Re: [R] passing arguments to simple plotting program.
Seems so simple when you explain it. Thanks very much. Gerard
> On May 9, 2017, at 9:40 AM, Ulrik Stervbo <ulrik.ster...@gmail.com> wrote:
>
> Hi Gerard,
> Quotation marks are used for strings. In you function body you try to use the
> strings "indata" and "fig_descrip" (the latter will work but is not what you
> want).
>
> In your current function call you pass the variable Figure as the value to
> the argument fig_descrip, followed by a lot of other stuff your function
> doesn't know what to do with.
>
> Remove the quotation marks around indata and fig_descrip in the function
> body, call your function with:
>
> plot_f1(indata=v5, n1=114, n2=119, n3=116, fig_descrip="Figure 2a\nChange in
> Composite Score at Visit 5 (Day 31)\nPer Protocol Population")
>
> and you should be fine.
>
> HTH
>
> Ulrik
>
> Gerard Smits <smits.gerar...@gmail.com <mailto:smits.gerar...@gmail.com>>
> schrieb am Di., 9. Mai 2017, 18:27:
> Hi Ulrik,
>
> If I can trouble you with one more question.
>
> Now trying to send a string to the main= . I was able to pass the data name
> in data=in_data, but same logic is not working in passion the main string.
>
>
> plot_f1 <-function(indata,n1,n2,n3,fig_descrip) {
> par(oma=c(2,2,2,2))
> boxplot(formula = d_comp ~ rx_grp,
> data="indata”,# <- worked fine here.
> main="fig_descrip",
> ylim=c(-10,5),
> names=c(paste0("Placebo(N=", n1, ")"),
> paste0("Low Dose(N=", n2, ")"),
> paste0("High Dose(N=", n3,")")),
> ylab='Change from Baseline')
> abline(h=c(0), col="lightgray")
> }
>
> plot_f1(indata=v5, n1=114, n2=119, n3=116, fig_descrip=Figure 2a\nChange in
> Composite Score at Visit 5 (Day 31)\nPer Protocol Population)
>
> Error Message: Error: unexpected numeric constant in "plot_f1(indata=v5,
> n1=114, n2=119, n3=116, fig_descrip=Figure 2”
>
> Even this call gives the same error: plot_f1(indata=v5, n1=114, n2=119,
> n3=116, fig_descrip=Figure)
>
>
> Thanks,
>
> Gerard
>
>
>
>
>
>
>> On May 8, 2017, at 11:40 PM, Ulrik Stervbo <ulrik.ster...@gmail.com
>> <mailto:ulrik.ster...@gmail.com>> wrote:
>>
>
>> HI Gerard,
>>
>> You get the literals because the variables are not implicitly expanded -
>> 'Placebo(N=n1) ' is just a string indicating the N = n1.
>>
>> What you want is to use paste() or paste0():
>> c(paste0("Placebo(N=", n1, ")"), paste0("Low Dose (N=", n2, ")"),
>> paste0("High Dose (N=", n3, ")"))
>> should do it.
>>
>> I was taught a long ago that attach() should be avoided to avoid name
>> conflicts. Also, it makes it difficult to figure out which data is actually
>> being used.
>>
>> HTH
>> Ulrik
>>
>> On Tue, 9 May 2017 at 06:44 Gerard Smits <smits.gerar...@gmail.com
>> <mailto:smits.gerar...@gmail.com>> wrote:
>> Hi All,
>>
>> I thought I’d try to get a function working instead of block copying code
>> and editing. My backorund is more SAS, so using a SAS Macro would be easy,
>> but not so lucky with R functions.
>>
>>
>> R being used on Mac Sierra 10.12.4:
>>
>> R version 3.3.1 (2016-06-21) -- "Bug in Your Hair"
>> Copyright (C) 2016 The R Foundation for Statistical Computing
>> Platform: x86_64-apple-darwin13.4.0 (64-bit)
>>
>>
>> resp<-read.csv("//users//gerard//gs//r_work//xyz.csv", header = TRUE)
>>
>> v5 <-subset(resp, subset=visit==5 & pp==1)
>>
>> plot_f1 <-function(n1,n2,n3) {
>> attach(v8)
>> par(oma=c(2,2,2,2))
>> boxplot(formula = d_comp ~ rx_grp,
>> main="Figure 2\nChange in Composite Score at Visit 5 (Day 31)\nPer
>> Protocol Population",
>> ylim=c(-10,5),
>> names=c('Placebo(N=n1) ',
>> 'Low Dose(N=n2) ',
>> 'High Dose(N=n3)'),
>> ylab='Change from Baseline')
>> abline(h=c(0), col="lightgray")
>> }
>>
>> plot_f1(n1=114, n2=119, n3=116)
>>
>> The above is a simplified example where I am trying to pass 3 arguments,
>> n1-n3, to be shown in the x-axis tables, Instead of the numbers, I get the
>> literal n1, n2, n3.
>>
>> Any help appreciated.
>>
>> Thanks,
>&g
Re: [R] passing arguments to simple plotting program.
Hi Ulrik,
If I can trouble you with one more question.
Now trying to send a string to the main= . I was able to pass the data name in
data=in_data, but same logic is not working in passion the main string.
plot_f1 <-function(indata,n1,n2,n3,fig_descrip) {
par(oma=c(2,2,2,2))
boxplot(formula = d_comp ~ rx_grp,
data="indata”,# <- worked fine here.
main="fig_descrip",
ylim=c(-10,5),
names=c(paste0("Placebo(N=", n1, ")"),
paste0("Low Dose(N=", n2, ")"),
paste0("High Dose(N=", n3,")")),
ylab='Change from Baseline')
abline(h=c(0), col="lightgray")
}
plot_f1(indata=v5, n1=114, n2=119, n3=116, fig_descrip=Figure 2a\nChange in
Composite Score at Visit 5 (Day 31)\nPer Protocol Population)
Error Message: Error: unexpected numeric constant in "plot_f1(indata=v5,
n1=114, n2=119, n3=116, fig_descrip=Figure 2”
Even this call gives the same error: plot_f1(indata=v5, n1=114, n2=119,
n3=116, fig_descrip=Figure)
Thanks,
Gerard
> On May 8, 2017, at 11:40 PM, Ulrik Stervbo <ulrik.ster...@gmail.com> wrote:
>
> HI Gerard,
>
> You get the literals because the variables are not implicitly expanded -
> 'Placebo(N=n1) ' is just a string indicating the N = n1.
>
> What you want is to use paste() or paste0():
> c(paste0("Placebo(N=", n1, ")"), paste0("Low Dose (N=", n2, ")"),
> paste0("High Dose (N=", n3, ")"))
> should do it.
>
> I was taught a long ago that attach() should be avoided to avoid name
> conflicts. Also, it makes it difficult to figure out which data is actually
> being used.
>
> HTH
> Ulrik
>
> On Tue, 9 May 2017 at 06:44 Gerard Smits <smits.gerar...@gmail.com
> <mailto:smits.gerar...@gmail.com>> wrote:
> Hi All,
>
> I thought I’d try to get a function working instead of block copying code and
> editing. My backorund is more SAS, so using a SAS Macro would be easy, but
> not so lucky with R functions.
>
>
> R being used on Mac Sierra 10.12.4:
>
> R version 3.3.1 (2016-06-21) -- "Bug in Your Hair"
> Copyright (C) 2016 The R Foundation for Statistical Computing
> Platform: x86_64-apple-darwin13.4.0 (64-bit)
>
>
> resp<-read.csv("//users//gerard//gs//r_work//xyz.csv", header = TRUE)
>
> v5 <-subset(resp, subset=visit==5 & pp==1)
>
> plot_f1 <-function(n1,n2,n3) {
> attach(v8)
> par(oma=c(2,2,2,2))
> boxplot(formula = d_comp ~ rx_grp,
> main="Figure 2\nChange in Composite Score at Visit 5 (Day 31)\nPer
> Protocol Population",
> ylim=c(-10,5),
> names=c('Placebo(N=n1) ',
> 'Low Dose(N=n2) ',
> 'High Dose(N=n3)'),
> ylab='Change from Baseline')
> abline(h=c(0), col="lightgray")
> }
>
> plot_f1(n1=114, n2=119, n3=116)
>
> The above is a simplified example where I am trying to pass 3 arguments,
> n1-n3, to be shown in the x-axis tables, Instead of the numbers, I get the
> literal n1, n2, n3.
>
> Any help appreciated.
>
> Thanks,
>
> Gerard
>
>
>
>
> [[alternative HTML version deleted]]
>
> __
> R-help@r-project.org <mailto:R-help@r-project.org> mailing list -- To
> UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
> <https://stat.ethz.ch/mailman/listinfo/r-help>
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> <http://www.r-project.org/posting-guide.html>
> and provide commented, minimal, self-contained, reproducible code.
[[alternative HTML version deleted]]
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] passing arguments to simple plotting program.
Hi Ulrik,
That worked perfectly. Thanks for your help. Much appreciated.
