[R] a function more appropriate than 'sapply'?
I'm wondering if I need to use a function other than sapply as the following line of code runs indefinitely (or 30 min so far) and uses up all 16Gb of memory on my machine for what seems like a very small dataset (data attached in a txt file wells.txt http://r.789695.n4.nabble.com/file/n4656723/wells.txt ). The R code is: wells-read.table(c:/temp/wells.txt,col.names=c(name,plc_hldr)) wells2-wells[sapply(wells[,1],function(x)length(strsplit(as.character(x), _)[[1]])==2),] The 2nd line of R code above gets bogged down and takes all my RAM with it: http://r.789695.n4.nabble.com/file/n4656723/memory_loss.png I'm simply trying to extract all of the lines of data that have a single _ in the first column and place them into a dataset called wells2. If that were to work, I then want to extract the lines of data that have two _ and put them into a separate dataset, say wells3. Is there a better way to do this than the one-liner above? -Eric -- View this message in context: http://r.789695.n4.nabble.com/a-function-more-appropriate-than-sapply-tp4656723.html Sent from the R help mailing list archive at Nabble.com. __ 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] resizing data
Undoubtedly this question has been asked before, I just can't seem to find the combination of search terms to produce it. I'm trying to resize a dataset that is pulled into R using read.table. However, I think the same problem can be produced using matrix: x-matrix(1:64,8) x # [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] #[1,]19 17 25 33 41 49 57 #[2,]2 10 18 26 34 42 50 58 #[3,]3 11 19 27 35 43 51 59 #[4,]4 12 20 28 36 44 52 60 #[5,]5 13 21 29 37 45 53 61 #[6,]6 14 22 30 38 46 54 62 #[7,]7 15 23 31 39 47 55 63 #[8,]8 16 24 32 40 48 56 64 The true order of data in the larger problem I'm working with is actually transposed, like so: x-t(x) x # [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] #[1,]12345678 #[2,]9 10 11 12 13 14 15 16 #[3,] 17 18 19 20 21 22 23 24 #[4,] 25 26 27 28 29 30 31 32 #[5,] 33 34 35 36 37 38 39 40 #[6,] 41 42 43 44 45 46 47 48 #[7,] 49 50 51 52 53 54 55 56 #[8,] 57 58 59 60 61 62 63 64 I'm trying to resize the data (in this example, a matrix) to say a 16 x 4 matrix while preserving the consecutive order of the individual elements in a left-to-right top-to-bottom fashion. The example below is wrong because the first row should be 1 2 3 4, how can I make this happen? It would also be nice to make a 4 x 16 matrix where the first row contains the values of x[1,1:8] followed by x[2,1:8]. I'm guessing there is a 1 liner of R code for this type of thing so I don't have to resort to nested for loops? y-matrix(x,nrow=16,ncol=4) y # [,1] [,2] [,3] [,4] # [1,]1357 # [2,]9 11 13 15 # [3,] 17 19 21 23 # [4,] 25 27 29 31 # [5,] 33 35 37 39 # [6,] 41 43 45 47 # [7,] 49 51 53 55 # [8,] 57 59 61 63 # [9,]2468 #[10,] 10 12 14 16 #[11,] 18 20 22 24 #[12,] 26 28 30 32 #[13,] 34 36 38 40 #[14,] 42 44 46 48 #[15,] 50 52 54 56 #[16,] 58 60 62 64 -- View this message in context: http://r.789695.n4.nabble.com/resizing-data-tp4656653.html Sent from the R help mailing list archive at Nabble.com. __ 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] resizing data
I played around with your example on the smaller dataset, and it seemed like it was doing what I wanted. However, applying it to the larger problem, I didn't get a resized 2D dataset that preserved the order I was hoping for. Hopefully the following illustrates the larger problem: x-matrix(0,nrow=29328,ncol=7) # Now set the 70,000th element to 1 to find out where it ends up? # Iterating on columns first, then rows, the 70,000th element is # at index [1,7] x[1,7]-1 y-t(matrix(x,nrow=546)) dim(y) # [1] 376 546 # Does 1 appear at index [129,112] as I expect # (128 complete rows x 546 cols = 69888 + 112 = 70,000) thus, row 129, col 112 y[129,112] # [1] 0 # No, so where is it? grep(1,y) # [1] 123293 Where is that? -- View this message in context: http://r.789695.n4.nabble.com/resizing-data-tp4656653p4656662.html Sent from the R help mailing list archive at Nabble.com. __ 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] setting off-diagonals to zero
The following 1460 x 1460 matrix can be throught of as 16 distinct 365 x 365
matrices. I'm trying to set off-diaganol terms in the 16 sub-matrices with
indices more than +/- 5 (days) from each other to zero using some for loops.
This works well for some, but not all, of the for loops. The R code Im
using follows. For some reason the third loop below zero's-out everything
in the sub-quadrant it is targeting, which is readily observable when
viewing the matrix (View(MAT)).
library(Matrix)
MAT-matrix(rnorm(1460*1460,mean=0,sd=1),nrow = 1460, ncol = 1460)
#works great
for (i in 1:365) {
SEQ - (i - 5):(i + 5)
SEQ - SEQ[SEQ 0 SEQ 366]
MAT[(1:365)[-SEQ], i] - 0
}
#works great
for (i in 1:365) {
SEQ - (i - 5):(i + 5)
SEQ - SEQ[SEQ 0 SEQ 366]
MAT[(1:365)[-SEQ], i + 365] - 0
}
#zero's out everything, including main-diagonal and near-main-diagonal
terms???
for (i in 731:1095) {
SEQ - (i - 5):(i + 5)
SEQ - SEQ[SEQ 730 SEQ 1096]
MAT[(731:1095)[-SEQ], i + 365] - 0
}
View(MAT)
I'm not sure why the third FOR loop above is not leaving the main-diagonal
and near-main-diagonal terms alone?
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Re: [R] setting off-diagonals to zero
I'm not following. Printing SEQ to the screen at the intermediate steps
using the following modified R code suggests that 'i' is fine and is not
getting reset to 1 as you suggest? My understanding, or rather my desired
output if someone else is able to weight-in, is that the values in the
second line of output (731 732 733 etc.) should not be appearing in the 3rd
line of output. The third line of output should be missing 731 thru 736.
Any suggestions on how to modify the R code are certainly welcome.
Suggested revisions will be substituted back into the third FOR loop in my
original post on this thread to prevent the main- and near-main-diagonal
terms from being set equal to zero.
for (i in 731:732) {
SEQ - (i - 5):(i + 5)
print(SEQ)
SEQ - SEQ[SEQ 730 SEQ 1096]
print(SEQ)
print((731:1095)[-SEQ])
}
# [1] 726 727 728 729 730 731 732 733 734 735 736
# [1] 731 732 733 734 735 736
# [1] 731 732 733 734 735 736 737 738 739 740 741 742 743 744
745 746 747 748 749 750 751 752 753 754 755 756...
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Re: [R] Filling a covariance matrix
Hi Joshua,
The code you put together is very helpful. I have run into a small issue,
however, and I am surprised you weren't getting similar error message when
you tried running the nested for loops? As an example, I've pared down the
for loop a bit to highlight the error message I'm getting (note also that I
had to modify xs[[i]]$Q to xs[[i]]$*x*Q):
for(i in seq_along(xs)) {
for(j in seq_along(xs)) {
xcov[paste0(Q,i),paste0(Q,j)] - xcov[paste0(Q,j),paste0(Q,i)]
- cov(xs[[i]]$xQ, xs[[j]]$xQ, use=complete.obs)
}
}
#Error in cov(xs[[i]]$xQ, xs[[j]]$xQ, use = complete.obs) :
# incompatible dimensions
#Some investigation...
i
# 1
j
# 95
length(xs[[i]]$xQ)
# 96
length(xs[[j]]$xQ)
# 92
xs[[j]]$xQ[1:10]
#xQ
#2004-04-04 00:00:00 674
#2004-04-04 00:15:00 669
#2004-04-04 00:30:00 664
#2004-04-04 00:45:00 664
#2004-04-04 01:00:00 669
#2004-04-04 01:15:00 659
#2004-04-04 01:30:00 674
#2004-04-04 01:45:00 669
#2004-04-04 03:00:00 664
#2004-04-04 03:15:00 674
xs[[i]]$xQ[1:10]
# xQ
#2004-01-01 00:00:00 0.43
#2004-01-01 00:15:00 0.43
#2004-01-01 00:30:00 0.43
#2004-01-01 00:45:00 0.43
#2004-01-01 01:00:00 0.57
#2004-01-01 01:15:00 0.57
#2004-01-01 01:30:00 0.57
#2004-01-01 01:45:00 0.43
#2004-01-01 02:00:00 0.43
#2004-01-01 02:15:00 0.57
I suppose the reason use=complete.obs, use=pair, or
use=pairwise.complete.obs won't work is because the date stamp is
different despite the fact there are similar time stamps during the
respective days. Thus, I'm wondering if there is a way to direct the cov
function to make the calculation using only the time stamp (and not the date
stamp) to determine pairs?
Thanks, Eric
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[R] Filling a covariance matrix
useRs –
I’m working with the attached data that contains one year’s worth of
sub-daily observations of flow (“Q”) and specific conductance (“SC”, a
surrogate for concentration) at a point in a stream. The R code posted
below shows the extent of data processing thus far. My goal is to create a
covariance matrix that takes on the following form:
Q1 Q2 … Q365 SC1 SC2 … SC365
Q1
Q2
…
Q365
SC1
SC2
...
SC365
Where the covariance between Q1 (flow on day 1) Q2 (flow on day 2) is
determined using the sub-daily data contained in the variable ‘x’ of the R
code below. Similarly, the covariance between Q1 SC1 (specific
conductance on day 1) would be made using the sub-daily observations of flow
and specific conductance. Covariance between observations that are more
than 5 days distant from one another are likely meaningless. Thus, the
covariance matrix should reflect this limitation with zeros. For example,
the covariance between Q1 Q6, or between Q1 SC6, or between SC359
(specific conductance on day 359) SC365 (specific conductance on day 365)
would be zero as these observations are more than 5 days apart. Here is the
R code that reads the attached files containing Q and SC and puts the
processed data into ‘x’:
07130500_BelowJM_q_2004.txt
http://r.789695.n4.nabble.com/file/n4647170/07130500_BelowJM_q_2004.txt
07130500_BelowJM_2004.txt
http://r.789695.n4.nabble.com/file/n4647170/07130500_BelowJM_2004.txt
library(xts)
Q_subDaily-read.table(C:/temp/07130500_BelowJM_q_2004.rdb,col.names=c('date','time','tz','Q','rating','unknown'),colClasses=c(character,character,character,numeric,character,character))
SC_subDaily-read.table(C:/temp/07130500_BelowJM_2004.rdb,col.names=c('date','time','tz','SC','rating','unknown'),colClasses=c(character,character,character,numeric,character,character))
Q_subDaily$datetime.str - paste(Q_subDaily$date, Q_subDaily$time)
SC_subDaily$datetime.str - paste(SC_subDaily$date, SC_subDaily$time)
fmt - %Y%m%d %H%M%S
xQ - xts(as.matrix(Q_subDaily[Q]), as.POSIXct(Q_subDaily$datetime.str,
format=fmt))
xSC - xts(as.matrix(SC_subDaily[SC]),
as.POSIXct(SC_subDaily$datetime.str, format=fmt))
x - merge(xQ,xSC)
And here’s where I’m stuck, I’m not sure how to create the covariance matrix
I’ve described above? I would appreciate and greatly benefit from the sort
of help often found in the useR community.
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Re: [R] calculate within-day correlations
Hi Joshua,
I was attempting to work with your code on the larger datasets, which I have
to read in with 'read.table', but I've apparently missed something. When I
attempt to massage the data.frame a bit in the code below, as.POSIXct drops
the time component which then precludes use of xts. I think you'll see what
I mean (the rdb file is a text file containing the data, which is attached
to the post):
library(xts)
Q_hourly-read.table(C:/temp/07130500_BelowJM_q.rdb,skip=59,col.names=c('date','time','tz','Q','rating','unknown'),colClasses=c(character,character,character,numeric,character,character))
notice that
Q_hourly[1,]
# date time tz Q rating unknown
#1 19981001 00 MDT 326 3 A
Q_hourly[2,]
# date time tz Q rating unknown
#2 19981001 001500 MDT 326 3 A
and
paste(strptime(Q_hourly[1,1],%Y%m%d),
,format(strptime(Q_hourly[1,2],%H%M%S),format=%H:%M:%S),sep='')
#[1] 1998-10-01 00:00:00
but for some reason, the time stamp is dropped in the following, which
breaks the call to xts (I think)
as.POSIXct(paste(strptime(Q_hourly[1,1],%Y%m%d),
,format(strptime(Q_hourly[1,2],%H%M%S),format=%H:%M:%S),sep=''),format=%Y-%m-%d
%H:%M:%S,tz=)
#[1] 1998-10-01 PDT
The Q_use data.frame I'm building here should come out exactly
http://r.789695.n4.nabble.com/file/n4643206/07130500_BelowJM_q.rdb
07130500_BelowJM_q.rdb the same as in my original post (above), but I can't
seem to seem to force the preservation of the time stamp even though I've
explicitly stated the format I want to be stored (...format=%Y-%m-%d
%H:%M:%S). Any ideas?
xQ - xts(Q_use[Q], Q_use$date)
#Error in `[.data.frame`(x, indx) : undefined columns selected
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[R] calculate within-day correlations
useRs, Here is some R-ready data for my question to follow. Of course this data is small snippet from a much larger dataset that is about a decade long. Q-read.table(textConnection(2002 3 28 15 77.38815 2002 3 28 30 77.09505 2002 3 28 45 76.80196 2002 3 28 60 76.50887 2002 3 28 75 76.50887 2002 3 28 90 76.50887 2002 3 28 105 76.50887 2002 3 28 120 76.50887 2002 3 28 135 76.80196 2002 3 28 150 77.09506 2002 3 28 165 77.38815 2002 3 28 180 77.68125 2002 3 28 195 77.68125 2002 3 28 210 77.68125 2002 3 28 225 77.68125 2002 3 28 240 77.68125 2002 3 28 255 77.38815 2002 3 28 270 77.09505 2002 3 28 285 76.80196 2002 3 28 300 76.50887 2002 3 28 315 76.80196 2002 3 28 330 77.09506 2002 3 28 345 77.38815 2002 3 28 360 77.68125 2002 3 28 375 77.68125 2002 3 28 390 77.68125 2002 3 28 405 77.68125 2002 3 28 420 77.68125 2002 3 28 435 77.68125 2002 3 28 450 77.68125 2002 3 28 465 77.68125 2002 3 28 480 77.68125 2002 3 28 495 77.97691 2002 3 28 510 78.27258 2002 3 28 525 78.56824 2002 3 28 540 78.86389 2002 3 28 555 78.86389 2002 3 28 570 78.86389 2002 3 28 585 78.86389 2002 3 28 600 78.86389 2002 3 28 615 78.86389 2002 3 28 630 78.86389 2002 3 28 645 78.86389 2002 3 28 660 78.86389 2002 3 28 675 78.86389 2002 3 28 690 78.86389 2002 3 28 705 78.86389 2002 3 28 720 78.86389 2002 3 28 735 79.16212 2002 3 28 750 79.46034 2002 3 28 765 79.75856 2002 3 28 780 80.05679 2002 3 28 795 79.75856 2002 3 28 810 79.46033 2002 3 28 825 79.16211 2002 3 28 840 78.86389 2002 3 28 855 78.56823 2002 3 28 870 78.27257 2002 3 28 885 77.97691 2002 3 28 900 77.68125 2002 3 28 915 77.68125 2002 3 28 930 77.68125 2002 3 28 945 77.68125 2002 3 28 960 77.68125 2002 3 28 975 77.68125 2002 3 28 990 77.29046 2002 3 28 1005 76.89966 2002 3 28 1020 76.50887 2002 3 28 1035 75.93033 2002 3 28 1050 75.35179 2002 3 28 1065 74.77325 2002 3 28 1080 74.19473 2002 3 28 1095 73.90929 2002 3 28 1110 73.62385 2002 3 28 1125 73.33841 2002 3 28 1140 73.05298 2002 3 28 1155 73.33842 2002 3 28 1170 73.62386 2002 3 28 1185 73.90929 2002 3 28 1200 74.19473 2002 3 28 1215 74.19473 2002 3 28 1230 74.19473 2002 3 28 1245 74.19473 2002 3 28 1260 74.19473 2002 3 28 1275 74.48272 2002 3 28 1290 74.77071 2002 3 28 1305 75.05871 2002 3 28 1320 75.3467 2002 3 28 1335 75.63725 2002 3 28 1350 75.92779 2002 3 28 1365 76.21832 2002 3 28 1380 76.50887 2002 3 28 1395 76.80196 2002 3 28 1410 77.09506 2002 3 28 1425 77.38815 2002 3 28 1440 77.68125 2002 3 29 15 77.38815 2002 3 29 30 77.09505 2002 3 29 45 76.80196 2002 3 29 60 76.50887 2002 3 29 75 76.21832 2002 3 29 90 75.92778 2002 3 29 105 75.63724 2002 3 29 120 75.3467 2002 3 29 135 75.63725 2002 3 29 150 75.92779 2002 3 29 165 76.21832 2002 3 29 180 76.50887 2002 3 29 195 76.21832 2002 3 29 210 75.92778 2002 3 29 225 75.63724 2002 3 29 240 75.3467 2002 3 29 255 75.3467 2002 3 29 270 75.3467 2002 3 29 285 75.3467 2002 3 29 300 75.3467 2002 3 29 315 75.3467 2002 3 29 330 75.3467 2002 3 29 345 75.3467 2002 3 29 360 75.3467 2002 3 29 375 75.63725 2002 3 29 390 75.92779 2002 3 29 405 76.21832 2002 3 29 420 76.50887 2002 3 29 435 76.50887 2002 3 29 450 76.50887 2002 3 29 465 76.50887 2002 3 29 480 76.50887 2002 3 29 495 76.80196 2002 3 29 510 77.09506 2002 3 29 525 77.38815 2002 3 29 540 77.68125 2002 3 29 555 77.97691 2002 3 29 570 78.27258 2002 3 29 585 78.56824 2002 3 29 600 78.86389 2002 3 29 615 79.16212 2002 3 29 630 79.46034 2002 3 29 645 79.75856 2002 3 29 660 80.05679 2002 3 29 675 80.05679 2002 3 29 690 80.05679 2002 3 29 705 80.05679 2002 3 29 720 80.05679 2002 3 29 735 80.35759 2002 3 29 750 80.65839 2002 3 29 765 80.9592 2002 3 29 780 81.26001 2002 3 29 795 81.26001 2002 3 29 810 81.26001 2002 3 29 825 81.26001 2002 3 29 840 81.26001 2002 3 29 855 81.26001 2002 3 29 870 81.26001 2002 3 29 885 81.26001
Re: [R] Problem with converting character vector to time
It's not entirely clear where your trying to go with your problem, but one 'solution' depending on your destination could be: time_obj-format(as.POSIXct(test,format=%H:%M:%S),format=%H:%M:%S) time_obj [1] 00:49:19 class(time_obj) [1] character But this is a fairly unsatisfactory solution since it simply undoes the application of as.POSIXct Perhaps you can share more of your problem so folks responding know where you're trying to get to. -- View this message in context: http://r.789695.n4.nabble.com/Problem-with-converting-character-vector-to-time-tp4634283p4634293.html Sent from the R help mailing list archive at Nabble.com. __ 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] extracting values from txt file that follow user-supplied quote
I'll summarize the results in terms of total run time for the suggestions
that have been made as well as post the code for those that come across this
post in the future. First the results (the code for which is provided
second):
What I tried to do using suggestions from Bert and Dan:
t1
# user system elapsed
# 208.211.68 210.34
Gabor's suggested code:
t2
# user system elapsed
# 51.120.63 51.75
Rui's suggested code:
t3a #(Get the number of lines)
# user system elapsed
# 45.13 11.08 56.23
t3b #(now perform the function
# user system elapsed
# 50.590.55 51.16
So in summary it appears that Gabor's and Rui's code are quite similar (in
terms of runtime) if a priori knowledge of the number of lines in the file
is known (e.g. t2 is roughly equal to t3b). It would seem Gabor's code is a
little more robust since it doesn't require the total number of lines in the
file be supplied. And here is the code used to get these times (note that
the file I used was the 1GB text file, not the reduced version attached to
the top post of this thread):
#
#modified attempt
#
library(gsubfn)
library(tcltk2)
p-[-0-9]\\S+
pd-numeric()
txt_con-file(description=D:/MCR_BeoPEST - Copy/MCR.out,open=r)
t1-system.time(
while (length(txt_line-readLines(txt_con,n=1))){
if (length(grep(DISCREPANCY = ,txt_line))) {
pd-c(pd,as.numeric(strapplyc(txt_line, p)[[1]]))
}
})
close(txt_con)
t1
# user system elapsed
# 208.211.68 210.34
#
#Suggested by G. Grothendieck
#
g-function(txt_con, string, from, to, ...) {
L - readLines(txt_con)
matched - grep(string, L, value = TRUE, ...)
as.numeric(substring(matched, from, to))
}
txt_con-file(description=D:/MCR_BeoPEST - Copy/MCR.out,open=r)
t2-system.time(
edm-g(txt_con, PERCENT DISCREPANCY = , 70, 78, fixed = TRUE)
)
close(txt_con)
t2
# user system elapsed
# 51.120.63 51.75
#-
#Suggested by Rui Barradas
#-
library(R.utils)
t3a-system.time(num_lines-countLines(D:/MCR_BeoPEST - Copy/MCR.out))
t3a
# user system elapsed
# 45.13 11.08 56.23
fun - function(con, pattern, nlines, n=5000L){
if(is.character(con)){
con - file(con, open=rt)
on.exit(close(con))
}
passes - nlines %/% n
remaining - nlines %% n
res - NULL
for(i in seq_len(passes)){
txt - readLines(con, n=n)
res - c(res, as.numeric(substr(txt[grepl(pattern, txt)],
70, 78)))
}
if(remaining){
txt - readLines(con, n=remaining)
res - c(res, as.numeric(substr(txt[grepl(pattern, txt)],
70, 78)))
}
res
}
txt_con-file(description=D:/MCR_BeoPEST - Copy/MCR.out,open=r)
pat-PERCENT DISCREPANCY =
num_lines - 14405247L
t3b - system.time(pd2 - fun(txt_con, pat, num_lines, 10L))
close(txt_con)
t3b
# user system elapsed
# 50.590.55 51.16
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and provide commented, minimal, self-contained, reproducible code.
