Re: [R] Calculating plotting a linear regression between two correlated variables
I know this isn't what you are asking, but have you considered examining the relationship between dA and the community density excluding dA? JulieV wrote Hi Josh, Thanks for your response ! Actually, I already tried to plot it with a classical regression and I know the relation is linear: dA = 0.765 * dCOM - 0.089 p(slope) 0.0001 p(intercept) = 0.0003 The fact is that I can not use these results as my variables dA and dCOM are correlated (as mentioned above, Eq.1). What I need to find out is which correction I should do on my data, and how, to be able to calculte the regression p-values correctly with Linear Mixed Models. I am interested in this because I know that my species decline at different rates when my community is declining. For example, with decreasing values of dCOM, dA reaches 0 before dB. Julie -- View this message in context: http://r.789695.n4.nabble.com/Calculating-plotting-a-linear-regression-between-two-correlated-variables-tp4319051p4321222.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] R not giving significance tests for coefficients/estimates?
3x4 Error: unexpected symbol in 3x4 R has no idea that you equate x as multiplication.. use an astrix 3*4 [1] 12 dominic wrote This is basically my code: library(MASS) lmsreg(formula = b0 ~ b1 + b3 + b1xb2, data=mydata) b1xb2 is an interaction but it was the centered value for a continuous variable times a categorical variable. I am used to using Mplus statistical software and SAS for robust regression, and they'll usually give you a t-test, p-value as well as the SE and/or CI. But I am not getting it here even though I have sent that is the correct code to get it on some websites that have posted their output. I I have also tried: library(MASS) lqs(formula = b0 ~ b1 + b3 + b1xb2, data=mydata, + method = c(lqs)) And rlm(formula ...) too, but neither provide tests of significance. Am I doing something wrong? Any help on this matter is greatly appreciated. Thanks, Dominic -- View this message in context: http://r.789695.n4.nabble.com/R-not-giving-significance-tests-for-coefficients-estimates-tp4321669p4321876.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] Automated Regressions
I don't know why you had c in a(1:2000))
c is a function see ?c ...
and you want a (the row number in SQL_Code) to change with each iteration
in the loop.
Perhaps this might work (I'm not saying this is the best option, just a
potential fix for what you have):
for (a in 1:2000) {
Dataset - sqlQuery(Database, SQL_Code[a,1])
print(summary(lm(Quant~ UPE + Mon, data = Dataset)))
}
ryanSt wrote
Hello R-Experts,
I've got a question, concerning the automation of a number of regressions
(lm) with the help of a loop (for i in ).
The situation is as follows (the code follows after that):
I have my data in an access database. I have historical data for 2000
parts, for each of this parts I want to do a regression analysis, so I
need to do 2000 regressions (just for one country, there are also more
countries). So this would be a lot of manual effort.
What I want to do is to automate this routine. I have already established
a database connection via RODBC.
So I can acces every part with its distinct code by a SQL Query. For every
part, the SQL Query has to be adapted for the parts name.
My idea was to generate the SQL Code in Excel for ervery part and save
this as an txt-file. So I can define the SQL-Codes as an object (SQL_Code
- load.table(...txt, header = FALSE).
In the next step I can build a loop, which tells R to go through the
object SQL_Code line for line, using the text for the SQL Query.
The code is as follows:
SQL Code in the sqlparts.txt. file: SELECT table.* FROM table WHERE
((table.part) = '2929AAD766') (2000 lines for all parts, each part has a
distinct code)
SQL_Code - load.table(sqlparts.txt, header = FALSE)
Loop:
for (c in a(1:2000)) {
Dataset - sqlQuery(Database, SQL_Code[a,1])
print(summary(lm(Quant~ UPE + Mon, data = Dataset)))
}
Unfortunately, this loop doesn't work. I think, it's because R does not
interpret the object SQL_Code as text, so the sqlQuery is incomplete.
Can anybody help me with that problem?
Thank you in advance.
Greets
Ryan
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Re: [R] Help with study guide for R exam
343GS You should consider dropping out of college; I don't think you belong there. McDonalds is hiring. 343GS wrote Thanks for nothing, jerk! DENMARK IS POO -- View this message in context: http://r.789695.n4.nabble.com/Help-with-study-guide-for-R-exam-tp4173951p4178139.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] profile likelihood
Try posting this question at least 1 more time. plocq wrote Hi, I try to use the function profile() of the SpatialExtremes' package to obtain the profile likelihood of parameters for an extreme values fit based on Poisson process : fit-fpot(data, threshold, model=pp, npp=365). But when I call profile(fit), I obtain the following error (even if I precise others arguments of the function) : [1] profiling loc Erreur dans nlpot(p[1], p[2], ...) : argument(s) inutilisé(s) (loc = 9.58435562072) I don't understand what is unused and why. From where does this error come from? If someone could help me it would be nice...! -- View this message in context: http://r.789695.n4.nabble.com/profile-likelihood-tp4174090p4175254.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] Problem with xlsx package
If all else fails, open it in Excel... save as .csv read.csv() Nikhil Joshi wrote I am trying to read an xlsx spreadsheet (1506 rows, 501columns) all populated but getting the following error: Please advise as to how to get around this issue. res - read.xlsx(c:\\BSE_v2.xlsx,1) Error in .jcall(RJavaTools, Ljava/lang/Object;, invokeMethod, cl, : java.lang.OutOfMemoryError: Java heap space Here is the session info: sessionInfo() R version 2.12.1 (2010-12-16) Platform: i386-pc-mingw32/i386 (32-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] grid stats graphics grDevices utils datasets methods base other attached packages: [1] xlsx_0.2.4 xlsxjars_0.2.0 rJava_0.8-8 [4] quantmod_0.3-15 TTR_0.20-2 Defaults_1.1-1 [7] urca_1.2-4 RBGL_1.26.0 graph_1.28.0 [10] colorspace_1.0-1 spatstat_1.21-4 deldir_0.0-13 [13] mgcv_1.7-2 reshape_0.8.4plyr_1.4 [16] fExtremes_2100.77fTrading_2100.76 fGarch_2110.80 [19] fBasics_2110.79 timeSeries_2130.90 timeDate_2130.91 [22] mc2d_0.1-9 mvtnorm_0.9-96 mixtools_0.4.4 [25] boot_1.2-43 MASS_7.3-9 SuppDists_1.1-8 [28] ghyp_1.5.4 gplots_2.8.0 caTools_1.11 [31] bitops_1.0-4.1 gdata_2.8.1 gtools_2.6.2 [34] numDeriv_2010.11-1 xtable_1.5-6 PerformanceAnalytics_1.0.3.2 [37] sn_0.4-16mnormt_1.4-0 xts_0.7-5 [40] zoo_1.6-4RODBC_1.3-2 loaded via a namespace (and not attached): [1] lattice_0.19-13Matrix_0.999375-46 nlme_3.1-97tools_2.12.1 [[alternative HTML version deleted]] __ R-help@ 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. -- View this message in context: http://r.789695.n4.nabble.com/Problem-with-xlsx-package-tp3298470p4169793.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] similarity matrix
That really all depends on what you need; and I can't tell you what you need. set wrote I'm sorry, I made a mistake in my example. you're right. I don't really know how a similarity alogrithm worksbut I'm willing to try that...are there any good examples available? Thank you -- View this message in context: http://r.789695.n4.nabble.com/similarity-matrix-tp4157576p4160902.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] Why can't I figure this out? :S
Why can't you figure this out?
I think you already know the answer: I don't speak computer.
Time to learn, or get another job I suppose. I hope you at least speak
math-statistics if you are going to attempt to understand this.
The only way you are going to be able to figure that code out is to break it
up into pieces until you DO understand it. If that is too much work for
you, then
Good luck.