Gerard
> On May 8, 2017, at 11:40 PM, Ulrik Stervbo <ulrik.ster...@gmail.com> wrote:
>
> HI Gerard,
>
> You get the literals because the variables are not implicitly expanded -
> 'Placebo(N=n1) ' is just a string indicating the N = n1.
>
> What you want is to use paste() or paste0():
> c(paste0("Placebo(N=", n1, ")"), paste0("Low Dose (N=", n2, ")"),
> paste0("High Dose (N=", n3, ")"))
> should do it.
>
> I was taught a long ago that attach() should be avoided to avoid name
> conflicts. Also, it makes it difficult to figure out which data is actually
> being used.
>
> HTH
> Ulrik
>
> On Tue, 9 May 2017 at 06:44 Gerard Smits <smits.gerar...@gmail.com
> <mailto:smits.gerar...@gmail.com>> wrote:
> Hi All,
>
> I thought I’d try to get a function working instead of block copying code and
> editing. My backorund is more SAS, so using a SAS Macro would be easy, but
> not so lucky with R functions.
>
>
> R being used on Mac Sierra 10.12.4:
>
> R version 3.3.1 (2016-06-21) -- "Bug in Your Hair"
> Copyright (C) 2016 The R Foundation for Statistical Computing
> Platform: x86_64-apple-darwin13.4.0 (64-bit)
>
>
> resp<-read.csv("//users//gerard//gs//r_work//xyz.csv", header = TRUE)
>
> v5 <-subset(resp, subset=visit==5 & pp==1)
>
> plot_f1 <-function(n1,n2,n3) {
> attach(v8)
> par(oma=c(2,2,2,2))
> boxplot(formula = d_comp ~ rx_grp,
> main="Figure 2\nChange in Composite Score at Visit 5 (Day 31)\nPer
> Protocol Population",
> ylim=c(-10,5),
> names=c('Placebo(N=n1) ',
> 'Low Dose(N=n2) ',
> 'High Dose(N=n3)'),
> ylab='Change from Baseline')
> abline(h=c(0), col="lightgray")
> }
>
> plot_f1(n1=114, n2=119, n3=116)
>
> The above is a simplified example where I am trying to pass 3 arguments,
> n1-n3, to be shown in the x-axis tables, Instead of the numbers, I get the
> literal n1, n2, n3.
>
> Any help appreciated.
>
> Thanks,
>
> Gerard
>
>
>
>
> [[alternative HTML version deleted]]
>
> __
> R-help@r-project.org <mailto:R-help@r-project.org> mailing list -- To
> UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
> <https://stat.ethz.ch/mailman/listinfo/r-help>
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> <http://www.r-project.org/posting-guide.html>
> and provide commented, minimal, self-contained, reproducible code.
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
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.
[R] passing arguments to simple plotting program.
Hi All,
I thought I’d try to get a function working instead of block copying code and
editing. My backorund is more SAS, so using a SAS Macro would be easy, but not
so lucky with R functions.
R being used on Mac Sierra 10.12.4:
R version 3.3.1 (2016-06-21) -- "Bug in Your Hair"
Copyright (C) 2016 The R Foundation for Statistical Computing
Platform: x86_64-apple-darwin13.4.0 (64-bit)
resp<-read.csv("//users//gerard//gs//r_work//xyz.csv", header = TRUE)
v5 <-subset(resp, subset=visit==5 & pp==1)
plot_f1 <-function(n1,n2,n3) {
attach(v8)
par(oma=c(2,2,2,2))
boxplot(formula = d_comp ~ rx_grp,
main="Figure 2\nChange in Composite Score at Visit 5 (Day 31)\nPer
Protocol Population",
ylim=c(-10,5),
names=c('Placebo(N=n1) ',
'Low Dose(N=n2) ',
'High Dose(N=n3)'),
ylab='Change from Baseline')
abline(h=c(0), col="lightgray")
}
plot_f1(n1=114, n2=119, n3=116)
The above is a simplified example where I am trying to pass 3 arguments, n1-n3,
to be shown in the x-axis tables, Instead of the numbers, I get the literal
n1, n2, n3.
Any help appreciated.
Thanks,
Gerard
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
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.
[R] changing font size in Forest plot code.
Hi All,
I have pulled the following function (fplot) from the internet, and
unfortunately I do not see an author to whom I can give credit. It used grid
graphics and relies mostly on package rmeta by Thomas Lumley. I am trying to
make the font smaller in my labeltext, but dont see any references to font
size in the code. Digitize changes the number size on the x-axis, but dont
see a corresponding way of making the labeling size smaller.
Using R 3.0.2
Any suggestions appreciated.
Gerard Smits
fplot=function (labeltext, mean, lower, upper, align = NULL, is.summary =
FALSE,
clip = c(-Inf, Inf), xlab = , zero = 1, graphwidth = unit(3,inches),
col = meta.colors(), xlog = FALSE, xticks = NULL,
xlow=0, xhigh, digitsize, boxsize,
...)
{
require(grid) || stop(`grid' package not found)
require(rmeta) || stop(`rmeta' package not found)
drawNormalCI - function(LL, OR, UL, size)
{
size = 0.75 * size
clipupper - convertX(unit(UL, native), npc, valueOnly = TRUE) 1
cliplower - convertX(unit(LL, native), npc, valueOnly = TRUE) 0
box - convertX(unit(OR, native), npc, valueOnly = TRUE)
clipbox - box 0 || box 1
if (clipupper || cliplower)
{
ends - both
lims - unit(c(0, 1), c(npc, npc))
if (!clipupper) {
ends - first
lims - unit(c(0, UL), c(npc, native))
}
if (!cliplower) {
ends - last
lims - unit(c(LL, 1), c(native, npc))
}
grid.lines(x = lims, y = 0.5, arrow = arrow(ends = ends,
length = unit(0.05, inches)), gp = gpar(col = col$lines))
if (!clipbox)
grid.rect(x = unit(OR, native), width = unit(size,
snpc), height = unit(size, snpc), gp = gpar(fill =
col$box,
col = col$box))
}
else {
grid.lines(x = unit(c(LL, UL), native), y = 0.5,
gp = gpar(col = col$lines))
grid.rect(x = unit(OR, native), width = unit(size,
snpc), height = unit(size, snpc), gp = gpar(fill = col$box,
col = col$box))
if ((convertX(unit(OR, native) + unit(0.5 * size,
lines), native, valueOnly = TRUE) UL)
(convertX(unit(OR, native) - unit(0.5 * size,
lines), native, valueOnly = TRUE) LL))
grid.lines(x = unit(c(LL, UL), native), y = 0.5,
gp = gpar(col = col$lines))
}
}
drawSummaryCI - function(LL, OR, UL, size) {
grid.polygon(x = unit(c(LL, OR, UL, OR), native), y = unit(0.5 +
c(0, 0.5 * size, 0, -0.5 * size), npc), gp = gpar(fill =
col$summary,
col = col$summary))
}
plot.new()
widthcolumn - !apply(is.na(labeltext), 1, any)
nc - NCOL(labeltext)
labels - vector(list, nc)
if (is.null(align))
align - c(l, rep(r, nc - 1))
else align - rep(align, length = nc)
nr - NROW(labeltext)
is.summary - rep(is.summary, length = nr)
for (j in 1:nc) {
labels[[j]] - vector(list, nr)
for (i in 1:nr) {
if (is.na(labeltext[i, j]))
next
x - switch(align[j], l = 0, r = 1, c = 0.5)
just - switch(align[j], l = left, r = right, c = center)
labels[[j]][[i]] - textGrob(labeltext[i, j], x = x,
just = just, gp = gpar(fontface = if (is.summary[i]) bold
else plain, col = rep(col$text, length = nr)[i]))
}
}
colgap - unit(3, mm)
colwidths - unit.c(max(unit(rep(1, sum(widthcolumn)), grobwidth,
labels[[1]][widthcolumn])), colgap)
if (nc 1) {
for (i in 2:nc) colwidths - unit.c(colwidths, max(unit(rep(1,
sum(widthcolumn)), grobwidth, labels[[i]][widthcolumn])),
colgap)
}
colwidths - unit.c(colwidths, graphwidth)
pushViewport(viewport(layout = grid.layout(nr + 1, nc * 2 +
1, widths = colwidths, heights = unit(c(rep(1, nr), 0.5),
lines
cwidth - (upper - lower)
#xrange - c(max(min(lower, na.rm = TRUE), clip[1]), min(max(upper, na.rm =
TRUE), clip[2]))
xrange - c(xlow,xhigh)
info - 1/cwidth
info - info/max(info[!is.summary], na.rm = TRUE)
info[is.summary] - 1
if (!is.null(boxsize))
info - rep(boxsize, length = length(info))
for (j in 1:nc) {
for (i in 1:nr) {
if (!is.null(labels[[j]][[i]])) {
pushViewport(viewport(layout.pos.row = i, layout.pos.col = 2 *
j - 1))
grid.draw(labels[[j]][[i]])
popViewport()
}
}
}
pushViewport(viewport(layout.pos.col = 2 * nc + 1, xscale = xrange))
grid.lines(x = unit(zero, native), y = 0:1, gp = gpar(col = col$zero))
if (xlog) {
if (is.null(xticks
Re: [R] changing font size in Forest plot code.
