Re: [R] shading in overlap between two ranges
This worked perfectly. An example graphic is located here:
ftp://ftp.sfwmd.gov/pub/ggraves/ribbon.bmp
-Original Message-
From: Dennis Murphy [mailto:djmu...@gmail.com]
Sent: Thursday, June 02, 2011 12:11 PM
To: Graves, Gregory
Cc: r-help@r-project.org; Kemp, Susan K SAJ; patrick_pi...@fws.gov
Subject: Re: [R] shading in overlap between two ranges
Hi:
Here's one approach using geom_ribbon() in ggplot2 - the 'overlap' is
the change in color where the two ribbons intersect. Using your
example data with the same names and the 'one.month' variable removed,
library(ggplot2)
ggplot() +
geom_ribbon(data = target, aes(x = i.value, ymin = X25, ymax = X75,
fill = 'Target'), alpha = 0.4) +
geom_ribbon(data = observed, aes(x = i.value, ymin = X25, ymax = X75,
fill = 'Observed'), alpha = 0.4) +
scale_fill_manual(, c('Target' = 'blue', 'Observed' = 'orange')) +
opts(legend.position = c(0.88, 0.85),
legend.background = theme_rect(colour = 'transparent'),
legend.text = theme_text(size = 12)) +
labs(x = 'Month', y = 'Value')
There is a separate geom_ribbon() for each of target and observed. A
factor variable for fill color is generated on the fly with colors
specified in scale_fill_manual(). The opts() reposition the legend
inside the graphics region (the values represent proportions of the
total graphics area in each direction), make the legend background
transparent and slightly increase the size of the legend labels
(default size = 10 in theme_text).
Alpha transparency is used so that the overlap creates a blend of the
two colors; without it, one overwrites the other.
HTH,
Dennis
On Thu, Jun 2, 2011 at 8:04 AM, Graves, Gregory ggra...@sfwmd.gov wrote:
I have 2 datafiles 'target' and 'observed' as shown below (I will gladly
email these 2 small files to whomever). X25. And X75. Indicate the
value of 25th and 75th-percentile of the target ('what should be') and
the observed ('what is'). The i.value is simply the month.
target
X i.value X25. X75.
1 one.month 1 10.845225 17.87237
2 one.month 2 12.235813 19.74490
3 one.month 3 14.611749 23.44810
4 one.month 4 17.529332 28.09647
5 one.month 5 19.458738 30.56936
6 one.month 6 15.264505 28.29333
7 one.month 7 12.370369 23.35455
8 one.month 8 12.471224 21.82794
9 one.month 9 9.716685 17.28762
10 one.month 10 6.470568 12.49830
11 one.month 11 6.180560 14.24961
12 one.month 12 9.673738 15.79208
observed
X i.value X25. X75.
1 one.month 1 19.81000 27.63500
2 one.month 2 23.64062 30.09125
3 one.month 3 26.04865 35.99104
4 one.month 4 32.02625 41.50958
5 one.month 5 34.74479 47.75958
6 one.month 6 37.48885 46.56448
7 one.month 7 30.06740 40.10146
8 one.month 8 26.14917 39.49458
9 one.month 9 14.12521 32.39406
10 one.month 10 11.04125 23.55479
11 one.month 11 13.14917 23.56833
12 one.month 12 17.17938 27.02458
The following plots 4 lines on one graph. The area between the two red
lines represents the target 'zone', and the area between the two black
lines is the observed 'zone'.
with(target, plot(X25.~i.value,ylim=c(0,55),type='l',col='red'))
par(new=T)
with(target, plot(X75.~i.value,ylim=c(0,55),type='l',col='red'))
par(new=T)
with(observed, plot(X25.~i.value,ylim=c(0,55),type='l'))
par(new=T)
with(observed, plot(X75.~i.value,ylim=c(0,55),type='l'))
par(new=F)
Ideally, the target and the observed should overlap in every month -
they don't. The desire is to visually accentuate the amount of overlap
by shading in the area where these two zones overlap. How would you
do that? Note, that in some of these characterizations, the overlap
wanders in and out [I already have routines that calculate the percent
of overlap, but I have been requested to find a way to shade the
overlap.]
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[R] shading in overlap between two ranges
I have 2 datafiles 'target' and 'observed' as shown below (I will gladly
email these 2 small files to whomever). X25. And X75. Indicate the
value of 25th and 75th-percentile of the target ('what should be') and
the observed ('what is'). The i.value is simply the month.
target
Xi.valueX25. X75.
1 one.month 1 10.845225 17.87237
2 one.month 2 12.235813 19.74490
3 one.month 3 14.611749 23.44810
4 one.month 4 17.529332 28.09647
5 one.month 5 19.458738 30.56936
6 one.month 6 15.264505 28.29333
7 one.month 7 12.370369 23.35455
8 one.month 8 12.471224 21.82794
9 one.month 9 9.716685 17.28762
10 one.month 10 6.470568 12.49830
11 one.month 11 6.180560 14.24961
12 one.month 12 9.673738 15.79208
observed
X i.value X25. X75.
1 one.month 1 19.81000 27.63500
2 one.month 2 23.64062 30.09125
3 one.month 3 26.04865 35.99104
4 one.month 4 32.02625 41.50958
5 one.month 5 34.74479 47.75958
6 one.month 6 37.48885 46.56448
7 one.month 7 30.06740 40.10146
8 one.month 8 26.14917 39.49458
9 one.month 9 14.12521 32.39406
10 one.month 10 11.04125 23.55479
11 one.month 11 13.14917 23.56833
12 one.month 12 17.17938 27.02458
The following plots 4 lines on one graph. The area between the two red
lines represents the target 'zone', and the area between the two black
lines is the observed 'zone'.
with(target, plot(X25.~i.value,ylim=c(0,55),type='l',col='red'))
par(new=T)
with(target, plot(X75.~i.value,ylim=c(0,55),type='l',col='red'))
par(new=T)
with(observed, plot(X25.~i.value,ylim=c(0,55),type='l'))
par(new=T)
with(observed, plot(X75.~i.value,ylim=c(0,55),type='l'))
par(new=F)
Ideally, the target and the observed should overlap in every month -
they don't. The desire is to visually accentuate the amount of overlap
by shading in the area where these two zones overlap. How would you
do that? Note, that in some of these characterizations, the overlap
wanders in and out [I already have routines that calculate the percent
of overlap, but I have been requested to find a way to shade the
overlap.]
__
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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.
[R] how to eliminate first row in datafile created in do loop
I am trying to create a routine that would take a time series and
generate monthly 25%tile and 75%tile limits based on 12 calendar months.
I have succeeded to create a do loop to do this, but can't figure out
how to initiate the receiving datafile (in this case limit.list)
without sticking an initial set of zeroes in. I'd like to be able to
insert a new column i.value for each time i increments 1 thru 12, to
represent the 12 months, but I can't figure out how to join the value of
i to the lower and upper variables. Close but no cigar here.
lower=0
upper=0
limit.list-data.frame(lower,upper)
for(i in 1:12){
s=subset(paleo,month==i);
one.month=with(s, quantile(WBpaleo, c(.25, .75), na.rm=T));
limit.list-rbind(limit.list,one.month)
}
limit.list
which gives me this:
limit.list
lowerupper
1 0.0 0.0
2 25.51786 29.68823
3 26.09828 30.92770
4 27.75703 33.04016
5 29.53395 36.14273
6 30.89420 37.65727
7 29.27843 37.59690
8 27.50141 34.36265
9 26.40554 32.88533
10 25.13494 29.83829
11 23.48982 27.14612
12 22.88814 27.52782
13 24.51789 28.24871
How do I avoid having row 1 in my list?