[R] extracting values from txt with regular expression
Thanks for your suggestions. Bert, in your response you raised my awareness to regular expressions. Are regular expressions the same across various languages? Consider the following line of text: txt_line- PERCENT DISCREPANCY = 0.01 PERCENT DISCREPANCY = -0.05 It seems python uses the following line of code to extract the two values in txt_line and store them in a variable called v: v = re.findall([+-]? *(?:\d+(?:\.\d*)|\.\d+)(?:[eE][+-]?\d+)?, line) #v[0] 0.01 #v[1] -0.05 I tried something similar in R (but it didn't work) by using the same regular expression, but got an error: edm-grep([+-]? *(?:\d+(?:\.\d*)|\.\d+)(?:[eE][+-]?\d+)?,txt_line) #Error: '\d' is an unrecognized escape in character string starting [+-]? *(?:\d I'm not even sure which function in R most efficiently extracts the values from txt_line. Basically, I want to peel out the values and think I can use the decimal point to construct the regular expression, but don't know where to go from here? -- View this message in context: http://r.789695.n4.nabble.com/extracting-values-from-txt-file-that-follow-user-supplied-quote-tp4632558p4632724.html Sent from the R help mailing list archive at Nabble.com. __ 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] extracting values from txt with regular expression
Hi Dan and Rui, Thank you for the suggestions, both were very helpful. Rui's code was quite fast...there is one more thing I want to explore for my own edification, but first I need some help fixing the code below, which is a slight modification to Dan's suggestion. It'll no doubt be tough to beat the time Rui's code finished the task in, but I'm willing to try. First, I need to fix the following, which 'peels' the wrong bit of text from txt_line. Instead of extracting as it now does (shown below), can the code be modified to extract the values 0.01 and -0.05, and store them in the variable 'extracted'? txt_line- PERCENT DISCREPANCY = 0.01 PERCENT DISCREPANCY = -0.05 extracted - strsplit(gsub([+-]?(?:\\d+(?:\\.\\d*)|\\.\\d+)(?:[eE][+-]?\\d+)?,\\1%,txt_line),%) extracted #[1] PERCENT DISCREPANCY =PERCENT DISCREPANCY = -- View this message in context: http://r.789695.n4.nabble.com/extracting-values-from-txt-file-that-follow-user-supplied-quote-tp4632558p4632753.html Sent from the R help mailing list archive at Nabble.com. __ 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] extracting values from txt file that follow user-supplied quote
useRs-
I'm attempting to scan a more than 1Gb text file and read and store the
values that follow a specific key-phrase that is repeated multiple time
throughout the file. A snippet of the text file I'm trying to read is
attached. The text file is a dumping ground for various aspects of the
performance of the model that generates it. Thus, the location of
information I'm wanting to extract from the file is not in a fixed position
(i.e. it does not always appears in a predictable location, like line 1000,
or 2000, etc.). Rather, the desired values always follow a specific phrase:
PERCENT DISCREPANCY =
One approach I took was the following:
library(R.utils)
txt_con-file(description=D:/MCR_BeoPEST - Copy/MCR.out,open=r)
#The above will need to be altered if one desires to test code on the
attached txt file, which will run much quicker
system.time(num_lines-countLines(D:/MCR_BeoPEST - Copy/MCR.out))
#elapsed time on full 1Gb file took about 55 seconds on a 3.6Gh Xeon
num_lines
#14405247
system.time(
for(i in 1:num_lines){
txt_line-readLines(txt_con,n=1)
if (length(grep(PERCENT DISCREPANCY =,txt_line))) {
pd-c(pd,as.numeric(substr(txt_line,70,78)))
}
}
)
#Time took about 5 minutes
The inefficiencies in this approach arise due to reading the file twice
(first to get num_lines, then to step through each line looking for the
desired text).
Is there a way to speed this process up through the use of a ?scan ? I
wan't able to get anything working, but what I had in mind was scan through
the more than 1Gb file and when the keyphrase (e.g. PERCENT
DISCREPANCY = ) is encountered, read and store the next 13 characters
(which will include some white spaces) as a numeric value, then resume the
scan until the key phrase is encountered again and repeat until the
end-of-the-file marker is encountered. Is such an approach even possible or
is line-by-line the best bet?
http://r.789695.n4.nabble.com/file/n4632558/MCR.out MCR.out
--
View this message in context:
http://r.789695.n4.nabble.com/extracting-values-from-txt-file-that-follow-user-supplied-quote-tp4632558.html
Sent from the R help mailing list archive at Nabble.com.
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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] obtaining a true/false vector with combination of strsplit, length, unlist,
Hi, A pared down version of the dataset I'm working with: edm-read.table(textConnection(WELLIDX_GRID Y_GRID LAYER ROW COLUMN SPECIES CALCULATED OBSERVED w301_3 4428. 1389 2 6 18 1 3558 6490. w304_12 4836. 6627 2 27 20 1 3509 3228. 02_10_120803.6125E+04 13875 1 56145 1 2774 -999.0 02_10_120803.6125E+04 13875 1 56145 1 2774 -999.0 02_10_120813.6375E+04 13875 1 56146 1 3493 -999.0 02_10_120923.9125E+04 13875 1 56157 1 4736 -999.0 w305_12 2962. 7326 2 30 12 1 4575 5899.),header=T) closeAllConnections() I'm having a hard time coming up with the R code that would produce a TRUE/FALSE vector based on whether or not the first column of the data.frame edm has a length of 2 or 3? To show what I mean going row-by-row, I could do the following: length(strsplit(as.character(edm$WELLID),_)[[1]])==3 [1] FALSE length(strsplit(as.character(edm$WELLID),_)[[2]])==3 [1] FALSE length(strsplit(as.character(edm$WELLID),_)[[3]])==3 [1] TRUE length(strsplit(as.character(edm$WELLID),_)[[4]])==3 [1] TRUE length(strsplit(as.character(edm$WELLID),_)[[5]])==3 [1] TRUE length(strsplit(as.character(edm$WELLID),_)[[6]])==3 [1] TRUE length(strsplit(as.character(edm$WELLID),_)[[7]])==3 [1] FALSE I've fumbled around trying to come up with a line of R code that would create a vector that looks like: FALSE FALSE TRUE TRUE TRUE TRUE FALSE The final goal is to use this vector to create two new data.frames, where, for example, the first contains all the rows of edm in which the first column has a length of 2 when split using a _ character. The second data.frame would contain all the rows in which the first column has a length of 3 when split using a _ character. Thanks, Eric -- View this message in context: http://r.789695.n4.nabble.com/obtaining-a-true-false-vector-with-combination-of-strsplit-length-unlist-tp4380050p4380050.html Sent from the R help mailing list archive at Nabble.com. __ 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] legend, lheight, and alignment
Hello, A bit of fairly simple code, yet I don't seem to be able to manipulate it quite as much as I would like: 1) It would be nice if the objects appearing in the legend were aligned, and by aligned I mean the boxes are centered over the lines. Do I need to adjust the use of NA in the code below to accomplish this? Here's how it appears on my machine: http://r.789695.n4.nabble.com/file/n4129402/example.png 2) Because I'm using a call to 'expression' in the text contained in the legend, it would be nice to space the lines of text a bit more. My feeble attempt was to increase the lheight parameter in the hopes this would affect the legend...to no avail. Further, lheight cannot be added to the argument list of legend. I've been unable to track down another parameter applicable to legend, suggestions very much appreciated. par(lheight=2) plot(1,1,col=white) legend(center, legend=c(expression(paste(Observed ,italic(bar(EC)[e]))),expression(paste(Simulated ,italic(bar(EC)[e]))),test1,test2), lty = c(NA,NA,1,1), col = c(black,black,red,blue), density = c(NA,25,NA,NA), border=c(black,black,NA,NA), fill = c(grey,black,NA,NA), angle = c(NA,45,NA,NA), cex = 1.25, bty = n, xpd = TRUE) sessionInfo() R version 2.13.2 (2011-09-30) Platform: x86_64-pc-mingw32/x64 (64-bit) locale: [1] LC_COLLATE=English_United States.1252 [2] LC_CTYPE=English_United States.1252 [3] LC_MONETARY=English_United States.1252 [4] LC_NUMERIC=C [5] LC_TIME=English_United States.1252 attached base packages: [1] stats graphics grDevices utils datasets methods base loaded via a namespace (and not attached): [1] tools_2.13.2 -- View this message in context: http://r.789695.n4.nabble.com/legend-lheight-and-alignment-tp4129402p4129402.html Sent from the R help mailing list archive at Nabble.com. __ 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] subplot strange behavoir
Hello Dr. Snow, With regard to your response from earlier this month: When I copy and paste your code I get what is expected, the 2 subplots line up on the same y-value. What version of R are you using, which version of subplot? What platform? I'm still troubled by the fact that layout and subplot (from TeachingDemos) are not playing nicely together on my machine. sessionInfo(): sessionInfo() #R version 2.13.2 (2011-09-30) #Platform: x86_64-pc-mingw32/x64 (64-bit) #other attached packages: #[1] TeachingDemos_2.7 I'd really like to get this working on my machine as it seems to be working on yours. While I previously tried a simply example for the initial forum post, I'm curious if the real plot I'm trying to make works on your machine. Should you happen to have a spare moment and I'm not pushing my luck, I've attached 4 small data files, 1 text file containing the R commands I'm trying to run (including 'layout' and 'subplot' called R_Commands_Plot_MT3D_Analytical_Comparison_For_Paper.txt), and the incorrect tiff output I'm getting on my machine. I've directed all paths in the R code to c:/temp/ so everything should quickly work if files are dropped there. Should it work on your machine as we would expect, does anything come to mind for how to fix it on my machine? Very Respectfully, Eric http://r.789695.n4.nabble.com/file/n3926941/AnalyticalDissolvedSorbedConcAt20PoreVols.txt AnalyticalDissolvedSorbedConcAt20PoreVols.txt http://r.789695.n4.nabble.com/file/n3926941/AnalyticalEffluentConcentration.txt AnalyticalEffluentConcentration.txt http://r.789695.n4.nabble.com/file/n3926941/Conc_Breakthru_at_100cm.txt Conc_Breakthru_at_100cm.txt http://r.789695.n4.nabble.com/file/n3926941/Conc_Profile_20T.txt Conc_Profile_20T.txt http://r.789695.n4.nabble.com/file/n3926941/R_Commands_Plot_MT3D_Analytical_Comparison_For_Paper.txt R_Commands_Plot_MT3D_Analytical_Comparison_For_Paper.txt http://r.789695.n4.nabble.com/file/n3926941/NonEquilibrium_ForPaper.tif NonEquilibrium_ForPaper.tif -- View this message in context: http://r.789695.n4.nabble.com/subplot-strange-behavoir-tp3875917p3926941.html Sent from the R help mailing list archive at Nabble.com. __ 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] contouring x y scatter data
Hello, I'm almost positive R can do the following, I just haven't hit upon the right package or search terms, however. Here's what I'm after: I've got concentration output from two different models that I want to qualitatively compare in a contour plot (or some variant of a contour plot). The problem as I see it is that the data is not gridded is the usual regular fashion, and even if it were, the models are working with two different irregular computational/numerical grids at which concentrations are calculated. Running the code below will produce a plot that will hopefully illuminate what I'm attempting to describe. At each location where 'O' appears, there is a calculated concentration that I would like to contour. I'd like to do the same thing at each 'X' location and then compare, visually, how close the contours are (hopefully close). Both data.frames contain columns/fields titled ts181 and ts1825 which contain the concentrations at time step 181 days and time step 1825 days. Is it possible to contour x-y-value data if it is not laid out on a regular grid? X-c(0,100,200,300,400,500,600,700,800,900,1000,1100,1200,1300,1400,1500,1600,1700,1800,1900,2000,2100,2200,2300,2400,2500,2600,2700,2800,2900,3000,3100,3200,3300,3400,3500,3600,3700,3800,3900,4000,0,100,200,300,400,500,600,700,800,900,1000,1100,1200,1300,1400,1500,1600,1700,1800,1900,2000,2100,2200,2300,2400,2500,2600,2700,2800,2900,3000,3100,3200,3300,3400,3500,3600,3700,3800,3900,4000,0,100,200,300,400,500,600,700,800,900,1000,1100,1200,1300,1400,1500,1600,1700,1800,1900,2000,2100,2200,2300,2400,2500,2600,2700,2800,2900,3000,3100,3200,3300,3400,3500,3600,3700,3800,3900,4000,0,100,200,300,400,500,600,700,800,900,1000,1100,1200,1300,1400,1500,1600,1700,1800,1900,2000,2100,2200,2300,2400,2500,2600,2700,2800,2900,3000,3100,3200,3300,3400,3500,3600,3700,3800,3900,4000,0,100,200,300,400,500,600,700,800,900,1000,1100,1200,1300,1400,1500,1600,1700,1800,1900,2000,2100,2200,2300,2400,2500,2600,2700,2800,2900,3000,3100,3200,3300,3400,3500,3600,3700,3800,3900,4000,0,100,200,300,400,5! 00,600,700,800,900,1000,1100,1200,1300,1400,1500,1600,1700,1800,1900,2000,2100,2200,2300,2400,2500,2600,2700,2800,2900,3000,3100,3200,3300,3400,3500,3600,3700,3800,3900,4000,0,100,200,300,400,500,600,700,800,900,1000,1100,1200,1300,1400,1500,1600,1700,1800,1900,2000,2100,2200,2300,2400,2500,2600,2700,2800,2900,3000,3100,3200,3300,3400,3500,3600,3700,3800,3900,4000,0,100,200,300,400,500,600,700,800,900,1000,1100,1200,1300,1400,1500,1600,1700,1800,1900,2000,2100,2200,2300,2400,2500,2600,2700,2800,2900,3000,3100,3200,3300,3400,3500,3600,3700,3800,3900,4000,0,100,200,300,400,500,600,700,800,900,1000,1100,1200,1300,1400,1500,1600,1700,1800,1900,2000,2100,2200,2300,2400,2500,2600,2700,2800,2900,3000,3100,3200,3300,3400,3500,3600,3700,3800,3900,4000,0,100,200,300,400,500,600,700,800,900,1000,1100,1200,1300,1400,1500,1600,1700,1800,1900,2000,2100,2200,2300,2400,2500,2600,2700,2800,2900,3000,3100,3200,3300,3400,3500,3600,3700,3800,3900,4000,0,100,200,300,400,500,600,700,800,900,1000! ,1100,1200,1300,1400,1500,1600,1700,1800,1900,2000,2100,2200,2300,2400 ,2500,2600,2700,2800,2900,3000,3100,3200,3300,3400,3500,3600,3700,3800,3900,4000) Z-c(50,49.95,49.85,49.75,49.65,49.55,49.45,49.35,49.25,49.15,49.05,48.95,48.85,48.75,48.65,48.55,48.45,48.35,48.25,48.15,48.05,47.95,47.85,47.75,47.65,47.55,47.45,47.35,47.25,47.15,47.05,46.95,46.85,46.75,46.65,46.55,46.45,46.35,46.25,46.15,46.1,49.25,49.2,49.1,49,48.9,48.8,48.7,48.6,48.5,48.4,48.3,48.2,48.1,48,47.9,47.8,47.7,47.6,47.5,47.4,47.3,47.2,47.1,47,46.9,46.8,46.7,46.6,46.5,46.4,46.3,46.2,46.1,46,45.9,45.8,45.7,45.6,45.5,45.4,45.35,48.2,48.15,48.05,47.95,47.85,47.75,47.65,47.55,47.45,47.35,47.25,47.15,47.05,46.95,46.85,46.75,46.65,46.55,46.45,46.35,46.25,46.15,46.05,45.95,45.85,45.75,45.65,45.55,45.45,45.35,45.25,45.15,45.05,44.95,44.85,44.75,44.65,44.55,44.45,44.35,44.3,47,46.95,46.85,46.75,46.65,46.55,46.45,46.35,46.25,46.15,46.05,45.95,45.85,45.75,45.65,45.55,45.45,45.35,45.25,45.15,45.05,44.95,44.85,44.75,44.65,44.55,44.45,44.35,44.25,44.15,44.05,43.95,43.85,43.75,43.65,43.55,43.45,43.35,43.25,43.15,43.1,45.5,45.45,45.35,45.25,45.15,45.05,44.95,44.85,44.75,44.6!