Jasmine007 wrote
Hi, so I don't speak computer and I have no idea what this code is telling
the program to do, but I apparently need to be able to find and isolate
influencial observations. Problem, I have no idea what the error means and
where it may be from in the code.
error I get is below the code
{
## OLS results
NameC- lm(gpanew~female+female:lastinit+agenew+canadian+mom_ed+yearstudy)
## default: choose psamp = quantile
n - length(residuals(NameC))
k - length(coef(NameC))
if(is.null(quantile)) quantile - c(floor((n + k + 1)/2),
floor((n + k)/2))
quantile - rep(quantile, length.out = 2)
if(is.null(psamp)) psamp - quantile[1]
## LTS results with robust residuals
NameC_lts -
lqs(gpanew~female+female:lastinit+agenew+canadian+mom_ed+yearstudy,
quantile = quantile[1], psamp = psamp, nsamp = nsamp)
rr - residuals(NameC_lts)/NameC_lts$scale[2]
rr_nok - abs(rr) critval[1]
## robust leverage via MCD (or MVE)
X - model.matrix(NameC)[,-1]
cv - cov.rob(X, method = method,
quantile = quantile[2], nsamp = dist_nsamp)
rd - sqrt(mahalanobis(X, cv$center, cv$cov))
rd_nok - rd critval[2]
## ROBUST results
nok - rr_nok rd_nok
NameC_rob - lm(formula, data[!nok,])
rval - list(ols = NameC, lts = NameC_lts, robust = NameC_rob,
cov.rob = cv, robresid = rr, robdist = rd,
high_residuals = rr[rr_nok], high_leverage = rd[rd_nok],
bad_leverage = nok, psamp = psamp, method = method,
nsamp = list(lts = nsamp, dist = dist_nsamp),
quantile = list(lts = quantile[1], dist = quantile[2]))
return(rval)
}
Error: object of type 'closure' is not subsettable
Any ideas, In plain non-computer language.. I have a headache from trying
to figure out if this question had been answered before.
Thank you
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Re: [R] frequency table?
Set, This is the same post as your Similarity Matrix post. I'm not trying to be a smart ass here, but ... ?Can you fit a square peg in a round hole?... yes, but it doesn't mean it belongs there. I suggest you get a piece of paper and a pencil and figure out 1) what you are trying to do and why, and 2) how you would do that by hand. After you do that, you should be able to figure this out on your own or formulate a question to post that someone will be able to answer. Reading might help as well. set wrote Hello R-users, I've got a file with individuals as colums and the clusters where they occur in as rows. And I wanted a table which tells me how many times each individual occurs with another. I don't really know how such a table is called...it is not a frequency tableMy eventual goal is to make Venn-diagrams from the occurence of my individuals. So I've this: cluster ind1 ind2 ind3 etc. 10 1 2 23 01 31 1 1 And I want to go to this: ind1 ind2 ind3 ind1 0 42 ind2 4 04 ind3 2 4 1 is there a way to do this? Thank you for your help -- View this message in context: http://r.789695.n4.nabble.com/frequency-table-tp4156422p4163192.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] Programming Statistical Functions
As Stephen pointed out, this is easy to do. The word file the OP posted has all the necessary formulae. Now you just need to learn how to convert those formulae into R functions Stephen gave you an example of how to create a function for CV. Now run with it. perhaps something like this is what you need: http://cran.r-project.org/doc/contrib/Short-refcard.pdf It is time for you to do some reading on your own. Good luck. gvjones wrote Hello, Did you find anything helpful for calculating the statistical functions in your list. I would like to also calculate these and have been looking for some code to do so. Thanks -- View this message in context: http://r.789695.n4.nabble.com/Programming-Statistical-Functions-tp2307880p4157101.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] problem merging data with different shapes
Please use dput() to post your example data sets. dput(Adata) dput(Bdata) **then copy and paste the results of each so that we can play around with it easily. Miriam -2 wrote I have been trying to merge datasets, one of which has a long format (Adata) and one has a (different) long format (Bdata): Adata Bdata subject order bpm subject order trial agegroup gender 1 1 70.21 1 3 2 1 1 1 69.51 2 1 2 1 1 1 68.81 3 2 2 1 1 2 69.12 1 2 1 2 1 270 2 2 3 1 2 1 2 70.52 3 1 1 2 1 3 70.2... 1 3 1 3 2 1 2 1 ... ... In the end I would like to have a dataset that contains A unchanged with the additional information from B added. subject order bpm trial agegroup gender 1 1 70.2 3 2 1 1 1 69.5 3 2 1 1 1 68.8 3 2 1 ... I have tried: newdataframe - merge(Adata,Bdata, by= c(subject, order), sort = FALSE) For some reason, the trial column is not matched to the subject and order information, despite them being identified as key-variables for the merge. (The same is true for other variables, the actual dataset has more variables and trials, but this is essentially the problem.) So it results in something like: subject order bpm trial agegroup gender 1 1 70.2 3 2 1 1 1 69.5 2 2 1 1 1 68.8 1 2 1 What could be my mistake? Thank you VERY much. Miriam -- -- __ R-help@ 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. -- View this message in context: http://r.789695.n4.nabble.com/problem-merging-data-with-different-shapes-tp4157137p4157442.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] Export cols to single csv files
just use indexing. without doing it all for you... df - structure(list(AA = c(0.3, 0.1, 0.6), BB = c(0.9, 0.4, 0.2), CC = c(1, 0.8, 0.6), DD = c(0.7, 0.5, 0.5)), .Names = c(AA, BB, CC, DD), class = data.frame, row.names = c(NA, -3L )) write.csv(df[,1], paste(colnames(df[1]), csv, sep=.)) Chega wrote Hi I am trying to batch export the columns of a numeric matrix to separate csv files by naming them according to the column names. So my matrix in R looks like this: AA BB CC DD etc. 1: 0.3 0.9 1.0 0.7 ... 2: 0.1 0.4 0.8 0.5 ... 3: 0.6 0.2 0.6 0.5 ... etc. Now I am looking for a way to get these files (file names in quotes): AA.csv BB.csv CC.csvetc. 1: 0.31: 0.91: 1.0 2: 0.12: 0.42: 0.8 3: 0.63: 0.23: 0.6 etc. etc. etc. As I understand this may be done using write.csv and a loop with the column names, but I have no idea how to export single columns. Thanks i.a. for help! Chega -- View this message in context: http://r.789695.n4.nabble.com/Export-cols-to-single-csv-files-tp4157484p4157537.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] similarity matrix
I apologize to the list and you if I am mis-understanding something, but... As an example: ind2 occurs with ind1 only in cluster#3, so why does it get a value of 4 in your similarity matrix? Also, if this isn't a recognized similarity algorithm, perhaps you should at the very least put quotes around similarity. Again, sorry if I am confused here. set wrote Hello R-users, I've got a file with individuals as colums and the clusters where they occur in as rows. And I wanted a similarity matrix which tells me how many times each individual occurs with another. My eventual goal is to make Venn-diagrams from the occurence of my individuals. So I've this: cluster ind1 ind2 ind3 etc. 10 1 2 23 01 31 1 1 And I want to go to this: ind1 ind2 ind3 ind1 0 42 ind2 4 04 ind3 2 4 1 is there a way to do this? Thank you for your help -- View this message in context: http://r.789695.n4.nabble.com/similarity-matrix-tp4157576p4157840.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] Iteration in R
Hi Michael, How would you do this with lapply to return a list? I can't seem to get that to work (I haven't used these much and am trying to learn). Thanks Brad Michael Weylandt wrote ? replicate or a for loop or do all one hundred simulations at once x - matrix(rnorm(100^2, 1, 2), 100) It's going to depend on what you want to do with the numbers. Michael On Sat, Dec 3, 2011 at 1:10 PM, Martin Zonyrah lt;martin2005z@gt; wrote: Hi, I need help. I am trying to iterate this command x - rnorm(100, 1.0, 2.0) one hundred times in R but I don't seem to have a clue. Can anyone help? Your help is very much appreciated. Martin [[alternative HTML version deleted]] __ R-help@ mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@ 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. -- View this message in context: http://r.789695.n4.nabble.com/Iteration-in-R-tp4154433p4154479.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] Iteration in R
