Vertical bwplot and stripplot

Hi,

I'd like to change the default orientation of bwplot() and stripplot()
so the plots are displayed vertically.  Passing horizontal=FALSE into
stripplot in the simple code below doesn't seem to be the answer.

library(lattice);
x <- rnorm(100);
y <- as.factor(sapply(1:100, function(k) sample(c("A","B","C"), 1,
prob=c(1/2, 1/3, 1/6))));
my.df <- data.frame(x=x, y=y);
stripplot(~x | y, data=my.df, as.table=TRUE, layout=c(1,3), hor);

On Apr 23, 2011, at 9:26 AM, David Neu wrote:

Hi,

I'd like to change the default orientation of bwplot() and stripplot()
so the plots are displayed vertically. ?Passing horizontal=FALSE into
stripplot in the simple code below doesn't seem to be the answer.

library(lattice);
x <- rnorm(100);
y <- as.factor(sapply(1:100, function(k) sample(c("A","B","C"), 1,
prob=c(1/2, 1/3, 1/6))));
my.df <- data.frame(x=x, y=y);
stripplot(~x | y, data=my.df, as.table=TRUE, layout=c(1,3), hor);

A) hor is not defined
B) it doesn't make sense to me to have the continuous variable as the
independent variable here, despite if being named `x`.

Try:
stripplot(x~y , data=my.df, as.table=TRUE, layout=c(1,3), horizontal=FALSE);

(I didn't recognize the as.table argument, but experimentation seems to
produce a top-down order to the plots.)

--
David Winsemius, MD
West Hartford, CT

A) hor is not defined

B) it doesn't make sense to me to have the continuous variable as the
independent variable here, despite if being named `x`.

Try:
stripplot(x~y , data=my.df, as.table=TRUE, layout=c(1,3), horizontal=FALSE);

On Sat, Apr 23, 2011 at 9:47 AM, David Winsemius <dwinsemius at comcast.net 

wrote:

On Apr 23, 2011, at 9:26 AM, David Neu wrote:

Hi,

I'd like to change the default orientation of bwplot() and  
stripplot()
so the plots are displayed vertically.  Passing horizontal=FALSE  
into
stripplot in the simple code below doesn't seem to be the answer.

library(lattice);
x <- rnorm(100);
y <- as.factor(sapply(1:100, function(k) sample(c("A","B","C"), 1,
prob=c(1/2, 1/3, 1/6))));
my.df <- data.frame(x=x, y=y);
stripplot(~x | y, data=my.df, as.table=TRUE, layout=c(1,3), hor);

A) hor is not defined
B) it doesn't make sense to me to have the continuous variable as the
independent variable here, despite if being named `x`.

Try:
stripplot(x~y , data=my.df, as.table=TRUE, layout=c(1,3),  
horizontal=FALSE);

(I didn't recognize the as.table argument, but experimentation  
seems to
produce a top-down order to the plots.)

--
David Winsemius, MD
West Hartford, CT

Many thanks for your reply!

A) hor is not defined

Ugggh, cut and paste mistake.

B) it doesn't make sense to me to have the continuous variable as the
independent variable here, despite if being named `x`.

I have data from related experiments in that involves two variables
conditioned on a third.  This data is displayed in an xyplot.  The
reason I'm trying to get the vertical orientation in the stripplot is
that in some experiments the variable plotted on the horizontal axis
is invariant and in these cases for consistency I'd like the variable
that is plotted on the vertical axis to continue to appear vertically.

For example in non-lattice graphics the following works:
stripchart(rnorm(100), vert=TRUE).

Try:
stripplot(x~y , data=my.df, as.table=TRUE, layout=c(1,3),  
horizontal=FALSE);

Yes, that's moving closer, but the strips containing the conditioning
info are missing.

On Apr 23, 2011, at 10:13 AM, David Neu wrote:

On Sat, Apr 23, 2011 at 9:47 AM, David Winsemius <dwinsemius at comcast.net>
wrote:

On Apr 23, 2011, at 9:26 AM, David Neu wrote:

Hi,

I'd like to change the default orientation of bwplot() and stripplot()
so the plots are displayed vertically. ?Passing horizontal=FALSE into
stripplot in the simple code below doesn't seem to be the answer.

library(lattice);
x <- rnorm(100);
y <- as.factor(sapply(1:100, function(k) sample(c("A","B","C"), 1,
prob=c(1/2, 1/3, 1/6))));
my.df <- data.frame(x=x, y=y);
stripplot(~x | y, data=my.df, as.table=TRUE, layout=c(1,3), hor);

A) hor is not defined
B) it doesn't make sense to me to have the continuous variable as the
independent variable here, despite if being named `x`.