Re: [R] contouring x y scatter data
Hi Carlos, Thanks for the response. The plot you suggested was not in line with what I'm trying to produce in R. The last plot found at: http://www.advsofteng.com/gallery_contour.html is more along the lines of what I'm looking for. Notice the scatter points (shown by x) are not laid out in a regular/gridded fashion, yet the plotting software was still able to use some kind of interpolation and plot contours. All the contouring options I'm finding in R require the spatial locations of data points to be in a 'regular' grid. That is, data in each row must have the same y coordinate and data in each column must have the same x coordinate. I'm seeking flexibility in the location of the points on which the contours are based. -- View this message in context: http://r.789695.n4.nabble.com/contouring-x-y-scatter-data-tp3912828p3913544.html Sent from the R help mailing list archive at Nabble.com. __ 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] contouring x y scatter data
Thank you David, I'm moving forward again. I was not aware of that website, I was recently at http://www.r-project.org/ and clicked on the 'mailing list' link and didn't see it there...with 8k+ posts, maybe its worth adding? -- View this message in context: http://r.789695.n4.nabble.com/contouring-x-y-scatter-data-tp3912828p3913617.html Sent from the R help mailing list archive at Nabble.com. __ 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] Irregular 3d objects with rgl
Hello, While exploring if rgl is along the lines of what I need, I checked out demo(rgl) and didn't find quite what I'm looking for, and am therefore seeking additional help/suggestions. The application is geared towards displaying a 3D rendering of a contaminant plume in the subsurface with the following highlights: Once the plume was rendered as a 3D object, a pie-like wedge could be removed (or cut away) exposing the higher concentrations within the plume as 'hotter' colors. About the closest example I could find is here: http://mclaneenv.com/graphicdownloads/plume.jpg Whereas this particular rendering shows a bullet-like object where 3/4 of the object is removed, I would like to try and show something where 3/4 of the object remains, and where the object has been cut away the colors would show concentrations within the plume, just as in the example referenced above. It would seem most software capable of this type of thing is proprietary (and perhaps for good reason if it is a difficult problem to solve). I've put together a very simple 6x6x6 cube with non-zero values internal to it representing the plume. I wondering if an isosurface where conc = 0.01 can be rendered in 3D and then if a bite or wedge can be removed from the 3d object exposing the higher concentrations inside as discussed above? x-c(1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6) y-c(1,1,1,1,1,1,2,2,2,2,2,2,3,3,3,3,3,3,4,4,4,4,4,4,5,5,5,5,5,5,6,6,6,6,6,6,1,1,1,1,1,1,2,2,2,2,2,2,3,3,3,3,3,3,4,4,4,4,4,4,5,5,5,5,5,5,6,6,6,6,6,6,1,1,1,1,1,1,2,2,2,2,2,2,3,3,3,3,3,3,4,4,4,4,4,4,5,5,5,5,5,5,6,6,6,6,6,6,1,1,1,1,1,1,2,2,2,2,2,2,3,3,3,3,3,3,4,4,4,4,4,4,5,5,5,5,5,5,6,6,6,6,6,6,1,1,1,1,1,1,2,2,2,2,2,2,3,3,3,3,3,3,4,4,4,4,4,4,5,5,5,5,5,5,6,6,6,6,6,6,1,1,1,1,1,1,2,2,2,2,2,2,3,3,3,3,3,3,4,4,4,4,4,4,5,5,5,5,5,5,6,6,6,6,6,6) z-c(1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6) conc-c(0,0,0,0,0,0,0,0,0.1,0.1,0,0,0,0.1,1,1,0.1,0,0,0.1,0.5,1,0.1,0,0,0,0.2,0.2,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0.05,0.1,0,0,0,0.05,0.8,0.8,0.05,0,0,0.05,0.4,0.8,0.05,0,0,0,0.1,0.1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0.05,0,0,0,0,0.6,0.6,0.02,0,0,0,0.2,0.5,0.02,0,0,0,0.05,0.05,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0.05,0.2,0,0,0,0,0,0.05,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0.02,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0) plume-data.frame(cbind(x=x,y=y,z=z,conc=conc)) if it helps to view the concentrations in layer by layer tabular form: Layer 1 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.10 0.10 0.00 0.00 0.00 0.10 1.00 0.50 0.20 0.00 0.00 0.10 1.00 1.00 0.20 0.00 0.00 0.00 0.10 0.10 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 Layer 2 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.05 0.05 0.00 0.00 0.00 0.05 0.80 0.40 0.10 0.00 0.00 0.10 0.80 0.80 0.10 0.00 0.00 0.00 0.05 0.05 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 Layer 3 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.60 0.20 0.05 0.00 0.00 0.05 0.60 0.50 0.05 0.00 0.00 0.00 0.02 0.02 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 Layer 4 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.05 0.00 0.00 0.00 0.00 0.00 0.20 0.05 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 Layer 5 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.02 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 Layer 6 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 -- View this message in context: http://r.789695.n4.nabble.com/Irregular-3d-objects-with-rgl-tp3906573p3906573.html Sent from the R help mailing list archive at Nabble.com. __ 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] subplot strange behavoir
Hello,
Below is some example code that should reproduce an error I'm encountering
while trying to create a tiff plot with two subplots. If I run just the
following bit of code through the R GUI the result is what I'd like to have
appear in the saved tiff image:
x-seq(0:20)
y-c(1,1,2,2,3,4,5,4,3,6,7,1,1,2,2,3,4,5,4,3,6)
plot(x,y,type=l,las=1,ylim=c(0,12))
subplot(edm.sub(x[seq(1:5)],y[seq(1:5)]),x=4,y=9,size=c(1,1.5))
subplot(edm.sub(x[seq(15,20,by=1)],y[seq(15,20,by=1)]),x=17,y=9,size=c(1,1.5))
However, if expanding on this code with:
edm.sub-function(x,y){plot(x,y,col=red,frame.plot=F,
las=1,xaxs=i,yaxs=i,type=b,
ylim=c(0,6),xlab=,ylab=)}
png(c:/temp/lookat.tif,res=120,height=600,width=1200)
layout(matrix(c(1,2),2,2,byrow=TRUE),c(1.5,2.5),respect=TRUE)
plot(seq(1:10),seq(1:10),type=l,las=1,col=blue)
plot(x,y,type=l,las=1,ylim=c(0,12))
subplot(edm.sub(x[seq(1:5)],y[seq(1:5)]),x=4,y=9,size=c(1,1.5))
subplot(edm.sub(x[seq(15,20,by=1)],y[seq(15,20,by=1)]),x=17,y=9,size=c(1,1.5))
dev.off()
One will notice the second subplot is out of position (notice the
y-coordinate is the same for both subplots...y=9):
http://r.789695.n4.nabble.com/file/n3875917/lookat.png
If I try to 'guess' a new y-coordinate for the second subplot, say y=10:
png(c:/temp/lookat.tif,res=120,height=600,width=1200)
layout(matrix(c(1,2),2,2,byrow=TRUE),c(1.5,2.5),respect=TRUE)
plot(seq(1:10),seq(1:10),type=l,las=1,col=blue)
plot(x,y,type=l,las=1,ylim=c(0,12))
subplot(edm.sub(x[seq(1:5)],y[seq(1:5)]),x=4,y=9,size=c(1,1.5))
subplot(edm.sub(x[seq(15,20,by=1)],y[seq(15,20,by=1)]),x=17,y=10,size=c(1,1.5))
dev.off()
R kicks back the following message
Error in plot.new() : plot region too large
Am I mis-using subplot?
Thanks, Eric
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Re: [R] subplot strange behavoir
Hello Greg, Session info is below. Running Win7 64-bit. I just upgraded my version of R and tried rerunning the code and got the same odd result. I, too, get an expected result when I create the plot in the R GUI. The problem crops up only when I try and create the plot in png() or tiff(). Perhaps I need to try par(new=T), or some other trick-of-the-trade? If you're unable to recreate the problem, I suppose this is nearly a dead-end? sessionInfo() R version 2.13.2 (2011-09-30) Platform: x86_64-pc-mingw32/x64 (64-bit) locale: [1] LC_COLLATE=English_United States.1252 LC_CTYPE=English_United States.1252 [3] LC_MONETARY=English_United States.1252 LC_NUMERIC=C [5] LC_TIME=English_United States.1252 attached base packages: [1] splines stats graphics grDevices utils datasets methods base other attached packages: [1] TeachingDemos_2.7 Hmisc_3.8-3 survival_2.36-9 loaded via a namespace (and not attached): [1] cluster_1.14.0 grid_2.13.2 lattice_0.19-33 tools_2.13.2 -- View this message in context: http://r.789695.n4.nabble.com/subplot-strange-behavoir-tp3875917p3876178.html Sent from the R help mailing list archive at Nabble.com. __ 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] subplot strange behavoir
I tried this trick, and clearly things are not going in the right direction. It seems 'layout' is at the root of my frustration, so I can make two plots and marge them in adobe illustrator (or something similar). png(c:/temp/lookat.png,res=120,height=600,width=1200) layout(matrix(c(1,2),2,2,byrow=TRUE),c(1.5,2.5),respect=TRUE) plot(seq(1:10),seq(1:10),type=l,las=1,col=blue) plot(x,y,type=l,las=1,ylim=c(0,12)) subplot(edm.sub(x[seq(1:5)],y[seq(1:5)]),x=4,y=9,size=c(1,1.5)) par(new=T) plot(x,y,las=1,ylim=c(0,12),type=n,ann=F,las=1,col=blue) subplot(edm.sub(x[seq(15,20,by=1)],y[seq(15,20,by=1)]),x=17,y=9,size=c(1,1.5)) dev.off() http://r.789695.n4.nabble.com/file/n3876285/lookat.png -- View this message in context: http://r.789695.n4.nabble.com/subplot-strange-behavoir-tp3875917p3876285.html Sent from the R help mailing list archive at Nabble.com. __ 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] animation error
Hello,
While running the following code I encountered a bit of a funny error.
First the code:
library(animation)
ani.options(convert = shQuote('C:/Program
Files/ImageMagick-6.7.0-Q16/convert.exe'))
plot.ani-function() {
for (ts in 1:730){
dat-read.table(paste(c:/temp/ConcProfile_,ts,.txt,sep=),
header=T,sep='\t')
#tiff(paste(c:/temp/Plot_ConcProfile_sp,ts,.tif,sep=),height=800,width=600,res=130)
plot(dat$Conc[1:200],rev(dat$Depth[1:200]),type=l,xlab=Conc.,ylab=Depth,
yaxt=n,col=blue,lwd=2,xlim=c(0,120))
axis(side=2,at=seq(0,200,by=25),labels=rev(seq(0,200,by=25)),las=1)
axis(side=2,at=seq(0,200,by=5),labels=F,tcl=-0.25)
text(x=110,y=200,paste(SP ,ts,sep=))
ani.pause()
#dev.off()
}
}
saveMovie(plot.ani(),movie.name=concSlug.gif,img.name=ConcProfile,interval=0.15,outdir=getwd())
#Executing:
#C:/Program Files/ImageMagick-6.7.0-Q16/convert.exe -loop 0
#-delay 15
#C:/Users/emorway/AppData/Local/Temp/2/RtmpwH5TwA/ConcProfile1.png
#C:/Users/emorway/AppData/Local/Temp/2/RtmpwH5TwA/ConcProfile2.png
...
#C:/Users/emorway/AppData/Local/Temp/2/RtmpwH5TwA/ConcProfile113.png
#C:/Users/emorway/AppData/Local/Temp/2/RtmpwH5TwA/ConcProfile114.png
#There seems to be an error in the conversion...
The code runs without issue when I dump individual tiff images to the temp
directory (that is, I run only the for loop). Is there a limit to the
number of frames that can be added? If I shorten the for loop to 100 the
gif is created, if I try to run it all the way out to 730 days (as coded
above) it terminates with the rather non-descriptive error above. Is there
an option I've overlooked?
Thanks...
Eric
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Re: [R] Help with filled.contour()
Michael,
Although this is a rather old post I'm responding to, I recently came across
it and have a suggestions for getting rid of the legend. Simply modify the
code associated with the function and stuff it into a new function,
edm-function (x = seq(0, 1, length.out = nrow(z)), y = seq(0, 1,
length.out = ncol(z)), z, xlim = range(x, finite = TRUE),
ylim = range(y, finite = TRUE), zlim = range(z, finite = TRUE),
levels = pretty(zlim, nlevels), nlevels = 20, color.palette = cm.colors,
col = color.palette(length(levels) - 1), plot.title, plot.axes,
key.title, key.axes, asp = NA, xaxs = i, yaxs = i, las = 1,
axes = TRUE, frame.plot = axes, ...)
{
if (missing(z)) {
if (!missing(x)) {
if (is.list(x)) {
z - x$z
y - x$y
x - x$x
}
else {
z - x
x - seq.int(0, 1, length.out = nrow(z))
}
}
else stop(no 'z' matrix specified)
}
else if (is.list(x)) {
y - x$y
x - x$x
}
if (any(diff(x) = 0) || any(diff(y) = 0))
stop(increasing 'x' and 'y' values expected)
mar.orig - (par.orig - par(c(mar, las, mfrow)))$mar
on.exit(par(par.orig))
w - (3 + mar.orig[2L]) * par(csi) * 2.54
#layout(matrix(c(2, 1), ncol = 2L), widths = c(1, lcm(w)))
par(las = las)
mar - mar.orig
mar[4L] - mar[2L]
mar[2L] - 1
par(mar = mar)
#plot.new()
#plot.window(xlim = c(0, 1), ylim = range(levels), xaxs = i,
#yaxs = i)
#rect(0, levels[-length(levels)], 1, levels[-1L], col = col)
#if (missing(key.axes)) {
#if (axes)
#axis(4)
#}
#else key.axes
#box()
#if (!missing(key.title))
#key.title
#mar - mar.orig
#mar[4L] - 1
#par(mar = mar)
plot.new()
plot.window(xlim, ylim, , xaxs = xaxs, yaxs = yaxs, asp = asp)
if (!is.matrix(z) || nrow(z) = 1L || ncol(z) = 1L)
stop(no proper 'z' matrix specified)
if (!is.double(z))
storage.mode(z) - double
.Internal(filledcontour(as.double(x), as.double(y), z,
as.double(levels),
col = col))
if (missing(plot.axes)) {
if (axes) {
title(main = , xlab = , ylab = )
Axis(x, side = 1)
Axis(y, side = 2)
}
}
else plot.axes
if (frame.plot)
box()
if (missing(plot.title))
title(...)
else plot.title
invisible()
}
Then call this new function,
edm(x, y, z, axes = F, frame.plot = F, asp = 1,
col = palette(gray(seq(0, 0.9, len = 25))), nlevels = 25)
Eric
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[R] Plotting Finite Element and Finite Difference Results together
As a simplification of a much larger problem, I'm using the following simple example to explore ways of getting the finite difference results to plot over the same extents as the finite element solution (see figure). I haven't discovered (if it exist) a way of getting the finite difference data to extrapolate, perhaps I'm using the wrong plotting tool? FD.dat-matrix(c(1,3.5,3.5,1, 1,2.5,2.5,1, 1,1.5,1.5,1, 1,1,1,1), nrow=4,ncol=4,byrow=TRUE) FE.dat-matrix(c(1,1,4,1,1, 1,1,3.5,1,1, 1,1,2.5,1,1, 1,1,1.5,1,1, 1,1,1,1,1), nrow=5,ncol=5,byrow=TRUE) FD.x.coords-c(0.5,1.5,2.5,3.5) FE.x.coords-c(0,1,2,3,4) FD.y.coords-FD.x.coords FE.y.coords-FE.x.coords contour(FD.x.coords,FD.y.coords,FD.dat,xlim=c(0,4),ylim=c(0,4)) contour(FE.x.coords,FE.y.coords,FE.dat,xlim=c(0,4),ylim=c(0,4),add=T,col=red) http://r.789695.n4.nabble.com/file/n3388485/Slide1.png -- View this message in context: http://r.789695.n4.nabble.com/Plotting-Finite-Element-and-Finite-Difference-Results-together-tp3388485p3388485.html Sent from the R help mailing list archive at Nabble.com. __ 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] Using 'contour' to compare concentration profiles on the same plot
Hello, Using the data and code below I've been using R to compare output from two different solute transport codes, the red contours are from verified code while the blue contours come from some modified code I've been working on. The goal is for the contours to match, and although there is currently a slight discrepancy this would be expected. The plot shows a pulse of infiltrating water with an arbitrary concentration moving downward. I'm curious if there is a clever way to plot the labels so that they don't interfere with one another? Getting even more fancy, I'm also wondering if spaces can be created and left empty in one set of contours (blue) to account for values from the other contours (say red) ? In other words, labels from the blue (or red) contours won't be drawn over the red's (or blue's) contour line. Thanks, Eric vs2d-matrix(c(0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0, 0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0, 0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0, 0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0, 0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0, 0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0, 0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0, 0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0, 0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0, 0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.14,0.15,0.15,0.16,0.15,0.15,0.14,0.14,0.14,0.14,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0, 0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.13,0.029600,0.029800,0.03,0.030100,0.03,0.029800,0.029600,0.029500,0.029400,0.029400,0.04,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0, 0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.71,0.156000,0.156000,0.156000,0.156000,0.156000,0.156000,0.156000,0.155000,0.155000,0.155000,0.19,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0, 0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.000108,0.289000,0.289000,0.289000,0.288000,0.289000,0.289000,0.289000,0.289000,0.289000,0.289000,0.29,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0, 0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.000109,0.436000,0.436000,0.436000,0.436000,0.436000,0.436000,0.436000,0.436000,0.436000,0.436000,0.29,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0, 0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.90,0.593000,0.593000,0.593000,0.592000,0.593000,0.593000,0.593000,0.593000,0.593000,0.593000,0.24,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0, 0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.61,0.742000,0.742000,0.741000,0.741000,0.741000,0.742000,0.742000,0.742000,0.742000,0.742000,0.17,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0, 0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.36,0.862000,0.862000,0.862000,0.861000,0.862000,0.862000,0.862000,0.862000,0.862000,0.862000,0.10,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0, 0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.16,0.942000,0.942000,0.942000,0.942000,0.942000,0.942000,0.942000,0.942000,0.942000,0.942000,0.04,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0, 0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.05,0.983000,0.983000,0.983000,0.983000,0.983000,0.983000,0.983000,0.983000,0.983000,0.983000,0.01,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0, 0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.01,0.997000,0.997000,0.997000,0.997000,0.997000,0.997000,0.997000,0.997000,0.997000,0.997000,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0) ,nrow=20,ncol=40,byrow=T) mt3d-matrix(c(0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,
Re: [R] Fitting an Inverse Gamma Distribution
Hello David, Thank you for pointing me to the GeneralizedHyperbolic package. I've been playing around with hyperbFit with my data, but get: Warning Messages: In besselK(zeta, nu = 1) : value out of range in 'bessel_k' 10 Next, just to see if I could get something to plot I tried the following to no avail: hist(iniSal_US_forHist,breaks=seq(1.1,21,by=0.625),col=grey,freq=F,xlim=c(0,21)) curve(dgig(x,param=c(11,0,-4)),col=blue,from=0,to=20,add=T) Could I trouble you for a bit more guidance? Also, does the fitting function also deal with shifts? In other words, if 500 values were sampled from an inv. gamma and and then had 4 added to them, would the fitting function account for this? Thanks, Eric -- View this message in context: http://r.789695.n4.nabble.com/Fitting-an-Inverse-Gamma-Distribution-tp3216865p3218474.html Sent from the R help mailing list archive at Nabble.com. __ 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] Grid drawing over the filled.contour's legend
Hello, For a reason I can't seem to figure out (have searched posts on this forum for filled.contour grid), the grid (in the code below) is plotting over the legend that accompanies the filled.contour. The dataset has 40 columns and 20 rows. Where have I gone wrong? How can I draw a grid with 40 columns and 20 rows that is constrained to the plotted region and doesn't draw over the legend? vs2d_conc20-read.table(textConnection(0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00
[R] Fitting an Inverse Gamma Distribution
http://r.789695.n4.nabble.com/file/n3216865/Inverse_Gamma.png
Hello,
I am seeking help in estimating the parameters of an inverse gamma
distribution (from the 'actuar' package) using a function like 'fitdistr'.
Unfortunately I haven't found such a package using findFn('fit Inverse
Gamma') from the 'sos' package and was therefore hoping someone might be
aware of such a function?