Interesting and thank you; I'm confused as to why this doesn't work with: lapply(rep(1,6), FUN=rnorm, n=10, mean=1.0, sd=1) andrija djurovic wrote Hi Brad. Maybe something like this: lapply(rep(1,6), function(x) rnorm(10,0,1)) Andrija On Sat, Dec 3, 2011 at 8:21 PM, B77S lt;bps0002@gt; wrote: Hi Michael, How would you do this with lapply to return a list? I can't seem to get that to work (I haven't used these much and am trying to learn). Thanks Brad Michael Weylandt wrote ? replicate or a for loop or do all one hundred simulations at once x - matrix(rnorm(100^2, 1, 2), 100) It's going to depend on what you want to do with the numbers. Michael On Sat, Dec 3, 2011 at 1:10 PM, Martin Zonyrah lt;martin2005z@gt; wrote: Hi, I need help. I am trying to iterate this command x - rnorm(100, 1.0, 2.0) one hundred times in R but I don't seem to have a clue. Can anyone help? Your help is very much appreciated. Martin [[alternative HTML version deleted]] __ R-help@ mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@ 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. -- View this message in context: http://r.789695.n4.nabble.com/Iteration-in-R-tp4154433p4154479.html Sent from the R help mailing list archive at Nabble.com. __ R-help@ 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. [[alternative HTML version deleted]] __ R-help@ 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. -- View this message in context: http://r.789695.n4.nabble.com/Iteration-in-R-tp4154433p4154559.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] Moving column averaging
I don't know the answer, but would suppose not. You could test this for yourself using: system.time() example: system.time(rnorm(10,0,1)) Chega wrote This solved my problem - Thanks a lot for your help! Please allow me one more question: Works zoo's rollapply on a plain matrix faster than on a zoo object? I am asking since I wanted to apply the provided averaging code to a larger matrix (2500 rows x 200 cols), which is quite time consuming... Thanks again! Chega From: B77S [via R] lt;ml-node+s789695n4143909h3@.nabblegt; To: Chega lt;chegaga@gt; Sent: Friday, December 2, 2011 6:16 AM Subject: Re: Moving column averaging Sorry for that, and thanks Gabor, I could have sworn that it wouldn't. Gabor Grothendieck wrote On Thu, Dec 1, 2011 at 7:13 PM, B77S [hidden email] wrote: # need zoo to use rollapply() # your data (I called df) df - structure(list(a = 1:2, b = 2:3, c = c(5L, 9L), d = c(9L, 6L), e = c(1L, 5L), f = c(4, 7)), .Names = c(a, b, c, d, e, f), class = data.frame, row.names = c(NA, -2L)) # transpose and make a zoo object df2 - zoo(t(df)) #rollapply to get means and transpose back means - t(rollapply(df2, width=2, by=2, FUN=mean)) # adding the combined column names you requested colnames(means) - apply(matrix(names(df), nrow=2), 2, paste, collapse=, ) Note that zoo's rollapply also works on plain matrices and vectors. -- Statistics Software Consulting GKX Group, GKX Associates Inc. tel: 1-877-GKX-GROUP email: ggrothendieck at gmail.com __ [hidden email] 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. If you reply to this email, your message will be added to the discussion below: http://r.789695.n4.nabble.com/Moving-column-averaging-tp4130179p4143909.html To unsubscribe from Moving column averaging, click here. NAML -- View this message in context: http://r.789695.n4.nabble.com/Moving-column-averaging-tp4130179p4154770.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] Data Analysis for Gas Prices
use a ? to get help on a function; example: ?read.table If you do this you will see an option called header... use header=T if your top row contains column names. Learn how to read these help pages. Also, read thru a few beginner R manuals and see this website: http://www.statmethods.net/interface/io.html As for the rest of your questions. Perhaps this is your sign to begin reading a intro-stats book, or inquire on a different forum. see: http://stats.stackexchange.com/questions good luck inferno846 wrote Hi there, I'm looking to analyze a set of data on local gas prices for a single day. I'm wondering what kind of questions I should be looking to ask and how to find and answer to them with R. Examples would be: Do prices differ between brands? Does location affect (NE, NW, SE, SW) price? Does the number of nearby (within .25 miles) competitors affect price? Do gas stations near shopping centers or highways have different prices? Also, could anyone help me figure out how to import a data table to R? When I try to create a .txt file from a word document and read it in R, the format of the first column always messes up. Any/all help is appreciated. -- View this message in context: http://r.789695.n4.nabble.com/Data-Analysis-for-Gas-Prices-tp4155078p4155185.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] How to test for Poisson?
A simple way to determine if it is NOT is to see if the mean (the single parameter of a poisson: lambda) and variance are the same. This really has nothing to do with R (other than the data source), and since it is homework, you will likely get no further help here. Good luck. RToss wrote Hi! I am sitting with a school assignment, but I got stuck on this one. I am suppose to test if my data is Poisson-distributed. The data I´m using is the studie Bids, found in the Ecdat-package, and the variable of interest is the dependent numbids. How do I practically perform a test for this? Kind regards/ Richard -- View this message in context: http://r.789695.n4.nabble.com/How-to-test-for-Poisson-tp4147356p4147519.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] Summarizing elements of a list
Someone is bound to know a better way, but...
subset(unlist(Version1_), subset=names(unlist(Version1_))==First)
LCOG1 wrote
Hi everyone,
I looked around the list for a while but couldn't find a solution to my
problem. I am storing some results to a simulation in a list and for each
element i have two separate vectors(is that what they are called, correct
my vocab if necessary). See below
Version1_-list()
for(i in 1:5){
Version1_[[i]]-list(First=rnorm(1),Second=rnorm(1))
}
What I want is to put all of the elements' 'First' vectors into a single
list to box plot. But whats a more elegant solution to the below?
c(Version1_[[1]]$First,Version1_[[2]]$First,Version1_[[3]]$First,Version1_[[4]]$First,Version1_[[5]]$First)
since i have 50 or more simulations this is impractical and sloppy. Do I
need to store my data differently or is their a solution on the back end?
Thanks all.
Josh
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Re: [R] Moving column averaging
# need zoo to use rollapply() # your data (I called df) df - structure(list(a = 1:2, b = 2:3, c = c(5L, 9L), d = c(9L, 6L), e = c(1L, 5L), f = c(4, 7)), .Names = c(a, b, c, d, e, f), class = data.frame, row.names = c(NA, -2L)) # transpose and make a zoo object df2 - zoo(t(df)) #rollapply to get means and transpose back means - t(rollapply(df2, width=2, by=2, FUN=mean)) # adding the combined column names you requested colnames(means) - apply(matrix(names(df), nrow=2), 2, paste, collapse=, ) HTH -- View this message in context: http://r.789695.n4.nabble.com/Moving-column-averaging-tp4130179p4143329.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] calculate mean of multiple rows in a data frame
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Re: [R] Moving column averaging
Sorry for that, and thanks Gabor, I could have sworn that it wouldn't. Gabor Grothendieck wrote On Thu, Dec 1, 2011 at 7:13 PM, B77S lt;bps0002@gt; wrote: # need zoo to use rollapply() # your data (I called df) df - structure(list(a = 1:2, b = 2:3, c = c(5L, 9L), d = c(9L, 6L), e = c(1L, 5L), f = c(4, 7)), .Names = c(a, b, c, d, e, f), class = data.frame, row.names = c(NA, -2L)) # transpose and make a zoo object df2 - zoo(t(df)) #rollapply to get means and transpose back means - t(rollapply(df2, width=2, by=2, FUN=mean)) # adding the combined column names you requested colnames(means) - apply(matrix(names(df), nrow=2), 2, paste, collapse=, ) Note that zoo's rollapply also works on plain matrices and vectors. -- Statistics Software Consulting GKX Group, GKX Associates Inc. tel: 1-877-GKX-GROUP email: ggrothendieck at gmail.com __ R-help@ 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. -- View this message in context: http://r.789695.n4.nabble.com/Moving-column-averaging-tp4130179p4143909.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] I cannot get species scores to plot with site scores in MDS when I use a distance matrix as input. Problems with NA's?