Try:
stripplot(x~y , data=my.df, as.table=TRUE, layout=c(1,3),
horizontal=FALSE);

(I didn't recognize the as.table argument, but experimentation seems to
produce a top-down order to the plots.)

--
David Winsemius, MD
West Hartford, CT

Many thanks for your reply!

A) hor is not defined

Ugggh, cut and paste mistake.

B) it doesn't make sense to me to have the continuous variable as the
independent variable here, despite if being named `x`.

I have data from related experiments in that involves two variables
conditioned on a third. ?This data is displayed in an xyplot. ?The
reason I'm trying to get the vertical orientation in the stripplot is
that in some experiments the variable plotted on the horizontal axis
is invariant and in these cases for consistency I'd like the variable
that is plotted on the vertical axis to continue to appear vertically.

For example in non-lattice graphics the following works:
stripchart(rnorm(100), vert=TRUE).

Try:
stripplot(x~y , data=my.df, as.table=TRUE, layout=c(1,3),
horizontal=FALSE);

Yes, that's moving closer, but the strips containing the conditioning
info are missing.

You only offered two variables, so It's unclear what sort of "conditioning
info" you imagine. Your stripchart() ?example that you say "works" told me
nothing.

Unless perhaps you are trying for:
stripplot( x~1 | y , data=my.df, ?layout=c(1,3), horizontal=FALSE)



--

David Winsemius, MD
West Hartford, CT

You only offered two variables, so It's unclear what sort of "conditioning
info" you imagine.

Your stripchart()  example that you say "works" told me
nothing.

Unless perhaps you are trying for:
stripplot( x~1 | y , data=my.df,  layout=c(1,3), horizontal=FALSE)

On Sat, Apr 23, 2011 at 9:47 AM, David Winsemius<dwinsemius at comcast.net>  wrote:

On Apr 23, 2011, at 9:26 AM, David Neu wrote:

Hi,

I'd like to change the default orientation of bwplot() and stripplot()
so the plots are displayed vertically.  Passing horizontal=FALSE into
stripplot in the simple code below doesn't seem to be the answer.

library(lattice);
x<- rnorm(100);
y<- as.factor(sapply(1:100, function(k) sample(c("A","B","C"), 1,
prob=c(1/2, 1/3, 1/6))));
my.df<- data.frame(x=x, y=y);
stripplot(~x | y, data=my.df, as.table=TRUE, layout=c(1,3), hor);

A) hor is not defined
B) it doesn't make sense to me to have the continuous variable as the
independent variable here, despite if being named `x`.

Try:
stripplot(x~y , data=my.df, as.table=TRUE, layout=c(1,3), horizontal=FALSE);

(I didn't recognize the as.table argument, but experimentation seems to
produce a top-down order to the plots.)

--
David Winsemius, MD
West Hartford, CT

Many thanks for your reply!

A) hor is not defined

Ugggh, cut and paste mistake.

B) it doesn't make sense to me to have the continuous variable as the
independent variable here, despite if being named `x`.

I have data from related experiments in that involves two variables
conditioned on a third.  This data is displayed in an xyplot.  The
reason I'm trying to get the vertical orientation in the stripplot is
that in some experiments the variable plotted on the horizontal axis
is invariant and in these cases for consistency I'd like the variable
that is plotted on the vertical axis to continue to appear vertically.

For example in non-lattice graphics the following works:
stripchart(rnorm(100), vert=TRUE).

Try:
stripplot(x~y , data=my.df, as.table=TRUE, layout=c(1,3), horizontal=FALSE);

Yes, that's moving closer, but the strips containing the conditioning
info are missing.

On 2011-04-23 07:13, David Neu wrote:

On Sat, Apr 23, 2011 at 9:47 AM, David Winsemius<dwinsemius at comcast.net>
?wrote:

On Apr 23, 2011, at 9:26 AM, David Neu wrote:

Hi,

I'd like to change the default orientation of bwplot() and stripplot()
so the plots are displayed vertically. ?Passing horizontal=FALSE into
stripplot in the simple code below doesn't seem to be the answer.

library(lattice);
x<- rnorm(100);
y<- as.factor(sapply(1:100, function(k) sample(c("A","B","C"), 1,
prob=c(1/2, 1/3, 1/6))));
my.df<- data.frame(x=x, y=y);
stripplot(~x | y, data=my.df, as.table=TRUE, layout=c(1,3), hor);

A) hor is not defined
B) it doesn't make sense to me to have the continuous variable as the
independent variable here, despite if being named `x`.