Hi David,
That worked perfectly. I had tried something like that, but obviously messed
up the change.
Thanks for your help. Much appreciated.
Gerard
On Jan 19, 2014, at 2:16 PM, David Winsemius dwinsem...@comcast.net wrote:
On Jan 19, 2014, at 1:13 PM, Gerard Smits wrote:
Hi All,
I have pulled the following function (fplot) from the internet, and
unfortunately I do not see an author to whom I can give credit. It used
grid graphics and relies mostly on package rmeta by Thomas Lumley. I am
trying to make the font smaller in my labeltext, but don‚t see any
references to font size in the code. Digitize changes the number size on
the x-axis, but don‚t see a corresponding way of making the labeling size
smaller.
Wouldn't it just be needed to specify grid parameters (as exemplified several
other places in that code) in the code where 'labels' are created?
...
labels[[j]][[i]] - textGrob(labeltext[i, j], x = x,
just = just, gp = gpar(fontsize=8, fontface = if
(is.summary[i]) bold
else plain, col = rep(col$text, length = nr)[i]))
...
Seems to succeed (once the errant and quite strange double comma character
'„' is removed and replaced with a proper double quote.) If you are doing
this on a word processor, then you should convert to a programming text
editor.
--
David.
Using R 3.0.2
Any suggestions appreciated.
Gerard Smits
fplot=function (labeltext, mean, lower, upper, align = NULL, is.summary =
FALSE,
clip = c(-Inf, Inf), xlab = , zero = 1, graphwidth = unit(3,inches),
col = meta.colors(), xlog = FALSE, xticks = NULL,
xlow=0, xhigh, digitsize, boxsize,
...)
{
require(grid) || stop(`grid' package not found)
require(rmeta) || stop(`rmeta' package not found)
drawNormalCI - function(LL, OR, UL, size)
{
size = 0.75 * size
clipupper - convertX(unit(UL, native), npc, valueOnly = TRUE) 1
cliplower - convertX(unit(LL, native), npc, valueOnly = TRUE) 0
box - convertX(unit(OR, native), npc, valueOnly = TRUE)
clipbox - box 0 || box 1
if (clipupper || cliplower)
{
ends - both
lims - unit(c(0, 1), c(npc, npc))
if (!clipupper) {
ends - first
lims - unit(c(0, UL), c(npc, native))
}
if (!cliplower) {
ends - last
lims - unit(c(LL, 1), c(native, npc))
}
grid.lines(x = lims, y = 0.5, arrow = arrow(ends = ends,
length = unit(0.05, inches)), gp = gpar(col = col$lines))
if (!clipbox)
grid.rect(x = unit(OR, native), width = unit(size,
snpc), height = unit(size, snpc), gp = gpar(fill =
col$box,
col = col$box))
}
else {
grid.lines(x = unit(c(LL, UL), native), y = 0.5,
gp = gpar(col = col$lines))
grid.rect(x = unit(OR, native), width = unit(size,
snpc), height = unit(size, snpc), gp = gpar(fill =
col$box,
col = col$box))
if ((convertX(unit(OR, native) + unit(0.5 * size,
lines), native, valueOnly = TRUE) UL)
(convertX(unit(OR, native) - unit(0.5 * size,
lines), native, valueOnly = TRUE) LL))
grid.lines(x = unit(c(LL, UL), native), y = 0.5,
gp = gpar(col = col$lines))
}
}
drawSummaryCI - function(LL, OR, UL, size) {
grid.polygon(x = unit(c(LL, OR, UL, OR), native), y = unit(0.5 +
c(0, 0.5 * size, 0, -0.5 * size), npc), gp = gpar(fill =
col$summary,
col = col$summary))
}
plot.new()
widthcolumn - !apply(is.na(labeltext), 1, any)
nc - NCOL(labeltext)
labels - vector(list, nc)
if (is.null(align))
align - c(l, rep(r, nc - 1))
else align - rep(align, length = nc)
nr - NROW(labeltext)
is.summary - rep(is.summary, length = nr)
for (j in 1:nc) {
labels[[j]] - vector(list, nr)
for (i in 1:nr) {
if (is.na(labeltext[i, j]))
next
x - switch(align[j], l = 0, r = 1, c = 0.5)
just - switch(align[j], l = left, r = right, c = center)
labels[[j]][[i]] - textGrob(labeltext[i, j], x = x,
just = just, gp = gpar(fontface = if (is.summary[i]) bold
else plain, col = rep(col$text, length = nr)[i]))
}
}
colgap - unit(3, mm)
colwidths - unit.c(max(unit(rep(1, sum(widthcolumn)), grobwidth,
labels[[1]][widthcolumn])), colgap)
if (nc 1) {
for (i in 2:nc) colwidths - unit.c(colwidths, max(unit(rep(1,
sum(widthcolumn)), grobwidth, labels[[i]][widthcolumn])),
colgap)
}
colwidths - unit.c(colwidths, graphwidth)
pushViewport(viewport(layout = grid.layout(nr + 1, nc * 2 +
1, widths = colwidths, heights = unit(c(rep(1
[R] problem with newline using bquote(paste())
Hi All, This is a variant of a problem I posted yesterday (see below) where I found I had a large gap between my N= and he number I had evaluated using .(x). I seem to have trouble with newlines in a main title. I find now that all works as expected (no unsightly gap between my N= and the value, if all of the title is put on the same line. Whenever I try the newline, I run into problems. Below, I have one example that gives me no syntax errors, but simply does not print the information after the \n (the N= xxx) part. Any help appreciated. Thanks, Gerard PS using R 3.0.0 ss-n(m18_das28*b_dascore) par(oma=c(2,2,2,2)) scatterplot(m18_das28~b_score, jitter=list(x=1, y=1), grid=F, smooth=F, las=1, pch=c(1), col='blue', main=as.expression(bquote(paste(Baseline xyz with Month 18 DAS28\n)), bquote(paste((N=,.(ss),, xlab=Baseline xyz, ylab=Month 18 DAS28, legend.plot=F) Prior, related post: On Aug 29, 2013, at 2:00 PM, Gerard Smits g_sm...@verizon.net wrote: Hi All, I'm using R 3.0.0. I'm trying to add the sample size of the paired data (calculated by a function n(), which returns a value of 70, correctly). My main title works fine except that the '70' appears far to the right on the line as in: at Month 18 (N= 70) Is there a way of left justifying the result of .(ss)? or some other way of removing with whitespace between n= and 70?. Thanks for any suggestions. Gerard library (car) data-read.csv(//users//smits//r_work//data.csv, header = TRUE) attach(data); ## ss-n(m18_das28*b_score) scatterplot(m18_das28~b_score, jitter=list(x=1, y=1), grid=F, smooth=F, las=1, pch=c(1), col='blue', main=bquote(paste(Hypothesis 9.4.1\nBaseline XYZ with Disease Activity (DAS28)\nat Month 18 (N=,.(ss),))), xlab=Baseline XYZ, ylab=Month 18 DAS28, legend.plot=F) __ R-help@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.
Re: [R] problem with newline using bquote(paste())
That explains it. Thanks for the info. Gerard On Aug 30, 2013, at 8:42 AM, Marc Schwartz marc_schwa...@me.com wrote: On Aug 30, 2013, at 10:26 AM, Gerard Smits g_sm...@verizon.net wrote: Hi All, This is a variant of a problem I posted yesterday (see below) where I found I had a large gap between my N= and he number I had evaluated using .(x). I seem to have trouble with newlines in a main title. I find now that all works as expected (no unsightly gap between my N= and the value, if all of the title is put on the same line. Whenever I try the newline, I run into problems. Below, I have one example that gives me no syntax errors, but simply does not print the information after the \n (the N= xxx) part. Any help appreciated. Thanks, Gerard PS using R 3.0.0 ss-n(m18_das28*b_dascore) par(oma=c(2,2,2,2)) scatterplot(m18_das28~b_score, jitter=list(x=1, y=1), grid=F, smooth=F, las=1, pch=c(1), col='blue', main=as.expression(bquote(paste(Baseline xyz with Month 18 DAS28\n)), bquote(paste((N=,.(ss),, xlab=Baseline xyz, ylab=Month 18 DAS28, legend.plot=F) Prior, related post: On Aug 29, 2013, at 2:00 PM, Gerard Smits g_sm...@verizon.net wrote: Hi All, I'm using R 3.0.0. I'm trying to add the sample size of the paired data (calculated by a function n(), which returns a value of 70, correctly). My main title works fine except that the '70' appears far to the right on the line as in: at Month 18 (N= 70) Is there a way of left justifying the result of .(ss)? or some other way of removing with whitespace between n= and 70?. Thanks for any suggestions. Gerard library (car) data-read.csv(//users//smits//r_work//data.csv, header = TRUE) attach(data); ## ss-n(m18_das28*b_score) scatterplot(m18_das28~b_score, jitter=list(x=1, y=1), grid=F, smooth=F, las=1, pch=c(1), col='blue', main=bquote(paste(Hypothesis 9.4.1\nBaseline XYZ with Disease Activity (DAS28)\nat Month 18 (N=,.(ss),))), xlab=Baseline XYZ, ylab=Month 18 DAS28, legend.plot=F) You cannot use newlines in plotmath expressions. You will need to create each line of the plot title text separately using ?mtext instead of specifying 'main' in the plot call. Regards, Marc Schwartz __ R-help@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.