Secondly, is there a way to shift the pdf (code below) to the right (rather
than the data to the left)? I tried:
par(new=T)
shift-1
hist(iniSal_US_forHist,breaks=seq(1.1,21,by=0.625),col=grey,freq=F,xlim=c(0-shift,21-shift),plot=F)
curve(dinvgamma(x,scale=11.835,shape=4.4242),from=0,to=20,add=T,col=purple,lwd=2)
but this failed in shifting the curve. More broadly, the data plotted in
the histogram represents a calibration target for output generated by a
finite difference solute transport model. The values that will be generated
by the finite difference model would ideally fit the shifted (if that can be
figured out) inverse gamma pdf. To the extent that fit is deemed poor, the
parameter estimation software associated with the finite difference model
will adjust parameters until a better fit is found. I will try to use a
goodness of fit test to determine if one set of parameter values in the
finite difference model produce output that fits the inv. gamma curve herein
better than another. But first I need to establish the curve the finite
difference model output should target and would greatly appreciate any leads
on how it might be shifted and/or fit more precisely by a fitdistr-like
function.
library(MASS)
library(actuar)
iniSal_US_forHist-c(2.368000,3.532614,3.064330,3.347069,3.066333,4.233636,3.465650,2.858553,
2.946731,2.945417,2.415000,2.873019,5.521000,5.788148,5.314630,5.509672,6.032840,6.009310,
4.110833,6.073182,5.652833,4.425733,6.481852,4.076857,3.289310,4.524000,3.985811,5.399714,
4.490606,6.956729,5.270933,8.099107,5.058250,6.394500,5.644000,5.202459,5.67,3.152680,
3.220952,2.777381,3.115467,3.642759,3.488333,3.022439,2.610290,2.618571,3.218000,3.417634,
10.327317,7.344270,6.886154,4.015800,3.063103,6.832292,4.600238,2.939000,5.999027,7.894878,
4.411538,2.384762,6.816154,2.782500,2.475333,2.799138,2.739063,2.619917,2.892545,2.468167,
2.577079,2.821875,2.502500,2.969032,2.046023,3.073077,4.408000,3.411774,3.50,4.283607,
4.284000,4.276714,3.228103,2.639875,3.453194,2.821200,3.838723,1.714253,2.273750,2.611882,
2.321781,2.567500,2.557045,1.288875,2.175211,1.736000,2.250781,7.433366,7.033553,5.47,
7.132727,8.505937,9.174545,6.554487,7.060286,6.617160,8.210986,4.404045,6.062381,5.149625,
2.972105,5.358889,3.910968,3.715873,1.728966,2.843667,4.413906,3.016346,7.168636,3.839394,
3.930141,7.019882,3.459429,5.050250,3.492714,3.226667,3.987667,2.770227,3.661167,1.553000,
2.867391,2.897193,2.611707,2.577167,2.904697,2.733077,2.507241,11.044865,6.425484,8.567222,
8.552344,7.493396,4.807381,9.697869,9.471333,6.783175,4.563571,8.059649,9.448679,5.803778,
4.769423,4.424634,7.586042,4.451556,3.622373,6.390152,4.424375,4.135806,5.025400,5.410635,
7.012292,2.961071,3.192188,2.989643,3.471429,2.867966,1.980541,3.172344,2.574783,2.958983,
1.708140,3.604853,3.479000,2.845000,2.742603,2.923968,3.620308,2.452500,2.721375,3.166333,
2.742162,2.793000,3.337000,5.192025,5.365875,3.079000,8.415970,6.612277,6.734706,4.856857,
5.164783,7.743667,6.894151,4.666538,9.227167,8.077581,6.109833,6.621724,18.098182,12.705600,
15.490784,17.394750,12.422364,14.832727,8.326000,11.352400,3.431429,2.658261,3.219773,3.605185,
4.030299,3.262241,3.503250,3.522763,2.847312,2.996618,3.075769,3.387731,3.066923,3.078200,
2.466957,3.214167,2.707778,3.384839,2.283556,2.912258,3.378000,2.726750,2.95,2.195000,
4.819063,3.604578,3.694906,5.068000,4.676582,3.028831,4.261042,3.593235,4.501224,2.880317,
5.750333,3.257833,3.967458,2.522292,2.725738,2.549231,2.591389,2.990488,2.681222,2.685854,
2.284750,2.585938,2.432824,3.108875,2.611340,3.916667,2.418095,2.476406,2.801235,3.278000,
2.434921,2.617826,3.133939,2.774321,4.196173,3.764286,3.555833,5.317361,3.970800,4.136400,
4.487013,3.746393,4.754000,3.854316,3.742353,3.044079,2.817821,3.995179,3.643134,3.642593,
3.604533,2.935902,4.088310,5.344407,3.076883,3.287105,3.720870,2.032258,2.872593,5.787313,
6.017838,5.425205,4.880600,3.582295,4.90,3.489016,4.603030,5.344407,6.184286,4.047083,
4.788304,4.661325,4.815938,4.056790,3.765595,5.348772,5.200222,4.906311,3.900147,3.782897,
3.767313,3.417732,3.725455,2.888750,2.552333,2.521613,2.531522,2.510833,2.710208,2.445273,
2.619750,2.094737,2.399355,2.758000,2.317077,2.247755,3.594333,4.607805,2.69,3.084706,
3.529000,2.326200,3.309851,2.647805,2.802250,2.778667,3.231235,2.418065,3.134545,3.807843,
2.988372,2.709792,3.084035,3.633279,2.958750,2.170081,2.67,2.640706,2.841600,3.399219,
2.561373,2.574824,3.046447,2.647500,3.554875,1.865000,2.858333,2.355000,2.508082,2.376833,
Re: [R] Strange behavior when trying to piggyback off of fitdistr
Avi, Its been nearly a year since you made this post, but I'm curious if you were able to find a solution to your problem? Your inquiry is closely related to a problem I'm having. Thanks, Eric -- View this message in context: http://r.789695.n4.nabble.com/Strange-behavior-when-trying-to-piggyback-off-of-fitdistr-tp1012536p3209887.html Sent from the R help mailing list archive at Nabble.com. __ 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] Fitting an Inverse Gamma Distribution to Survey Data
Hello, I've been attempting to fit the data below with an inverse gamma distribution. The reason for this is outside proprietary software (@Risk) kicked back a Pearson5 (inverse gamma) as the best fitting distribution with a Chi-Sqr goodness-of-fit roughly 40% better than with a log-normal fit. Looking up Inverse gamma on this forum led me the following post: http://r.789695.n4.nabble.com/Inverse-gamma-td825481.html#a825482 But I think I'm misunderstanding the suggestion made in that post. Is there way to estimate the shape and rate parameters for an inverse-gamma and then plot the PDF as I have done below using other more readily available pdf's in R? Thanks, Eric library(MASS) iniSal_US_forHist-c(2.368000,3.532614,3.064330,3.347069,3.066333,4.233636,3.465650,2.858553, 2.946731,2.945417,2.415000,2.873019,5.521000,5.788148,5.314630,5.509672,6.032840,6.009310, 4.110833,6.073182,5.652833,4.425733,6.481852,4.076857,3.289310,4.524000,3.985811,5.399714, 4.490606,6.956729,5.270933,8.099107,5.058250,6.394500,5.644000,5.202459,5.67,3.152680, 3.220952,2.777381,3.115467,3.642759,3.488333,3.022439,2.610290,2.618571,3.218000,3.417634, 10.327317,7.344270,6.886154,4.015800,3.063103,6.832292,4.600238,2.939000,5.999027,7.894878, 4.411538,2.384762,6.816154,2.782500,2.475333,2.799138,2.739063,2.619917,2.892545,2.468167, 2.577079,2.821875,2.502500,2.969032,2.046023,3.073077,4.408000,3.411774,3.50,4.283607, 4.284000,4.276714,3.228103,2.639875,3.453194,2.821200,3.838723,1.714253,2.273750,2.611882, 2.321781,2.567500,2.557045,1.288875,2.175211,1.736000,2.250781,7.433366,7.033553,5.47, 7.132727,8.505937,9.174545,6.554487,7.060286,6.617160,8.210986,4.404045,6.062381,5.149625, 2.972105,5.358889,3.910968,3.715873,1.728966,2.843667,4.413906,3.016346,7.168636,3.839394, 3.930141,7.019882,3.459429,5.050250,3.492714,3.226667,3.987667,2.770227,3.661167,1.553000, 2.867391,2.897193,2.611707,2.577167,2.904697,2.733077,2.507241,11.044865,6.425484,8.567222, 8.552344,7.493396,4.807381,9.697869,9.471333,6.783175,4.563571,8.059649,9.448679,5.803778, 4.769423,4.424634,7.586042,4.451556,3.622373,6.390152,4.424375,4.135806,5.025400,5.410635, 7.012292,2.961071,3.192188,2.989643,3.471429,2.867966,1.980541,3.172344,2.574783,2.958983, 1.708140,3.604853,3.479000,2.845000,2.742603,2.923968,3.620308,2.452500,2.721375,3.166333, 2.742162,2.793000,3.337000,5.192025,5.365875,3.079000,8.415970,6.612277,6.734706,4.856857, 5.164783,7.743667,6.894151,4.666538,9.227167,8.077581,6.109833,6.621724,18.098182,12.705600, 15.490784,17.394750,12.422364,14.832727,8.326000,11.352400,3.431429,2.658261,3.219773,3.605185, 4.030299,3.262241,3.503250,3.522763,2.847312,2.996618,3.075769,3.387731,3.066923,3.078200, 2.466957,3.214167,2.707778,3.384839,2.283556,2.912258,3.378000,2.726750,2.95,2.195000, 4.819063,3.604578,3.694906,5.068000,4.676582,3.028831,4.261042,3.593235,4.501224,2.880317, 5.750333,3.257833,3.967458,2.522292,2.725738,2.549231,2.591389,2.990488,2.681222,2.685854, 2.284750,2.585938,2.432824,3.108875,2.611340,3.916667,2.418095,2.476406,2.801235,3.278000, 2.434921,2.617826,3.133939,2.774321,4.196173,3.764286,3.555833,5.317361,3.970800,4.136400, 4.487013,3.746393,4.754000,3.854316,3.742353,3.044079,2.817821,3.995179,3.643134,3.642593, 3.604533,2.935902,4.088310,5.344407,3.076883,3.287105,3.720870,2.032258,2.872593,5.787313, 6.017838,5.425205,4.880600,3.582295,4.90,3.489016,4.603030,5.344407,6.184286,4.047083, 4.788304,4.661325,4.815938,4.056790,3.765595,5.348772,5.200222,4.906311,3.900147,3.782897, 3.767313,3.417732,3.725455,2.888750,2.552333,2.521613,2.531522,2.510833,2.710208,2.445273, 2.619750,2.094737,2.399355,2.758000,2.317077,2.247755,3.594333,4.607805,2.69,3.084706, 3.529000,2.326200,3.309851,2.647805,2.802250,2.778667,3.231235,2.418065,3.134545,3.807843, 2.988372,2.709792,3.084035,3.633279,2.958750,2.170081,2.67,2.640706,2.841600,3.399219, 2.561373,2.574824,3.046447,2.647500,3.554875,1.865000,2.858333,2.355000,2.508082,2.376833, 2.728710,1.752833,1.571967,1.800333,2.265455,2.353226,2.568028,2.359500,2.472639,1.675385, 2.667386,2.49,2.482632,2.593452,2.695510,2.466941,2.624211,3.835645,3.519667,2.661940, 3.516167,3.146585,3.420462,4.809833,3.028500,3.192297,4.256333,2.516897,3.033846,2.793359, 6.700714,5.441250,6.872500,4.528333,7.490328,4.673788,6.376885,3.023167,4.429167,4.317625, 16.729231,8.372500,6.279828,10.805098,8.359452,8.277576,8.008846,8.742308,12.155942,5.975063, 3.317333,2.883021,3.310822,3.119219,3.921190,3.552986,3.647162,4.017667,3.895342,3.381096, 2.769412,3.225294,4.169333,3.733919,2.859492,3.674875,3.401017,3.160267,4.109545,4.347867, 3.867000,3.672763,4.364844,3.804250,2.613000,4.289909,4.212059,4.785797,4.149687,6.299444, 5.640135,5.713448,4.766438,7.032027,5.610533,3.126111,6.322029,4.417692,6.392532,2.753175, 2.549655,3.456533,3.199000,3.852338,3.387549,3.098033,3.271733,3.679180,2.655484,6.858167, 5.808033,7.55,8.388387,5.108732,7.895152,5.223580,5.741714,8.026250,4.928421,2.797292,
Re: [R] optim and singularity
Thanks Berend, setting lower=c(0,0.01) seems to have solved the problem. I have a dataset with 200+ cropped fields with associated data and wanted to generate each plot in a way that minimized the visual residuals between the green squares and blue circles. I have inserted a resulting plot so you can see how your suggestion has helped me accomplish this end. Eric http://r.789695.n4.nabble.com/file/n3169658/Field__18_.png -- View this message in context: http://r.789695.n4.nabble.com/optim-and-singularity-tp3168722p3169658.html Sent from the R help mailing list archive at Nabble.com. __ 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] optim and singularity
Hello, I was unable to find clues to my problem in ?optim. Using the data and code below, I get an error (system is exactly singular) when a particular line of code is left in, but have found that 'optim' works when I comment it out. The line of code in question is after the closeAllConnections() line of code and contains a call to na.approx from the zoo package. What's confusing to me is why optim works without it versus with it? Any clues would be very helpful in moving forward. The overall goal of the code below is to minimize the residuals between the blue circles and green squares by adjusting the limits of the secondary y-axis. Thanks, Eric library(zoo) temp.dat-read.table(textConnection(Well Meas_Date WTD ECgw GW_Project_Area Region Avg.EM.Survey.Value Avg.Soil.Paste.EC SD_EM_Survey 1 4/12/1999 1.75 2.27 LOWER-UPSTREAM US NA NA NA 1 5/11/1999 1.24 5.04 LOWER-UPSTREAM US NA NA NA 1 5/27/1999 1.27 4.45 LOWER-UPSTREAM US NA NA NA 1 6/3/1999 1.27 4.09 LOWER-UPSTREAM US 3.347069 3.126667 0.6347013 1 6/11/1999 1.52 2.84 LOWER-UPSTREAM US NA NA NA 1 6/18/1999 1.19 2.34 LOWER-UPSTREAM US NA NA NA 1 6/24/1999 1.21 2.48 LOWER-UPSTREAM US NA NA NA 1 6/30/1999 1.19 2.45 LOWER-UPSTREAM US NA NA NA 1 7/7/1999 1.59 2.47 LOWER-UPSTREAM US NA NA NA 1 7/14/1999 1.75 2.73 LOWER-UPSTREAM US NA NA NA 1 7/21/1999 1.75 2.33 LOWER-UPSTREAM US NA NA NA 1 7/28/1999 1.78 2.63 LOWER-UPSTREAM US NA NA NA 1 8/4/1999 1.37 2.25 LOWER-UPSTREAM US NA NA NA 1 8/12/1999 0.08 2.37 LOWER-UPSTREAM US 2.858553 NA 0.3560363 1 8/18/1999 2.29 2.48 LOWER-UPSTREAM US NA NA NA 1 9/4/1999 0.02 2.48 LOWER-UPSTREAM US NA NA NA 1 9/16/1999 1.93 2.45 LOWER-UPSTREAM US NA NA NA 1 10/1/1999 2.01 2.49 LOWER-UPSTREAM US NA NA NA 1 11/13/1999 1.91 2.32 LOWER-UPSTREAM US NA NA NA 1 1/4/2000 1.83 2.59 LOWER-UPSTREAM US NA NA NA 1 2/24/2000 1.93 2.83 LOWER-UPSTREAM US NA NA NA 1 3/25/2000 1.73 2.71 LOWER-UPSTREAM US NA NA NA 1 4/15/2000 1.47 3.12 LOWER-UPSTREAM US NA NA NA 1 4/28/2000 1.98 2.76 LOWER-UPSTREAM US NA NA NA 1 5/17/2000 1.80 2.92 LOWER-UPSTREAM US NA NA NA 1 5/30/2000 NA NA NA US 3.064330 NA 0.3970829 1 6/1/2000 1.58 2.83 LOWER-UPSTREAM US NA NA NA 1 6/6/2000 1.65 2.97 LOWER-UPSTREAM US NA NA NA 1 6/15/2000 1.85 2.65 LOWER-UPSTREAM US NA NA NA 1 6/23/2000 1.90 2.78 LOWER-UPSTREAM US NA NA NA 1 6/29/2000 1.80 2.67 LOWER-UPSTREAM US NA NA NA 1 7/6/2000 2.00 2.69 LOWER-UPSTREAM US NA NA NA 1 7/14/2000 1.97 3.46 LOWER-UPSTREAM US NA NA NA 1 7/20/2000 1.69 2.57 LOWER-UPSTREAM US NA NA NA 1 7/27/2000 1.95 3.16 LOWER-UPSTREAM US NA NA NA 1 8/2/2000 2.03 3.12 LOWER-UPSTREAM US NA NA NA 1 8/8/2000 NA NA NA US 3.185567 2.522467 0.3765025 1 8/11/2000 2.00 2.65 LOWER-UPSTREAM US NA NA NA 1 8/19/2000 1.95 2.50 LOWER-UPSTREAM US NA NA NA 1 9/29/2000 2.11 2.11 LOWER-UPSTREAM US NA NA NA 1 10/20/2000 2.16 2.75 LOWER-UPSTREAM US NA NA NA 1 12/1/2000 1.83 2.81 LOWER-UPSTREAM US NA NA NA 1 1/19/2001 2.01 3.01 LOWER-UPSTREAM US NA NA NA 1 3/5/2001 2.01 2.35 LOWER-UPSTREAM US NA NA NA 1 3/30/2001 2.13 2.10 LOWER-UPSTREAM US NA NA NA 1 4/21/2001 1.87 2.36 LOWER-UPSTREAM US NA NA NA 1 5/15/2001 2.00 2.37 LOWER-UPSTREAM US NA NA NA 1 5/29/2001 1.60 2.59 LOWER-UPSTREAM US NA NA NA 1 6/4/2001 NA NA NA US 4.233636 NA 1.3738540 1 6/7/2001 1.71 2.55 LOWER-UPSTREAM US NA NA NA 1 6/14/2001 1.83 2.75 LOWER-UPSTREAM US NA NA NA 1 6/20/2001 1.93 3.02 LOWER-UPSTREAM US NA NA NA 1 6/29/2001 1.80 2.74 LOWER-UPSTREAM US NA NA NA 1 7/6/2001 1.91 2.70 LOWER-UPSTREAM US NA NA NA 1 7/12/2001 2.04 2.46 LOWER-UPSTREAM US NA NA NA 1 7/18/2001 1.87 2.55 LOWER-UPSTREAM US NA NA NA 1 7/25/2001 1.96 2.67 LOWER-UPSTREAM US NA NA NA 1 8/2/2001 2.02 2.42 LOWER-UPSTREAM US NA NA NA 1 8/8/2001 NA NA NA US 2.946731 NA 0.3748240 1 8/10/2001 1.94 2.36 LOWER-UPSTREAM US NA NA NA 1 8/17/2001 1.94 0.75 LOWER-UPSTREAM US NA NA NA 1 9/1/2001 2.12 2.22 LOWER-UPSTREAM US NA NA NA 1 9/15/2001 2.15 2.35 LOWER-UPSTREAM US NA NA NA 1 9/29/2001 1.95 2.20 LOWER-UPSTREAM US NA NA NA 1 10/27/2001 1.08 1.72 LOWER-UPSTREAM US NA NA NA 1 11/17/2001 1.99 2.21 LOWER-UPSTREAM US NA NA NA 1 1/10/2002 2.01 2.31 LOWER-UPSTREAM US NA NA NA 1 2/16/2002 1.94 1.74 LOWER-UPSTREAM US NA NA NA 1 3/23/2002 2.03 2.46 LOWER-UPSTREAM US NA NA NA 1 4/13/2002 2.11 2.50 LOWER-UPSTREAM US NA NA NA 1 5/11/2002 2.11 2.54 LOWER-UPSTREAM US NA NA NA 1 5/21/2002 NA NA NA US 2.368000 NA 0.3235852 1 5/30/2002 2.07 2.56 LOWER-UPSTREAM US NA NA NA 1 6/6/2002 2.11 2.55 LOWER-UPSTREAM US NA NA NA 1 6/13/2002 2.03 2.66 LOWER-UPSTREAM US NA NA NA 1 6/27/2002 2.19 2.55 LOWER-UPSTREAM US NA NA NA 1 7/3/2002 2.22 2.68 LOWER-UPSTREAM US NA NA NA 1 7/11/2002 2.06 2.61 LOWER-UPSTREAM US NA NA NA 1 7/18/2002 2.23 2.58 LOWER-UPSTREAM US NA NA NA 1 7/25/2002 2.23 2.62 LOWER-UPSTREAM US NA NA NA 1 8/1/2002 2.21 2.60 LOWER-UPSTREAM US NA NA NA 1 8/8/2002 2.29 2.60 LOWER-UPSTREAM US NA NA NA 1 8/14/2002 NA NA NA US 1.863548 NA 0.2400621 1 8/15/2002 2.26 2.61 LOWER-UPSTREAM US NA NA NA 1 8/21/2002 2.34 2.51 LOWER-UPSTREAM US NA NA NA 1 9/6/2002 2.31 2.51