I can delete the first row with:
limit.list = limit.list[-1,]
but that messes up my month variable
limit.list
lowerupper
2 25.51786 29.68823
3 26.09828 30.92770
4 27.75703 33.04016
5 29.53395 36.14273
6 30.89420 37.65727
7 29.27843 37.59690
8 27.50141 34.36265
9 26.40554 32.88533
10 25.13494 29.83829
11 23.48982 27.14612
12 22.88814 27.52782
13 24.51789 28.24871
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Re: [R] how to eliminate first row in datafile created in do loop
Certainly closer, and I can get by with this ... but ... would like to
not have the one.months .
limit.list-NULL
for(i in 1:12){
s=subset(paleo,month==i);
one.month=with(s, quantile(WBpaleo, c(.25, .75), na.rm=T));
one.month=c(as.character(i),one.month);
limit.list-rbind(limit.list,one.month)
}
limit.list
25% 75%
one.month 1 25.51786389 29.68823492
one.month 2 26.09827974 30.9276985
one.month 3 27.757034405 33.040163045
one.month 4 29.53395155 36.1427293975
one.month 5 30.894195075 37.657271835
one.month 6 29.27843098 37.59689852
one.month 7 27.5014142975 34.36265367
one.month 8 26.405542532.88533117
one.month 9 25.1349430375 29.838291435
one.month 10 23.48982439 27.14611846
one.month 11 22.88814497 27.52782293
one.month 12 24.5178851225 28.248711925
-Original Message-
From: Ian Gow [mailto:iand...@gmail.com]
Sent: Tuesday, May 24, 2011 8:58 AM
To: Graves, Gregory; Nungesser, Martha; r-help@r-project.org
Cc: Kemp, Susan K SAJ; patrick_pi...@fws.gov
Subject: Re: [R] how to eliminate first row in datafile created in do
loop
Gregory: Would setting limit.list - NULL at the start do the trick? See
example below:
Analogue of your example:
lower - 0
upper - 0
limit.list-data.frame(lower,upper)
for(i in 1:12) {
some.data - rnorm(2)
lower - min(some.data)
upper - max(some.data)
one.month - data.frame(lower, upper)
limit.list - rbind(limit.list, one.month)
}
limit.list
Approach that doesn't give zeroes at the start:
limit.list - NULL
for(i in 1:12) {
some.data - rnorm(2)
lower - min(some.data)
upper - max(some.data)
one.month - data.frame(lower, upper)
limit.list - rbind(limit.list, one.month)
}
limit.list
-Ian
On 5/24/11 7:30 AM, Graves, Gregory ggra...@sfwmd.gov wrote:
I am trying to create a routine that would take a time series and
generate monthly 25%tile and 75%tile limits based on 12 calendar
months.
I have succeeded to create a do loop to do this, but can't figure out
how to initiate the receiving datafile (in this case limit.list)
without sticking an initial set of zeroes in. I'd like to be able to
insert a new column i.value for each time i increments 1 thru 12,
to
represent the 12 months, but I can't figure out how to join the value
of
i to the lower and upper variables. Close but no cigar here.
lower=0
upper=0
limit.list-data.frame(lower,upper)
for(i in 1:12){
s=subset(paleo,month==i);
one.month=with(s, quantile(WBpaleo, c(.25, .75), na.rm=T));
limit.list-rbind(limit.list,one.month)
}
limit.list
which gives me this:
limit.list
lowerupper
1 0.0 0.0
2 25.51786 29.68823
3 26.09828 30.92770
4 27.75703 33.04016
5 29.53395 36.14273
6 30.89420 37.65727
7 29.27843 37.59690
8 27.50141 34.36265
9 26.40554 32.88533
10 25.13494 29.83829
11 23.48982 27.14612
12 22.88814 27.52782
13 24.51789 28.24871
How do I avoid having row 1 in my list?
I can delete the first row with:
limit.list = limit.list[-1,]
but that messes up my month variable
limit.list
lowerupper
2 25.51786 29.68823
3 26.09828 30.92770
4 27.75703 33.04016
5 29.53395 36.14273
6 30.89420 37.65727
7 29.27843 37.59690
8 27.50141 34.36265
9 26.40554 32.88533
10 25.13494 29.83829
11 23.48982 27.14612
12 22.88814 27.52782
13 24.51789 28.24871
__
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-help@r-project.org mailing list
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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] tapply output as a dataframe
On Mon, Apr 13, 2009 at 12:41 PM, Dan Dube ddube-at-advisen.com wrote: i use tapply and by often, but i always end up banging my head against the wall with the output. The proposed solution of Dan's problem posted on R-help was: do.call(rbind,a) When I use this 'solution' I get 'ERROR: second argument must be a list'. So head on wall continues. My tapply output is generated as follows: a=tapply(value,list(sampling.date,station.code),mean) which gives me this (in part): A B C D E F G H I J K 1/15/2008 0.004 0.027 0.019 0.015 0.035 0.022 0.007 0.038 0.042 0.045 0.0350 1/15/2009 0.027 0.027 0.031 0.015 0.008 0.021 0.007 0.027 0.026 0.029 0.0210 1/15/2010 0.016 0.020 0.015 0.022 0.015 0.013 0.007 0.014 0.019 0.019 0.0180 10/15/2007 0.052 0.051 0.032 0.024 0.017 0.044 0.015 0.058 0.063 0.061 0.0640 10/15/2008 0.042 0.054 0.030 0.017 0.024 0.030 0.019 0.044 0.047 0.051 0.0390 10/15/2009 0.047 0.035 0.031 0.020 0.012 0.039 0.019 0.051 0.055 0.054 0.0350 The only way I can figure out how to resolve this, such that I can, for example, plot station A against date, is to export the tapply output as a csv, and then reimport. Suggestions? I couldn't find a solution to this likely SIMPLE problem in Crawley or multiple searches of R help. Gregory A. Graves, Lead Scientist Everglades REstoration COoordination and VERification (RECOVER) Wetland Watershed Sciences / Restoration Sciences Department South Florida Water Management District Phones: DESK: 561 / 682 - 2429 CELL: 561 / 719 - 8157 __ 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] tapply output as a dataframe
Yes, as far as I can tell, sampling.date is a character vector of the format 1/15/2008. It resides in the leftmost column of the tapply output. station.code are the A, B, C column headers which refer actual water quality station locations, and the values below those headers correspond to the sampling.date when samples were taken. Actually what I have done is to take the mid-point of each month and calculated its mean to deal with multiple samples taken in one month, and to generate NAs where no sample was taken by purposefully not adding the na.rm=T to the tapply command. Normally I would do this: rdate-as.POSIXct(strptime(date,format=%m/%d/%Y)) #convert sampling.date to date R can handle plot(A~rdate) If I just submit station.code like A I get all the values for Station A. It is in converting the sampling.date to an rdate that has me stumped. One reason being that in the tapply output the character vector representing date has no column name. I can't access that column. Gregory A. Graves, Lead Scientist Everglades REstoration COoordination and VERification (RECOVER) Wetland Watershed Sciences / Restoration Sciences Department South Florida Water Management District Phones: DESK: 561 / 682 - 2429 CELL: 561 / 719 - 8157 -Original Message- From: David Winsemius [mailto:dwinsem...