[R] spacing problem in main title using car package scatterplot
Hi All, I'm using R 3.0.0. I'm trying to add the sample size of the paired data (calculated by a function n(), which returns a value of 70, correctly). My main title works fine except that the '70' appears far to the right on the line as in: at Month 18 (N= 70) Is there a way of left justifying the result of .(ss)? or some other way of removing with whitespace between n= and 70?. Thanks for any suggestions. Gerard library (car) data-read.csv(//users//smits//r_work//data.csv, header = TRUE) attach(data); ## ss-n(m18_das28*b_score) scatterplot(m18_das28~b_score, jitter=list(x=1, y=1), grid=F, smooth=F, las=1, pch=c(1), col='blue', main=bquote(paste(Hypothesis 9.4.1\nBaseline XYZ with Disease Activity (DAS28)\nat Month 18 (N=,.(ss),))), xlab=Baseline XYZ, ylab=Month 18 DAS28, legend.plot=F) __ R-help@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.
[R] Problem using a function to produce multiple scatterplots (cars package)
Hi All,
I have a number of plots to run and was hoping to use a function instead of
bock copying my code and retyping my parameters.
I am using R 3.0.0 on a mac.
My code is as follows:
library (car)
cres-read.csv(//users//smits//r_work//cres.csv, header = TRUE)
attach(cres);
run_plots - functon()
{
scatterplot(y~x|z,
jitter=list(x=1, y=1),
grid=F,
smooth=F,
las=1,
pch=c(1,2),
col=c('red','blue'),
main=Baseline Clinical CRP by RAMRIS Synovitis at Day 113,
xlab=Baseline CRP,
ylab=Day 113 Synovitis,
legend.plot=F)
legend(bottomright,
box.lty=0, # line type to surround the legend box (0 for none)
legend=c(Active+MTX,Placebo+MTX),
col=c('red','blue'),
pch=c(1,2),
pt.cex=c(1,1,1,1))
}
then I was hoping to execute the function by passing 3 parameters:
run_plots(b_crp, t_synos113, treatment)
This approach follows the way I usually do it with a SAS macro, were I can
repeat the innovation over and over using different variables, etc.
Right off the bat, you can see that I have a problem finding the function
function. Then it does not recognize the first parameter it comes across.
Any help appreciated.
Thanks,
Gerard
cres-read.csv(//users//smits//r_work//cres.csv, header = TRUE)
attach(cres);
run_plots - functon()
Error: could not find function functon
{
+ scatterplot(y~x|z,
+ jitter=list(x=1, y=1),
+ grid=F,
+ smooth=F,
+ las=1,
+ pch=c(1,2),
+ col=c('red','blue'),
+ main=Baseline Clinical CRP by RAMRIS Synovitis at Day 113,
+ xlab=Baseline CRP,
+ ylab=Day 113 Synovitis,
+ legend.plot=F)
+ legend(bottomright,
+ box.lty=0, # line type to surround the legend box (0 for none)
+ legend=c(Active+MTX,Placebo+MTX),
+ col=c('red','blue'),
+ pch=c(1,2),
+ pt.cex=c(1,1,1,1))
+ }
Error in eval(expr, envir, enclos) : object 'y' not found
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Problem using a function to produce multiple scatterplots (cars package)
Dumb error. Thanks for letting me know. Gerard
On Aug 15, 2013, at 11:38 AM, Richard M. Heiberger r...@temple.edu wrote:
typo
run_plots - functon()
On Thu, Aug 15, 2013 at 1:10 PM, Gerard Smits g_sm...@verizon.net wrote:
Hi All,
I have a number of plots to run and was hoping to use a function instead of
bock copying my code and retyping my parameters.
I am using R 3.0.0 on a mac.
My code is as follows:
library (car)
cres-read.csv(//users//smits//r_work//cres.csv, header = TRUE)
attach(cres);
run_plots - functon()
{
scatterplot(y~x|z,
jitter=list(x=1, y=1),
grid=F,
smooth=F,
las=1,
pch=c(1,2),
col=c('red','blue'),
main=Baseline Clinical CRP by RAMRIS Synovitis at Day 113,
xlab=Baseline CRP,
ylab=Day 113 Synovitis,
legend.plot=F)
legend(bottomright,
box.lty=0, # line type to surround the legend box (0 for none)
legend=c(Active+MTX,Placebo+MTX),
col=c('red','blue'),
pch=c(1,2),
pt.cex=c(1,1,1,1))
}
then I was hoping to execute the function by passing 3 parameters:
run_plots(b_crp, t_synos113, treatment)
This approach follows the way I usually do it with a SAS macro, were I can
repeat the innovation over and over using different variables, etc.
Right off the bat, you can see that I have a problem finding the function
function. Then it does not recognize the first parameter it comes across.
Any help appreciated.
Thanks,
Gerard
cres-read.csv(//users//smits//r_work//cres.csv, header = TRUE)
attach(cres);
run_plots - functon()
Error: could not find function functon
{
+ scatterplot(y~x|z,
+ jitter=list(x=1, y=1),
+ grid=F,
+ smooth=F,
+ las=1,
+ pch=c(1,2),
+ col=c('red','blue'),
+ main=Baseline Clinical CRP by RAMRIS Synovitis at Day 113,
+ xlab=Baseline CRP,
+ ylab=Day 113 Synovitis,
+ legend.plot=F)
+ legend(bottomright,
+ box.lty=0, # line type to surround the legend box (0 for none)
+ legend=c(Active+MTX,Placebo+MTX),
+ col=c('red','blue'),
+ pch=c(1,2),
+ pt.cex=c(1,1,1,1))
+ }
Error in eval(expr, envir, enclos) : object 'y' not found
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and provide commented, minimal, self-contained, reproducible code.
[[alternative HTML version deleted]]
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and provide commented, minimal, self-contained, reproducible code.
[R] question on axis labels
Hi All, I'm trying to label my plot axis with times (HH:MM) that correspond to a numeric index (values 0:6) for my time variable. I'd like to plot 08:00, 12:00, and so on, instead of 0 through 6. I have used the following line of code: axis(1, 0:6, labels=c(08:00, 12:00, 16:00, 20:00, 24:00, 04:00, 08:00), cex=0.8) [I've used both the 0:6 and at=c(0:6), with no effect.] My labels come out with a 0 - 6, location dependent, superimposed over my colon in my HH:MM string. So 08:00 looks like 08000, 12:00 looks like 12100. Any way of suppressing the at locations? I'm using version 2.14.0 on a mac My program pulls in the following packages (not sure relevant): require (Hmisc) require (lattice) require (gplots) Thanks for any suggestions, Gerard __ R-help@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.
Re: [R] question on axis labels
Worked like a charm! Thanks for your help. Gerard On Feb 20, 2012, at 3:52 PM, Sarah Goslee wrote: This works for me: plot(0:6, runif(7), xaxt=n) axis(1, at=0:6, labels=c(08:00, 12:00, 16:00, 20:00, 24:00, 04:00, 08:00), cex=0.8) You need the xaxt=n in the plot statement, and the correct form is at=0:6 Sarah On Mon, Feb 20, 2012 at 6:39 PM, Gerard Smits g_sm...@verizon.net wrote: Hi All, I'm trying to label my plot axis with times (HH:MM) that correspond to a numeric index (values 0:6) for my time variable. I'd like to plot 08:00, 12:00, and so on, instead of 0 through 6. I have used the following line of code: axis(1, 0:6, labels=c(08:00, 12:00, 16:00, 20:00, 24:00, 04:00, 08:00), cex=0.8) [I've used both the 0:6 and at=c(0:6), with no effect.] My labels come out with a 0 - 6, location dependent, superimposed over my colon in my HH:MM string. So 08:00 looks like 08000, 12:00 looks like 12100. Any way of suppressing the at locations? I'm using version 2.14.0 on a mac My program pulls in the following packages (not sure relevant): require (Hmisc) require (lattice) require (gplots) Thanks for any suggestions, Gerard -- Sarah Goslee http://www.functionaldiversity.org __ R-help@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.
[R] question of elimination drawn lines on competing risk graph
Hi All,
I am using the package, cmprisk, to plot competing risks. In my case, I have
four lines showing risk of going on dialysis (by a lab test [fgf-23] in
quartiles), where the 4 lower lines are for the competing risk of death.
I am trying to edit the function to just plot the upper curves (and not show
the lower (competing) risk death curves. I am not especially facile in
editing this code, so would appreciate any help as to where I should be making
changes.