-- View this message in context: http://r.789695.n4.nabble.com/I-cannot-get-species-scores-to-plot-with-site-scores-in-MDS-when-I-use-a-distance-matrix-as-input-Pr-tp4103699p4104295.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] I cannot get species scores to plot with site scores in MDS when I use a distance matrix as input. Problems with NA's?
Try the daisy() function from the package cluster, it seems to be able to handle NAs and non-dummy coded character variables metaMDS(daisy(df, metric=gower)) Edwin Lebrija Trejos wrote Hi, First I should note I am relatively new to R so I would appreciate answers that take this into account. I am trying to perform an MDS ordination using the function “metaMDS” of the “vegan” package. I want to ordinate species according to a set of functional traits. “Species” here refers to “sites” in traditional vegetation analyses while “traits” here correspond to “species” in such analyses. My data looks like this: Trait1 Trait2 Trait3 Trait4 Trait5 Trait… Species1 228.44 16.56 1.66 13.22 1 short Species2 150.55 28.07 0.41 0.60 1 mid Species3 NA 25.89 NA 0.55 0 large Species4 147.70 17.65 0.42 1.12 NA large Species… 132.68 NA 1.28 2.75 0 short Because the traits have different variable types, different measurement scales, and also missing values for some species, I have calculated the matrix of species distances using the Gower coefficient of similarity available in Package “FD” (which allows missing values). My problem comes when I create a bi-plot of species and traits. As I have used a distance matrix in function “metaMDS” there are no species scores available. This is given as a warning in R: NMDSgowdis- metaMDS(SpeciesGowdis) plot(NMDSgowdis, type = t) Warning message:In ordiplot(x, choices = choices, type = type, display = display, :Species scores not available” I have read from internet resources that in principle I could obtain the trait (species) scores to plot them in the ordination but my attempts have been unsuccessful. I have tried using the function “wascores” in package vegan and “add.spec.scores” in package BiodiversityR. For this purpuse I have created a new species x traits table where factor traits were coded into dummy variables and all integer variables (including binary) were coerced to numeric variables. Here are the codes used and the error messages I have got: “ NMDSgowdis- metaMDS(SpeciesGowdis) NMDSpoints-postMDS(NMDSgowdis$points,SpeciesGowdis) NMDSwasc-wascores(NMDSpoints,TraitsNMDSdummies) Error in if (any(w 0) || sum(w) == 0) stop(weights must be non-negative and not all zero) : missing value where TRUE/FALSE needed” I imagine the problem is with the NA’s in the data. Alternatively, I have used the “add.spec.scores” function, method=”cor.scores”, found in package BiodiversityR. This seemed to work, as I got no error message, but the species scores were not returned. Here the R code and results: “ A-add.spec.scores(ordi=NMDSgowdis,comm=TraitsNMDSdummies,method=cor.scores,multi=1, Rscale=F,scaling=1) plot(A) Warning message:In ordiplot(x, choices = choices, type = type, display = display, :Species scores not available“ Can anyone guide me to get the trait (“species”) scores to plot together with my species (“site”) scores? Thanks in advance, Edwin __ R-help@ 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. -- View this message in context: http://r.789695.n4.nabble.com/I-cannot-get-species-scores-to-plot-with-site-scores-in-MDS-when-I-use-a-distance-matrix-as-input-Pr-tp4103699p4104406.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] Looping and paste
out - vector(list)
Ylab - for(i in 1:length(BndY))
{
out[i] - paste(BndY[i], to ,BndY[i],mN)
}
Ylab - do.call(c, out)
markm0705 wrote
Dear R helpers
I'm trying to make up some labels for plot from this vector
BndY-seq(from = 18900,to= 19700, by = 50)
using
Ylab-for(i in BndY) {c((paste(i, to ,i+50,mN)))}
but the vector created is NULL
However if i use
for(i in BndY) {print(c(paste(i, to ,i+50,mN)))}
I can see the for loop is making the labels I'm looking for but not sure
on my error in assigning them to a vector
Thanks in advance
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Re: [R] Removing rows in dataframe w'o duplicated values
This is ugly, but it gets what you want.
dat[which(dat[,1] %in% unique((dat[duplicated(dat[,1], fromLast = T),
1]))),]
AC Del Re wrote
Hi,
Is there an easy way to remove dataframe rows without duplicated values of
a specified column ('id')? e.g.,
dat - data.frame(id = c(1,1,1,2,3,3), value = c(5,6,7,4,5,4), value2 =
c(1,4,3,3,4,3))
dat
id value value2
1 1 5 1
2 1 6 4
3 1 7 3
4 2 4 3
5 3 5 4
6 3 4 3
This is sample data and the real data has hundreds of rows. In this
case, only row 4 does not have a duplicated id and I would like to
remove it without using:
dat$id[4] - NULL
Any help is appreciated!
AC
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Re: [R] return only pairwise correlations greater than given value
This is probably not the prettiest or most efficient function ever, but this
seems to do what I wanted.
spec.cor - function(dat, r, ...){
require(reshape)
d1 - data.frame(cor(dat))
d2 - melt(d1)
d2[,3] - rep(rownames(d1), nrow(d2)/length(unique(d2[,1])))
d2 - d2[,c(variable, V3, value)]
colnames(d2) - c(V1, V2, value)
d2 - d2[with(d2, which(V1 != V2, arr.ind=T)), ]
d2 - d2[which(d2[,3] =r | d2[,3] = -r, arr.ind=T),]
d2[,1:2] - t(apply(d2[,1:2], MARGIN=1, function(x) sort(x)))
d2 - unique(d2)
return(d2)
}
data(mtcars)
spec.cor(mtcars[,2:5], .6)
Using as id variables
V1 V2 value
2 cyl disp 0.9020329
3 cyl hp 0.8324475
4 cyl drat -0.6999381
7 disp hp 0.7909486
8 disp drat -0.7102139
I'm not sure how to make melt() quit giving the Using as id variables
warning, but I don't really care either.
B77S wrote:
Thanks Michael,
I just started on the following code (below), and realized I should ask,
as this likely exists already.
basically what I'd like is for the function to return (basically) what you
just suggested, plus the names of the two variables (I suppose pasted
together would be good).
I hope that is clear, and obviously I didn't get so far as to add the
names to the output.
#
sig.cor - function(dat, r, ...){
cv2 - data.frame(cor(dat))
var.names - rownames(cv2)
list.cv2 - which(cv2 =r | cv2 = -r, arr.ind=T)
cor.r - cv2[list.cv2[which(list.cv2 [,row]!=list.cv2 [,col]),]]
cor.names - var.names[list.cv2[which(list.cv2 [,row]!=list.cv2
[,col]),]]
return(cor.r)
}
data(mtcars)
sig.cor(mtcars[,2:5], .90)
# sig.cor(mtcars[,2:5], .90)
#[1] 0.9020329 0.9020329
# Ideally this would look likt this:
cyl-disp
0.9020329
Michael Weylandt wrote:
What exactly do you mean returns them? More generally I suppose,
what do you have in mind to do with this?
You could do something like this:
BigCorrelation - function(X){
return(which(abs(cor(X)) 0.9, arr.ind = T))
}
but it hardly seems worth its own function call.
On Thu, Nov 17, 2011 at 12:42 AM, B77S lt;bps0002@gt; wrote:
Hello,
I would like to find out if a function already exists that returns only
pairwise correlations above/below a certain threshold (e.g, -.90, .90)
Thank you.
--
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Re: [R] return only pairwise correlations greater than given value
Excellent; thanks Josh.
Joshua Wiley-2 wrote:
Hi Brad,
You do not really need to reshape the correlation matrix. This seems
to do what you want:
spec.cor - function(dat, r, ...) {
x - cor(dat, ...)
x[upper.tri(x, TRUE)] - NA
i - which(abs(x) = r, arr.ind = TRUE)
data.frame(matrix(colnames(x)[as.vector(i)], ncol = 2), value = x[i])
}
spec.cor(mtcars[, 2:5], .6)
Cheers,
Josh
On Wed, Nov 16, 2011 at 9:58 PM, B77S lt;bps0002@gt; wrote:
Thanks Michael,
I just started on the following code (below), and realized I should as as
this might exist.
basically what I'd like is for the function to return (basically) what
you
just suggested, plus the names of the two variables (I suppose pasted
together would be good).