@comcast.net] Sent: Thursday, February 03, 2011 12:50 PM To: Graves, Gregory Cc: r-help@r-project.org; Goodman, Patricia; Gorman, Patricia Subject: Re: [R] tapply output as a dataframe On Feb 3, 2011, at 11:29 AM, Graves, Gregory wrote: My tapply output is generated as follows: a=tapply(value,list(sampling.date,station.code),mean) Why not give us sampling.date (which is probably NOT really a date but rather a character vector) and station.code so we can show you how to create a more appropriate structure? which gives me this (in part): A B C D E F G H I J K 1/15/2008 0.004 0.027 0.019 0.015 0.035 0.022 0.007 0.038 0.042 0.045 0.0350 1/15/2009 0.027 0.027 0.031 0.015 0.008 0.021 0.007 0.027 0.026 0.029 0.0210 1/15/2010 0.016 0.020 0.015 0.022 0.015 0.013 0.007 0.014 0.019 0.019 0.0180 10/15/2007 0.052 0.051 0.032 0.024 0.017 0.044 0.015 0.058 0.063 0.061 0.0640 10/15/2008 0.042 0.054 0.030 0.017 0.024 0.030 0.019 0.044 0.047 0.051 0.0390 10/15/2009 0.047 0.035 0.031 0.020 0.012 0.039 0.019 0.051 0.055 0.054 0.0350 The only way I can figure out how to resolve this, such that I can, for example, plot station A against date, is to export the tapply output as a csv, and then reimport. Suggestions? I couldn't find a solution to this likely SIMPLE problem Perhaps. but we haven't really been told what the problem is, have we? in Crawley or multiple searches of R help. Gregory A. Graves, Lead Scientist David Winsemius, MD West Hartford, CT __ 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] tapply output as a dataframe
This works. Thanks. Gregory A. Graves, Lead Scientist Everglades REstoration COoordination and VERification (RECOVER) Wetland Watershed Sciences / Restoration Sciences Department South Florida Water Management District Phones: DESK: 561 / 682 - 2429 CELL: 561 / 719 - 8157 -Original Message- From: Phil Spector [mailto:spec...@stat.berkeley.edu] Sent: Thursday, February 03, 2011 12:41 PM To: Graves, Gregory Cc: r-help@r-project.org; Goodman, Patricia; Gorman, Patricia Subject: Re: [R] tapply output as a dataframe Try as.data.frame(as.table(a)) - Phil Spector Statistical Computing Facility Department of Statistics UC Berkeley spec...@stat.berkeley.edu On Thu, 3 Feb 2011, Graves, Gregory wrote: On Mon, Apr 13, 2009 at 12:41 PM, Dan Dube ddube-at-advisen.com wrote: i use tapply and by often, but i always end up banging my head against the wall with the output. The proposed solution of Dan's problem posted on R-help was: do.call(rbind,a) When I use this 'solution' I get 'ERROR: second argument must be a list'. So head on wall continues. My tapply output is generated as follows: a=tapply(value,list(sampling.date,station.code),mean) which gives me this (in part): A B C D E F G H I J K 1/15/2008 0.004 0.027 0.019 0.015 0.035 0.022 0.007 0.038 0.042 0.045 0.0350 1/15/2009 0.027 0.027 0.031 0.015 0.008 0.021 0.007 0.027 0.026 0.029 0.0210 1/15/2010 0.016 0.020 0.015 0.022 0.015 0.013 0.007 0.014 0.019 0.019 0.0180 10/15/2007 0.052 0.051 0.032 0.024 0.017 0.044 0.015 0.058 0.063 0.061 0.0640 10/15/2008 0.042 0.054 0.030 0.017 0.024 0.030 0.019 0.044 0.047 0.051 0.0390 10/15/2009 0.047 0.035 0.031 0.020 0.012 0.039 0.019 0.051 0.055 0.054 0.0350 The only way I can figure out how to resolve this, such that I can, for example, plot station A against date, is to export the tapply output as a csv, and then reimport. Suggestions? I couldn't find a solution to this likely SIMPLE problem in Crawley or multiple searches of R help. Gregory A. Graves, Lead Scientist Everglades REstoration COoordination and VERification (RECOVER) Wetland Watershed Sciences / Restoration Sciences Department South Florida Water Management District Phones: DESK: 561 / 682 - 2429 CELL: 561 / 719 - 8157 __ 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-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] reference text variables as column name to plot [SOLVED]
The following works nicely. Thank you.
plot(z, sf[[s]], type=l) #where the dataframe name in my case is sf
Sorry - I meant to say there were 24 rows in each column, so z=1:24
Gregory A. Graves, Lead Scientist
Everglades REstoration COoordination and VERification (RECOVER)
Restoration Sciences Department
South Florida Water Management District
Phones: DESK: 561 / 682 - 2429
CELL: 561 / 719 - 8157
-Original Message-
From: David Winsemius [mailto:dwinsem...@comcast.net]
Sent: Tuesday, November 30, 2010 11:50 AM
To: Graves, Gregory
Cc: r-help@r-project.org
Subject: Re: [R] reference text variables as column name to plot
On Nov 30, 2010, at 11:40 AM, Graves, Gregory wrote:
Having given myself carpal tunnel looking for answer to this ...
I have a dataset
Named what?
each column of which has 12 rows in it. I created a
variable 'z' as follows:
z=1:24 s
Why did you put twice as many elements in z as there are in a column?
Since I have a large number of these plots to make, and they are a bit
complex, I want to want to reference the column I want to plot via a
variable containing the name of that column. As follows:
similar='1978'
s=paste('Y',similar,sep='')
variable s now contains 'Y1978' which is the name of one of the
columns.
However, when I try to plot
plot(z,s,type='l')
Try this ... assuming that there really are 12 item length columns in
a dataframe named, dfm.
plot(1:12, dfm[[s]], type=l)
dataframes are lists that can be accessed by the names of columns
which are interpreted. Don't assume that you can get such
interpretation with the $ operator.
I get a 'x and y lengths differ' error because variable s is being
recognized as 'Y1978' length=1, rather than the contents of the column
Y1978 length=12.
I tried all the usual tricks I know like s.
Huh? is a logical operator.
How do you get R to
reference a variable as a column name?
Thank you.
Gregory A. Graves, Lead Scientist
Everglades REstoration COoordination and VERification (RECOVER)
Restoration Sciences Department
South Florida Water Management District
Phones: DESK: 561 / 682 - 2429
CELL: 561 / 719 - 8157
[[alternative HTML version deleted]]
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and provide commented, minimal, self-contained, reproducible code.
David Winsemius, MD
West Hartford, CT
__
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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.
[R] reference text variables as column name to plot
Having given myself carpal tunnel looking for answer to this ...
I have a dataset each column of which has 12 rows in it. I created a
variable 'z' as follows:
z=1:24
Since I have a large number of these plots to make, and they are a bit
complex, I want to want to reference the column I want to plot via a
variable containing the name of that column. As follows:
similar='1978'
s=paste('Y',similar,sep='')
variable s now contains 'Y1978' which is the name of one of the columns.
However, when I try to plot
plot(z,s,type='l')
I get a 'x and y lengths differ' error because variable s is being
recognized as 'Y1978' length=1, rather than the contents of the column
Y1978 length=12.
I tried all the usual tricks I know like s. How do you get R to
reference a variable as a column name?
Thank you.
Gregory A. Graves, Lead Scientist
Everglades REstoration COoordination and VERification (RECOVER)
Restoration Sciences Department
South Florida Water Management District
Phones: DESK: 561 / 682 - 2429
CELL: 561 / 719 - 8157
[[alternative HTML version deleted]]
__
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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.