Hope OK to include the function code.
Thanks.
Gerard
require(cmprsk)
#
# #
# CUMULATIVE INCIDENCE CURVES IN R #
# #
# Written by Luca Scrucca #
# #
# Reference:#
# Scrucca L., Santucci A., Aversa F. (2007) Competing risks analysis using #
# R: an easy guide for clinicians. Bone Marrow Transplantation, 40, #
# 381--387. #
#
# ver. 1.1 Feb 2008
# - allow group to be missing
# - if t is provided both computation and plots use t as time points
# - allow col, lwd to be used for curves with confidence bands
# - fix some bugs in the legend
# - added help on source code
# ver. 1.0 May 2007
# - Version appearing in the BMT paper
#
#
# Usage:
#
# CumIncidence(ftime, fstatus, group, t, strata, rho = 0, cencode = 0,
#subset, na.action = na.omit, level,
#xlab = Time, ylab = Probability,
#col, lty, lwd, digits = 4)
#
# Arguments:
#
# ftime = failure time variable.
# fstatus = variable with distinct codes for different causes of
# failure and also a distinct code for censored observations.
# group = estimates will be calculated within groups given by distinct
# values of this variable. Tests will compare these groups. If
# missing then treated as all one group (no test statistics).
# t = a vector of time points where the cumulative incidence function
# should be evaluated.
# strata = stratification variable. Has no effect on estimates. Tests
# will be stratified on this variable. (all data in 1 stratum,
# if missing).
# rho = power of the weight function used in the tests. By default is
# set to 0.
# cencode = value of fstatus variable which indicates the failure time
# is censored.
# subset = a logical vector specifying a subset of cases to include in
# the analysis.
# na.action=a function specifying the action to take for any cases
# missing any of ftime, fstatus, group, strata, or subset.
# By default missing cases are omitted.
# level = a value in the range [0,1] specifying the level for pointwise
# confidence bands.
# xlab =text for the x-axis label.
# ylab =text for the y-axis label.
# col = color(s) used for plotting curves (see plot.default).
# lty = line type(s) used for plotting curves (see plot.default).
# lwd = line width(s) used for plotting curves (see plot.default).
# digits = number of significant digits used for printing values. By
# default set at 4.
#
#
CumIncidence - function(ftime, fstatus, group, t, strata, rho = 0,
cencode = 0, subset, na.action = na.omit,
level,
xlab = Time, ylab = Probability,
col, lty, lwd, digits = 4)
{
# check for the required package
if(!require(cmprsk))
{ stop(Package `cmprsk' is required and must be installed.\n
See help(install.packages) or write the following command at prompt
and then follow the instructions:\n
install.packages(\cmprsk\)) }
# collect data
mf - match.call(expand.dots = FALSE)
mf[[1]] - as.name(list)
mf$t - mf$digits - mf$col - mf$lty - mf$lwd - mf$level -
mf$xlab - mf$ylab - NULL
mf - eval(mf, parent.frame())
g - max(1, length(unique(mf$group)))
s - length(unique(mf$fstatus))
if(missing(t))
{ time - pretty(c(0, max(mf$ftime)), 6)
ttime - time - time[time max(mf$ftime)] }
else { ttime - time - t }
# fit model and estimates at time points
fit - do.call(cuminc, mf)
tfit - timepoints(fit, time)
# print result
cat(\n+, paste(rep(-, 67), collapse=), +, sep =)
cat(\n|
[R] problem with abline
Hi All, I am running a scatter plot and trying to add a best fit line. I use an abline function, but get no line drawn over the points. I also get no error. I arm using V 2.10.0 on Windows 7. Here is my code, including the SAS transport file import: require (foreign) require (chron) require (Hmisc) require (lattice) clin - sasxport.get(y:\\temp\\subset.xpt) attach(clin) plot.new() xyplot(jitter(b.lvef)~jitter(log.fgf), main=Scatter Plot of Baseline Ejection Fraction\nby Log10 FGF-23, ylim=c(10,90), ylab=Ejection Fraction, xlab=Log10 FGF-23) abline(lm(b.lvef~log.fgf)) Any suggestions appreciated. Thanks, Gerard [[alternative HTML version deleted]] __ R-help@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.
Re: [R] problem with abline
Hi David,
Just changed to the standard plot from xyplot and it worked fine. I'll check
out panel.lmline.
Thanks for your help.
Gerard
On Oct 6, 2010, at 3:27 PM, David Winsemius wrote:
On Oct 6, 2010, at 5:56 PM, Gerard Smits wrote:
Hi All,
I am running a scatter plot and trying to add a best fit line. I use an
abline function, but get no line drawn over the points.
xyplot is Lattice
abline is base graphics
I also get no error. I arm using V 2.10.0 on Windows 7.
Here is my code, including the SAS transport file import:
Nope. Read the Posting Guide about attachments to the list. Nice try, though.
require (foreign)
require (chron)
require (Hmisc)
require (lattice)
clin - sasxport.get(y:\\temp\\subset.xpt)
attach(clin)
plot.new()
xyplot(jitter(b.lvef)~jitter(log.fgf),
main=Scatter Plot of Baseline Ejection Fraction\nby Log10 FGF-23,
ylim=c(10,90),
ylab=Ejection Fraction, xlab=Log10 FGF-23)
Instead consider:
?panel.lmline
Perhaps (untested in absence of data):
plot.new()
xyplot(jitter(b.lvef)~jitter(log.fgf), panel = function(x, y) {
panel.xyplot(x, y)
panel.lmline(x,y)
},
main=Scatter Plot of Baseline Ejection Fraction\nby Log10 FGF-23,
ylim=c(10,90),
ylab=Ejection Fraction, xlab=Log10 FGF-23)
abline(lm(b.lvef~log.fgf))
Any suggestions appreciated.
Thanks,
Gerard
[[alternative HTML version deleted]]
David Winsemius, MD
West Hartford, CT
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and provide commented, minimal, self-contained, reproducible code.
[R] special symbols into a text string
All, I am trying to put a few special symbols into a string to place on a graph (e.g., = and superscript 2). This clearly works, but does not look good: text-c(x2, A=B) I tried pasting in the = symbol, but just comes out as an =. I have tried expression() with no luck: text-c(x2, expression(x^2)) Any help appreciated. Thanks, Gerard [[alternative HTML version deleted]] __ R-help@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.
Re: [R] Forestplot () box size question
Yes, my R is a few versions old. I did not realize that the package version was dependent on the R version. Thanks. Gerard PS was able to apply the code suggested by David W. to the 2.14 version and got it to work. At 12:32 AM 3/22/2009, Thomas Lumley wrote: On Sat, 21 Mar 2009, Gerard Smits wrote: I have tried several sites (2 in Ca and also Australia) and find only version 2.14. Which site did you pull 2.15 from? I have checked half a dozen sites in various countries, and they all have 2.15 (even in the Southern Hemisphere -- it's not that updates flow the wrong way south of the equator). The CRAN mirror status page says that no mirrors are more than a couple of days out of date. Mostly likely you have an old version of R. CRAN binaries are built only for the current version of R. -thomas Thomas Lumley Assoc. Professor, Biostatistics tlum...@u.washington.eduUniversity of Washington, Seattle [[alternative HTML version deleted]] __ R-help@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.
[R] Forestplot () box size question
Hi All,
I have been able to modify the x-axis to start at zero by adding xlow
and xhigh parameters; that was pretty simple. I have been unable to
find the location of the code that would turn off the information
weighting of the box size (I have smaller randomized trials getting
less weight than a much larger non-randomized trial). The function
is forestplot() from rmeta.
Thanks for any help.
Gerard
Slightly modified working function with data and a call follows:
fplot=function (labeltext, mean, lower, upper, align = NULL,
is.summary = FALSE,
clip = c(-Inf, Inf), xlab = , zero = 1, graphwidth = unit(3,inches),
col = meta.colors(), xlog = FALSE, xticks = NULL,
xlow=0, xhigh, digitsize,
...)