I hope that is clear.
#
sig.cor - function(dat, r, ...){
cv2 - data.frame(cor(dat))
var.names - rownames(cv2)
list.cv2 - which(cv2 =r | cv2 = -r, arr.ind=T)
cor.r - cv2[list.cv2[which(list.cv2 [,row]!=list.cv2 [,col]),]]
cor.names - var.names[list.cv2[which(list.cv2 [,row]!=list.cv2
[,col]),]]
return(cor.r)
}
Michael Weylandt wrote:
What exactly do you mean returns them? More generally I suppose,
what do you have in mind to do with this?
You could do something like this:
BigCorrelation - function(X){
return(which(abs(cor(X)) 0.9, arr.ind = T))
}
but it hardly seems worth its own function call.
On Thu, Nov 17, 2011 at 12:42 AM, B77S lt;bps0002@gt; wrote:
Hello,
I would like to find out if a function already exists that returns
only
pairwise correlations above/below a certain threshold (e.g, -.90, .90)
Thank you.
--
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--
Joshua Wiley
Ph.D. Student, Health Psychology
Programmer Analyst II, ATS Statistical Consulting Group
University of California, Los Angeles
https://joshuawiley.com/
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Re: [R] R forum for only Statistics
In addition to getting help from others, I always find that seeking answers through reading to be helpful. I would suggest a basic stats book and this paper regarding issues indices like species richness: http://www.amazon.com/Primer-Ecological-Statistics-Nicholas-Gotelli/dp/0878932690 Here is one you should probably read as well: http://onlinelibrary.wiley.com/doi/10.1046/j.1461-0248.2001.00230.x/full Happy reading. geeknick wrote: Hi there I need some http://calculate-conditional-and-distributio.blogspot.com/2011/11/blog-post.html Statistics Help !! What statistical tests should i use/consider (pref in SPSS) in regards to an invertebrate survey i have carried out comparing urban street trees to urban park trees. I am looking at species richness and abundance between the two sites and also other variables such as tree species, tree width at base, distance from road/tarmac, tree base substrate and weather at time of collection. any help would be greatly appreciated! i tried on this link and i found this blog very usefull. -- View this message in context: http://r.789695.n4.nabble.com/R-forum-for-only-Statistics-tp3480715p4076599.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] return only pairwise correlations greater than given value
Hello, I would like to find out if a function already exists that returns only pairwise correlations above/below a certain threshold (e.g, -.90, .90) Thank you. -- View this message in context: http://r.789695.n4.nabble.com/return-only-pairwise-correlations-greater-than-given-value-tp4079028p4079028.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] split list of characters in groups of 2
hi, If i have a list of things, like this var.names - c(a, b, c, d, e, f) how can i get this: a, b, c, d, e, f thanks ahead of time. -- View this message in context: http://r.789695.n4.nabble.com/split-list-of-characters-in-groups-of-2-tp4079031p4079031.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] return only pairwise correlations greater than given value
Thanks Michael,
I just started on the following code (below), and realized I should as as
this might exist.
basically what I'd like is for the function to return (basically) what you
just suggested, plus the names of the two variables (I suppose pasted
together would be good).
I hope that is clear.
#
sig.cor - function(dat, r, ...){
cv2 - data.frame(cor(dat))
var.names - rownames(cv2)
list.cv2 - which(cv2 =r | cv2 = -r, arr.ind=T)
cor.r - cv2[list.cv2[which(list.cv2 [,row]!=list.cv2 [,col]),]]
cor.names - var.names[list.cv2[which(list.cv2 [,row]!=list.cv2
[,col]),]]
return(cor.r)
}
Michael Weylandt wrote:
What exactly do you mean returns them? More generally I suppose,
what do you have in mind to do with this?
You could do something like this:
BigCorrelation - function(X){
return(which(abs(cor(X)) 0.9, arr.ind = T))
}
but it hardly seems worth its own function call.
On Thu, Nov 17, 2011 at 12:42 AM, B77S lt;bps0002@gt; wrote:
Hello,
I would like to find out if a function already exists that returns only
pairwise correlations above/below a certain threshold (e.g, -.90, .90)
Thank you.
--
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Re: [R] pairs(), no axis labels/values for upper panel?
Steffen,
Did you ever have luck getting rid of the tick marks?...
I like your idea and have modified it, but yes, the tick marks need to go.
Steffen Fleischer wrote:
Dear all,
I want to draw a graph that contains the scatterplot matrix in the lower
panel and coefficients in the upper panel. I used and adapted the example
for the function pairs but cannot figure out how to get no values and
ticks in the upper panel (the values should only be in the lower panel).
The upper panel looks odd to me this way. Any hints?
Thanks in advance
Steffen
Here is an example what the graph looks like:
#
data(mtcars)
panel.cor - function(a, b, digits=2, ...)
{
usr - par(usr); on.exit(par(usr))
par(usr = c(0, 1, 0, 1))
x-cbind(a,b)
x-na.omit(x)
n - nrow(x)
pp - c(0.025, 0.975)
corx - cor(x,method=s)[1, 2]
CI-c(tanh(atanh(corx) + qnorm(pp)/sqrt((n - 3)/1.06)))
txt1 - paste(rho =,format(c(corx, 0.123456789),
digits=digits)[1])
txt2-paste((,format(c(CI,0.123456789)[1],digits=digits),;
,format(c(CI,0.123456789)[2],digits=digits),),sep=)
txt3-paste(N =,round(n,0))
txt - paste(txt1,\n,95%KI ,txt2,\n,txt3, sep=)
text(0.5, 0.5, txt,cex=.8)
}
diag.cor-function(a,b, ...)