Re: [R] optim and singularity
Hi Tom, I followed you suggestion and found the following: class(temp.dat$WTD) #[1] numeric temp.dat$WTD-na.approx(temp.dat$WTD,as.numeric(as.Date(temp.dat$Meas_Date,%m/%d/%Y))) class(temp.dat$WTD) #[1] numeric Have I misunderstood you? If not, then my original confusion still stands... Respectfully, Eric -- View this message in context: http://r.789695.n4.nabble.com/optim-and-singularity-tp3168722p3168955.html Sent from the R help mailing list archive at Nabble.com. __ 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] adjust secondary y-axis bounds to minimize visual residuals
Hello, I'm plotting two sets of data referenced to either the left or right y-axes. The first, water table depth (blue circles), is plotted on the left y-axis in reverse order (0 at the top) as this is more intuitive when thinking in terms of depth. The second is electrical conductance (a surrogate for salinity), and is referenced to the right y-axis. The data and plot commands follow below. I've been trying to figure out a way to minimize the variable called min_this by changing the values in ylim_2 only, with one constraint: ylim_2[1]0. I'm unsure of how to stuff the code that follows the plot commands into a loop that adjusts ylim_2 so as to minimize min_this? I'm guessing there is a function to do this? Thanks, Eric library(zoo) #example data and plotting commands follow temp.dat-read.table(textConnection(Well Meas_Date WTD ECgw GW_Project_Area Region Avg.EM.Survey.Value Avg.Soil.Paste.EC SD_EM_Survey 1 4/12/1999 1.75 2.27 LOWER-UPSTREAM US NA NA NA 1 5/11/1999 1.24 5.04 LOWER-UPSTREAM US NA NA NA 1 5/27/1999 1.27 4.45 LOWER-UPSTREAM US NA NA NA 1 6/3/1999 1.27 4.09 LOWER-UPSTREAM US 3.347069 3.126667 0.6347013 1 6/11/1999 1.52 2.84 LOWER-UPSTREAM US NA NA NA 1 6/18/1999 1.19 2.34 LOWER-UPSTREAM US NA NA NA 1 6/24/1999 1.21 2.48 LOWER-UPSTREAM US NA NA NA 1 6/30/1999 1.19 2.45 LOWER-UPSTREAM US NA NA NA 1 7/7/1999 1.59 2.47 LOWER-UPSTREAM US NA NA NA 1 7/14/1999 1.75 2.73 LOWER-UPSTREAM US NA NA NA 1 7/21/1999 1.75 2.33 LOWER-UPSTREAM US NA NA NA 1 7/28/1999 1.78 2.63 LOWER-UPSTREAM US NA NA NA 1 8/4/1999 1.37 2.25 LOWER-UPSTREAM US NA NA NA 1 8/12/1999 0.08 2.37 LOWER-UPSTREAM US 2.858553 NA 0.3560363 1 8/18/1999 2.29 2.48 LOWER-UPSTREAM US NA NA NA 1 9/4/1999 0.02 2.48 LOWER-UPSTREAM US NA NA NA 1 9/16/1999 1.93 2.45 LOWER-UPSTREAM US NA NA NA 1 10/1/1999 2.01 2.49 LOWER-UPSTREAM US NA NA NA 1 11/13/1999 1.91 2.32 LOWER-UPSTREAM US NA NA NA 1 1/4/2000 1.83 2.59 LOWER-UPSTREAM US NA NA NA 1 2/24/2000 1.93 2.83 LOWER-UPSTREAM US NA NA NA 1 3/25/2000 1.73 2.71 LOWER-UPSTREAM US NA NA NA 1 4/15/2000 1.47 3.12 LOWER-UPSTREAM US NA NA NA 1 4/28/2000 1.98 2.76 LOWER-UPSTREAM US NA NA NA 1 5/17/2000 1.80 2.92 LOWER-UPSTREAM US NA NA NA 1 5/30/2000 NA NA NA US 3.064330 NA 0.3970829 1 6/1/2000 1.58 2.83 LOWER-UPSTREAM US NA NA NA 1 6/6/2000 1.65 2.97 LOWER-UPSTREAM US NA NA NA 1 6/15/2000 1.85 2.65 LOWER-UPSTREAM US NA NA NA 1 6/23/2000 1.90 2.78 LOWER-UPSTREAM US NA NA NA 1 6/29/2000 1.80 2.67 LOWER-UPSTREAM US NA NA NA 1 7/6/2000 2.00 2.69 LOWER-UPSTREAM US NA NA NA 1 7/14/2000 1.97 3.46 LOWER-UPSTREAM US NA NA NA 1 7/20/2000 1.69 2.57 LOWER-UPSTREAM US NA NA NA 1 7/27/2000 1.95 3.16 LOWER-UPSTREAM US NA NA NA 1 8/2/2000 2.03 3.12 LOWER-UPSTREAM US NA NA NA 1 8/8/2000 NA NA NA US 3.185567 2.522467 0.3765025 1 8/11/2000 2.00 2.65 LOWER-UPSTREAM US NA NA NA 1 8/19/2000 1.95 2.50 LOWER-UPSTREAM US NA NA NA 1 9/29/2000 2.11 2.11 LOWER-UPSTREAM US NA NA NA 1 10/20/2000 2.16 2.75 LOWER-UPSTREAM US NA NA NA 1 12/1/2000 1.83 2.81 LOWER-UPSTREAM US NA NA NA 1 1/19/2001 2.01 3.01 LOWER-UPSTREAM US NA NA NA 1 3/5/2001 2.01 2.35 LOWER-UPSTREAM US NA NA NA 1 3/30/2001 2.13 2.10 LOWER-UPSTREAM US NA NA NA 1 4/21/2001 1.87 2.36 LOWER-UPSTREAM US NA NA NA 1 5/15/2001 2.00 2.37 LOWER-UPSTREAM US NA NA NA 1 5/29/2001 1.60 2.59 LOWER-UPSTREAM US NA NA NA 1 6/4/2001 NA NA NA US 4.233636 NA 1.3738540 1 6/7/2001 1.71 2.55 LOWER-UPSTREAM US NA NA NA 1 6/14/2001 1.83 2.75 LOWER-UPSTREAM US NA NA NA 1 6/20/2001 1.93 3.02 LOWER-UPSTREAM US NA NA NA 1 6/29/2001 1.80 2.74 LOWER-UPSTREAM US NA NA NA 1 7/6/2001 1.91 2.70 LOWER-UPSTREAM US NA NA NA 1 7/12/2001 2.04 2.46 LOWER-UPSTREAM US NA NA NA 1 7/18/2001 1.87 2.55 LOWER-UPSTREAM US NA NA NA 1 7/25/2001 1.96 2.67 LOWER-UPSTREAM US NA NA NA 1 8/2/2001 2.02 2.42 LOWER-UPSTREAM US NA NA NA 1 8/8/2001 NA NA NA US 2.946731 NA 0.3748240 1 8/10/2001 1.94 2.36 LOWER-UPSTREAM US NA NA NA 1 8/17/2001 1.94 0.75 LOWER-UPSTREAM US NA NA NA 1 9/1/2001 2.12 2.22 LOWER-UPSTREAM US NA NA NA 1 9/15/2001 2.15 2.35 LOWER-UPSTREAM US NA NA NA 1 9/29/2001 1.95 2.20 LOWER-UPSTREAM US NA NA NA 1 10/27/2001 1.08 1.72 LOWER-UPSTREAM US NA NA NA 1 11/17/2001 1.99 2.21 LOWER-UPSTREAM US NA NA NA 1 1/10/2002 2.01 2.31 LOWER-UPSTREAM US NA NA NA 1 2/16/2002 1.94 1.74 LOWER-UPSTREAM US NA NA NA 1 3/23/2002 2.03 2.46 LOWER-UPSTREAM US NA NA NA 1 4/13/2002 2.11 2.50 LOWER-UPSTREAM US NA NA NA 1 5/11/2002 2.11 2.54 LOWER-UPSTREAM US NA NA NA 1 5/21/2002 NA NA NA US 2.368000 NA 0.3235852 1 5/30/2002 2.07 2.56 LOWER-UPSTREAM US NA NA NA 1 6/6/2002 2.11 2.55 LOWER-UPSTREAM US NA NA NA 1 6/13/2002 2.03 2.66 LOWER-UPSTREAM US NA NA NA 1 6/27/2002 2.19 2.55 LOWER-UPSTREAM US NA NA NA 1 7/3/2002 2.22 2.68 LOWER-UPSTREAM US NA NA NA 1 7/11/2002 2.06 2.61 LOWER-UPSTREAM US NA NA NA 1 7/18/2002 2.23 2.58 LOWER-UPSTREAM US NA NA NA 1 7/25/2002 2.23 2.62 LOWER-UPSTREAM US NA NA NA 1 8/1/2002 2.21 2.60 LOWER-UPSTREAM US NA NA NA 1 8/8/2002 2.29 2.60 LOWER-UPSTREAM US NA NA NA 1 8/14/2002 NA NA NA US 1.863548 NA 0.2400621
[R] Formatting 'names.arg' in barplot
Hello, I've been looking through ?phantom and ?expression and this forum for examples of how I might be able to manipulate some of the names that appear on the y-axis of the barplot below. For example, the gw in ECgw would appear as a subscript...or qr would be the theta symbol followed by subscript r. My attempts haven't even come close to what I'm after. I could switch to ?text or ?mtext, but thought for positioning purposes it would be easier to do it with names.arg in the barplot function call. Thanks, Eric base.dat.sel2-read.table(textConnection(base.dat.Covariate base.dat.US.Number.Observations base.dat.US.Num.Obs.to.Achieve.Starting.Residual base.dat.US.Percent.Reduction.in.EM38.Surveys base.dat.DS.Number.Observations base.dat.DS.Num.Obs.to.Achieve.Starting.Residual base.dat.DS.Percent.Reduction.in.EM38.Surveys Baseline 391 391 NA 281 281 NA WTD 391 315 0.194 281 210 0.253 ECgw 391 362 0.074 281 280 0.004 SM 391 349 0.107 281 280 0.004 WTD_ECgw 391 300 0.233 281 209 0.256 WTD_SM 391 286 0.269 281 210 0.253 ECgw_SM 391 318 0.187 281 280 0.004 WTD_ECgw_SM 391 269 0.312 281 210 0.253 Sand 391 359 0.082 281 279 0.007 Silt 391 369 0.056 281 280 0.001 Clay 391 370 0.054 281 269 0.043 WTD_ECgw_SM_Sand 391 263 0.327 281 181 0.356 qr 391 360 0.079 281 270 0.038 Ks 391 370 0.054 281 276 0.017 qr_Ks 391 358 0.084 281 271 0.035 WTD_ECgw_SM_qr 391 261 0.332 281 188 0.331 WTD_ECgw_SM_Ks 391 260 0.335 281 205 0.270),header=T) closeAllConnections() par(mar=c(3,15,2,1)) barplot(base.dat.sel2$base.dat.US.Num.Obs.to.Achieve.Starting.Residual[order(base.dat.sel2$base.dat.US.Num.Obs.to.Achieve.Starting.Residual,decreasing=F)], names.arg=base.dat.sel2$base.dat.Covariate[order(base.dat.sel2$base.dat.US.Num.Obs.to.Achieve.Starting.Residual,decreasing=F)], col=grey,horiz=T,las=1,xlim=c(0,450),yaxs=i,cex.axis=1.2,cex.names=1.1) -- View this message in context: http://r.789695.n4.nabble.com/Formatting-names-arg-in-barplot-tp3078861p3078861.html Sent from the R help mailing list archive at Nabble.com. __ 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] boxplot: reverse y-axis order
Hello, Searching this forum has enabled me to get pretty far in what I'm trying to do. However, there is one more manipulation I would like to make and I haven't found a solution. Using the data and code below, I generate the plot produced by the last command. If possible I would like to reverse the order of the y-axis (bearing in mind horizontal=T) so that 0 is plotted at the upper most part of the y-axis and 3 at the axis intercepts. I've tried the experiment approach to no avail, meaning I've placed rev(...) around various arguments but with wrong results. Thanks, Eric df-read.table(textConnection(Field Date AvgWTD Region variable value hole depth depthM 204 17-Aug-00 2.897428989 US R1 NA R 1 0 205 17-Aug-00 4.017264366 US R1 1.01 R 1 0 206 10-Jun-03 NA US R1 1.19 R 1 0 207 26-May-04 NA US R1 1.6 R 1 0 207 29-May-03 NA US R1 1.23 R 1 0 207 17-Aug-00 NA US R1 1.82 R 1 0 21 03-Jun-99 1.896872044 US R1 NA R 1 0 22 10-Aug-00 1.546097994 US R1 1.3 R 1 0 22 17-Aug-99 1.639824152 US R1 1.11 R 1 0 22 02-Jun-99 1.202943921 US R1 1.44 R 1 0 23 22-May-02 1.52449 US R1 1.58 R 1 0 23 09-Aug-00 1.938527942 US R1 2.2 R 1 0 23 17-Aug-99 0.93040204 US R1 1.36 R 1 0 23 02-Jun-99 1.479804039 US R1 2.65 R 1 0 24 09-Jun-03 NA US R1 1.08 R 1 0 24 04-Jun-01 1.484376073 US R1 NA R 1 0 24 09-Aug-00 1.439671993 US R1 2.41 R 1 0 24 03-Jun-99 1.481328011 US R1 2.36 R 1 0 26A 16-Aug-00 3.128010035 US R1 2.44 R 1 0 26A 06-Apr-99 NA US R1 1.81 R 1 0 26B 10-Aug-00 2.090928078 US R1 1.62 R 1 0 27 26-May-04 NA US R1 1.6 R 1 0 27 29-May-03 NA US R1 NA R 1 0 27 04-Jun-01 NA US R1 2.24 R 1 0 27 09-Aug-00 5.324855804 US R1 9.79 R 1 0 27 10-Jun-99 NA US R1 1.25 R 1 0 28 10-Aug-04 3.565550327 US R1 1.01 R 1 0 28 10-Aug-00 2.169414043 US R1 1.79 R 1 0 28 10-Jun-99 1.633728027 US R1 2.1 R 1 0 29 09-Jun-03 1.847087979 US R1 3.01 R 1 0 29 22-May-02 1.950719953 US R1 0.981 R 1 0 29 08-Jun-01 1.647139192 US R1 1.8 R 1 0 29 15-Aug-00 1.946148038 US R1 3.01 R 1 0 29 04-Jun-99 1.824735999 US R1 0.845 R 1 0 30 28-May-03 NA US R1 1.09 R 1 0 30 10-Aug-00 3.727704048 US R1 0.76 R 1 0 30 17-Aug-99 NA US R1 NA R 1 0 30 03-Jun-99 NA US R1 1.87 R 1 0 31 15-Aug-00 1.594104052 US R1 2.32 R 1 0 31 17-Aug-99 0.961643994 US R1 0.99 R 1 0 31 03-Jun-99 0.907288015 US R1 1.48 R 1 0 33 04-Jun-01 5.030724049 US R1 1.23 R 1 0 33 15-Aug-00 4.110228062 US R1 2.24 R 1 0 33 12-Aug-99 1.383792043 US R1 NA R 1 0 33 04-Jun-99 1.542287946 US R1 1.42 R 1 0 35 14-Aug-00 1.918715954 US R1 2.1 R 1 0 35 17-Aug-99 1.495044112 US R1 1.52 R 1 0 35 06-Apr-99 NA US R1 2.02 R 1 0 37 22-May-02 4.803647995 US R1 2.12 R 1 0 37 06-Aug-01 3.499104261 US R1 1.91 R 1 0 37 16-Aug-00 3.473958015 US R1 2.41 R 1 0 37 08-Jun-99 NA US R1 0.648 R 1 0 38 26-May-04 NA US R1 1.94 R 1 0 38 04-Aug-03 NA US R1 2.02 R 1 0 38 16-Aug-01 2.645663977 US R1 4.97 R 1 0 38 08-Jun-99 1.240535975 US R1 1.06 R 1 0 39 16-Aug-00 1.674571276 US R1 3.25 R 1 0 39 08-Jun-99 0.774192035 US R1 7.41 R 1 0 40 09-Aug-04 NA US R1 18 R 1 0 40 26-May-04 NA US R1 7.03 R 1 0 40 29-May-03 NA US R1 3.61 R 1 0 40 04-Jun-01 0.669035971 US R1 3.92 R 1 0 40 16-Aug-00 0.518160045 US R1 2.81 R 1 0 40 11-Aug-99 0.232257605 US R1 NA R 1 0 41 09-Aug-05 0.704087973 US R1 1.65 R 1 0 41 26-May-04 1.865375996 US R1 1.62 R 1 0 41 04-Aug-03 1.544319987 US R1 0.968 R 1 0 41 02-Jun-03 2.767584085 US R1 1.02 R 1 0 41 11-Aug-00 1.138427973 US R1 0.98 R 1 0 41 12-Aug-99 0.924763203 US R1 1.64 R 1 0 41 09-Jun-99 0.964184046 US R1 1.24 R 1 0 42 11-Aug-00 1.082802057 US R1 1.09 R 1 0 43 09-Jun-03 1.775460005 US R1 1.14 R 1 0 43 30-May-01 1.38074398 US R1 2.38 R 1 0 43 16-Aug-00 0.61214 US R1 2.23 R 1 0 43 12-Aug-99 0.583996832 US R1 4.97 R 1 0 43 09-Jun-99 0.838199973 US R1 3.41 R 1 0 45 06-Jun-01 5.977128029 US R1 1.72 R 1 0 45 10-Aug-00 5.331714153 US R1 1.57 R 1 0 45 10-Jun-99 NA US R1 1.13 R 1 0 48 09-Aug-04 3.940454483 US R1 0.886 R 1 0 48 28-May-03 NA US R1 1.15 R 1 0 48 30-May-01 2.782824039 US R1 2.79 R 1 0 48 15-Aug-00 2.336292028 US R1 7.22 R 1 0 48 12-Aug-99 1.379220009 US R1 NA R 1 0 49 16-Aug-00 2.613964796 US R1 2.1 R 1 0 49 31-May-99 NA US R1 NA R 1 0 5 08-Aug-00 NA US R1 1.19 R 1 0 50 16-Aug-00 2.816961765 US R1 1.02 R 1 0 50 11-Aug-99 2.66755 US R1 1.45 R 1 0 50 31-May-99 NA US R1 1.54 R 1 0 51 02-Jun-03 NA US R1 4.15 R 1 0 51 16-Aug-00 NA US R1 4.68 R 1 0 51 11-Aug-99 NA US R1 NA R 1 0 51 09-Jun-99 5.42076 US R1 2.37 R 1 0 52 04-Aug-03 2.141219854 US R1 1.88 R 1 0 52 09-Jun-03 1.549908042 US R1 1.13 R 1 0 52 04-Jun-01 2.020823956 US R1 3.13 R 1 0 52 10-Aug-99 2.029206038 US R1 NA R 1 0 52 09-Jun-99 1.862328053 US R1 1.56 R 1 0 53 12-Aug-99 NA US R1 1.37 R 1 0 53 09-Jun-99 NA US R1 3.77 R 1 0 57 11-Aug-99 1.907286048 US R1 4.58 R 1 0 57 09-Jun-99 0.930655956 US R1 11.1 R 1 0 6 28-May-03 3.97459197 US R1 1.35 R 1 0 6 31-May-01 4.736591816 US R1 2.69 R 1 0 6 08-Aug-00 5.587999821 US R1 1.43 R 1 0 60A 11-Jun-99 NA US R1 1.56 R 1 0 61 22-May-02 2.862071991 US R1 1.57 R 1 0 61 06-Aug-01 2.178558111 US R1 1.69 R 1 0 61
[R] barplot3d cutting off labels