{
require(grid) || stop(`grid' package not found)
require(rmeta) || stop(`rmeta' package not found)
drawNormalCI - function(LL, OR, UL, size)
{
size = 0.75 * size
clipupper - convertX(unit(UL, native), npc, valueOnly = TRUE) 1
cliplower - convertX(unit(LL, native), npc, valueOnly = TRUE) 0
box - convertX(unit(OR, native), npc, valueOnly = TRUE)
clipbox - box 0 || box 1
if (clipupper || cliplower)
{
ends - both
lims - unit(c(0, 1), c(npc, npc))
if (!clipupper) {
ends - first
lims - unit(c(0, UL), c(npc, native))
}
if (!cliplower) {
ends - last
lims - unit(c(LL, 1), c(native, npc))
}
grid.lines(x = lims, y = 0.5, arrow = arrow(ends = ends,
length = unit(0.05, inches)), gp = gpar(col = col$lines))
if (!clipbox)
grid.rect(x = unit(OR, native), width = unit(size,
snpc), height = unit(size, snpc), gp =
gpar(fill = col$box,
col = col$box))
}
else {
grid.lines(x = unit(c(LL, UL), native), y = 0.5,
gp = gpar(col = col$lines))
grid.rect(x = unit(OR, native), width = unit(size,
snpc), height = unit(size, snpc), gp = gpar(fill
= col$box,
col = col$box))
if ((convertX(unit(OR, native) + unit(0.5 * size,
lines), native, valueOnly = TRUE) UL)
(convertX(unit(OR, native) - unit(0.5 * size,
lines), native, valueOnly = TRUE) LL))
grid.lines(x = unit(c(LL, UL), native), y = 0.5,
gp = gpar(col = col$lines))
}
}
drawSummaryCI - function(LL, OR, UL, size) {
grid.polygon(x = unit(c(LL, OR, UL, OR), native), y = unit(0.5 +
c(0, 0.5 * size, 0, -0.5 * size), npc), gp = gpar(fill
= col$summary,
col = col$summary))
}
plot.new()
widthcolumn - !apply(is.na(labeltext), 1, any)
nc - NCOL(labeltext)
labels - vector(list, nc)
if (is.null(align))
align - c(l, rep(r, nc - 1))
else align - rep(align, length = nc)
nr - NROW(labeltext)
is.summary - rep(is.summary, length = nr)
for (j in 1:nc) {
labels[[j]] - vector(list, nr)
for (i in 1:nr) {
if (is.na(labeltext[i, j]))
next
x - switch(align[j], l = 0, r = 1, c = 0.5)
just - switch(align[j], l = left, r = right, c = center)
labels[[j]][[i]] - textGrob(labeltext[i, j], x = x,
just = just, gp = gpar(fontface = if (is.summary[i]) bold
else plain, col = rep(col$text, length = nr)[i]))
}
}
colgap - unit(3, mm)
colwidths - unit.c(max(unit(rep(1, sum(widthcolumn)), grobwidth,
labels[[1]][widthcolumn])), colgap)
if (nc 1) {
for (i in 2:nc) colwidths - unit.c(colwidths, max(unit(rep(1,
sum(widthcolumn)), grobwidth, labels[[i]][widthcolumn])),
colgap)
}
colwidths - unit.c(colwidths, graphwidth)
pushViewport(viewport(layout = grid.layout(nr + 1, nc * 2 +
1, widths = colwidths, heights = unit(c(rep(1, nr), 0.5),
lines
cwidth - (upper - lower)
#xrange - c(max(min(lower, na.rm = TRUE), clip[1]),
min(max(upper, na.rm = TRUE), clip[2]))
xrange - c(xlow,xhigh)
info - 1/cwidth
info - info/max(info[!is.summary], na.rm = TRUE)
info[is.summary] - 1
for (j in 1:nc) {
for (i in 1:nr) {
if (!is.null(labels[[j]][[i]])) {
pushViewport(viewport(layout.pos.row = i,
layout.pos.col = 2 *
j - 1))
grid.draw(labels[[j]][[i]])
popViewport()
}
}
}
pushViewport(viewport(layout.pos.col = 2 * nc + 1, xscale = xrange))
grid.lines(x = unit(zero, native), y = 0:1, gp = gpar(col = col$zero))
if (xlog) {
if (is.null(xticks)) {
ticks - pretty(exp(xrange))
Re: [R] Forestplot () box size question
Hi David,
I just checked to make sure I had the latest version. I see no
boxsize option in the forestplot function parameters or any other
place in the code.
function (labeltext, mean, lower, upper, align = NULL, is.summary = FALSE,
clip = c(-Inf, Inf), xlab = , zero = 0, graphwidth = unit(2,
inches), col = meta.colors(), xlog = FALSE, xticks = NULL,
...)
{
Thanks,
Gerard
At 10:32 AM 3/21/2009, David Winsemius wrote:
if (!is.null(boxsize))
info - rep(boxsize, length = length(info))
[[alternative HTML version deleted]]
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Forestplot () box size question
Hi David,
I did a general package update, but it must not have been quite as
complete as I had hoped.
I will make sure I download 2.15 and go from there.
Thanks for you help.
Gerard
At 11:10 AM 3/21/2009, David Winsemius wrote:
Latest version is a tad non-specific. The help page for (my) package
rmeta, version 2.15 (download and compiled today) says:
Usage
forestplot(labeltext, mean, lower, upper, align = NULL, is.summary =
FALSE, clip = c(-Inf, Inf), xlab = , zero = 0, graphwidth = unit(2,
inches), col = meta.colors(), xlog = FALSE, xticks=NULL,
boxsize=NULL,...)
--
sessionInfo()
R version 2.8.1 Patched (2009-01-19 r47650)
i386-apple-darwin9.6.0
locale:
en_US.UTF-8/en_US.UTF-8/C/C/en_US.UTF-8/en_US.UTF-8
attached base packages:
[1] grid stats graphics grDevices utils datasets
methods base
other attached packages:
[1] rmeta_2.15 zoo_1.5-5
loaded via a namespace (and not attached):
[1] lattice_0.17-20 tools_2.8.1
--
David Winsemius, MD
Heritage Laboratories
West Hartford, CT
On Mar 21, 2009, at 2:01 PM, Gerard Smits wrote:
Hi David,
I just checked to make sure I had the latest version. I see no
boxsize option in the forestplot function parameters or any other
place in the code.
function (labeltext, mean, lower, upper, align = NULL, is.summary =
FALSE,
clip = c(-Inf, Inf), xlab = , zero = 0, graphwidth = unit(2,
inches), col = meta.colors(), xlog = FALSE, xticks = NULL,
...)
{
Thanks,
Gerard
At 10:32 AM 3/21/2009, David Winsemius wrote:
if (!is.null(boxsize))
info - rep(boxsize, length = length(info))
[[alternative HTML version deleted]]
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Forestplot () box size question
I have tried several sites (2 in Ca and also Australia) and find only
version 2.14.
Which site did you pull 2.15 from?
Thanks
At 11:10 AM 3/21/2009, David Winsemius wrote:
Latest version is a tad non-specific. The help page for (my) package
rmeta, version 2.15 (download and compiled today) says:
Usage
forestplot(labeltext, mean, lower, upper, align = NULL, is.summary =
FALSE, clip = c(-Inf, Inf), xlab = , zero = 0, graphwidth = unit(2,
inches), col = meta.colors(), xlog = FALSE, xticks=NULL,
boxsize=NULL,...)
--
sessionInfo()
R version 2.8.1 Patched (2009-01-19 r47650)
i386-apple-darwin9.6.0
locale:
en_US.UTF-8/en_US.UTF-8/C/C/en_US.UTF-8/en_US.UTF-8
attached base packages:
[1] grid stats graphics grDevices utils datasets
methods base
other attached packages:
[1] rmeta_2.15 zoo_1.5-5
loaded via a namespace (and not attached):
[1] lattice_0.17-20 tools_2.8.1
--
David Winsemius, MD
Heritage Laboratories
West Hartford, CT
On Mar 21, 2009, at 2:01 PM, Gerard Smits wrote:
Hi David,
I just checked to make sure I had the latest version. I see no
boxsize option in the forestplot function parameters or any other
place in the code.
function (labeltext, mean, lower, upper, align = NULL, is.summary =
FALSE,
clip = c(-Inf, Inf), xlab = , zero = 0, graphwidth = unit(2,
inches), col = meta.colors(), xlog = FALSE, xticks = NULL,
...)
{
Thanks,
Gerard
At 10:32 AM 3/21/2009, David Winsemius wrote:
if (!is.null(boxsize))
info - rep(boxsize, length = length(info))
[[alternative HTML version deleted]]
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] SAS Institute Adding Support for R
It seems that SAS is willing to offer this interface if you are willing to purchase one of their expensive add-on packages. Not surprising. Gerard Smits [[alternative HTML version deleted]] __ R-help@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.
[R] Modifying forestplot function in rmeta
All,
I am using the forestplot function in rmeta.
I was able to modify the x axis range by commenting out one line and
feeding it two new parameters (I wanted to set zero as the axis start point).
#xrange - c(max(min(lower, na.rm = TRUE), clip[1]),
min(max(upper, na.rm = TRUE), clip[2]))
#new line
xrange - c(xlow,xhigh)
Now I am trying to modify the text font size for the elements in
labeltext and also trying to remove the variability in OR box size,
which is some function of the width of the lower and upper confidence
interval boundaries or the OR.
I am a reasonable programer (other than with R), but have not been
able to decipher what I should be modifying. Any suggestions will be
appreciated.
Thanks,
Gerard Smits
At 02:32 PM 9/7/2007, Gerard Smits wrote:
Hi R users,
I have a test dataframe (file1, shown below) for which I am trying
to create a flag for the first and last ID record (equivalent to SAS
first.id and last.id variables.