{
usr - par(usr); on.exit(par(usr))
par(usr = c(0, 1, 0,1))
rect(0,0,1,1,col=grey)
}
pairs(mtcars[1:4],upper.panel=panel.cor,diag.panel=diag.cor,label.pos=0.5)
#
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Re: [R] multivariate modeling codes
Not sure if this helps, but did you try Google? http://www.jstatsoft.org/v35/i09/paper http://cran.r-project.org/web/packages/lcmm/lcmm.pdf http://www.warwick.ac.uk/statsdept/useR-2011/abstracts/010411-liquetbenoit.pdf yurirouge wrote: HI, I am relatively new to R and would appreciate some help or directions for this. I am trying to model 3 longitudinal outcomes jointly and to identify some predictors for these 3 joint outcomes (all continuous). I am trying to find some codes that I may modify to do this but cannot seem to find anything. -- View this message in context: http://r.789695.n4.nabble.com/multivariate-modeling-codes-tp4032624p4032643.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] Interpreting Multiple Linear Regression Summary
Please see ?dput use dput(your data) and paste the output into a reply, thanks. This way we know what you are working with. Rich Shepard wrote: I would appreciate pointers on what I should read to understand this output: summary(lm(TDS ~ Cond + Ca + Cl + Mg + Na + SO4)) Call: lm(formula = TDS ~ Cond + Ca + Cl + Mg + Na + SO4) Residuals: ALL 1 residuals are 0: no residual degrees of freedom! Coefficients: (6 not defined because of singularities) Estimate Std. Error t value Pr(|t|) (Intercept) 125 NA NA NA Cond NA NA NA NA CaNA NA NA NA ClNA NA NA NA MgNA NA NA NA NaNA NA NA NA SO4 NA NA NA NA Residual standard error: NaN on 0 degrees of freedom (63 observations deleted due to missingness) When I look at the summary for the data frame used for this model I do not see an excessive number of missing values or indications why there are no residual degrees of freedom. The same model applied to 8 other data frames did not produce similar results. Puzzled, Rich __ R-help@ 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. -- View this message in context: http://r.789695.n4.nabble.com/Interpreting-Multiple-Linear-Regression-Summary-tp4020516p4020567.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] Interpreting Multiple Linear Regression Summary
There is only one row with a complete set of observations; I think lm() is throwing out the rest. Rich Shepard wrote: On Wed, 9 Nov 2011, John C Frain wrote: As far as I know if there is an NA in any variable in an observation the default is to drop the entire observation. Thus there are no observations in your calculation John, Hadn't realized that. I know there are NA's in other data frames that yield model results. Perhaps it is the excessive numbers in this set that are the problem. Thanks, Rich __ R-help@ 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. -- View this message in context: http://r.789695.n4.nabble.com/Interpreting-Multiple-Linear-Regression-Summary-tp4020516p4021352.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] Interpreting Multiple Linear Regression Summary
This is the output of dput(your data) structure(list(Ca = c(NA, NA, 24.4, NA, 21.4, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, 28, 32, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, 34.7, NA, 42.5, NA, 26, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, 0.6, 21.4, NA, 48.3, 63.5, NA, NA, 28.7, NA, NA, NA, NA, 64.3, 23), Cl = c(1.58, 5.6, 3, NA, 1, 5, 1.2, 4, 4, 8.4, 1, 1.4, 4.9, 1.7, 2, 1.6, 3.3, 2.2, 9, 1, 2, 1, 1, 5, 4, 3, 2.27, 1.76, 5.81, 4.23, 4.23, 6.25, 6.72, 4, NA, 5, 5.8, 5.8, 2.2, 5.4, 5.4, 4.8, 8, 1, 4.8, 5.9, 5.9, 13, 5.6, 1.2, NA, NA, NA, 3, 7, NA, NA, 2, NA, NA, NA, NA, 7, 4.1), Cond = c(NA, NA, 190, 187, 184, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, 248, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, 304, 354, 379, NA, 300, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, 2.2, 187, 285, 378, 533, 207, 262, 244, 238, 280, 380, 402, 636, 300), Mg = c(NA, NA, 10, NA, 9.1, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, 11, 12, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, 17.4, NA, 21.1, NA, 24, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, 9.5, NA, 22.1, 29.9, NA, NA, 12.6, NA, NA, NA, NA, 32.4, 21), Na = c(NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, 4L, 4L, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA), SO4 = c(9.4, 6.5, 9, NA, 7, 55, 6.8, 105, 15.6, 8.4, 8.8, 19.4, 37, 12, 10, 9.1, 34, 11, 69, 18, 9, 13, 9, 7, 6, 5, 7.8, 7.8, 7.5, 6, 7.3, 7, 7.5, 6, 7, 7, 5.6, 5.6, 5.4, 11, 10.5, 9.9, 11.7, 8.4, 12.1, 16, 20, 7.6, 17, 6.5, NA, 8, 22, 24, 44, NA, 13, 13, 12, 18, 23, 23, 73, 4), TDS = c(105L, 181L, 112L, 144L, 114L, 308L, 96L, 430L, 108L, 108L, 125L, 129L, 360L, 140L, 95L, 120L, 280L, 130L, 352L, 148L, 107L, 125L, 139L, 188L, 201L, 178L, 197L, 187L, 182L, 165L, 186L, 191L, 190L, 176L, 175L, 220L, 163L, 163L, 152L, 221L, 171L, 204L, 174L, 190L, 174L, 210L, 190L, 180L, 200L, 180L, NA, 120L, 135L, 228L, 14L, NA, 156L, 140L, 128L, 160L, 215L, 230L, 316L, 163L)), .Names = c(Ca, Cl, Cond, Mg, Na, SO4, TDS), class = data.frame, row.names = c(NA, -64L)) B77S wrote: Please see ?dput use dput(your data) and paste the output into a reply, thanks. This way we know what you are working with. Rich Shepard wrote: I would appreciate pointers on what I should read to understand this output: summary(lm(TDS ~ Cond + Ca + Cl + Mg + Na + SO4)) Call: lm(formula = TDS ~ Cond + Ca + Cl + Mg + Na + SO4) Residuals: ALL 1 residuals are 0: no residual degrees of freedom! Coefficients: (6 not defined because of singularities) Estimate Std. Error t value Pr(|t|) (Intercept) 125 NA NA NA Cond NA NA NA NA CaNA NA NA NA ClNA NA NA NA MgNA NA NA NA NaNA NA NA NA SO4 NA NA NA NA Residual standard error: NaN on 0 degrees of freedom (63 observations deleted due to missingness) When I look at the summary for the data frame used for this model I do not see an excessive number of missing values or indications why there are no residual degrees of freedom. The same model applied to 8 other data frames did not produce similar results. Puzzled, Rich __ R-help@ 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. -- View this message in context: http://r.789695.n4.nabble.com/Interpreting-Multiple-Linear-Regression-Summary-tp4020516p4021355.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] Correlation analysis
I would start by reading one or more of the introduction manuals available here: http://mirrors.ibiblio.org/pub/mirrors/CRAN/ wizi wrote: Hi everyone, I am new to R-project. I did search through the list for my problem but i can't find it. I am sorry if this question has been asked. I would like to perform a correlation analysis between a hiv data and gene expression. Basically, i have a file that contains: hiv_name, start_position, end_position, chromosome. I would like to see if these data has anything to do with the location of our genes (I also have another file contains gene_name, start_position, end_position, chromosome). What functions that allow me to do this? I am very new to R and hopefully someone can guide me to the right direction. Thank you very much, -- View this message in context: http://r.789695.n4.nabble.com/Correlation-analysis-tp3996877p3996961.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] help with unequal variances