Hello, Using the function that is available at: http://addictedtor.free.fr/graphiques/sources/source_161.R The following command leads to a plot with labels that are cut off: barplot3d(c(4.01,4.71,2.03,1.28,1.42,1.76,0, 6.58,3.25,3.11,2.10,2.05,1.19,1.28, 6.44,5.50,3.69,4.53,3.33,1.70,2.57, 6.01,5.36,5.90,6.61,4.67,2.62,3.83, 10.69,5.90,5.62,6.64,5.71,5.35,3.18, 8.98,7.30,7.73,7.23,14.10,8.65,4.00, 13.39,10.91,5.57,7.34,6.03,5.44,3.24), rows=7, theta = 55, phi = 22, expand=0.9, col.lab=c(GWTD 1,1 = GWTD 2,2 = GWTD 3,3 = GWTD 4,4 = GWTD 5,5 = GWTD 6,GWTD 6), row.lab=c(GW ECe 1,1 = GW ECe 2,2 = GW ECe 3,3 = GW ECe 4,4 = GW ECe 5,5 = GW ECe 6,GW ECe 6), col.bar=c(#FF6633,#33,#99CCFF,#9933FF,#44ff58,#CC,#FF), z.lab=Soil ECe) I've also tried to use par(mar=c(5,5,5,5)) or par(oma=c(5,5,5,5)), but to no avail. Any ideas? Thanks, E -- View this message in context: http://r.789695.n4.nabble.com/barplot3d-cutting-off-labels-tp3041309p3041309.html Sent from the R help mailing list archive at Nabble.com. __ 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] histograms resulting from call to soil.texture
Hello, Using the dataset below I produce a soil.texture plot (R code for this included at the end). One will notice I have plotted the points based on whether or not they are in a region called 'upstream' or 'downstream'. I'm curious if there is a way to somehow extract counts of the number of points (red points and blue points) falling within each classification (e.g. silty loam, sandy clay, etc.)? Eric soil.dat-read.table(textConnection(Field_Num Region Region_Num Region_Sym Region_Col Date Location Location_Num sand silt clay 1 US 1 1 4 6/3/1999 Surface 1 26.25 52.5 21.25 1 US 1 1 4 8/8/2000 Surface 1 27.5 52.5 20 1 US 1 1 4 8/8/2000 One_Foot 2 27.5 50 22.5 1 US 1 1 4 8/8/2000 Two_Foot 3 80 20 0 1 US 1 1 4 8/8/2000 Three_Foot 4 67.5 22.5 10 1 US 1 1 4 8/8/2000 Four_Foot 5 95 0 5 2 US 1 1 4 8/8/2000 Surface 1 55 12.5 32.5 2 US 1 1 4 6/4/2001 Surface 1 48.75 43.75 7.5 2 US 1 1 4 6/1/1999 Surface 1 28 34 38 2 US 1 1 4 8/8/2000 One_Foot 2 62.5 7.5 30 2 US 1 1 4 6/4/2001 One_Foot 2 65 16.25 18.75 2 US 1 1 4 8/8/2000 Two_Foot 3 55 0 45 2 US 1 1 4 6/4/2001 Two_Foot 3 50 21.25 28.75 2 US 1 1 4 8/8/2000 Three_Foot 4 47.5 5 47.5 2 US 1 1 4 6/4/2001 Three_Foot 4 77.5 15 7.5 2 US 1 1 4 8/8/2000 Four_Foot 5 62.5 0 37.5 2 US 1 1 4 6/4/2001 Four_Foot 5 75 15 10 5 US 1 1 4 8/1/2000 Surface 1 25 36.25 38.75 5 US 1 1 4 6/9/1999 Surface 1 48.75 31.25 20 5 US 1 1 4 8/1/2000 One_Foot 2 25 40 35 5 US 1 1 4 8/1/2000 Two_Foot 3 21.13 30.99 47.89 5 US 1 1 4 8/1/2000 Three_Foot 4 37.5 25 37.5 5 US 1 1 4 8/1/2000 Four_Foot 5 37.5 25 37.5 6 US 1 1 4 5/28/2003 Surface 1 37.5 37.5 25 6 US 1 1 4 8/8/2000 Surface 1 37.5 22.5 40 6 US 1 1 4 5/31/2001 Surface 1 21.25 35 43.75 6 US 1 1 4 5/28/2003 One_Foot 2 77.5 10 12.5 6 US 1 1 4 8/8/2000 One_Foot 2 82.5 5 12.5 6 US 1 1 4 5/31/2001 One_Foot 2 61.25 21.25 17.5 6 US 1 1 4 5/28/2003 Two_Foot 3 82.5 6.25 11.25 6 US 1 1 4 8/8/2000 Two_Foot 3 85 5 10 6 US 1 1 4 5/31/2001 Two_Foot 3 76.25 15 8.75 6 US 1 1 4 5/28/2003 Three_Foot 4 85 5 10 6 US 1 1 4 8/8/2000 Three_Foot 4 72.5 7.5 20 6 US 1 1 4 5/31/2001 Three_Foot 4 68.75 16.25 15 6 US 1 1 4 5/28/2003 Four_Foot 5 67.5 18.75 13.75 6 US 1 1 4 8/8/2000 Four_Foot 5 67.5 7.5 25 6 US 1 1 4 5/31/2001 Four_Foot 5 76.25 13.75 10 6 US 1 1 4 5/28/2003 Six_Foot 6 61.25 23.75 15 7 US 1 1 4 8/4/2003 Surface 1 40 42.5 17.5 7 US 1 1 4 6/2/2003 Surface 1 41.25 46.25 12.5 7 US 1 1 4 8/8/2000 Surface 1 52.5 12.5 35 7 US 1 1 4 6/1/2001 Surface 1 40 27.5 32.5 7 US 1 1 4 8/4/2003 One_Foot 2 42.5 40 17.5 7 US 1 1 4 6/2/2003 One_Foot 2 40 48.75 11.25 7 US 1 1 4 8/8/2000 One_Foot 2 47.5 10 42.5 7 US 1 1 4 6/1/2001 One_Foot 2 47.5 26.25 26.25 7 US 1 1 4 8/4/2003 Two_Foot 3 65 20 15 7 US 1 1 4 6/2/2003 Two_Foot 3 40 55 5 7 US 1 1 4 8/8/2000 Two_Foot 3 45 5 50 7 US 1 1 4 6/1/2001 Two_Foot 3 57.5 17.5 25 7 US 1 1 4 8/4/2003 Three_Foot 4 77.5 11.25 11.25 7 US 1 1 4 6/2/2003 Three_Foot 4 85 12.5 2.5 7 US 1 1 4 8/8/2000 Three_Foot 4 60 7.5 32.5 7 US 1 1 4 6/1/2001 Three_Foot 4 65 18.75 16.25 7 US 1 1 4 8/4/2003 Four_Foot 5 22.5 60 17.5 7 US 1 1 4 6/2/2003 Four_Foot 5 86.25 6.25 7.5 7 US 1 1 4 8/8/2000 Four_Foot 5 25 20 55 7 US 1 1 4 6/1/2001 Four_Foot 5 37.5 35 27.5 7 US 1 1 4 8/4/2003 Six_Foot 6 52.5 33.75 13.75 7 US 1 1 4 6/2/2003 Six_Foot 6 56.25 36.25 7.5 8 US 1 1 4 8/8/2000 Surface 1 40 15 45 8 US 1 1 4 6/3/1999 Surface 1 32.5 45 22.5 8 US 1 1 4 8/8/2000 One_Foot 2 45 21.25 33.75 8 US 1 1 4 8/8/2000 Two_Foot 3 33.75 20 46.25 8 US 1 1 4 8/8/2000 Three_Foot 4 38.75 18.75 42.5 8 US 1 1 4 8/8/2000 Four_Foot 5 32.5 23.75 43.75 11 US 1 1 4 8/4/2003 Surface 1 28.75 45 26.25 11 US 1 1 4 5/29/2003 Surface 1 27.5 47.5 25 11 US 1 1 4 8/9/2000 Surface 1 32.5 20 47.5 11 US 1 1 4 5/31/2001 Surface 1 25 30 45 11 US 1 1 4 6/2/1999 Surface 1 26 34 40 11 US 1 1 4 8/4/2003 One_Foot 2 20 50 30 11 US 1 1 4 5/29/2003 One_Foot 2 25 47.5 27.5 11 US 1 1 4 8/9/2000 One_Foot 2 25 25 50 11 US 1 1 4 5/31/2001 One_Foot 2 12.5 37.5 50 11 US 1 1 4 8/4/2003 Two_Foot 3 20 20 60 11 US 1 1 4 5/29/2003 Two_Foot 3 20 57.5 22.5 11 US 1 1 4 8/9/2000 Two_Foot 3 27.5 30 42.5 11 US 1 1 4 5/31/2001 Two_Foot 3 10 37.5 52.5 11 US 1 1 4 8/4/2003 Three_Foot 4 22.5 60 17.5 11 US 1 1 4 5/29/2003 Three_Foot 4 26.25 51.25 22.5 11 US 1 1 4 8/9/2000 Three_Foot 4 30 22.5 47.5 11 US 1 1 4 5/31/2001 Three_Foot 4 15 22.5 62.5 11 US 1 1 4 8/4/2003 Four_Foot 5 30 47.5 22.5 11 US 1 1 4 5/29/2003 Four_Foot 5 35 43.75 21.25 11 US 1 1 4 8/9/2000 Four_Foot 5 30 22.5 47.5 11 US 1 1 4 5/31/2001 Four_Foot 5 20 25 55 11 US 1 1 4 8/4/2003 Six_Foot 6 51.25 18.75 30 11 US 1 1 4 5/29/2003 Six_Foot 6 56.25 17.5 26.25 12 US 1 1 4 8/4/2003 Surface 1 28.75 55 16.25 12 US 1 1 4 6/9/2003 Surface 1 32.5 36.25 31.25 12 US 1 1 4 8/9/2000 Surface 1 30 27.5 42.5 12 US 1 1 4 6/9/1999 Surface 1 37.5 38.75 23.75 12 US 1 1 4 8/4/2003 One_Foot 2 18.75 45 36.25 12 US 1 1 4 6/9/2003 One_Foot 2 21.25 43.75 35 12 US 1 1 4 8/9/2000 One_Foot 2 15 38.75 46.25 12 US 1 1 4 8/4/2003 Two_Foot 3 15 45 40 12 US 1 1 4 6/9/2003 Two_Foot 3 12.5 38.75 48.75 12 US 1 1
[R] hdiffplot
Hello, I'm trying to track down more information on hdiffplot than what is supplied in ?hdiffplot. More specifically, the example code found at the bottom ?hdiffplot (##Compare *three* of them:) is something I'm very interested in using for my own analysis. However, I don't see how to add a legend or even how to dig into the bnlst object that is created for determining what the displayed colors mean. The following example helps illustrate what I mean: library(hexbin) x1 - rnorm(1) y1 - rnorm(1) x2 - rnorm(1,mean=1) y2 - rnorm(1,mean=1) x3 - rnorm(1,mean=1) y3 - rnorm(1,mean=0) xbnds - range(x1,x2,x3) ybnds - range(y1,y2,y3) bin1 - hexbin(x1,y1,xbnds=xbnds,ybnds=ybnds) bin2 - hexbin(x2,y2,xbnds=xbnds,ybnds=ybnds) bin3 - hexbin(x3,y3,xbnds=xbnds,ybnds=ybnds) erodebin1 - erode.hexbin(smooth.hexbin(bin1)) erodebin2 - erode.hexbin(smooth.hexbin(bin2)) erodebin3 - erode.hexbin(smooth.hexbin(bin3)) bnlst - list(b1=erodebin1, b2=erodebin2, b3=erodebin3) hdiffplot(bnlst) Can 'grid.hexlegend' be used to add a legend somehow? -Eric -- View this message in context: http://r.789695.n4.nabble.com/hdiffplot-tp3000644p3000644.html Sent from the R help mailing list archive at Nabble.com. __ 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] spplot cuts
Hello-
After looking through ?spplot, I would expect that I could specify the
values of the cuts:
...‘cuts’ number of cuts or the actual cuts to use...
So in the following command,
spplot(lzm.krige.dir[var1.pred], scales=list(draw=TRUE),
xlab=Easting,ylab=Northing,
cuts=seq(0.0,0.4,by=0.01),key.space=right,cex=1.1,col.regions=terrain.colors(41),
main=Specific Yield Layer 1,sp.layout=list(pts))
I get the following warning messages and the plot is a solid color:
Warning messages:
1: In if (length.out == 0L) integer(0L) else if (One) seq_len(length.out)
else if (missing(by)) { :
the condition has length 1 and only the first element will be used
2: In if (length.out 2L) if (from == to) rep.int(from, length.out) else
as.vector(c(from, :
the condition has length 1 and only the first element will be used
Looking at Edzer's response to a post:
http://r.789695.n4.nabble.com/using-spplot-sp-package-with-5-quantiles-td836245.html#a836246
it reinforces to me the idea that I should be able to specify where the cuts
should occur. I'm not sure what to do with the warning message and my
google searches have been unfruitful.
Some supplemental information that might be helpful:
range(data.frame(lzm.krige.dir[var1.pred])[,1])
[1] 0.1277067 0.2933876
I give the range of the plotted values because it varies between datasets
and I would like the color scale to remain fixed between images rather than
changing image to image making visual comparison's more difficult.
Any suggestions would be greatly appreciated,
Eric
--
View this message in context:
http://r.789695.n4.nabble.com/spplot-cuts-tp2716237p2716237.html
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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] R's Data Dredging Philosophy for Distribution Fitting
Forum, I'm a grad student in Civil Eng, took some Stats classes that required students learn R, and I have since taken to R and use it for as much as I can. Back in my lab/office, many of my fellow grad students still use proprietary software at the behest of advisers who are familiar with the recommended software (Statistica, @Risk (Excel Add-on), etc). I have spent a lot of time learning R and am confident it can generally out-process, out-graph, or more simply stated, out-perform most of these other software packages. However, one area my view has been humbled in is distribution fitting. I started by reading through http://cran.r-project.org/doc/contrib/Ricci-distributions-en.pdf After that I started digging around on this forum and found posts like this one http://r.789695.n4.nabble.com/Fitting-usual-distributions-td80.html#a80 that are close to what I'm after. That is, given an observation dataset, I would like to call a function that cycles through numerous distributions (common or not) and then ranks them for me based on Chi-Square, Kolmogorov-Smirnov and/or Anderson-Darling, for example. This question was asked back in 2004: http://finzi.psych.upenn.edu/R/Rhelp02a/archive/37053.html but the response was that this kind of thing wasn't in R nor in proprietary software to the best of the responding author's memory. In 2010, however, this is no longer true as @Risk's (http://www.palisade.com/risk/?gclid=CKvblPSM7KICFZQz5wodDRI2fg) Distribution Fitting function does this very thing. And it is here that my R pride has taken a hit. Based on the first response to the question posed here http://r.789695.n4.nabble.com/Which-distribution-best-fits-the-data-td859448.html#a859448 is it fair to say that the R community (I realize this is only 1 view) would take exception to this kind of data mining? Unless I've missed a discussion of a package that does this very thing, it seems as though I would need to code something up using fitdistr() and do all the ranking myself. Undoubtedly that would be a good exercise for me, but its hard for me to believe R would be a runner-up to something like distribution fitting in @Risk. Eric -- View this message in context: http://r.789695.n4.nabble.com/R-s-Data-Dredging-Philosophy-for-Distribution-Fitting-tp2289508p2289508.html Sent from the R help mailing list archive at Nabble.com. __ 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] Average 2 Columns when possible, or return available value
Forum, Using the following data: DF-read.table(textConnection(A B 22.60 NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA 102.00 NA 19.20 NA 19.20 NA NA NA NA NA NA NA 11.80 NA 7.62 NA NA NA NA NA NA NA NA NA NA NA 75.00 NA NA NA 18.30 18.2 NA NA NA NA 8.44 NA 18.00 NA NA NA 12.90 NA),header=T) closeAllConnections() The second column is a duplicate reading of the first column, and when two values are available, I would like to average column 1 and 2 (example code below). But if there is only one reading, I would like to retain it, but I haven't found a good way to exclude NA's using the following code: t(as.matrix(aggregate(t(as.matrix(DF)),list(rep(1:1,each=2)),mean)[,-1])) Currently, row 24 is the only row with a returned value. I'd like the result to return column A if it is the only available value, and average where possible. Of course, if both columns are NA, NA is the only possible result. The result I'm after would look like this (row 24 is an avg): 22.60 NA NA NA NA NA NA NA 102.00 19.20 19.20 NA NA NA 11.80 7.62 NA NA NA NA NA 75.00 NA 18.25 NA NA 8.44 18.00 NA 12.90 This is a small example from a much larger data frame, so if you're wondering what the deal is with list(), that will come into play for the larger problem I'm trying to solve. Respectfully, Eric -- View this message in context: http://r.789695.n4.nabble.com/Average-2-Columns-when-possible-or-return-available-value-tp2269049p2269049.html Sent from the R help mailing list archive at Nabble.com. __ 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] barplot: invalid 'xlim' value
Forum,
I am attempting to plot weekly horizontal histograms, something very similar
to:
http://addictedtor.free.fr/graphiques/RGraphGallery.php?graph=109
The part of the code used to create this graphic that interest me:
# Draw the histograms
for (i in 1:length(ages)){
par(new = TRUE)
xmin - -ages[i] + xlim[1]
xmax - xlim[2] - ages[i]
ser - freqs[, i+1]
ser - ser/max(ser) * barscale
barplot(ser, horiz = TRUE, axes = FALSE, xlim = c(xmin, xmax),
ylim = ylim, col = barcol, space = 0)
}
When I try something similar with the code below, I get the error:
Error in plot.window(xlim, ylim, log = log, ...) : invalid 'xlim' value
However, I can successfully use the the xlim and ylim data to plot points,
no R complaints (see code below that creates purple dots). I can't figure
out why barplot doesn't like xlim. Assuming that it is something small and
obvious, I'm also curious if someone could explain what is going on with
xmin and xmax in the code snippet above? When I investigated the ages
array, its value increases at each i index, but this leads to xmin steadily
decreasing. xmax has similar behavoir. Although I know what is going on, I
don't know why? It seems very counter intuitive. Also, I did look at the
post:
http://r.789695.n4.nabble.com/multiple-graphs-td830743.html#a830744
But subplot disconnects the data from the existing coordinate system.
