Change
geom_point(aes(y = tmax_mean, color = "blue"))
to
geom_point(aes(y = tmax_mean), color = "blue")
if you want blue points.
aes(color = ) does not set the color of the points.
aes(color = ) takes a column (best if it is a factor) and uses that for
different colors.
/Martin
On Tue, Sep
ncludes both min and max
> > > thresholds.
> > >
> > > All I have done is provide an example where min and max could have
> > > a real world use. I use max(temp) over some interval and then
> > > update an accumulated thermal units variable based on the outcome.
t the date (say Day). Group_by the day and apply a max
function to the grouped data. Then plot the result.
Tim
-Original Message-
From: Kevin Zembower
Sent: Wednesday, September 13, 2023 3:26 PM
To: Ebert,Timothy Aaron ; Richard O'Keefe
Cc: r-help@r-project.org
Subject: Re: [R] Hel
rom: Kevin Zembower
> Sent: Wednesday, September 13, 2023 2:05 PM
> To: Ebert,Timothy Aaron ; Richard O'Keefe
>
> Cc: r-help@r-project.org
> Subject: Re: [R] Help with plotting and date-times for climate data
>
> [External Email]
>
> Well, I looked for this, on bo
; Richard O'Keefe
Cc: r-help@r-project.org
Subject: Re: [R] Help with plotting and date-times for climate data
[External Email]
Well, I looked for this, on both the NWS and WeatherUnderground, but couldn't
find what I was looking for. Didn't check Weather.com, but if you can find
depends on how far away the trip will be.
>
> -Original Message-
> From: Kevin Zembower
> Sent: Wednesday, September 13, 2023 1:22 PM
> To: Richard O'Keefe ; Ebert,Timothy Aaron
>
> Cc: r-help@r-project.org
> Subject: Re: [R] Help with plotting and date-times
: Wednesday, September 13, 2023 1:22 PM
To: Richard O'Keefe ; Ebert,Timothy Aaron
Cc: r-help@r-project.org
Subject: Re: [R] Help with plotting and date-times for climate data
[External Email]
Tim, Richard, y'all are reading too much into this. I believe that TMAX is the
high temperature
Rui, thanks so much for your clear explanation, solution to my problem,
and additional help with making the graph come out exactly as I was
hoping. I learned a lot from your solution. Thanks, again, for your
help.
-Kevin
On Tue, 2023-09-12 at 23:06 +0100, Rui Barradas wrote:
> Às 21:50 de 12/09/2
l is not evident in the original request.
> >
> > Tim
> >
> > -Original Message-
> > From: R-help On Behalf Of Richard
> > O'Keefe
> > Sent: Wednesday, September 13, 2023 9:58 AM
> > To: Kevin Zembower
> > Cc: r-help@r-project.org
&g
;
> Tim
>
> -Original Message-
> From: R-help On Behalf Of Richard O'Keefe
> Sent: Wednesday, September 13, 2023 9:58 AM
> To: Kevin Zembower
> Cc: r-help@r-project.org
> Subject: Re: [R] Help with plotting and date-times for climate data
>
> [External Ema
On Behalf Of Richard O'Keefe
Sent: Wednesday, September 13, 2023 9:58 AM
To: Kevin Zembower
Cc: r-help@r-project.org
Subject: Re: [R] Help with plotting and date-times for climate data
[External Email]
Off-topic, but what is a "mean temperature max"
and what good would it do you
Off-topic, but what is a "mean temperature max"
and what good would it do you to know you if you did?
I've been looking at a lot of weather station data
and for no question I've ever had (except "would the
newspapers get excited about this") was "max" (or min)
the answer. Considering the way that
Às 21:50 de 12/09/2023, Kevin Zembower via R-help escreveu:
Hello,
I'm trying to calculate the mean temperature max from a file of climate
date, and plot it over a range of days in the year. I've downloaded the
data, and cleaned it up the way I think it should be. However, when I
plot it, the ge
Hello,
I'm trying to calculate the mean temperature max from a file of climate
date, and plot it over a range of days in the year. I've downloaded the
data, and cleaned it up the way I think it should be. However, when I
plot it, the geom_smooth line doesn't show up. I think that's because
my x ax
Dear R helpers,
I am trying to create a xyplot similar to this one:
https://stat.ethz.ch/pipermail/r-help/2008-June/164968.html
I can plot the data correctly when I only have one Y. However, I need to add
another Y (Y2) to the plot and each Y must have their own error bars and also
different s
Amit Patel writes:
>>plot(BHPLS1, "loadings", comps = 1:2, legendpos = "topleft", labels =
>>"numbers",
>>xlab = "nm")
>
> Error in loadingplot.default(x, ...) :
> Could not convert variable names to numbers.
>
>
> str(BHPLS1_Loadings)
> loadings [1:8892, 1:60] -0.00717 0.00414 0.02611 0.0
Hi
I am attempting to do a loadings plot from a plsr object. I have managed to do
this using the gasoline data that comes with the pls package. However when I
conduct this on my dataset i get the following error message.
>plot(BHPLS1, "loadings", comps = 1:2, legendpos = "topleft", labels =
On 2011-03-28 09:33, Pam Allen wrote:
Thank you Jim and David for your help.
The 'levels' call is not a misdirection, in my actual dataset it is
necessary because the flows aren't symmetrical. So while your solution is
quite elegant David, it doesn't apply to my actual data, just the example.
Thank you Jim and David for your help.