[R] stacking consecutive columns
I have a file, each column of which is a separate year, and each row of each column is mean precipitation for that month. Looks like this (except it goes back to 1964). monthX2000 X2001 X2002 X2003 X2004 X2005 X2006 X2007 X2008 X2009 11.600 1.010 4.320 2.110 0.925 3.275 3.460 0.675 1.315 2.920 22.960 3.905 3.230 2.380 2.720 1.880 2.430 1.380 2.480 2.380 31.240 1.815 1.755 1.785 1.250 3.940 10.025 0.420 2.845 2.460 43.775 1.350 2.745 0.170 0.710 2.570 0.255 0.425 4.470 1.250 54.050 1.385 5.650 1.515 12.005 6.895 7.020 4.060 7.725 2.775 68.635 8.900 15.715 12.680 16.270 12.605 7.095 7.025 10.465 7.345 75.475 7.955 7.880 6.670 7.955 7.355 5.475 5.650 7.255 7.985 88.435 5.525 7.120 6.250 7.150 7.610 5.525 6.500 6.275 10.405 95.855 7.830 7.250 7.355 9.715 7.850 6.385 7.960 4.485 7.250 10 7.965 11.915 6.735 8.125 7.855 10.465 4.340 6.165 2.400 3.240 11 1.705 1.525 0.905 1.670 1.840 2.100 0.255 2.830 4.425 1.645 12 2.335 0.840 0.795 1.890 0.145 1.700 0.260 2.160 2.300 2.220 What I want to do is to stack 2008 data underneath 2009 data, 2007 under 2008, 2006 under 2007, etc. I have so far figured out that I can do this with the following clumsy approach: a=stack(yearmonth,select=c(X2009,X2008)) b=stack(yearmonth,select=c(X2008,X2007)) x=as.data.frame(c(a,b)) write.table(x,clipboard,sep= ,col.names=NA) #then paste this back into Excel to get this values ind values.1ind.1 1 0.275 X2009 1.285 X2008 2 0.41X2009 3.85X2008 3 1.915 X2009 3.995 X2008 4 1.25X2009 3.845 X2008 5 8.76X2009 2.095 X2008 6 8.65X2009 8.29X2008 7 7.175 X2009 9.405 X2008 8 7.19X2009 13.44 X2008 9 8.13X2009 7.245 X2008 10 1.46X2009 5.645 X2008 11 2.56X2009 0.535 X2008 12 5.01X2009 1.225 X2008 13 1.285 X2008 0.72X2007 14 3.85X2008 1.89X2007 15 3.995 X2008 1.035 X2007 16 3.845 X2008 2.86X2007 17 2.095 X2008 3.785 X2007 18 8.29X2008 9.62X2007 19 9.405 X2008 9.245 X2007 20 13.44 X2008 5.595 X2007 21 7.245 X2008 8.4 X2007 22 5.645 X2008 6.705 X2007 23 0.535 X2008 1.47X2007 24 1.225 X2008 1.665 X2007 Is there an easier, cleaner way to do this? Thanks. Gregory A. Graves, Lead Scientist Everglades REstoration COoordination and VERification (RECOVER) Restoration Sciences Department South Florida Water Management District Phones: DESK: 561 / 682 - 2429 CELL: 561 / 719 - 8157 __ 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] stacking consecutive columns
Elegant or not ... This approach worked as intended, the objective being to
create individual 'years' each of which was 24 months long (not 12) such that
each 'year' reflected the antecedent rainfall condition. THANKS.
I needed to do this to help us identify years that were similar to one another
with respect to resultant water-level stage pattern in the Everglades. Thus
the need to examine each year in respect to its preceeding year.
Gregory A. Graves, Lead Scientist
Everglades REstoration COoordination and VERification (RECOVER)
Restoration Sciences Department
South Florida Water Management District
Phones: DESK: 561 / 682 - 2429
CELL: 561 / 719 - 8157
-Original Message-
From: Patrick Hausmann [mailto:patrick.hausm...@uni-bremen.de]
Sent: Wednesday, November 17, 2010 9:49 AM
To: r-help@r-project.org; Graves, Gregory
Subject: Re: [R] stacking consecutive columns
Hi Gregory,
is this what you want? Ok, not the most elegant way...
# using 'melt' from the 'reshape' package
library(reshape)
Data - data.frame(month = 1:12,
x2002 = runif(12),
x2003 = runif(12),
x2004 = runif(12),
x2005 = runif(12))
v - NULL
for(i in 2:4) {
kk - melt(Data[, c(1, i, i+1)], id.vars=month, variable_name=year)
v[[i-1]] - kk[order(kk$year, decreasing=TRUE),]
}
out - do.call(cbind, v)
HTH
Patrick
Am 17.11.2010 15:03, schrieb Graves, Gregory:
I have a file, each column of which is a separate year, and each row of each
column is mean precipitation for that month. Looks like this (except it goes
back to 1964).
monthX2000 X2001 X2002 X2003 X2004 X2005 X2006 X2007 X2008
X2009
11.600 1.010 4.320 2.110 0.925 3.275 3.460 0.675 1.315
2.920
22.960 3.905 3.230 2.380 2.720 1.880 2.430 1.380 2.480
2.380
31.240 1.815 1.755 1.785 1.250 3.940 10.025 0.420 2.845
2.460
43.775 1.350 2.745 0.170 0.710 2.570 0.255 0.425 4.470
1.250
54.050 1.385 5.650 1.515 12.005 6.895 7.020 4.060 7.725
2.775
68.635 8.900 15.715 12.680 16.270 12.605 7.095 7.025 10.465
7.345
75.475 7.955 7.880 6.670 7.955 7.355 5.475 5.650 7.255
7.985
88.435 5.525 7.120 6.250 7.150 7.610 5.525 6.500 6.275
10.405
95.855 7.830 7.250 7.355 9.715 7.850 6.385 7.960 4.485
7.250
10 7.965 11.915 6.735 8.125 7.855 10.465 4.340 6.165 2.400
3.240
11 1.705 1.525 0.905 1.670 1.840 2.100 0.255 2.830 4.425
1.645
12 2.335 0.840 0.795 1.890 0.145 1.700 0.260 2.160 2.300
2.220
What I want to do is to stack 2008 data underneath 2009 data, 2007 under
2008, 2006 under 2007, etc. I have so far figured out that I can do this
with the following clumsy approach:
a=stack(yearmonth,select=c(X2009,X2008))
b=stack(yearmonth,select=c(X2008,X2007))
x=as.data.frame(c(a,b))
write.table(x,clipboard,sep= ,col.names=NA) #then paste this back
into Excel to get this
values ind values.1ind.1
1 0.275 X2009 1.285 X2008
2 0.41X2009 3.85X2008
3 1.915 X2009 3.995 X2008
4 1.25X2009 3.845 X2008
5 8.76X2009 2.095 X2008
6 8.65X2009 8.29X2008
7 7.175 X2009 9.405 X2008
8 7.19X2009 13.44 X2008
9 8.13X2009 7.245 X2008
101.46X2009 5.645 X2008
112.56X2009 0.535 X2008
125.01X2009 1.225 X2008
131.285 X2008 0.72X2007
143.85X2008 1.89X2007
153.995 X2008 1.035 X2007
163.845 X2008 2.86X2007
172.095 X2008 3.785 X2007
188.29X2008 9.62X2007
199.405 X2008 9.245 X2007
2013.44 X2008 5.595 X2007
217.245 X2008 8.4 X2007
225.645 X2008 6.705 X2007
230.535 X2008 1.47X2007
241.225 X2008 1.665 X2007
Is there an easier, cleaner way to do this? Thanks.
Gregory A. Graves, Lead Scientist
Everglades REstoration COoordination and VERification (RECOVER)
Restoration Sciences Department
South Florida Water Management District
Phones: DESK: 561 / 682 - 2429
CELL: 561 / 719 - 8157
__
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-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] stacking consecutive columns
Follows is the exact solution to this:
v - NULL
#note that decreasing is FALSE so preceding year preceeds
for(i in 2:46) {
kk - melt(yearmonth[, c(1, i, i+1)], id.vars=month, variable_name=year)
v[[i-1]] - kk[order(kk$year, decreasing=FALSE),]
}
x - do.call(cbind, v)
write.table(x,clipboard,sep= ,col.names=NA) #export to Excell via Ctrl-V
Gregory A. Graves, Lead Scientist
Everglades REstoration COoordination and VERification (RECOVER)
Restoration Sciences Department
South Florida Water Management District
Phones: DESK: 561 / 682 - 2429
CELL: 561 / 719 - 8157
-Original Message-
From: Patrick Hausmann [mailto:patrick.hausm...@uni-bremen.de]
Sent: Wednesday, November 17, 2010 9:49 AM
To: r-help@r-project.org; Graves, Gregory
Subject: Re: [R] stacking consecutive columns
Hi Gregory,
is this what you want? Ok, not the most elegant way...