I've prepared my data to be easily uploaded, but the paste command into R
took about 10-20 sec. on my machine to complete (I did pare down the dataset
by 80% or so):
library(zoo)
#create a time series
ttime.80.2-seq(as.POSIXlt(2000-06-12 02:00,%Y-%m-%d
%H:%M,tz=),as.POSIXlt(2000-07-17 01:00,%Y-%m-%d
%H:%M,tz=),by=hour)
#Load in the data for the base plot
well.80.2-read.table(textConnection(row~ID~date~Reading~WellID~week
19578~Well_80-2~6/12/2000 2:00~-202.034~80-2~0
19579~Well_80-2~6/12/2000 3:00~-201.018~80-2~0
19580~Well_80-2~6/12/2000 4:00~-199.494~80-2~0
19581~Well_80-2~6/12/2000 5:00~-197.716~80-2~0
19582~Well_80-2~6/12/2000 6:00~-190.858~80-2~0
19583~Well_80-2~6/12/2000 7:00~-181.460~80-2~0
19584~Well_80-2~6/12/2000 8:00~-170.030~80-2~0
19585~Well_80-2~6/12/2000 9:00~-155.552~80-2~0
19586~Well_80-2~6/12/2000 10:00~-118.468~80-2~0
19587~Well_80-2~6/12/2000 11:00~-96.116~80-2~0
19588~Well_80-2~6/12/2000 12:00~-75.542~80-2~0
19589~Well_80-2~6/12/2000 13:00~-65.128~80-2~0
19590~Well_80-2~6/12/2000 14:00~-59.032~80-2~0
19591~Well_80-2~6/12/2000 15:00~-68.684~80-2~0
19592~Well_80-2~6/12/2000 16:00~-78.844~80-2~0
19593~Well_80-2~6/12/2000 17:00~-87.480~80-2~0
19594~Well_80-2~6/12/2000 18:00~-95.354~80-2~0
19595~Well_80-2~6/12/2000 19:00~-102.974~80-2~0
19596~Well_80-2~6/12/2000 20:00~-109.070~80-2~0
19597~Well_80-2~6/12/2000 21:00~-113.388~80-2~0
19598~Well_80-2~6/12/2000 22:00~-117.706~80-2~0
19599~Well_80-2~6/12/2000 23:00~-121.008~80-2~0
19600~Well_80-2~6/13/2000 0:00~-123.802~80-2~0
19601~Well_80-2~6/13/2000 1:00~-125.326~80-2~0
19602~Well_80-2~6/13/2000 2:00~-126.342~80-2~0
19603~Well_80-2~6/13/2000 3:00~-127.104~80-2~0
19604~Well_80-2~6/13/2000 4:00~-127.866~80-2~0
19605~Well_80-2~6/13/2000 5:00~-128.882~80-2~0
19606~Well_80-2~6/13/2000 6:00~-128.882~80-2~0
19607~Well_80-2~6/13/2000 7:00~-129.644~80-2~0
19608~Well_80-2~6/13/2000 8:00~-130.660~80-2~0
19609~Well_80-2~6/13/2000 9:00~-132.184~80-2~0
19610~Well_80-2~6/13/2000 10:00~-132.184~80-2~0
19611~Well_80-2~6/13/2000 11:00~-133.962~80-2~0
19612~Well_80-2~6/13/2000 12:00~-135.740~80-2~0
19613~Well_80-2~6/13/2000 13:00~-138.280~80-2~0
19614~Well_80-2~6/13/2000 14:00~-139.296~80-2~0
19615~Well_80-2~6/13/2000 15:00~-140.820~80-2~0
19616~Well_80-2~6/13/2000 16:00~-141.836~80-2~0
19617~Well_80-2~6/13/2000 17:00~-143.360~80-2~0
19618~Well_80-2~6/13/2000 18:00~-145.138~80-2~0
19619~Well_80-2~6/13/2000 19:00~-145.138~80-2~0
19620~Well_80-2~6/13/2000 20:00~-146.154~80-2~0
19621~Well_80-2~6/13/2000 21:00~-146.916~80-2~0
19622~Well_80-2~6/13/2000 22:00~-147.678~80-2~0
19623~Well_80-2~6/13/2000 23:00~-147.678~80-2~0
19624~Well_80-2~6/14/2000 0:00~-148.694~80-2~0
19625~Well_80-2~6/14/2000 1:00~-148.694~80-2~0
19626~Well_80-2~6/14/2000 2:00~-148.694~80-2~0
19627~Well_80-2~6/14/2000 3:00~-148.694~80-2~0
19628~Well_80-2~6/14/2000 4:00~-148.694~80-2~0
19629~Well_80-2~6/14/2000 5:00~-149.456~80-2~0
19630~Well_80-2~6/14/2000 6:00~-149.456~80-2~0
19631~Well_80-2~6/14/2000 7:00~-149.456~80-2~0
19632~Well_80-2~6/14/2000 8:00~-150.472~80-2~0
19633~Well_80-2~6/14/2000 9:00~-150.472~80-2~0
19634~Well_80-2~6/14/2000 10:00~-150.472~80-2~0
19635~Well_80-2~6/14/2000 11:00~-150.472~80-2~0
19636~Well_80-2~6/14/2000 12:00~-149.456~80-2~0
19637~Well_80-2~6/14/2000 13:00~-148.694~80-2~0
19638~Well_80-2~6/14/2000 14:00~-148.694~80-2~0
19639~Well_80-2~6/14/2000 15:00~-148.694~80-2~0
19640~Well_80-2~6/14/2000 16:00~-148.694~80-2~0
19641~Well_80-2~6/14/2000 17:00~-148.694~80-2~0
19642~Well_80-2~6/14/2000
Re: [R] barplot: invalid 'xlim' value
Josh, I wonder if you try pulling the data from the post directly it will work better for you: http://r.789695.n4.nabble.com/barplot-invalid-xlim-value-td2217919.html#a2217919 As for your response to the other code I was asking about...that code works perfectly well as it is written, so there is no typo. I just can't figure out why it needs to be this way for it to work. As I mentioned, it is counter intuitive. Eric -- View this message in context: http://r.789695.n4.nabble.com/barplot-invalid-xlim-value-tp2217919p2217953.html Sent from the R help mailing list archive at Nabble.com. __ 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] barplot: invalid 'xlim' value
Josh, Thanks for looking into this further. Picking up on your hint I tried to coerce the xlim values as well: barplot(coldat,horiz=TRUE,axes=FALSE,xlim=c(as.numeric(strptime(well.80.2$date[1],%m/%d/%Y %H:%M)),as.numeric(strptime(well.80.2$date[length(well.80.2$date)],%m/%d/%Y %H:%M))),ylim=ylim,col=8,space=0) It doesn't complain at this point, but it doesn't appear to do anything either. Hopefully someone else will spot the problem. Eric -- View this message in context: http://r.789695.n4.nabble.com/barplot-invalid-xlim-value-tp2217919p2218029.html Sent from the R help mailing list archive at Nabble.com. __ 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] Add 1 week to date with hours included for subset
For those whose search lands on this post, here is the solution I found that worked well.80.2-data.frame(well.80.2,week=floor(as.numeric(difftime(strptime(well.80.2$date,%m/%d/%Y %H:%M,tz=),strptime(well.80.2$date[1],%m/%d/%Y %H:%M,tz=),tz=,units=hours))/168)) -- View this message in context: http://r.789695.n4.nabble.com/Add-1-week-to-date-with-hours-included-for-subset-tp2216062p2216710.html Sent from the R help mailing list archive at Nabble.com. __ 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] insert values based on common ID
Forum, with the datasets a and b below, I'm trying to establish a relationship based on the common column week and insert the value from the column weekAvg in b to the column weekAvg in a. The dataset a is several thousand lines long. I've tried looking at 'match', writing functions, 'rbind.fill' and various search terms in this forum to no avail. Thanks... a-read.table(textConnection(row ID date Reading WellID week weekAvg 1 Well_80-2 6/12/2000 2:00 -202.034 80-2 0 NA 2 Well_80-2 6/12/2000 3:00 -201.018 80-2 0 NA 3 Well_80-2 6/12/2000 4:00 -199.494 80-2 0 NA 4 Well_80-2 6/12/2000 5:00 -197.716 80-2 0 NA 5 Well_80-2 6/12/2000 6:00 -190.858 80-2 0 NA 6 Well_80-2 6/12/2000 7:00 -181.460 80-2 0 NA 7 Well_80-2 6/19/2000 10:00 -166.728 80-2 1 NA 8 Well_80-2 6/19/2000 11:00 -167.490 80-2 1 NA 9 Well_80-2 6/19/2000 12:00 -167.490 80-2 1 NA 10 Well_80-2 6/19/2000 13:00 -167.490 80-2 1 NA 11 Well_80-2 6/19/2000 14:00 -168.506 80-2 1 NA 12 Well_80-2 6/19/2000 15:00 -168.506 80-2 1 NA),header=T) closeAllConnections() b-read.table(textConnection(week weekAvg 0 -147.3726 1 -181.3429 2 -151.7208 3 -188.8653 4 -163.7465 5 -161.6873 6 -158.5168 7 -146.6136 8 -175.4351 9 -100.9450 10 -151.3655 11 -125.8975 12 -162.5993),header=T) closeAllConnections() -- View this message in context: http://r.789695.n4.nabble.com/insert-values-based-on-common-ID-tp2216735p2216735.html Sent from the R help mailing list archive at Nabble.com. __ 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] Add 1 week to date with hours included for subset
Forum, I've got hourly data for roughly 5 months of time that I want to subset into successive weeks. The first two elements of the dataset are well.80.2$date[1] [1] 6/12/2000 2:00 9256 Levels: 10/1/2001... well.80.2$date[2] [1] 6/12/2000 3:00 9256 Levels: 10/1/2001 and so on until mid-october. I've been able to add 1 week the first element with the following: as.POSIXlt(paste(as.character(as.Date(well.80.2$date[1],%m/%d/%Y %H:%M)+7),as.character(format(strptime(well.80.2$date[1],%m/%d/%Y %H:%M),%H:%M)),sep= )) [1] 2000-06-19 02:00:00 What I've been unable to do is then use this for comparative purposes in the subset command, something to the effect of: contin.80.2-subset(well.80.2,well.80.2$date=well.80.2$date[1]) well.80.2$date as.POSIXlt(paste(as.character(as.Date(well.80.2$date[1],%m/%d/%Y %H:%M)+7),as.character(format(strptime(well.80.2$date[1],%m/%d/%Y %H:%M),%H:%M)),sep= )) If there is an easier way to parse the dataset into weekly bins taking into consideration the hours, I'm open to suggestions. Thank you. -- View this message in context: http://r.789695.n4.nabble.com/Add-1-week-to-date-with-hours-included-for-subset-tp2216062p2216062.html Sent from the R help mailing list archive at Nabble.com. __ 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] Polygon Shading Based on Relative Line Position
Forum, I have two time series that I plot up using the data/code below. The x-axis shows weekly time step starting on 6/3/1999 (SP=10) and ending on 10/25/2007 (SP=448). I'm trying to shade between the two lines in a similar fashion as the graph found here: http://addictedtor.free.fr/graphiques/RGraphGallery.php?graph=122 Can the polygon function be used to do this? There are lots of posts pertaining to shading between two lines on this forum, but none with alternating shading. I did look over the ?polygon discussion, specifically the brownian motion plot, but the shading is constant regardless of which line is greater than/above the other. Due to the 20 or so times the lines cross each other I wasn't sure how to modify the last example in ?plot DS.Load.filtered-read.table(textConnection(SP JMLoad CoolidgeLoad 10 6287371153.3 5383351151 11 5016438684.4 5149840395 12 4020238617.8 4097757349 18 1681570096.3 2330476184 19 2011084698.2 1568941725 21 1145032462.8959165782 22 1063307043.8645435740 24 925120077.1667723509 25 587189249.9707324096 26 541535137.0639181098 27 421481931.6692405402 28 268198706.4599814619 30 490052206.1509190477 32 429672101.9518539938 33 479360154.8518084664 34 332705173.5627963150 407437921.6520362947 417272780.2511155928 44 189312657.1461023436 46 159935695.6834850739 48 580467416.1909478429 49 711678748.9825542943 54 820999693.4602141144 55 1281245122.3539526986 56 1234525377.3607948868 57 1107369351.4761832777 59 930102499.8513626069 60 934106161.7525646818 61 1074012858.8436276835 62 1856920332.5660276684 64 2564136121.9 1610652560 65 2404760972.5 1465864246 66 2035806517.6 1591773461 68 1886124760.8 1611344996 71 2287805156.8 1382811156 73 1376909678.5791595277 74 1265209973.9638691715 75 777429216.0404560430 76 595882924.8315035344 77 521183273.0253026361 78 533726296.6245560652 82 267965540.5295694926 943645170.3334233598 953644636.1338526100 963616835.0357388387 973478475.0360407980 983395990.1383170168 993424680.2375567185 1037706454.2338718083 1047484685.1341367271 105 12100851.7350978464 106 43374657.9326563936 109 1172149909.6263931016 110 1281917018.7345653054 113 835813366.3682986351 114 524765450.1717050861 115 220652819.6948602823 116 1147516084.1349630240 117 2205837663.4750724733 118 2506424170.0 1178190964 119 2265639020.8 1075464253 120 2068263612.8 1413273356 121 2156640167.4 1066482651 123 1910796858.1 1037178735 124 1611203132.5 1073785193 125 778788726.3542689397 126 833341620.3306049833 127 761683183.2237702159 128 719063683.5195396505 129 1146047977.4285866547 130 801138377.3316385292 131 657429028.1213995657 133 118871474.0211083899 134 204384729.1192625209 135 260090209.9199517710 136 24201037.7194351081 1378417831.9198857821 138 15899802.8191076815 1399772533.4199805343 1403616601.1203149621 1413893325.1207562895 1462789497.9256905051 1472624087.4239137572 1482320129.9233871812 1492375061.3244789361 1502623139.9230084008 1512679009.1235954773 1522623771.6221495933 1532679186.6222735406 1552651789.1172518801 1562679171.0161670825 157 13314442.4144901548 158 56080331.1125125488 159 786640402.5184918943 160 1069335727.1260721047 161 1077616331.7124086059 163 183185160.3118822561 164 143488200.9 69403023 165 200176394.9 76669321 166 229339546.2 57984637 167 270128506.1 58218605 168 794854222.2 63732496 169 1043519643.8429644515 170 987053037.5480691194 173 196554934.9 52094336 174 208620907.2 45293829 175 79198798.3 33293628 176 91447590.0 29101572 181 511657497.8143139470 187 90245277.8101068632 188 15889549.8116258690 1893182084.1112911409 1903042679.0113186554 1912903819.8113216345 1922903116.0114596143 1932710229.9116519095 1942710557.5116534186 1952599733.3110767298 1962516739.6111859396 1972543807.5116030474 1982322890.6116630304 2012681845.9114896789 2022488111.8115222450 2032626004.0117142088 2042238467.7113530229 2052293570.7119880375 2062266600.2113747430 2072570052.8104537669 2082515095.9100642204 2102432006.7 70423676 211 158252693.6 67914034 212 339920353.6 59404637 213 337973820.1 74727730 214 403331802.8 75101010 219 1278541396.5244833157 220 1134609769.0204191503 221 1137461413.1255674658 222 1101354706.1117538411 223 509565793.1 40451505 224 147178862.1 21781015 225
Re: [R] Polygon Shading Based on Relative Line Position
The last sentence should read: Due to the 20 or so times the lines cross each other I wasn't sure how to modify the last example in ?polygon -- View this message in context: http://r.789695.n4.nabble.com/Polygon-Shading-Based-on-Relative-Line-Position-tp2132718p2132720.html Sent from the R help mailing list archive at Nabble.com. __ 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] melt, remove rows with common id if NA encountered in value
forum, After melting a data.frame I get some 18,000+ rows (some rows below). I want to filter/remove all rows that have an SP variable value in common with a row containing NA in the value column. So in the example rows below, all rows with SP=425 variable=CoolidgeLoad would be removed because NA was encountered in at least one of the rows (in this case 2 of the rows). Continuing the example, rows with SP=426 variable=CoolidgeLoad would be retained since NA doesn't appear in the value column of any row containing both of these values. Rows with SP=361 variable=JMLoad would also survive. After performing the filter I plan to cast the data. -Eric SP variable value 425 CoolidgeLoad 6.044483e+07 425 CoolidgeLoad 5.314102e+07 425 CoolidgeLoad 5.287761e+07 425 CoolidgeLoad 4.592003e+07 425 CoolidgeLoad 4.314002e+07 425 CoolidgeLoad NA 425 CoolidgeLoad NA 426 CoolidgeLoad 5.412465e+07 426 CoolidgeLoad 6.122925e+07 426 CoolidgeLoad 6.514926e+07 426 CoolidgeLoad 6.605055e+07 426 CoolidgeLoad 6.525006e+07 426 CoolidgeLoad 6.166645e+07 426 CoolidgeLoad 6.167510e+07 427 CoolidgeLoad 5.372381e+07 427 CoolidgeLoad 4.551596e+07 427 CoolidgeLoad 4.246508e+07 427 CoolidgeLoad NA 427 CoolidgeLoad NA 427 CoolidgeLoad NA 427 CoolidgeLoad NA 359JMLoad 3.036887e+05 359JMLoad NA 359JMLoad NA 359JMLoad NA 359JMLoad NA 359JMLoad 3.313669e+05 359JMLoad 3.037101e+05 360JMLoad 3.036887e+05 360JMLoad 2.761196e+05 360JMLoad NA 360JMLoad 3.037101e+05 360JMLoad 3.036887e+05 360JMLoad 3.036672e+05 360JMLoad 3.036458e+05 361JMLoad 3.036029e+05 361JMLoad 3.035814e+05 361JMLoad 3.312733e+05 361JMLoad 3.037101e+05 361JMLoad 3.313435e+05 361JMLoad 3.037315e+05 361JMLoad 3.313669e+05 -- View this message in context: http://r.789695.n4.nabble.com/melt-remove-rows-with-common-id-if-NA-encountered-in-value-tp2131770p2131770.html Sent from the R help mailing list archive at Nabble.com. __ 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] different forms of nls recommendations
Hello,
Using this data:
http://n4.nabble.com/file/n1676330/US_Final_Values.txt US_Final_Values.txt
and the following code i got the image at the end of this message:
US.final.values-read.table(c:/tmp/US_Final_Values.txt,header=T,sep= )
US.nls.1-nls(US.final.values$ECe~a*US.final.values$WTD^b+c,data=US.final.values,start=list(a=2.75,b=-0.95,c=0.731),trace=TRUE)
f.US1-function(x){coef(US.nls.1)[a]*x^coef(US.nls.1)[b]+coef(US.nls.1)[c]}
xvals.US1-seq(min(US.final.values$WTD),max(US.final.values$WTD),length.out=75)
yvals.US1-f.US1(xvals.US1)
Rsq.nls.1-sum((predict(US.nls.1)-mean(US.final.values$ECe))^2/sum((US.final.values$ECe-mean(US.final.values$ECe))^2))
plot(US.final.values$WTD,US.final.values$ECe,col=red,pch=19,cex=.75)
lines(xvals.US1,yvals.US1,col=blue)
but the r^2 wasn't so hot.