Try:
stripplot(x~y , data=my.df, as.table=TRUE, layout=c(1,3),
horizontal=FALSE);

(I didn't recognize the as.table argument, but experimentation seems to
produce a top-down order to the plots.)

--
David Winsemius, MD
West Hartford, CT

Many thanks for your reply!

A) hor is not defined

Ugggh, cut and paste mistake.

B) it doesn't make sense to me to have the continuous variable as the
independent variable here, despite if being named `x`.

I have data from related experiments in that involves two variables
conditioned on a third. ?This data is displayed in an xyplot. ?The
reason I'm trying to get the vertical orientation in the stripplot is
that in some experiments the variable plotted on the horizontal axis
is invariant and in these cases for consistency I'd like the variable
that is plotted on the vertical axis to continue to appear vertically.

For example in non-lattice graphics the following works:
stripchart(rnorm(100), vert=TRUE).

Try:
stripplot(x~y , data=my.df, as.table=TRUE, layout=c(1,3),
horizontal=FALSE);

Yes, that's moving closer, but the strips containing the conditioning
info are missing.

You can define a 'phantom' single-level factor

?my.df$fac <- rep("", 100)
?stripplot(x ~ fac | y, data = my.df, layout = c(1, 3))

and I'd consider 'jitter'.

BTW, your method of generating 'y' seems overly complicated:

?y <- sample(c("A","B","C"), 100,
? ? ? ? ? ? ?replace=TRUE,
? ? ? ? ? ? ?prob=c(1/2, 1/3, 1/6))

Peter Ehlers

On Sat, Apr 23, 2011 at 10:37 AM, Peter Ehlers<ehlers at ucalgary.ca>  wrote:

On 2011-04-23 07:13, David Neu wrote:

On Sat, Apr 23, 2011 at 9:47 AM, David Winsemius<dwinsemius at comcast.net>
  wrote:

On Apr 23, 2011, at 9:26 AM, David Neu wrote:

Hi,

I'd like to change the default orientation of bwplot() and stripplot()
so the plots are displayed vertically.  Passing horizontal=FALSE into
stripplot in the simple code below doesn't seem to be the answer.

library(lattice);
x<- rnorm(100);
y<- as.factor(sapply(1:100, function(k) sample(c("A","B","C"), 1,
prob=c(1/2, 1/3, 1/6))));
my.df<- data.frame(x=x, y=y);
stripplot(~x | y, data=my.df, as.table=TRUE, layout=c(1,3), hor);

A) hor is not defined
B) it doesn't make sense to me to have the continuous variable as the
independent variable here, despite if being named `x`.

Try:
stripplot(x~y , data=my.df, as.table=TRUE, layout=c(1,3),
horizontal=FALSE);

(I didn't recognize the as.table argument, but experimentation seems to
produce a top-down order to the plots.)

--
David Winsemius, MD
West Hartford, CT

Many thanks for your reply!

A) hor is not defined

Ugggh, cut and paste mistake.

B) it doesn't make sense to me to have the continuous variable as the
independent variable here, despite if being named `x`.

I have data from related experiments in that involves two variables
conditioned on a third.  This data is displayed in an xyplot.  The
reason I'm trying to get the vertical orientation in the stripplot is
that in some experiments the variable plotted on the horizontal axis
is invariant and in these cases for consistency I'd like the variable
that is plotted on the vertical axis to continue to appear vertically.

For example in non-lattice graphics the following works:
stripchart(rnorm(100), vert=TRUE).

Try:
stripplot(x~y , data=my.df, as.table=TRUE, layout=c(1,3),
horizontal=FALSE);

Yes, that's moving closer, but the strips containing the conditioning
info are missing.

You can define a 'phantom' single-level factor

  my.df$fac<- rep("", 100)
  stripplot(x ~ fac | y, data = my.df, layout = c(1, 3))

and I'd consider 'jitter'.

BTW, your method of generating 'y' seems overly complicated:

  y<- sample(c("A","B","C"), 100,
              replace=TRUE,
              prob=c(1/2, 1/3, 1/6))

Peter Ehlers

Ahh, that's nice.

BTW, for my understanding, could you please explain why you suggested
the use of 'jitter'?  I'm thinking it's to aid in the visualization.