# using 'melt' from the 'reshape' package
library(reshape)
Data - data.frame(month = 1:12,
x2002 = runif(12),
x2003 = runif(12),
x2004 = runif(12),
x2005 = runif(12))
v - NULL
for(i in 2:4) {
kk - melt(Data[, c(1, i, i+1)], id.vars=month, variable_name=year)
v[[i-1]] - kk[order(kk$year, decreasing=TRUE),]
}
out - do.call(cbind, v)
HTH
Patrick
Am 17.11.2010 15:03, schrieb Graves, Gregory:
I have a file, each column of which is a separate year, and each row of each
column is mean precipitation for that month. Looks like this (except it goes
back to 1964).
monthX2000 X2001 X2002 X2003 X2004 X2005 X2006 X2007 X2008
X2009
11.600 1.010 4.320 2.110 0.925 3.275 3.460 0.675 1.315
2.920
22.960 3.905 3.230 2.380 2.720 1.880 2.430 1.380 2.480
2.380
31.240 1.815 1.755 1.785 1.250 3.940 10.025 0.420 2.845
2.460
43.775 1.350 2.745 0.170 0.710 2.570 0.255 0.425 4.470
1.250
54.050 1.385 5.650 1.515 12.005 6.895 7.020 4.060 7.725
2.775
68.635 8.900 15.715 12.680 16.270 12.605 7.095 7.025 10.465
7.345
75.475 7.955 7.880 6.670 7.955 7.355 5.475 5.650 7.255
7.985
88.435 5.525 7.120 6.250 7.150 7.610 5.525 6.500 6.275
10.405
95.855 7.830 7.250 7.355 9.715 7.850 6.385 7.960 4.485
7.250
10 7.965 11.915 6.735 8.125 7.855 10.465 4.340 6.165 2.400
3.240
11 1.705 1.525 0.905 1.670 1.840 2.100 0.255 2.830 4.425
1.645
12 2.335 0.840 0.795 1.890 0.145 1.700 0.260 2.160 2.300
2.220
What I want to do is to stack 2008 data underneath 2009 data, 2007 under
2008, 2006 under 2007, etc. I have so far figured out that I can do this
with the following clumsy approach:
a=stack(yearmonth,select=c(X2009,X2008))
b=stack(yearmonth,select=c(X2008,X2007))
x=as.data.frame(c(a,b))
write.table(x,clipboard,sep= ,col.names=NA) #then paste this back
into Excel to get this
values ind values.1ind.1
1 0.275 X2009 1.285 X2008
2 0.41X2009 3.85X2008
3 1.915 X2009 3.995 X2008
4 1.25X2009 3.845 X2008
5 8.76X2009 2.095 X2008
6 8.65X2009 8.29X2008
7 7.175 X2009 9.405 X2008
8 7.19X2009 13.44 X2008
9 8.13X2009 7.245 X2008
101.46X2009 5.645 X2008
112.56X2009 0.535 X2008
125.01X2009 1.225 X2008
131.285 X2008 0.72X2007
143.85X2008 1.89X2007
153.995 X2008 1.035 X2007
163.845 X2008 2.86X2007
172.095 X2008 3.785 X2007
188.29X2008 9.62X2007
199.405 X2008 9.245 X2007
2013.44 X2008 5.595 X2007
217.245 X2008 8.4 X2007
225.645 X2008 6.705 X2007
230.535 X2008 1.47X2007
241.225 X2008 1.665 X2007
Is there an easier, cleaner way to do this? Thanks.
Gregory A. Graves, Lead Scientist
Everglades REstoration COoordination and VERification (RECOVER)
Restoration Sciences Department
South Florida Water Management District
Phones: DESK: 561 / 682 - 2429
CELL: 561 / 719 - 8157
__
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-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] failure to access packages
I am helping a fellow worker get R up and running, and he has run into a
peculiar problem I've not encountered in previous install situations.
From the Rconsole menu choice, he can set CRAN mirror to USA CA2, but when he
selects load packages a very truncated list of packages appears (many
packages don't appear at all).
If he issues chooseCRANmirror() from the prompt, he can still select USA
CA2, but still can't see all the packages that are there.
If he issues
local({pkg - select.list(sort(.packages(all.available = TRUE)),graphics=TRUE)
+ if(nchar(pkg)) library(pkg, character.only=TRUE)})
he gets this error:
Error in select.list(sort(.packages(all.available = TRIE)), graphics = TRUE :
Unused argument(s) (graphics = TRUE)
When I issue same command from prompt, I get the expected list of packages.
Has anyone encountered this problem? I've searched CRAN help, but got nowhere.
Gregory A. Graves, Lead Scientist
Everglades REstoration COoordination and VERification (RECOVER)
Restoration Sciences Department
South Florida Water Management District
Phones: DESK: 561 / 682 - 2429
CELL: 561 / 719 - 8157
__
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] failure to access packages
The problem R ver is 2.10.1 - I am using 2.11.1
Running windows xp sp3 - same system as what I use
The suggested solution is for the user to upgrade to 2.11.1 to resolve package
incompatibilities
Gregory A. Graves, Lead Scientist
Everglades REstoration COoordination and VERification (RECOVER)
Restoration Sciences Department
South Florida Water Management District
Phones: DESK: 561 / 682 - 2429
CELL: 561 / 719 - 8157
-Original Message-
From: Graves, Gregory
Sent: Wednesday, September 22, 2010 11:45 AM
To: 'r-help@r-project.org'
Cc: King, Christopher
Subject: RE: failure to access packages
I am helping a fellow worker get R up and running, and he has run into a
peculiar problem I've not encountered in previous install situations.
From the Rconsole menu choice, he can set CRAN mirror to USA CA2, but when he
selects load packages a very truncated list of packages appears (many
packages don't appear at all).
If he issues chooseCRANmirror() from the prompt, he can still select USA
CA2, but still can't see all the packages that are there.
If he issues
local({pkg - select.list(sort(.packages(all.available = TRUE)),graphics=TRUE)
+ if(nchar(pkg)) library(pkg, character.only=TRUE)})
he gets this error:
Error in select.list(sort(.packages(all.available = TRIE)), graphics = TRUE :
Unused argument(s) (graphics = TRUE)
When I issue same command from prompt, I get the expected list of packages.
Has anyone encountered this problem? I've searched CRAN help, but got nowhere.