Rsq.nls.1
[1] 0.2377306
So I wanted to try a different equation of the general form a/(b+c*x^d)
US.nls.2-nls(US.final.values$ECe~(a/(b+c*US.final.values$WTD^d)),data=US.final.values,start=list(a=100.81,b=73.7299,c=0.0565,d=-6.043),trace=TRUE,algorithm=port)
but that ended with Convergence failure: false convergence (8). I tried
relaxing the convergence criteria to no avail. Assuming the form of the
equation I'm trying to use is the problem, I've been unable to track down a
source that shows the shapes of various non-linear equations that I might be
able to try as alternatives. Any suggestions?
http://n4.nabble.com/file/n1676330/nls_image.jpg
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Re: [R] adding infrequent date labels to x-axis
Hi Hrishi, With regard to the post you helped me out with, I've got my graph almost dialed in the way I would like it. There was one more small thing my searches have been unable to turn up. I've tried searching with stagger labels offset Labels alternate labels to no avail. the pertinent lines of code I'm basing my question on are: m2-ticks.lab==Feb|ticks.lab==May|ticks.lab==Aug|ticks.lab==Nov Axis(x,at=ticks.at[m2],side=1,labels=ticks.lab[m2],las=1,cex.axis=0.75,tick=F,padj=-2) I would prefer to use cex.axis=1, but because the labels crowd each other it doesn't place them all. Is there a command I can use that would essentially stagger the labels?, like so... | | | (Graph Area) +__ May Nov May Nov Aug Feb Aug Feb http://n4.nabble.com/file/n1566433/TimeSeries_Example1.jpg Respectfully, Eric -- View this message in context: http://n4.nabble.com/adding-infrequent-date-labels-to-x-axis-tp1564804p1566433.html Sent from the R help mailing list archive at Nabble.com. __ 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] adding infrequent date labels to x-axis
I'm sure there is a clever way to do the following, but I've been unable to find it on this forum or by writing my own functions. I have 8 years worth of weekly data but would like to restrict the labels on the x-axis to months only. I've included the first year's worth of data below. My line of thought has been along these lines x-seq(as.Date(1999-04-01),as.Date(2007-10-25),by=1 month) y-as.POSIXlt(x)$mon+1 months-month.name month.names-months[as.numeric(y)] month.names-substr(month.names,1,3) plot(cropped.cast1$date,cropped.cast1$Frac_ET_Satsfd_mean,xaxt='n') mtext(text=month.names,side=1,at=??? I'm not sure how to tie month.names back to their true location on the x-axis? I appreciate any insights, Eric Also, class(cropped.cast1$date) [1] POSIXt POSIXct cropped.cast1 date Frac_ET_Satsfd_mean Frac_ET_Satsfd_sd 1999-04-08 0.83448850.13545515 1999-04-15 0.83554660.12810387 1999-04-22 0.85955790.11259251 1999-04-29 0.89972250.09611060 1999-05-06 0.87143640.09527164 1999-05-13 0.85302030.11088544 1999-05-20 0.84378660.12689882 1999-05-27 0.83100030.13985307 1999-06-03 0.80312030.15851422 1999-06-10 0.82885050.12827027 1999-06-17 0.82511300.13051783 1999-06-24 0.82276390.14227501 1999-07-01 0.79146890.15892716 1999-07-08 0.80509290.14465413 1999-07-15 0.83701410.11843615 1999-07-22 0.84486970.10823010 1999-07-29 0.85619250.10694348 1999-08-05 0.85207900.09953065 1999-08-12 0.84299250.10545427 1999-08-19 0.83979660.11629002 1999-08-26 0.83679530.12363411 1999-09-02 0.82194790.13870596 1999-09-09 0.82181930.13617427 1999-09-16 0.82175840.13346997 1999-09-23 0.82168340.13304117 1999-09-30 0.81110050.14367143 1999-10-07 0.80908130.14967750 1999-10-14 0.82651880.13484263 1999-10-21 0.83913330.11873929 1999-10-28 0.84261020.11215439 1999-11-04 0.84318130.11007485 1999-11-11 0.83941400.11206864 1999-11-18 0.83506500.11042384 1999-11-25 0.83600820.11011926 1999-12-02 0.83621290.10834491 1999-12-09 0.83775120.10519698 1999-12-16 0.83673390.10176535 1999-12-23 0.83386210.10273662 1999-12-30 0.83170940.10470654 -- View this message in context: http://n4.nabble.com/adding-infrequent-date-labels-to-x-axis-tp1564804p1564804.html Sent from the R help mailing list archive at Nabble.com. __ 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] adding infrequent date labels to x-axis
Hello Hrishi, The command you suggested plotted the years on the x-axis only. It was the same exact plot as the one I included in the original post. Respectfully, Eric -- View this message in context: http://n4.nabble.com/adding-infrequent-date-labels-to-x-axis-tp1564804p1564875.html Sent from the R help mailing list archive at Nabble.com. __ 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] adding infrequent date labels to x-axis
Hello Hrishi, That worked great, and in the process I learned some new ways of going about writing R code. Thank you very much for helping me out! Eric -- View this message in context: http://n4.nabble.com/adding-infrequent-date-labels-to-x-axis-tp1564804p1564943.html Sent from the R help mailing list archive at Nabble.com. __ 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] formatting do.call output
Hello,
With a dataset that is about 1.4e6 rows long
my.dat[seq(1:20),]
date Cell_ID Classification WT_Depth Frac_ET_Satsfd
1999-04-08 974 3 3.5850830.244561400
1999-04-081188 3 3.5260010.123484700
1999-04-081189 3 3.1870120.215916700
1999-04-081403 3 3.3950200.163972900
1999-04-081409 3 2.3680420.415165500
1999-04-081617 3 4.0399170.003224544
1999-04-081618 3 3.4578860.148585900
1999-04-081619 3 2.1489260.475924300
1999-04-081620 2 1.5239260.633190900
1999-04-081621 3 2.1979980.469294300
1999-04-151622 7 2.7590330.325698400
1999-04-151623 3 2.8020020.313719600
1999-04-151624 3 3.0629880.243275900
1999-04-151833 2 3.4830320.141840700
1999-04-151834 2 3.1280520.232235400
1999-04-151835 7 3.3540040.176209000
1999-04-151836 3 2.6910400.341234700
1999-04-151837 3 3.1409910.228083800
1999-04-151838 3 2.3929440.413379300
1999-04-152048 2 3.7120360.084534560
.
.
.
I use
edm.func-function(x){c(mu=mean(x),var=var(x))}
output-do.call(rbind,tapply(my.dat$Frac_ET_Satsfd,list(my.dat$date,my.dat$Classification),edm.func))
data.frame(output)
muvar
1 0.7980007 0.03446669
2 0.7947966 0.03429280
3 0.8240736 0.02482441
.
.
3573 0.4509044 0.03821251
3574 0.4484108 0.03856110
3575 0.4519150 0.03889944
There are 447 dates and 8 classifications (1-8). What is the best way to
include the corresponding date and classification that goes with each row?
Thanks Eric
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[R] boxplot label cleanup
The boxplot below is close to what I would like, but needs some clean up.
I've been unable to address the following:
1) How can I use the full extent of the plot space (there is white space on
either side of the data)?
2) How can I reduce the # of month labels (e.g. every 3rd month Jan Mar
Jun...) but still keep them staggered?
3) The year labels at the top of the plot have been positioned using the
following line
mtext(year.observed,3,at=seq(3,99,by=12),padj=-1)
I would prefer to position the label using the actual value contained in the
date column of the data.frame (which is a date variable type). Fixing the
position as I have doesn't seem...pure?
4) The grid is meant to partition the data from year to year and was
generated using
grid(nx=9,ny=0,col=gray60,lty=1)
but it is clearly off. Is there a way to draw the grid lines exactly
between december and January? I wasn't sure how much code I should include,
so here it all is: (images follow)
my.dat-read.table(C:/Eric/Programs/UtilityProgram_Summarize_UNSAT_Zone_Results_8Class/Process_UZF_Output/bin/R_Data_fracETSat_US.txt,header=TRUE,sep=\t)
attach(my.dat)
dates-strptime(as.character(my.dat$date),%m/%e/%Y)
my.dat=my.dat[,2:5]
my.dat=data.frame(date=dates,my.dat)
detach(my.dat)
daterange=c(as.POSIXlt(min(my.dat$date)),as.POSIXlt(max(my.dat$date)))
rm(dates)
attach(my.dat)
cropped=subset(my.dat,Classification==5)
natveg=subset(my.dat,Classification==7)
cropped[c(1,2,3),] #Example data
date Cell_ID Classification WT_Depth Frac_ET_Satsfd
94 1999-04-084395 5 1.572998 0.7596927
95 1999-04-084396 5 1.923096 0.7588058
101 1999-04-084403 5 2.290039 0.7911426
month.year - function (x) {12*(as.POSIXlt(x)$year-99)+as.POSIXlt(x)$mon+1}
month-month.year(my.dat$date)
months-month.name
cropped.month-month.year(cropped$date)
mean.cropped.frac-tapply(cropped$Frac_ET_Satsfd,cropped.month,mean)
month.observed - numeric(length(mean.cropped.frac))
for(i in
1:length(mean.cropped.frac)){month.observed[i]=as.numeric(names(mean.cropped.frac[i]))%%12}
month.observed[month.observed==0]-12 #make the decembers 12 instead of 0
month.names-months[as.numeric(month.observed)]
month.names-substr(month.names,1,3)
year.vector-function (x) {as.POSIXlt(x)$year+1900}
cropped.year-year.vector(cropped$date)
mean.cropped.frac.yr-tapply(cropped$Frac_ET_Satsfd,cropped.year,mean)
year.observed-numeric(length(mean.cropped.frac.yr))
for(i in
1:length(mean.cropped.frac.yr)){year.observed[i]=as.numeric(names(mean.cropped.frac.yr[i]))}
boxplot(Frac_ET_Satsfd~month.year(cropped$date),data=cropped,col=lightblue,outline=F,
range=0.0001,ylim=c(0,1),xaxt=n)
axis(1,at=1:103,labels=F)
mtext(month.names,1,at=1:103,line=rep(c(1,2),2))
mtext(year.observed,3,at=seq(3,99,by=12),padj=-1)
grid(nx=9,ny=0,col=gray60,lty=1)
http://n4.nabble.com/file/n1476343/timeSeriesMessy.jpg
Thanks, Eric
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[R] object 'xxx' not found
The following line of code seems fairly straight forward, yet it is kicking
back an error message:
for (i in
1:length(mean.natveg.frac)){month.observed[i]=as.numeric(names(mean.natveg.frac[i]))%%12}
Error message:
Error in month.observed[i] = as numeric(names(mean.natveg.frac[i]))%%12 :
object 'month.observed' not found
Seems like its not found because I'm trying to define it. Also
length(mean.natveg.frac) = 103
Eric
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and provide commented, minimal, self-contained, reproducible code.
[R] log-normal overlay
Hello, Using the following lines of code, I created the following graph: d.resid-c(-5.63,4.25,4.38,3.05,-1.85,1.40,1.80,-0.80,4.20,3.20,-2.70,-7.20,-0.10,-2.50,0.60,1.20,1.30,1.20,1.30,1.27,6.91,8.55,7.18,8.73,1.82,-1.45,5.91,5.45,-0.82,-4.55,0.82,82.91,1.73,3.09,5.64,-1.73,-9.55,4.27,17.45,9.64,-18.33,21.77,-31.56,-0.65,-13.11,-14.75,-18.75,-15.56,-3.11,-1.75,-1.84,-2.02,-8.93,-7.47,0.89,10.77,-15.93,-16.93,-13.33,-9.05,-2.25,-3.53,4.95,5.13,69.38,61.76,71.18,20.27,60.51,-5.38,-14.52,-5.04,3.02,-13.19,-14.70,-14.53,-8.81,4.30,-7.89,-6.37,60.72,84.11,25.01,91.55,95.65,96.13,92.23,93.22,91.74,92.12,90.79,85.30,73.77,72.44,65.51,49.31,45.50,-10.31,-22.50,-20.21,-3.20,-11.32,-9.29,-1.75,-0.91,-1.93,-0.23,-6.32,-0.84,-3.89,-5.04,1.71,-18.29,-36.63,-36.75,-24.63,-24.95,-27.85,-26.60,-21.20,-14.80,-29.80,-22.80,-21.70,-26.20,-25.10,-37.50,-34.40,-34.80,-33.70,-34.80,-33.70,-19.73,-18.09,-23.45,-23.82,-19.27,-20.18,-24.45,-25.09,-23.55,-25.82,-21.55,-24.18,-34.09,-27.27,-25.91,-26.36,-28.73,-23.55,-32.73,-29.55,-28.36,-30.41,-125.31,-15.64,-24.73,-19.1! 9,-19.83,-17.83,-19.64,-22.19,-21.83,-22.92,-51.10,-27.01,-27.55,-26.19,-17.01,-5.63,-0.75,-5.63,-2.95,9.15,4.40,6.80,-0.80,-0.80,9.20,10.30,12.80,5.90,6.50,-0.40,-6.80,-4.70,-6.80,-4.70,7.27,7.91,3.55,13.18,1.73,3.82,6.55,6.91,16.45,9.18,5.45,11.82,-5.09,13.73,15.09,20.64,9.27,15.45,10.27,5.45,2.64,-3.41,7.69,13.36,7.27,12.81,7.17,17.17,19.36,14.81,12.17,24.08,19.90,21.99,19.45,15.81,-67.31,24.99,24.99,14.59,11.64,15.38,14.10,16.49,27.00,-58.88,-89.97,-59.18,-65.28,-62.50,-69.03,-75.57,-89.66,-58.22,-49.96,-61.24,-68.06,-41.82,-80.30,-60.49,-70.98,-79.45,-77.85,-127.19,-62.80,-103.81,-119.91,-102.65,-113.82,-105.53,19.62,17.05,15.38,12.25,35.38,36.05,33.15,30.40,38.80,35.20,17.20,32.20,59.30,61.80,15.90,35.50,28.60,10.20,14.30,10.20,14.30,33.27,22.91,10.55,4.18,10.73,31.82,32.55,13.91,15.45,21.18,26.45,25.82,25.91,22.73,16.09,17.64,18.27,35.45,31.27,22.45,26.64,26.59,40.69,39.36,47.27,44.81,33.17,32.17,38.36,39.81,35.17,33.08,40.90,44.99,47.45,23.81,33.69,41.99,38.99,44.59! ,46.82,39.90,43.19,33.89,13.57, 27.35,-51.71,34.38,33.25,19.38,33.05,46.65,45.40,51.80,39.20,38.20,47.20,33.30,32.80,58.90,57.50,57.60,54.20,57.30,54.20,57.30,38.27,38.91,48.55,46.18,30.73,25.82,25.55,39.91,50.45,35.18,30.45,30.82,21.91,44.73,50.09,45.64,37.27,39.45,60.27,62.45,64.64,65.59,90.69,80.36,54.27,59.81,64.17,66.17,53.36,38.81,32.17,53.08,37.90,33.99,37.45,40.81,56.69,35.99,34.99,43.59,53.92,60.72,64.01,50.71,54.15,56.38,3.05,23.52,43.00,40.03,29.59,46.00,46.30,51.52,69.93,53.73,43.37,46.90,60.85,77.61,68.17,66.64,77.04,81.18,31.68,78.85,73.41,89.58,64.58,50.15,53.52,40.95,59.24,47.83,35.71,33.65,41.31,46.26,61.64,34.87,47.56,34.87,23.34,18.96,34.29,24.79,30.10,16.40,27.16,11.92,-11.71,31.22,23.09,28.17,38.76,39.60,44.68,40.28,44.86,47.29,51.87,56.81,60.51,42.97,48.51,33.85,24.10,4.79,4.72,-16.47,-14.64,-14.41,-3.89,1.14,-2.22,-8.31,-14.71,-0.63,-1.75,-14.13,-6.95,9.65,3.40,12.80,-1.80,-1.80,9.20,-5.70,-5.20,16.90,14.50,17.60,21.20,22.30,21.20,22.30,-0.73,-1.09,12.55,7.18,-5.27,-8.18,-9.45,3.91,1! 0.45,-4.82,-3.55,-8.18,-32.09,6.73,11.09,1.64,3.27,1.45,22.27,24.45,30.64,-32.41,-34.31,-19.64,-50.73,-45.19,-36.83,-34.83,-53.64,-62.19,-66.83,-50.92,-62.10,-67.01,-63.55,-59.19,-44.31,-59.01,-54.01,-55.41,-47.84,-41.96,-40.19,35.35,12.57,18.54,23.38,29.91,-10.36,31.24,29.30,103.89,1.66,-34.99,-17.01,-10.43,-14.63,-7.00,30.05,-57.81,-57.81,35.92,25.51,14.33,13.31,6.46,5.44,24.05,3.36,25.82,17.36,-4.63,-0.75,-12.63,-4.45,3.15,-1.60,3.80,-4.80,-2.80,1.20,-10.70,-12.20,5.90,5.50,6.60,12.20,11.30,12.20,11.30,-2.73,-2.09,4.55,13.18,-4.27,-8.18,-11.45,-0.09,6.45,-8.82,-8.55,-9.18,-12.09,0.73,3.09,0.64,3.27,-3.55,11.27,14.45,22.64,18.59,28.69,30.36,9.27,13.81,19.17,19.17,7.36,-1.19,-2.83,10.08,0.90,-6.01,-2.55,2.81,13.69,1.99,4.99,-2.41,6.25,13.04,13.29,26.85,-1.53,6.48,7.38,-27.86,-8.91,7.69,-5.34,-1.46,8.61,10.75,-23.28,-87.47,-24.73,-30.51,-26.38,-21.26,-3.59,-19.13,-17.61,-7.21,-0.02,81.98,-5.88,-13.45,-2.29,17.14,12.77,11.87,27.04,32.83,14.71,7.77,4.80,8.80,10.70,27.00,63.63! ,16.58,-0.69,-7.64,
[R] sppolot: fill below minimum legend value
In the plot below, there are some grid cells that have values below 10, which is the lowest cut value I have specified. Is there a way, without adjusting the number of cuts, to tell R to fill in those cells with the lowest possible color (in this case greeen)? There is a white hole in the image about a quarter of the way in from the left side, this is what I would like to correct. Thanks...Eric The code: pts-list(sp.points,K.dat,pch=3,col=black) cuts-c(10,20,30,40,50,60,70,80,90,100,200,300,400,500,600,700,800,900,1000) spplot(lzm.krige.dir[var1.pred],at=cuts,colorkey=list(at=log10(cuts),at=log10(cuts),labels=as.character(cuts)),scales=list(draw=TRUE), xlab=Easting,ylab=Northing,key.space=right,cex=1.1,col.regions=terrain.colors(30),main=Hydraulic Conductivity of Layer 2,sp.layout=list(pts)) The image: http://www.nabble.com/file/p25392472/Image3.jpeg -- View this message in context: http://www.nabble.com/sppolot%3A-fill-below-minimum-legend-value-tp25392472p25392472.html Sent from the R help mailing list archive at Nabble.com. __ 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] spplot modifications
http://www.nabble.com/file/p25336596/Conductivity1.jpeg I need a little help making modifications to the image included with this post. First, rather than using a linear color legend to display the output I would like to use a log-scale legend. Thus, the legend on the right would go from 1 to 1000, for example, with a classic log-scale gradation. What I hope to avoid is taking the log of the values and displaying the logged values using a linear scale. Second, at the top of the legend there is a random green bar that shouldn't be there. If you know what causes this or more importantly, how to get rid of it, please help. I've included the code and text files necessary to create this figure in the zip file. Thank you, Eric Morway http://www.nabble.com/file/p25336596/Morway_R_Qs.zip Morway_R_Qs.zip -- View this message in context: http://www.nabble.com/spplot-modifications-tp25336596p25336596.html Sent from the R help mailing list archive at Nabble.com. __ 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] assign point values as labels in spplot
In the code to follow, I'm trying to label points with their corresponding values but have been unsuccessful after many variations to the following code. The code below creates the plot I want, I simply cannot get the black points (+) to display the actual value. I'm guessing the problem is somewhere in the second to last line of code (starts with pts-...). I have attached the two text files needed to run the code. http://www.nabble.com/file/p21734824/R_Hk_Krig_File_Log.txt R_Hk_Krig_File_Log.txt http://www.nabble.com/file/p21734824/Predict_Location_XY.txt Predict_Location_XY.txt Eric library(gstat) K.dat-read.table(C:/temp/R_Hk_Krig_File_Log.txt, header = TRUE) my.dat-data.frame(K.dat) attach(my.dat) coordinates(my.dat)=~X+Y pred.dat-read.table(C:/temp/Predict_Location_XY.txt, header = FALSE) names(pred.dat)-c(x,y,p) my.pred.dat-data.frame(pred.dat) coordinates(my.pred.dat)-~x+y gridded(my.pred.dat) = TRUE #Set up 2 exponential variograms lzm.vgm.exp.noNug-vgm(1.1,Exp,2000,0) lzm.vgm.exp.Nug-vgm(0.7,Exp,3000,0.4) #Krige the exponential variograms lzm.krig.exp.noNug-krige(Kh_Log10~1,my.dat,my.pred.dat,model=lzm.vgm.exp.noNug) lzm.krig.exp.Nug-krige(Kh_Log10~1,my.dat,my.pred.dat,model=lzm.vgm.exp.Nug) lzm.krig.exp.noNug$var1.pred-10^lzm.krig.exp.noNug$var1.pred lzm.krig.exp.Nug$var1.pred-10^lzm.krig.exp.Nug$var1.pred library(sp) library(lattice) pts-list(sp.points,K.dat,pch=3,col=black,labels=as.character(K.dat$Kh)) spplot(lzm.krig.exp.Nug[var1.pred], scales=list(draw=TRUE), xlab=Easting,ylab=Northing,cuts=25,key.space=right,cex=1.1, col.regions=terrain.colors(25),main=Hydraulic Conductivity of Layer 2, sp.layout=list(pts)) -- View this message in context: http://www.nabble.com/assign-point-values-as-labels-in-spplot-tp21734824p21734824.html Sent from the R help mailing list archive at Nabble.com. __ 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.