Dump of file1:
file1
id rx week dv1
1 1 11 1
2 1 12 1
3 1 13 2
4 2 11 3
5 2 12 4
6 2 13 1
7 3 11 2
8 3 12 3
9 3 13 4
10 4 11 2
11 4 12 6
12 4 13 5
13 5 21 7
14 5 22 8
15 5 23 5
16 6 21 2
17 6 22 4
18 6 23 6
19 7 21 7
20 7 22 8
21 8 21 9
22 9 21 4
23 9 22 5
I have written code that correctly assigns the first.id and last.id variabes:
require(Hmisc) #for Lags
#ascending order to define first dot
file1- file1[order(file1$id, file1$week),]
file1$first.id - (Lag(file1$id) != file1$id)
file1$first.id[1]-TRUE #force NA to TRUE
#descending order to define last dot
file1- file1[order(-file1$id,-file1$week),]
file1$last.id - (Lag(file1$id) != file1$id)
file1$last.id[1]-TRUE #force NA to TRUE
#resort to original order
file1- file1[order(file1$id,file1$week),]
I am now trying to get the above code to work as a function, and am
clearly doing something wrong:
first.last - function (df, idvar, sortvars1, sortvars2)
+ {
+ #sort in ascending order to define first dot
+ df- df[order(sortvars1),]
+ df$first.idvar - (Lag(df$idvar) != df$idvar)
+ #force first record NA to TRUE
+ df$first.idvar[1]-TRUE
+
+ #sort in descending order to define last dot
+ df- df[order(-sortvars2),]
+ df$last.idvar - (Lag(df$idvar) != df$idvar)
+ #force last record NA to TRUE
+ df$last.idvar[1]-TRUE
+
+ #resort to original order
+ df- df[order(sortvars1),]
+ }
Function call:
first.last(df=file1, idvar=file1$id,
sortvars1=c(file1$id,file1$week), sortvars2=c(-file1$id,-file1$week))
R Error:
Error in as.vector(x, mode) : invalid argument 'mode'
I am not sure about the passing of the sort strings. Perhaps this
is were things are off. Any help greatly appreciated.
Thanks,
Gerard
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and provide commented, minimal, self-contained, reproducible code.
[R] product of vector elements
All, I would like to find a simper method that I now have to find the product of all elements in a vector: #get product of vector elements: (1,2,3,4,5) vec.product - exp(sum(log(c(1,2,3,4,5 I have not found a vector product function, if one has been written. Thanks, Gerard [[alternative HTML version deleted]] __ R-help@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.
Re: [R] product of vector elements
Thanks for cumprod() and prod(). I should have tried the help.search first. Next time Gerard At 04:55 PM 1/12/2008, Gabor Grothendieck wrote: help.search(product) On Jan 12, 2008 7:38 PM, Gerard Smits [EMAIL PROTECTED] wrote: All, I would like to find a simper method that I now have to find the product of all elements in a vector: #get product of vector elements: (1,2,3,4,5) vec.product - exp(sum(log(c(1,2,3,4,5 I have not found a vector product function, if one has been written. Thanks, Gerard [[alternative HTML version deleted]] __ R-help@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. [[alternative HTML version deleted]] __ R-help@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.
[R] if statement problem
Hi All,
I have a small dataset named das (43 cases) in which I am trying to
create a binary outcome (1/0) based on the following code:
if (das$age65 das$bmi30) {das$danger-1} else das$danger-0
I am setting a flag called 'danger' to 1 of the subject is over 65
and has a BMI 30.
I find that my statement evaluates the first record in the data.frame
and then carries that assignment for all. I detected this as I played
around with the values and found that the T/F status of the first
record was always carried dowqn. I have gotten this to work with an
elseif construction, but would like to know what is happening here.
Thanks,
Gerard
Using: Windows Vista and R 2.61
Code and output:
das- sasxport.get(c:\\personal\\r\\das.xpt)
if (das$age65 das$bmi30) {das$danger-1} else das$danger-0
attach(das)
das
The following object(s) are masked from das ( position 3 ) :
age bmi day id male sex status
id age sex bmi status day male danger
1 33001 35 M 27.5 0 3651 0
2 33002 29 M 34.9 1 221 0
3 33003 41 F 23.6 0 3650 0
4 33004 55 F 27.0 0 3650 0
5 42001 37 M 39.0 0 3651 0
6 42002 53 M 26.6 1 1241 0
7 42003 46 F 45.4 1 2870 0
8 42004 35 F 36.2 0 3650 0
9 42005 38 F 24.6 0 3650 0
10 42006 58 F 28.0 0 3650 0
11 42007 27 M 25.0 0 3651 0
12 42008 65 F 24.6 0 3650 0
13 42009 25 F 28.0 0 3650 0
14 43001 66 M 27.8 0 3651 0
15 43002 57 F 34.0 0 3650 0
16 43003 45 F 38.1 0 3650 0
17 43004 33 F 53.3 1 620 0
18 43005 56 F 36.5 0 3650 0
19 43006 31 F 22.4 1 10 0
20 43007 53 F 32.2 1 210 0
21 55001 51 M 29.2 0 3651 0
22 55002 33 F 18.7 0 3650 0
23 55003 40 F 30.3 0 3650 0
24 55004 67 M 31.9 0 3651 0 - Problem case should
=1 for danger
25 55005 41 F 35.0 0 3650 0
26 55006 44 F 37.3 0 3650 0
27 55007 67 M 28.4 1 11 0
28 55008 65 F 28.8 0 3650 0
29 55009 76 M 18.8 1 2251 0
30 55010 75 F 21.1 1 390 0
31 63001 30 F 24.9 0 3650 0
32 63002 36 F 47.2 1 3770 0
33 63003 45 F 32.0 0 3650 0
34 63004 49 F 32.3 0 3650 0
35 63005 41 F 20.2 0 3650 0
36 63006 60 F 28.2 0 3650 0
37 63007 33 F 24.5 0 3650 0
38 63008 36 F 28.4 1 560 0
39 63009 31 F 22.1 0 3650 0
40 63010 77 M 26.6 1 91 0
41 63011 41 F 32.0 0 3650 0
42 63012 40 F 38.5 1 920 0
43 63013 27 M 20.6 0 3651 0
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] if statement problem
Thanks, but I tried the single ampersand, but got a warning msg with
the same lack of correct assignment:
if (das$age65 das$bmi30) {das$danger-1} else das$danger-0
Warning message:
In if (das$age 65 das$bmi 30) { :
the condition has length 1 and only the first element will be used
attach(das)
The following object(s) are masked from das ( position 3 ) :
age bmi day id male sex status
das
id age sex bmi status day male danger
1 33001 35 M 27.5 0 3651 0
2 33002 29 M 34.9 1 221 0
3 33003 41 F 23.6 0 3650 0
4 33004 55 F 27.0 0 3650 0
5 42001 37 M 39.0 0 3651 0
6 42002 53 M 26.6 1 1241 0
7 42003 46 F 45.4 1 2870 0
8 42004 35 F 36.2 0 3650 0
9 42005 38 F 24.6 0 3650 0
10 42006 58 F 28.0 0 3650 0
11 42007 27 M 25.0 0 3651 0
12 42008 65 F 24.6 0 3650 0
13 42009 25 F 28.0 0 3650 0
14 43001 66 M 27.8 0 3651 0
15 43002 57 F 34.0 0 3650 0
16 43003 45 F 38.1 0 3650 0
17 43004 33 F 53.3 1 620 0
18 43005 56 F 36.5 0 3650 0
19 43006 31 F 22.4 1 10 0
20 43007 53 F 32.2 1 210 0
21 55001 51 M 29.2 0 3651 0
22 55002 33 F 18.7 0 3650 0
23 55003 40 F 30.3 0 3650 0
24 55004 67 M 31.9 0 3651 0 -
25 55005 41 F 35.0 0 3650 0
26 55006 44 F 37.3 0 3650 0
27 55007 67 M 28.4 1 11 0
28 55008 65 F 28.8 0 3650 0
29 55009 76 M 18.8 1 2251 0
30 55010 75 F 21.1 1 390 0
31 63001 30 F 24.9 0 3650 0
32 63002 36 F 47.2 1 3770 0
33 63003 45 F 32.0 0 3650 0
34 63004 49 F 32.3 0 3650 0
35 63005 41 F 20.2 0 3650 0
36 63006 60 F 28.2 0 3650 0
37 63007 33 F 24.5 0 3650 0
38 63008 36 F 28.4 1 560 0
39 63009 31 F 22.1 0 3650 0
40 63010 77 M 26.6 1 91 0
41 63011 41 F 32.0 0 3650 0
42 63012 40 F 38.5 1 920 0
43 63013 27 M 20.6 0 3651 0
At 02:11 PM 1/1/2008, Christos Hatzis wrote:
You need to use '' instead of '':
A shorter version of your code using ifelse:
das$danger - with(das, ifelse(age65 bmi30, 1, 0))
HTH
-Christos
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Gerard Smits
Sent: Tuesday, January 01, 2008 5:04 PM
To: r-help@r-project.org
Subject: [R] if statement problem
Hi All,
I have a small dataset named das (43 cases) in which I am
trying to create a binary outcome (1/0) based on the following code:
if (das$age65 das$bmi30) {das$danger-1} else das$danger-0
I am setting a flag called 'danger' to 1 of the subject is
over 65 and has a BMI 30.