Gregory A. Graves, Lead Scientist
Everglades REstoration COoordination and VERification (RECOVER)
Restoration Sciences Department
South Florida Water Management District
Phones: DESK: 561 / 682 - 2429
CELL: 561 / 719 - 8157
__
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] interpretation of stress in NMDS
Among those users of Primer, stress values greater than 0.3 are interpreted as questionable. Using both isoMDS and metaMDS (vegan package), the stress values returned are much higher using my own data and using examples provided in R Help. For example Rstress = 8.3, and the stressplot r2 = 0.99 indicating (to me) that the ordination is OK. I am guessing that the stress values across packages are not the same, and googling about has not returned a satisfactory answer ... thus this posting. My concern being that reporting a stress value of 8 for what I consider a satisfactory result may raise a few Primer-user's eyebrows. Gregory A. Graves, Lead Scientist Everglades REstoration COoordination and VERification (RECOVER) Restoration Sciences Department South Florida Water Management District Phones: DESK: 561 / 682 - 2429 CELL: 561 / 719 - 8157 [[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] Exporting NMDS distance matris to csv
If you submit these lines, you end up with variable vare.dis. I want to export vare.dis to csv. Stuck I am. library(vegan,logical.return = TRUE) #return=true verifies package is available library(MASS,logical.return=TRUE) #return=true verifies package is available data(varespec) #varespec is an example data file in the vegan package vare.dis - vegdist(varespec) vare.dis #this is the triangular matrix I'd like to export x=format.data.frame(vare.dis) #nope, don't work x=format(vare.dis) #nope x=data.frame(vare.dis) #ditto x=unknown.function(vare.dis) #??? What is the unknown bit here to do this? #this is what I want to be able to do, i.e., write the matrix x out as a csv write.csv(x, file = temp.csv) Gregory A. Graves, Lead Scientist Everglades REstoration COoordination and VERification (RECOVER) Restoration Sciences Department South Florida Water Management District Phones: DESK: 561 / 682 - 2429 CELL: 561 / 719 - 8157 [[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] nls model fitting errors
What am I failing to understand here? The script below works fine if the dataset being used is DNase1 - DNase[ DNase$Run == 1, ] per the example given in help(nlrob). Obviously, I am trying to understand how to use nls and nlrob to fit curves to data using R. #package=DAAG attach(codling) plot(pobs~dose) #next command returns 'step factor reduced below min factor error' m.nls - nls( pobs ~ a/(1 + exp(( b - log(dose) )/c ) ), data = codling, start = list( a = 3, b = 0, c = 1 ), trace = TRUE ) s-seq(min(pobs), max(pobs), .01) p.nls-predict(m.nls,list(pobs=s)) lines(s,p.nls,col='blue') #generates 'x and y lengths differ' error Gregory A. Graves, Lead Scientist Everglades REstoration COoordination and VERification (RECOVER) Restoration Sciences Department South Florida Water Management District Phones: DESK: 561 / 682 - 2429 CELL: 561 / 719 - 8157 [[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] shading between two smoothed curves
I will attempt to distill my problem down to a simpler one, the solution of which will hopefully lead me to the nirvana of complete understanding, i.e., inserting a polygon beneath an irregular line terminated by the x axis. I 'know' that polygon will work -provided- I knew the xy coordinates of the line(s) beneath which I desire to create the polygon. Unfortunately, I don't know how to get at that info. Below is an example of a irregular line for which the xy coordinates are not known (i.e., at least not by me). I assume my problem boils down to finding out how to extract the xy from that line. attach(cars) plot(dist ~ speed) scatter.smooth(speed, dist,family = gaussian,span = .2) Gregory A. Graves Lead Scientist Everglades REstoration COoordination and VERification (RECOVER) Watershed Division South Florida Water Management District Phones: DESK: 561 / 682 - 2429 CELL: 561 / 719 - 8157 From: Debabrata Midya [mailto:debabrata.mi...@commerce.nsw.gov.au] Sent: Thursday, August 13, 2009 6:36 PM To: Graves, Gregory Cc: r-help@r-project.org Subject: Re: [R] shading between two smoothed curves Hi Gregory, I can plot out two nice curves using the following code, but I can't figure out how to shade in the area lying between the two curves. Use polygon. ?polygon Graves, Gregory ggra...@sfwmd.gov 14/08/2009 4:17 am I have a set of 52 weekly values, one is the desired high lake stage for a week, and the other is the desired low lake stage for each week. It looks like this: week High Low 1 16 14.5 2 16 14.5 3 15.95 14.45 4 15.84 14.34 5 15.73 14.23 6 15.61 14.11 7 15.5 14 8 15.38 13.88 9 15.25 13.75 10 15.13 13.63 52 15 13.5 I can plot out two nice curves using the following code, but I can't figure out how to shade in the area lying between the two curves. I also can't figure out how to adjust the line weight as lwd doesn't seem to affect scatter.smooth. Thanks in advance. l=11 #lower ylim h=17 #upper ylim par(new=F) good - complete.cases(week, high) #get rid of the NA error and only using non-missing pairs par(new=T) scatter.smooth(week[good], high[good],family = gaussian,span = .2, ylim=c(l,h), type='n', lwd=2, color=blue, ylab=, xlab=) good - complete.cases(week, low) par(new=T) scatter.smooth(week[good], low[good],family = gaussian,span = .2, ylim=c(l,h), type='n', lwd=2, color=red, ylab=Stage, feet, xlab=Week of Year) #loess, lower span makes more irregular par(new=F) Gregory A. Graves Lead Scientist Everglades REstoration COoordination and VERification (RECOVER) Watershed Division South Florida Water Management District Phones: DESK: 561 / 682 - 2429 CELL: 561 / 719 - 8157 [[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. ** This email message, including any attached files, is confidential and intended solely for the use of the individual or entity to whom it is addressed. The NSW Department of Commerce prohibits the right to publish, copy, distribute or disclose any information contained in this email, or its attachments, by any party other than the intended recipient. If you have received this email in error please notify the sender and delete it from your system. No employee or agent is authorised to conclude any binding agreement on behalf of the NSW Department of Commerce by email. The views or opinions presented in this email are solely those of the author and do not necessarily represent those of the Department, except where the sender expressly, and with authority, states them to be the views of NSW Department of Commerce. The NSW Department of Commerce accepts no liability for any loss or damage arising from the use of this email and recommends that the recipient check this email and any attached files for the presence of viruses. ** __ 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] shading between two smoothed curves
