On Aug 26, 2011, at 11:31 AM, heverkuhn wrote:
No attachment. No code.
I apology for that. I provided below the code and the vectors,
Whatever that mean. Perhaps it means "below"?
Yes, below, I mean out of the axis...
And I think, that as you suggested, the xpd is what I was looking for.
I need something like this
http://www.springerimages.com/Images/Biomedicine/1-10.1385_0-89603-217-5_315-2
http://www.springerimages.com/Images/Biomedicine/1-10.1385_0-89603-217-5_315-2
but with the vertical bars below but with just the axis of the
cumulative
responses.
That looks like what I have seen called a "rug plot" and I think there
is coverage of such plotting in existing graphics packages. Try this:
?rug # and be sure to run the example
--
David
Thanks guys for your time and your patience.
Claudio
here it is the code:
plot( activeT,activeR, pch="", type="s",
ylim=c(-20, tail(activeR,1)),
xlim=c(0, breakpT),
main=c("Subj", DATA[7]),
ylab="Cumulative Responses",
xlab="Time (sec)",
font.lab=2
)
segments(reinfT$activeT, reinfR$activeR, reinfT$activeT-150,
reinfR$activeR+3, lwd=2)
points(inactT, inactR, pch="|")
segments( 0, -10, max(activeT), -10)
segments(breakpT+50,breakpR-2,breakpT-50,breakpR+2, lwd=4)
and here the vectors:
activeT
[1] 2.6 34.1 37.6 45.9 46.6 53.2 93.5
116.3 172.1
[10] 616.7 651.3 711.4 722.5 725.4 772.8 796.4
863.0 880.3
[19] 918.6 945.3 1013.2 1066.9 1132.6 1148.6 1150.8
1171.1 1172.6
[28] 1345.0 1346.5 1596.5 1597.9 1632.4 1682.7 1695.9
1780.9 1904.8
[37] 2267.3 2326.8 3259.6 3314.8 4671.7 10671.7
activeR
[1] 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21
22 23 24
25
[26] 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 41
breakpT
[1] 10671.7
breakpR
[1] 41
reinfT
activeT
5 46.6
12 711.4
22 1066.9
36 1904.8
reinfR
activeR
5 5
12 12
22 22
36 36
*
inactT
[1] 31.6 65.6 637.8 809.8 809.9 1075.5 1144.0 1546.3 2247.4
2260.0
[11] 2793.6 2794.4 3075.7 3250.3 3250.6 3252.2 3308.2 5143.6
inactR
[1] "-10" "-10" "-10" "-10" "-10" "-10" "-10" "-10" "-10" "-10"
"-10" "-10"
[13] "-10" "-10" "-10" "-10" "-10" "-10"
David Winsemius wrote:
On Aug 25, 2011, at 2:06 PM, Claudio Zanettini wrote:
---------- Forwarded message ----------
From: Claudio Zanettini <claudio.zanett...@gmail.com>
Date: 2011/8/25
Subject: Re: [R] Segment out of the Graph
To: David Winsemius <dwinsem...@comcast.net>
Thanks David and Michael,
In attachment there is one of the graph.
No attachment. No code.
the line below the graph is not related to the y label.
I draw it at y= -10 just becouse it was the only way I know for draw
a line
parallel
to the x axis.
But it shoud be "out" of the xy axis.
Whatever that mean. Perhaps it means "below"?
It rappresents another variable related to the one shown in the
graph...
No graph.
Sorry for not being very clear :)
Please read the Posting Guide. You probably want the segments
function
with xpd=NA or xpd =TRUE
plot(1:2,c(1, 500), ylim=c(0,1000), xlim=c(0,3))
segments(0,-10, 3,-10, xpd=TRUE, col="red")
?par
?segments
Claudio
2011/8/25 David Winsemius <dwinsem...@comcast.net>
On Aug 25, 2011, at 1:30 PM, Claudio Zanettini wrote:
I tried setting ylim=(0, 1000)
but the segment that I have draw is at y=-10
so if I set the y origin to 0 I don t have the segment,
if a l leave it at -10 I have the segment but the axis start from
-10.
I would like to have both.
So to have a graph with ylim=( 0, 1000)
and under it a segment parallel to the x axis at -10.
If you really want that then you will need to look at the xpd
parameter for
par.
I am sorry it is contorted thing :)
Claudio
If instead you want to plot at y= -10 without changing xpd, then
set
axes=FALSE in the plot command and then construct your x and y axes
separately to your specifications. Any more specific comments will
require
that you present code (as well as a better description) that
constructs an
example.
--
David.
2011/8/25 R. Michael Weylandt <michael.weyla...@gmail.com>
"lim" is not the argument: "ylim" is.
You put in a vector of length 2 comprising the min and max y you
wish:
consider this:
x = -5:5; y = x^2; z = rep(-5,11);
layout(1:2)
plot(x,y,type="b"); lines(x,z,col=2)
plot(x,y,ylim = c(-8,max(y)+3),type="b"); lines(x,z,col=2)
For your work, you'd need ylim = c(0, 1.03*max(y)) or something
similar.
Michael Weylandt
On Thu, Aug 25, 2011 at 1:12 PM, Claudio Zanettini <
claudio.zanett...@gmail.com> wrote:
Yes I tried but if I set the lim to 0 then it will not displayed
the
line
that is at -10, right?
2011/8/25 R. Michael Weylandt
<michael.weyla...@gmail.com>
Look at ylim, as an optional argument to plot.
Michael
On Thu, Aug 25, 2011 at 1:07 PM, Claudio Zanettini <
claudio.zanett...@gmail.com> wrote:
Hello everyone,
I have a graph and a segment parallel to the x axis at
y=-10, x=0,
and
bars on it.
Now the question is,
Is there a way to leave the segment there but let the graph
axis
start
from
the origin?
In this way the segment will be out of the graph
Thanks
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David Winsemius, MD
West Hartford, CT
______________________________________________
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