The 'levels' call is not a misdirection, in my actual dataset it is
necessary because the flows aren't symmetrical. So while your solution is
quite elegant David, it doesn't apply to my actual data, just the example.
Too bad, it's quite nice!
I do think
On 03/26/2011 07:19 AM, Pam Allen wrote:
Hello again,
I wrote an example that better represents my data, since the coloured points
are actually consecutive, but with variable lengths:
date=as.Date(c(1:300))
flow=sin(2*pi/53*c(1:300))
levels=c(rep(c("high","med","low"),100))
data=cbind.data.fram
Hello again,
I wrote an example that better represents my data, since the coloured points
are actually consecutive, but with variable lengths:
date=as.Date(c(1:300))
flow=sin(2*pi/53*c(1:300))
levels=c(rep(c("high","med","low"),100))
data=cbind.data.frame(date, flow, levels)
library(zoo)
z <- z
On Mar 25, 2011, at 3:23 PM, Pam Allen wrote:
Hello Baptiste and others,
I tried your example with my dataset, and for a few days I thought
it worked
for me. But I realized yesterday that the result wasn't quite what
I hoped
for. In my actual data the flows aren't perfectly sinusoidal, a
Hello Baptiste and others,
I tried your example with my dataset, and for a few days I thought it worked
for me. But I realized yesterday that the result wasn't quite what I hoped
for. In my actual data the flows aren't perfectly sinusoidal, and I used a
series of ifelse queries to code the flow
On Mar 17, 2011, at 6:33 PM, Pamela Allen wrote:
Hi All,
I'm trying to plot data that is a time series of flows that are
associated
with a specific level, and I would like each level to represent a
colour
in a line plot. Here is some data that approximates what I'm using:
date=c(1:300)
Indeed, I forgot about the segments function.
with(d,plot(date,flow,type="n"))
with(d,segments(start,start.y,end,end.y,col=colour))
> Hi,
>
> because each colour is defined on non-consecutive points, you'll
> probably need to cut the intervals to define segments around each
> point. One approa
Hi,
because each colour is defined on non-consecutive points, you'll
probably need to cut the intervals to define segments around each
point. One approach might be the following,
d = transform(data, start = date - c(0, diff(date)/2), end = date +
c(0, diff(date)/2) )
d$start.y = approx(d$date, d
Hi!
Not an elegant solution, but seems to work:
date <- c(1:300)
flow <- sin(2*pi/53*c(1:300))
levels <- factor(rep(c("high","med","low"),100))
data <- cbind.data.frame(date, flow, levels)
colours <- as.numeric(levels)+1
# interpolate
resolution <- 0.001
appres <- approx(date,flow,seq(min(date
Hi All,
I'm trying to plot data that is a time series of flows that are associated
with a specific level, and I would like each level to represent a colour
in a line plot. Here is some data that approximates what I'm using:
date=c(1:300)
flow=sin(2*pi/53*c(1:300))
levels=c(rep(c("
Dear all,
I recently started using the kohonen package for my thesis project. I
have a very simple question which I cannot figure out by myself:
When I execute the following example code, from the paper of Wehrens
and Buydens (http://www.jstatsoft.org/v21/i05/paper):
R> library("kohonen")
Loadin
Hi:
Does this work for you?
df <- read.table(textConnection("
pool age age2 density body_length body_length2
11 7.4 11.3 25 0.887 1.322550
22 7.4 11.3 100 0.921 1.152000
33 7.4 11.3 250 0.896 1.136300
44 7.4 11.3 75 0.723
Hello everyone,
I'm trying to plot the initial and final size of some tadpoles and look at
their growth rates (slope of line). They are divided into 6 densities, and
so far I have plotted the initial and final sizes (these are means for a
pool), and color coded for the 6 densities, but I want a l
Hi Edward,
On Sep 11, 2009, at 1:47 PM, Edward Chen wrote:
HI all,
raw_urine =
read.table("Z:\\bruce.9.3.09.sample.stability.analysis\\urine\\mz.spot.sam.dat.new
",
header = TRUE )
pvalue =
read.table("Z:\\bruce.9.3.09.sample.stability.analysis\\urine\\all.urine.features.t.test.result
",
he
HI all,
raw_urine =
read.table("Z:\\bruce.9.3.09.sample.stability.analysis\\urine\\mz.spot.sam.dat.new",
header = TRUE )
pvalue =
read.table("Z:\\bruce.9.3.09.sample.stability.analysis\\urine\\all.urine.features.t.test.result",
header = TRUE )
library(compositions)
p = function(a,b){
y = pvalue[,a]
On Sun, 26 Apr 2009, Uwe Ligges wrote:
Works for me with
library(MASS)
plot(lda(Species~., data=iris))
hence you may want to profide the data to enable us to reproduce your
problem...
He is trying to plot the results from a cross-validation. As the help
page clearly states, that is a list
Works for me with
library(MASS)
plot(lda(Species~., data=iris))
hence you may want to profide the data to enable us to reproduce your
problem...
Uwe Ligges
pgseye wrote:
Hi,
I've performed an lda and obtained a classification table for some of my
data:
efa.dfa<-lda(groups~.,efa.scores.
Hi,
I've performed an lda and obtained a classification table for some of my
data:
> efa.dfa<-lda(groups~.,efa.scores.8,CV=T)
> str(efa.dfa)
List of 5
$ class: Factor w/ 2 levels "1","2": 1 2 1 2 1 1 2 2 1 2 ...
$ posterior: num [1:160, 1:2] 0.99083 0.00852 0.93983 0.23186 0.85931 ...
..
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