Here is the solution for shading under a curve attach(cars) #example dataset scatter.smooth(speed, dist,family = gaussian,span = .3, xlim=c(3,25), type='n') #plot a curve (smoothed line) gg-loess.smooth(speed,dist,family = gaussian,span = .3) #put coordinates of smoothed line into 'gg' DF-data.frame(gg) #convert gg into a dataframe; this dataframe has 50 rows DF[51,]-c(25,0) #add a 51st row with the desired bottom right corner of the polygon under the curve DF[52,]-c(3,0) #add a 52nd row with the desired bottom left corner of the polygon under the curve polygon(c(DF), col=gray, border = red) # the shading Here is NOT the solution for shading between 2 curves [I am going to use a crayon]. attach(cars) #put 2 curves on the graph scatter.smooth(speed, dist+25,family = gaussian,span = .3,xlim=c(3,25),type='n',ylim=c(0,120)) par(new=T) scatter.smooth(speed, dist-5,family = gaussian,span = .3,xlim=c(3,25),type='n',ylim=c(0,120)) #get the xy coords of the 2 curves gg1-loess.smooth(speed,dist+25,family = gaussian,span = .3) gg2-loess.smooth(speed,dist-5,family = gaussian,span = .3) #stick the xy's into a dataframe and combine the 2 dataframes DF1-data.frame(gg1) DF2-data.frame(gg2) DFnew-do.call(rbind, list(DF1, DF2)) #Arrrgh polygon(c(DFnew), col=gray, border = red) Gregory A. Graves Lead Scientist Everglades REstoration COoordination and VERification (RECOVER) Watershed Division South Florida Water Management District Phones: DESK: 561 / 682 - 2429 CELL: 561 / 719 - 8157 -Original Message- From: Jim Lemon [mailto:j...@bitwrit.com.au] Sent: Friday, August 14, 2009 8:47 AM To: Graves, Gregory Cc: r-help@r-project.org Subject: Re: [R] shading between two smoothed curves Graves, Gregory wrote: I will attempt to distill my problem down to a simpler one, the solution of which will hopefully lead me to the nirvana of complete understanding, i.e., inserting a polygon beneath an irregular line terminated by the x axis. I 'know' that polygon will work -provided- I knew the xy coordinates of the line(s) beneath which I desire to create the polygon. Unfortunately, I don't know how to get at that info. Below is an example of a irregular line for which the xy coordinates are not known (i.e., at least not by me). I assume my problem boils down to finding out how to extract the xy from that line. attach(cars) plot(dist ~ speed) scatter.smooth(speed, dist,family = gaussian,span = .2) Hi Gregory, That's right, just call the underlying function (loess.smooth) and use str on the result. There should be x and y components that can be passed to lines. Jim __ 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] shading between two smoothed curves
Here is the solution for a shading between two curves. [much thanks to W.D.) attach(cars) #put 2 curves on the graph scatter.smooth(speed, dist+25,family = gaussian,span = .3,xlim=c(3,25),type='n',ylim=c(0,120)) par(new=T) scatter.smooth(speed, dist-5,family = gaussian,span = .3,xlim=c(3,25),type='n',ylim=c(0,120)) #get the xy coords of the 2 curves gg1-loess.smooth(speed,dist+25,family = gaussian,span = .3) gg2-loess.smooth(speed,dist-5,family = gaussian,span = .3) #stick the xy's into a dataframe and combine the 2 dataframes DF1-data.frame(gg1) DF2-data.frame(gg2) DFnew-do.call(rbind, list(DF1, DF2)) #add the shading polygon(rbind(DF1, DF2[nrow(DF2):1,]), col=gray, border = red) Gregory A. Graves Lead Scientist Everglades REstoration COoordination and VERification (RECOVER) Watershed Division South Florida Water Management District Phones: DESK: 561 / 682 - 2429 CELL: 561 / 719 - 8157 -Original Message- From: Graves, Gregory Sent: Friday, August 14, 2009 1:38 PM To: r-help@r-project.org Subject: RE: [R] shading between two smoothed curves Here is the solution for shading under a curve attach(cars) #example dataset scatter.smooth(speed, dist,family = gaussian,span = .3, xlim=c(3,25), type='n') #plot a curve (smoothed line) gg-loess.smooth(speed,dist,family = gaussian,span = .3) #put coordinates of smoothed line into 'gg' DF-data.frame(gg) #convert gg into a dataframe; this dataframe has 50 rows DF[51,]-c(25,0) #add a 51st row with the desired bottom right corner of the polygon under the curve DF[52,]-c(3,0) #add a 52nd row with the desired bottom left corner of the polygon under the curve polygon(c(DF), col=gray, border = red) # the shading Here is NOT the solution for shading between 2 curves [I am going to use a crayon]. attach(cars) #put 2 curves on the graph scatter.smooth(speed, dist+25,family = gaussian,span = .3,xlim=c(3,25),type='n',ylim=c(0,120)) par(new=T) scatter.smooth(speed, dist-5,family = gaussian,span = .3,xlim=c(3,25),type='n',ylim=c(0,120)) #get the xy coords of the 2 curves gg1-loess.smooth(speed,dist+25,family = gaussian,span = .3) gg2-loess.smooth(speed,dist-5,family = gaussian,span = .3) #stick the xy's into a dataframe and combine the 2 dataframes DF1-data.frame(gg1) DF2-data.frame(gg2) DFnew-do.call(rbind, list(DF1, DF2)) #Arrrgh polygon(c(DFnew), col=gray, border = red) Gregory A. Graves Lead Scientist Everglades REstoration COoordination and VERification (RECOVER) Watershed Division South Florida Water Management District Phones: DESK: 561 / 682 - 2429 CELL: 561 / 719 - 8157 -Original Message- From: Jim Lemon [mailto:j...@bitwrit.com.au] Sent: Friday, August 14, 2009 8:47 AM To: Graves, Gregory Cc: r-help@r-project.org Subject: Re: [R] shading between two smoothed curves Graves, Gregory wrote: I will attempt to distill my problem down to a simpler one, the solution of which will hopefully lead me to the nirvana of complete understanding, i.e., inserting a polygon beneath an irregular line terminated by the x axis. I 'know' that polygon will work -provided- I knew the xy coordinates of the line(s) beneath which I desire to create the polygon. Unfortunately, I don't know how to get at that info. Below is an example of a irregular line for which the xy coordinates are not known (i.e., at least not by me). I assume my problem boils down to finding out how to extract the xy from that line. attach(cars) plot(dist ~ speed) scatter.smooth(speed, dist,family = gaussian,span = .2) Hi Gregory, That's right, just call the underlying function (loess.smooth) and use str on the result. There should be x and y components that can be passed to lines. Jim __ 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] shading between two smoothed curves
I have a set of 52 weekly values, one is the desired high lake stage for a week, and the other is the desired low lake stage for each week. It looks like this: week High Low 1 16 14.5 2 16 14.5 3 15.95 14.45 4 15.84 14.34 5 15.73 14.23 6 15.61 14.11 7 15.5 14 8 15.38 13.88 9 15.25 13.75 10 15.13 13.63 52 15 13.5 I can plot out two nice curves using the following code, but I can't figure out how to shade in the area lying between the two curves. I also can't figure out how to adjust the line weight as lwd doesn't seem to affect scatter.smooth. Thanks in advance. l=11 #lower ylim h=17 #upper ylim par(new=F) good - complete.cases(week, high) #get rid of the NA error and only using non-missing pairs par(new=T) scatter.smooth(week[good], high[good],family = gaussian,span = .2, ylim=c(l,h), type='n', lwd=2, color=blue, ylab=, xlab=) good - complete.cases(week, low) par(new=T) scatter.smooth(week[good], low[good],family = gaussian,span = .2, ylim=c(l,h), type='n', lwd=2, color=red, ylab=Stage, feet, xlab=Week of Year) #loess, lower span makes more irregular par(new=F) Gregory A. Graves Lead Scientist Everglades REstoration COoordination and VERification (RECOVER) Watershed Division South Florida Water Management District Phones: DESK: 561 / 682 - 2429 CELL: 561 / 719 - 8157 [[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] how to use a list to create a plot
I issued the following command to obtain the std dev for each month. psd-numSummary(Sal, groups=month, statistics=c(sd)) which resulted in psd sd n NA 1 6.930340 9367 2319 2 7.847003 10827 1008 3 5.962308 12988 404 4 3.632105 12576 384 5 3.328189 13030 362 6 10.101336129555 7 11.75958513071 321 8 10.27756612286 1105 9 8.561922 12286 674 10 9.252890 13343 27 11 9.568852 12591 369 12 9.622851 12020 946 typeof(psd) [1] list I want plot sd versus column 1 (month of year) of the list psd. The only way I can figure out how to do this is to manually copy the table into Excel, reformat, and then export that back out as a new csv. There has to be a better way? Gregory A. Graves Lead Scientist REstoration COoordination and VERification (RECOVER) Watershed Division South Florida Water Management District Phones: DESK: 561 / 682 - 2429 CELL: 561 / 719 - 8157 [[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] increase number of ticks on x axis of dates