the following is a more appropriate forum for your question, seeing as this has nothing to do with R (per se). http://stats.stackexchange.com/questions good luck. peak99 wrote: Hello, I have some patient data for my masters thesis with three groups (n=16, 19 20) I have completed compiling the results of 7 tests, for which one of these tests the variances are unequal. I wish to perform an ANOVA between the three groups but for the one test with unequal variance (0.001 by both bartlett and levene's test) I am not sure what to do. I thought i would run ANOVA with bonferonni post-test for groups with equal variances, then for the test with unequal variance i would use the welch correction and games-howell post-test. Does this sound reasonable? Someone has also recommended to me to use Kruskal-wallis ANOVA, then use Wilcoxon sign rank test pairwise to determine which groups are significantly different (ON ALL DATA, both equal and unequal variance tests). I don't think this is right, for two reasons: 1) Kruskal-wallis is for non-gaussian data, and i have no reason to believe they are not normal. - I have run normality tests which say they are normal, although perhaps my sample sizes are too small for a normality test? 2) i believe running pairwise Wilcoxon sign rank test is not acceptable unless there is a post-test correction for multiple comparisons (i am not aware of one); also on the wiki page for this test one of the assumptions says Under the null hypothesis the distributions of both groups are equal which i read to say that the variances must be equal. So I think there recommendations were based more on sample size and normality, and not my issue with variance? Ultimately i would like to know if i am going about this right with my deduction (ANOVA/Bonferonni of the test results, but welch correction and games-howell for the test with significantly different variances). and if not why and/or what you think is a better option. most appreciated to any help received! -- View this message in context: http://r.789695.n4.nabble.com/help-with-unequal-variances-tp3964779p3965046.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] Subsampling-oversampling from a data frame
If no one has a better solution, split it, take a sample of size X from both and put it back together. hgwelec wrote: Dear members, Consider the following data frame (first 4 rows shown) age sex class 15 m low 20 f high 15 f low 10 m low in my original data set i have 1200 rows and a class distribution of low=0.3 and high=0.7 My question : how can i create a new data frame as the one shown above but with the 'high' class subsampled so that in the new data frame the class distribution is low=0.5 and high=0.5? I tried looking at the sample function and prob option but all examples i seen do not use an imbalanced class problem as the one shown above Thank you in advance Thank you in advance -- View this message in context: http://r.789695.n4.nabble.com/Subsampling-oversampling-from-a-data-frame-tp3965771p3965827.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] Subsampling-oversampling from a data frame
# Perhaps I misunderstand your original need, but ## I added a few lines to your data and used dput() to get the below data (I named df) df- structure(list(age = c(15L, 20L, 15L, 10L, 10L, 12L, 17L, 17L, 11L, 12L, 16L, 20L, 23L, 14L, 22L, 16L, 10L, 11L, 21L, 10L, 13L, 17L), sex = structure(c(2L, 1L, 1L, 2L, 2L, 2L, 1L, 1L, 2L, 1L, 2L, 1L, 1L, 2L, 2L, 2L, 1L, 1L, 2L, 1L, 2L, 1L), .Label = c(f, m), class = factor), class = structure(c(2L, 1L, 2L, 2L, 2L, 1L, 1L, 2L, 2L, 1L, 2L, 1L, 1L, 2L, 1L, 2L, 2L, 2L, 1L, 1L, 2L, 1L), .Label = c(high, low), class = factor)), .Names = c(age, sex, class), class = data.frame, row.names = c(NA, -22L )) ## the following line uses which(), sample(), and rbind(), along with some indexing to get a new dataframe; see ?which, ?sample, and ?rbind for more info # For the indexing, play with it, ... type in df[1:3,1:2] as an example new_df - rbind(df[sample(which(df$class==low), 4),], df[sample(which(df$class==high), 4),]) Now replace 4 with the the size of each you want. hgwelec wrote: Thank you for your answer. The problem is that i am learning R now, so i do not know how i could do this. I have found the following code but it does not work unfortunately (=create distribution 0.1 low class - 0.9 high) : data[c(rownames(data.df[data.df$class==high,]), sample(rownames(data[data.df$class==low]), 0.1)) , ] 2 posts This post has NOT been accepted by the mailing list yet. Dear members, Consider the following data frame (first 4 rows shown) age sex class 15 m low 20 f high 15 f low 10 m low in my original data set i have 1200 rows and a class distribution of low=0.3 and high=0.7 My question : how can i create a new data frame as the one shown above but with the 'high' class subsampled so that in the new data frame the class distribution is low=0.5 and high=0.5? I tried looking at the sample function and prob option but all examples i seen do not use an imbalanced class problem as the one shown above Thank you in advance Thank you in advance -- View this message in context: http://r.789695.n4.nabble.com/Subsampling-oversampling-from-a-data-frame-tp3965771p3971840.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] Syntax Help for xyplot()
Why not format the data like this: site sampledate SO4 TDS NA Mg Cond Cl Ca i.e. with a column for each parameter? It seems to me that you summary doesn't make any sense. Those quantiles are meaningless as they encompass all the parameters. Am I missing something? Rich Shepard wrote: Thanks to David's help I subset my large data set and produced a smaller one for a single stream and 7 factors of interest. The structure of this data frame is: str(burns.tds.anal) 'data.frame': 718 obs. of 4 variables: $ site: Factor w/ 143 levels BC-0.5,BC-1,..: 1 1 4 6 4 4 4 5 5 5 $ sampdate: Date, format: 1996-06-02 1996-06-02 ... $ param : Factor w/ 7 levels Ca,Cl,Cond,..: 6 7 3 6 3 3 3 3 3 3 $ quant : num 194 530 826 36 848 ... and a summary of it shows: summary(burns.tds.anal) sitesampdate param quant BC-3 :460 Min. :1992-03-27 Ca : 65 Min. : 1.00 BC-2 :107 1st Qu.:1994-09-21 Cl :148 1st Qu.: 14.03 BC-1 : 62 Median :1996-05-21 Cond: 94 Median : 64.40 BC-4 : 38 Mean :1998-11-19 Mg : 65 Mean : 189.47 BC-1.5 : 28 3rd Qu.:2002-10-31 Na : 34 3rd Qu.: 285.75 BC-0.5 : 12 Max. :2011-05-18 SO4 :155 Max. :2058.00 (Other): 11TDS :157 NA's : 8.00 a sample of the data in burns.tds.anal is: burns.tds.anal site sampdate param quant 82BC-0.5 1996-06-02 SO4 194.00 83BC-0.5 1996-06-02 TDS 530.00 6903BC-2 1994-07-25 Cond 826.00 6905BC-4 1996-08-23 SO4 36.00 6977BC-2 1994-10-19 Cond 848.00 6980BC-2 1995-03-16 Cond 1795.00 6983BC-2 1995-06-21 Cond 640.00 7833BC-3 1994-01-20 Cond 406.00 7838BC-3 1994-02-17 Cond 401.00 7847BC-3 1994-03-24 Cond 441.00 7854BC-3 1994-06-13 Cond 400.00 7866BC-3 1994-07-25 Cond 393.00 7871BC-3 1994-08-18 Cond 420.00 7877BC-3 1994-10-20 Cond 438.00 Perhaps because it's Monday I'm not successfully writing the xyplot() command to show the quant, for example, for TDS by site. What I need to do is plot the quant values for TDS vs. Cond, TDS vs. SO4, etc. Have I incorrectly subset the data? I know that I've missed something but cannot determine what it is. Rich __ R-help@ 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. -- View this message in context: http://r.789695.n4.nabble.com/Syntax-Help-for-xyplot-tp3934533p3934721.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] Syntax Help for xyplot()
The following might not be exactly the way to do this, but see the package
reshape and the line of code following your data:
df - structure(list(site = structure(c(1L, 1L, 2L, 4L, 2L, 2L, 2L,
3L, 3L, 3L, 3L, 3L, 3L, 3L), .Label = c(BC-0.5, BC-2, BC-3,
BC-4), class = factor), sampdate = structure(c(11L, 11L,
5L, 12L, 7L, 9L, 10L, 1L, 2L, 3L, 4L, 5L, 6L, 8L), .Label = c(1994-01-20,
1994-02-17, 1994-03-24, 1994-06-13, 1994-07-25, 1994-08-18,
1994-10-19, 1994-10-20, 1995-03-16, 1995-06-21, 1996-06-02,
1996-08-23), class = factor), param = structure(c(2L, 3L,
1L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = c(Cond,
SO4, TDS), class = factor), quant = c(194, 530, 826, 36,
848, 1795, 640, 406, 401, 441, 400, 393, 420, 438)), .Names = c(site,
sampdate, param, quant), class = data.frame, row.names = c(82,
83, 6903, 6905, 6977, 6980, 6983, 7833, 7838,
7847, 7854, 7866, 7871, 7877))
library(reshape)
cast(melt(df), site + sampdate ~ param )
Using site, sampdate, param as id variables
site sampdate Cond SO4 TDS
1 BC-0.5 1996-06-02 NA 194 530
2BC-2 1994-07-25 826 NA NA
3BC-2 1994-10-19 848 NA NA
4BC-2 1995-03-16 1795 NA NA
5BC-2 1995-06-21 640 NA NA
6BC-3 1994-01-20 406 NA NA
7BC-3 1994-02-17 401 NA NA
8BC-3 1994-03-24 441 NA NA
9BC-3 1994-06-13 400 NA NA
10 BC-3 1994-07-25 393 NA NA
11 BC-3 1994-08-18 420 NA NA
12 BC-3 1994-10-20 438 NA NA
13 BC-4 1996-08-23 NA 36 NA
Rich Shepard wrote:
On Mon, 24 Oct 2011, B77S wrote:
Why not format the data like this:
site sampledate SO4 TDS NA Mg Cond Cl Ca
Because I don't know how to reformat the base data frame (chemdata) to
achieve this.
It seems to me that you summary doesn't make any sense. Those quantiles
are meaningless as they encompass all the parameters. Am I missing
something?
One site/sampdate/param has one value associated with it.
