
> John
>
> -----------------------------
> John Fox, Professor Emeritus
> McMaster University
> Hamilton, Ontario, Canada
> Web: http::/socserv.mcmaster.ca/jfox
>
> > On Jul 10, 2020, at 1:50 PM, Jean-Louis Abitbol <abit...@sent.com> wrote:
> >
> > Dear All
> >
> > I have a character vector, representing histology stages, such as for
> > example:
> > xc <- c("1", "1a", "1b", "1c", "2", "2a", "2b", "2c")
> >
> > and this goes on to 3, 3a etc in various order for each patient. I do have
> > of course a pre-established classification available which does change
> > according to the histology criteria under assessment.
> >
> > I would want to convert xc, for plotting reasons, to a numeric vector such
> > as
> >
> > xn <- c(1, 1.3, 1.5, 1.7, 2, 2.3, 2.5, 2.7)
> >
> > Unfortunately I have no clue on how to do that.
> >
> > Thanks for any help and apologies if I am missing the obvious way to do it.
> >
> > JL
> > --
> > Verif30042020
> >
> > ______________________________________________
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> >
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> >
> > and provide commented, minimal, self-contained, reproducible code.
>
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>
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