My x axis is a series of daily dates (e.g., 01/01/2000, 01/02/2000, etc.) from 2000 to end of 2008. The default only gives me 4 ticks. I want more. Why doesn't this work? sdate-as.POSIXct(strptime(date,format=%m/%d/%Y)) plot(ppt~sdate,type=l,ylim=c(0,47),col=1,lwd=1,pch=16,ylab=Salinity, psu,xlab=Year,las=1,main=Duck Key Salinity,xaxt=n) year.text=c(2001,2002,2003,2004, 2005, 2006, 2007,2008, 2009) axis(1, at=c(1:9),labels=year.text) Gregory A. Graves Lead Scientist REstoration COoordination and VERification (RECOVER) Watershed Division South Florida Water Management District Phones: DESK: 561 / 682 - 2429 CELL: 561 / 719 - 8157 __ 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] replacing default axis labels on a plot - SOLUTION
The original problem posed was: On 14/05/2009 7:31 AM, Graves, Gregory wrote: I have 3 columns: flow, month, and monthname, where month is 1-12, and monthname is name of month. I can't get the plot to replace the 1-12 with monthname using ticks.lab. What am I doing wrong? plot(flow~factor(month),xlab=Month,ylab=Total Flow per Month, ylim=c(0,55000), ticks.lab=monthname) Here is the solution to this: # make a boxplot but suppress default labels on x axis with xaxt=n plot(flow~factor(month),xlab=Month,ylab=Total Flow per Month, ylim=c(0,55000), xaxt=n) #NOTE xaxt # create a vector containing month abbrevs with [[1]] suffix as follows month.name-list(c(Jan, Feb, Mar, Apr, May, Jun, Jul, Aug, Sep, Oct, Nov, Dec))[[1]] # place the 12 months on axis 1 (the x axis) as follows: axis(1, at=1:12, labels=month.name) Gregory A. Graves Lead Scientist REstoration COoordination and VERification (RECOVER) Watershed Division South Florida Water Management District Phones: DESK: 561 / 682 - 2429 CELL: 561 / 719 - 8157 __ 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] replacing default axis labels on a plot
I have 3 columns: flow, month, and monthname, where month is 1-12, and monthname is name of month. I can't get the plot to replace the 1-12 with monthname using ticks.lab. What am I doing wrong? plot(flow~factor(month),xlab=Month,ylab=Total Flow per Month, ylim=c(0,55000), ticks.lab=monthname) Thanks Gregory A. Graves Lead Scientist REstoration COoordination and VERification (RECOVER) Watershed Division South Florida Water Management District Phones: DESK: 561 / 682 - 2429 CELL: 561 / 719 - 8157 __ 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] replacing default axis labels on a plot
Thanks. But ... Still don't get it. Referring explicitly to the monthname column which had many entries across many years did not work. So taking your clue I created a new variable list of names of month month.name-list(c(Jan, Feb, Mar, Apr, May, Jun, Jul, Aug, Sep, Oct, Nov, Dec)) then issued this command plot(flow~factor(month),xlab=Month,ylab=Total Flow per Month, ylim=c(0,55000), xaxt=n, axis(1, at=1:12, labels=month.name)) which gave this error ERROR: 12, labels = month.name) Gregory A. Graves Lead Scientist REstoration COoordination and VERification (RECOVER) Watershed Division South Florida Water Management District Phones: DESK: 561 / 682 - 2429 CELL: 561 / 719 - 8157 -Original Message- From: Duncan Murdoch [mailto:murd...@stats.uwo.ca] Sent: Thursday, May 14, 2009 8:12 AM To: Graves, Gregory Cc: r-help@r-project.org Subject: Re: [R] replacing default axis labels on a plot On 14/05/2009 7:31 AM, Graves, Gregory wrote: I have 3 columns: flow, month, and monthname, where month is 1-12, and monthname is name of month. I can't get the plot to replace the 1-12 with monthname using ticks.lab. What am I doing wrong? plot(flow~factor(month),xlab=Month,ylab=Total Flow per Month, ylim=c(0,55000), ticks.lab=monthname) ticks.lab is not a parameter to the plot.formula function. You don't say where you found it described, but I'd suggest you read your source more closely, or just use xaxt=n in the plot call, and construct your own call to axis(), e.g. axis(1, at=1:12, labels=monthname). Duncan Murdoch Thanks Gregory A. Graves Lead Scientist REstoration COoordination and VERification (RECOVER) Watershed Division South Florida Water Management District Phones: DESK: 561 / 682 - 2429 CELL: 561 / 719 - 8157 __ 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-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] creating a matrix as input to lowess
I have a very large file with many rows and columns. I want to create a plot with lowess. If I try the following it works fine: data(PrecipGL) plot(PrecipGL) lines(lowess(time(PrecipGL),PrecipGL),lwd=3, col=2) In my file, 2 columns are nox and sdate, and are both typeof() = double. If I issue command Plot(nox~sdate) I can get a nice plot. However if I try lines(lowess(time(nox~sdate),nox~sdate),lwd=3, col=2) it returns an error that it is not a matrix if I try to extract these 2 columns into a matrix mdat - matrix(date,nox, byrow=TRUE) doesn't work, and search for help did not work, so am posting. Obviously I am a newbee here. Thanks for any help!! Gregory A. Graves Lead Scientist REstoration COoordination and VERification (RECOVER) Division Everglades Restoration Resource Area South Florida Water Management District Phones: DESK: 561 / 681 - 2563 x3730 CELL: 561 / 719 - 8157 [[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] sdate-as.POSIXct(strptime(date,format=%m/%d/%Y %H:%M))
Newbie here~ I've spent a ton of time on this, but have to admit I am stuck. I've tried various combinations of strptime and the package CHRON, but still can't get there. My simple task is that I want to plot salinity (S1 or S2) on y-axis and time on x-axis, but I want to use a lay-understandable representation of time (not Julian). If I use the following command, I can make a plot using plot(S1~sdate), where: sdate-as.POSIXct(strptime(date,format=%m/%d/%Y %H:%M)) ... however, every 00:00 time winds up being an NA. Here is an excerpt of the input data (this is a HUGE file): indexDate Julian S1 S2 24985 11/19/2006 00:00 39040.01 20.40 22.11 (this line winds up being an NA) 24986 11/19/2006 01:00 39040.05 20.42 22.22 24987 11/19/2006 02:00 39040.09 20.60 22.29 24988 11/19/2006 03:00 39040.14 20.62 22.21 24989 11/19/2006 04:00 39040.18 20.75 22.39 ... an excerpt of the variable sdate winds up looking like this: [35017] NA2007-09-08 01:00:00 EDT [35019] 2007-09-08 01:00:00 EDT 2007-09-08 10:00:00 EDT [35021] 2007-09-08 10:00:00 EDT 2007-09-08 11:00:00 EDT [35023] 2007-09-08 11:00:00 EDT 2007-09-08 12:00:00 EDT [35025] 2007-09-08 02:00:00 EDT 2007-09-08 02:00:00 EDT [35027] 2007-09-08 03:00:00 EDT 2007-09-08 03:00:00 EDT [35029] 2007-09-08 04:00:00 EDT 2007-09-08 04:00:00 EDT [35031] 2007-09-08 05:00:00 EDT 2007-09-08 05:00:00 EDT [35033] 2007-09-08 06:00:00 EDT 2007-09-08 06:00:00 EDT [35035] 2007-09-08 07:00:00 EDT 2007-09-08 07:00:00 EDT [35037] 2007-09-08 08:00:00 EDT 2007-09-08 08:00:00 EDT [35039] 2007-09-08 09:00:00 EDT 2007-09-08 09:00:00 EDT [35041] NA2007-09-09 01:00:00 EDT [35043] 2007-09-09 01:00:00 EDT 2007-09-09 10:00:00 EDT [35045] 2007-09-09 10:00:00 EDT 2007-09-09 11:00:00 EDT [35047] 2007-09-09 11:00:00 EDT 2007-09-09 12:00:00 EDT [35049] 2007-09-09 02:00:00 EDT 2007-09-09 02:00:00 EDT [35051] 2007-09-09 03:00:00 EDT 2007-09-09 03:00:00 EDT [35053] 2007-09-09 04:00:00 EDT 2007-09-09 04:00:00 EDT [35055] 2007-09-09 05:00:00 EDT 2007-09-09 05:00:00 EDT [35057] 2007-09-09 06:00:00 EDT 2007-09-09 06:00:00 EDT [35059] 2007-09-09 07:00:00 EDT 2007-09-09 07:00:00 EDT [35061] 2007-09-09 08:00:00 EDT 2007-09-09 08:00:00 EDT [35063] 2007-09-09 09:00:00 EDT 2007-09-09 09:00:00 EDT Gregory A. Graves Lead Scientist RECOVER Division Everglades Restoration Resource Area South Florida Water Management District Phones: DESK: 561 / 681 - 2563 x3730 CELL: 561 / 719 - 8157 [[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.