I find that my statement evaluates the first record in the
data.frame and then carries that assignment for all. I
detected this as I played around with the values and found
that the T/F status of the first record was always carried
dowqn. I have gotten this to work with an elseif
construction, but would like to know what is happening here.
Thanks,
Gerard
Using: Windows Vista and R 2.61
Code and output:
das- sasxport.get(c:\\personal\\r\\das.xpt)
if (das$age65 das$bmi30) {das$danger-1} else das$danger-0
attach(das)
das
The following object(s) are masked from das ( position 3 ) :
age bmi day id male sex status
id age sex bmi status day male danger
1 33001 35 M 27.5 0 3651 0
2 33002 29 M 34.9 1 221 0
3 33003 41 F 23.6 0 3650 0
4 33004 55 F 27.0 0 3650 0
5 42001 37 M 39.0 0 3651 0
6 42002 53 M 26.6 1 1241 0
7 42003 46 F 45.4 1 2870 0
8 42004 35 F 36.2 0 3650 0
9 42005 38 F 24.6 0 3650 0
10 42006 58 F 28.0 0 3650 0
11 42007 27 M 25.0 0 3651 0
12 42008 65 F 24.6 0 3650 0
13 42009 25 F 28.0 0 3650 0
14 43001 66 M 27.8 0 3651 0
15 43002 57 F 34.0 0 3650 0
16 43003 45 F 38.1 0 3650 0
17 43004 33 F 53.3 1 620 0
18 43005 56 F 36.5 0 3650 0
19 43006 31 F 22.4 1 10 0
20 43007 53 F 32.2 1 210 0
21 55001 51 M 29.2 0 3651 0
22 55002 33 F 18.7 0 3650 0
23 55003 40 F 30.3 0 3650 0
24 55004 67 M 31.9 0 3651 0
Re: [R] if statement problem
Hi Domenico,
I was incorrectly assuming it would use a vector of equal length to
my data. frame. Thanks for the clarification.
Also, thanks for the many alternate programming approaches provided by others.
Gerard
At 02:25 PM 1/1/2008, Domenico Vistocco wrote:
You should look for your answer using the help for the if statement (?if).
The cond argument should be a scalar (otherwise only the first
element is used).
?if
.
cond: A length-one logical vector that is not 'NA'. Conditions of
length greater than one are accepted with a warning, but only
the first element is used. Other types are coerced to
logical if possible, ignoring any class.
...
The ifelse statement check the condition for each element of your
input (returning
a value with the same shape).
domenico
Gerard Smits wrote:
Hi All,
I have a small dataset named das (43 cases) in which I am trying
to create a binary outcome (1/0) based on the following code:
if (das$age65 das$bmi30) {das$danger-1} else das$danger-0
I am setting a flag called 'danger' to 1 of the subject is over 65
and has a BMI 30.
I find that my statement evaluates the first record in the
data.frame and then carries that assignment for all. I detected
this as I played around with the values and found that the T/F
status of the first record was always carried dowqn. I have gotten
this to work with an elseif construction, but would like to know
what is happening here.
Thanks,
Gerard
Using: Windows Vista and R 2.61
Code and output:
das- sasxport.get(c:\\personal\\r\\das.xpt)
if (das$age65 das$bmi30) {das$danger-1} else das$danger-0
attach(das)
das
The following object(s) are masked from das ( position 3 ) :
age bmi day id male sex status
id age sex bmi status day male danger
1 33001 35 M 27.5 0 3651 0
2 33002 29 M 34.9 1 221 0
3 33003 41 F 23.6 0 3650 0
4 33004 55 F 27.0 0 3650 0
5 42001 37 M 39.0 0 3651 0
6 42002 53 M 26.6 1 1241 0
7 42003 46 F 45.4 1 2870 0
8 42004 35 F 36.2 0 3650 0
9 42005 38 F 24.6 0 3650 0
10 42006 58 F 28.0 0 3650 0
11 42007 27 M 25.0 0 3651 0
12 42008 65 F 24.6 0 3650 0
13 42009 25 F 28.0 0 3650 0
14 43001 66 M 27.8 0 3651 0
15 43002 57 F 34.0 0 3650 0
16 43003 45 F 38.1 0 3650 0
17 43004 33 F 53.3 1 620 0
18 43005 56 F 36.5 0 3650 0
19 43006 31 F 22.4 1 10 0
20 43007 53 F 32.2 1 210 0
21 55001 51 M 29.2 0 3651 0
22 55002 33 F 18.7 0 3650 0
23 55003 40 F 30.3 0 3650 0
24 55004 67 M 31.9 0 3651 0 - Problem case should
=1 for danger
25 55005 41 F 35.0 0 3650 0
26 55006 44 F 37.3 0 3650 0
27 55007 67 M 28.4 1 11 0
28 55008 65 F 28.8 0 3650 0
29 55009 76 M 18.8 1 2251 0
30 55010 75 F 21.1 1 390 0
31 63001 30 F 24.9 0 3650 0
32 63002 36 F 47.2 1 3770 0
33 63003 45 F 32.0 0 3650 0
34 63004 49 F 32.3 0 3650 0
35 63005 41 F 20.2 0 3650 0
36 63006 60 F 28.2 0 3650 0
37 63007 33 F 24.5 0 3650 0
38 63008 36 F 28.4 1 560 0
39 63009 31 F 22.1 0 3650 0
40 63010 77 M 26.6 1 91 0
41 63011 41 F 32.0 0 3650 0
42 63012 40 F 38.5 1 920 0
43 63013 27 M 20.6 0 3651 0
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and provide commented, minimal, self-contained, reproducible code.
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] applying math/stat functions to rows in data frame
Hi all, Thanks for the suggestions. I have not yet tried the apply () approach, but have tried to get the indexed version working, so far with limited success. I realize that a transpose, as suggested, would work, but want to avoid that for something simpler. To repeat, the task is to perform a function on N rows (all numeric) in a data frame. I can use rowMeans, rowSums, pmin, and pmax to successfully average, sum, and find the min/max var1-var4). But If I try to get the mean (var5) using, no doubt the incorrect syntax: mean(df[,c(4,5,6,7)], na.rm=T), I do not get the mean of 4 column. I am not sure what is being returned as a value. I summarize the code below with the output: any further suggestions are appreciated, Thanks Gerard Code with output: df - read.table(textConnection(id,workshop,gender,q1,q2,q3,q4 + 1,1,f,1,1,5,1 + 2,2,f,2,1,4,1 + 3,1,f,2,2,4,3 + 4,2,f,3,1, ,3 + 5,1,m,4,5,2,4 + 6,2,m,5,4,5,5 + 7,1,m,5,3,4,4 + 8,2,m,4,5,5,5), header=TRUE, sep=,) attach(df) df$var1-rowMeans(cbind(q1,q2,q3,q4),na.rm=T) df$var2-rowSums (cbind(q1,q2,q3,q4),na.rm=T) df$var3-pmin(q1,q2,q3,q4, na.rm=T) df$var4-pmax(q1,q2,q3,q4, na.rm=T) df$var5-mean(df[,c(4,5,6,7)],na.rm=T) #not doing what I want df$var6-sd (df[,c(4,5,6,5)],na.rm=T) #not doing what I want df$var7-min (df[,c(4,5,6,5)],na.rm=T) #not doing what I want df$var8-max (df[,c(4,5,6,5)],na.rm=T) #not doing what I want df output with problem vars underlined: id workshop gender q1 q2 q3 q4 var1 var2 var3 var4 var5 var6 var7 var8 1 11 f 1 1 5 1 2.00815 3.25 1.48804815 2 22 f 2 1 4 1 2.00814 2.75 1.75254915 3 31 f 2 2 4 3 2.75 1124 4.142857 1.06904515 4 42 f 3 1 NA 3 2.33713 3.25 1.75254915 5 51 m 4 5 2 4 3.75 1525 3.25 1.48804815 6 62 m 5 4 5 5 4.75 1945 2.75 1.75254915 7 71 m 5 3 4 4 4.00 1635 4.142857 1.06904515 8 82 m 4 5 5 5 4.75 1945 3.25 1.75254915 [[alternative HTML version deleted]] __ R-help@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.
[R] applying math/stat functions to rows in data frame
Hi All, There are a variety of functions that can be applied to a variable (column) in a data frame: mean, min, max, sd, range, IQR, etc. I am aware of only two that work on the rows, using q1-q3 as example variables: rowMeans(cbind(q1,q2,q3),na.rm=T) #mean of multiple variables rowSums (cbind(q1,q2,q3),na.rm=T) #sum of multiple variables Can the standard column functions (listed in the first sentence) be applied to rows, with the use of correct indexes to reference the columns of interest? Or, must these summary functions be programmed separately to work on a row? Thanks, Gerard [[alternative HTML version deleted]] __ R-help@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.