82BC-0.5 1996-06-02 SO4 194.00
83BC-0.5 1996-06-02 TDS 530.00
6903BC-2 1994-07-25 Cond 826.00
6905BC-4 1996-08-23 SO4 36.00
6977BC-2 1994-10-19 Cond 848.00
6980BC-2 1995-03-16 Cond 1795.00
6983BC-2 1995-06-21 Cond 640.00
7833BC-3 1994-01-20 Cond 406.00
7838BC-3 1994-02-17 Cond 401.00
7847BC-3 1994-03-24 Cond 441.00
7854BC-3 1994-06-13 Cond 400.00
7866BC-3 1994-07-25 Cond 393.00
7871BC-3 1994-08-18 Cond 420.00
7877BC-3 1994-10-20 Cond 438.00
I acknowledge that if the chemicals ('param') were in a vector, their
associated
concentrations ('quant') in another vector following it, and both
associated
with a site and sampdate life would be much better.
Thanks,
Rich
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Re: [R] how to delete rows by a list of rownames
here is one way df1 - data.frame(c(1:20), c(21:40), c(31:50)) list1 - c(3, 6, 20) df2 - df1[-list1,] hanansela wrote: Hello I have a list of row names that needs to be deleted from a data frame. How do i do that? one of the columns in the data frame contains the row names as numbers. I can also select by this column (will it be easier?). Thank you -- View this message in context: http://r.789695.n4.nabble.com/how-to-delete-rows-by-a-list-of-rownames-tp3930206p3930273.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] Working With Variables Having Different Lengths
I know in my experience Cond (conductivity??) doesn't vary much within a stream except for during high flow events, and I would imagine the same is true for TDS. If these are all low flow values, you could possibly determine a mean/median value to use for the missing data points. Obviously this is going to be different if you are sampling storm events. If you have stage data and lots of data points, you may be able to model the parameters as a function of stage. HTH Rich Shepard wrote: Because of regulatory requirement changes over several decades and weather conditions preventing site access the variables in my data set have different lengths. I'd like guidance on how to perform linear regressions and other models with these variables. For example, there are 2206 rows for the parameter TDS but only 1191 rows for the parameter Cond. Such discrepancies are common in these data. Is there a reference I can read to learn how to analyze such data? Rich __ R-help@ 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. -- View this message in context: http://r.789695.n4.nabble.com/Working-With-Variables-Having-Different-Lengths-tp3926023p3926158.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] p value in R - beginners question
This is just scientific notation, so 8.15e-01 is the same as: 8.15*10^-1 [1] 0.815 niki wrote: Dear all, i have done some regression analyses but i do not understand the p value. These are the results t-value p value geno.1 -0.229 0.978 -0.234 8.15e-01 geno.50.647 1.146 0.565 5.73e-01 stress:geno.5-1.337 1.022 -1.307 1.92e-01 how can i see if my results are significant ? (apart from looking at the t-value). Can someone explain the p values? thanks niki -- View this message in context: http://r.789695.n4.nabble.com/p-value-in-R-beginners-question-tp3915873p3916308.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] matrix multiplication
Your question as answered by Timothy in your previous thread http://r.789695.n4.nabble.com/Re-Creating-the-mean-using-algebra-matrix-td3895689.html flokke wrote: Dear all, Sorry to bother you with such a stupid question, but I just cannot find the solution to my problem. I'd like to use matrix multiplication for meanA and factorial 3. I use the command meanA%*%factorial 3. But everything I get is: Error in factorial3 %*% A : non-conformable arguments I know that the number of the columns of the first vector has to be the same number of rows of the second vector to be able to use matrix multiplication, but that is the case here. I also tried it with two columns for factorial 3 and that didnt work either. Can someone help me out with this?' these are my matrices: meanA [,1] [,2] [1,] 3.67 4.67 factorial3 [,1] [1,]1 Thank you so much! Cheers, maria -- View this message in context: http://r.789695.n4.nabble.com/matrix-multiplication-tp3895833p3895860.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] For loop for subset - repeating same over and over?
The first suggestion would be to use dput() to allow people on here to access
your data.
see ?dput
I think you will want to post the output from:
dput(QInflAvgbyPlot) ## or whatever object contains your data
kelseyann wrote:
Hello, I am using the following script to run an anova for numerous
species in a table that I have:
SiteSpp -
c(ADHALP,ADLCON,ADLARC,BDALAT,BDPARC,BDLCON,BDLARC,AWCAQU,AWERUS,AWEANG,AWDPSI,BWCSTA,BWHPAU,BWETRI,BWERUS,BWDFIS,BWPARC,BWLCON,BWLARC,BWJBIG)
n.SiteSpp - length(SiteSpp)
for (i in (1:n.SiteSpp)) {
QInfl.aov - aov(AvgOfnumResponse ~ strTrea*strYear,
data=QInflAvgbyPlot, subset=(SiteSpp == SiteSpp[i]))
print(summary(QInfl.aov))
}
The resulting summary just continuously prints the same anova results (For
ADHALP) over and over again Any suggestions?
Thanks,
Kelsey
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Re: [R] Data import
I see what you mean. Sorry and thanks for pointing that out to me Ben. bbolker wrote: B77S bps0002 at auburn.edu writes: I have never used that function, but I know that with read.csv() you can do the following to select only the columns you want: chosen_vars - read.csv(Workbook1.csv, header=T)[c(variable1, variable3)] This is not actually selectively importing: it's importing the whole thing and *then* selecting the relevant columns. If the original poster is trying to avoid importing the whole data set because (for example) it's got a gigantic number of columns and will be very slow and/or tax their system, then this idiom won't help. Ben Bolker __ 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. -- View this message in context: http://r.789695.n4.nabble.com/Data-import-tp3842196p3848802.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] Data import
I have never used that function, but I know that with read.csv() you can do the following to select only the columns you want: chosen_vars - read.csv(Workbook1.csv, header=T)[c(variable1, variable3)] HTH sassorauk wrote: Is it possible to import only certain variables from a SPSS file. I know that read.spss in the foreign library will bring the data into R but can I choose to important only chosen variables from the SPSS dataset to R? Thanks for your help. R -- View this message in context: http://r.789695.n4.nabble.com/Data-import-tp3842196p3842600.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] Help with Integration Output
It is not simply the answer, it is a list str(integrate(dnorm, -1.96, 1.96)) List of 5 $ value : num 0.95 $ abs.error : num 1.05e-11 $ subdivisions: int 1 $ message : chr OK $ call: language integrate(f = dnorm, lower = -1.96, upper = 1.96) - attr(*, class)= chr integrate Is this what you want (?): 1 + (integrate(dnorm, -1.96, 1.96)[[1]]) ChrisQ wrote: Hi, I need to do a calculation in R which involves adding the estimate of a numerical integration with another number - however, if I add the number to the output of the intergrate function (which when run on its own provides me with a number with error), I get the message 'non-numeric argument to binary operator'. How would I be able to avoid this and use the estimate? eg: I have the code integrate(dnorm, -1.96, 1.96), which gives the output 0.9500042 with absolute error 1.0e-11, but if I put in integrate(dnorm, -1.96, 1.96)+1, I don't get 1.9500042 - how would I? Thanks very much, Chris. -- View this message in context: http://r.789695.n4.nabble.com/Help-with-Integration-Output-tp3839360p3839360.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. -- View this message in context: http://r.789695.n4.nabble.com/Help-with-Integration-Output-tp3839532p3839552.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] Identifying Package for Function
Where is dropvalue(s) mentioned? ?subset subset: logical expression indicating elements or rows to keep: missing values are taken as false. select: expression, indicating columns to select from a data frame. drop: passed on to ‘[’ indexing operator. ...: further arguments to be passed to or from other methods. Details: This is a generic function, with methods supplied for matrices, data frames and vectors (including lists). Packages and users can add further methods. For ordinary vectors, the result is simply ‘x[subset !is.na(subset)]’. For data frames, the ‘subset’ argument works on the rows. Note that ‘subset’ will be evaluated in the data frame, so columns can be referred to (by name) as variables in the expression (see the examples). The ‘select’ argument exists only for the methods for data frames and matrices. It works by first replacing column names in the selection expression with the corresponding column numbers in the data frame and then using the resulting integer vector to index the columns. This allows the use of the standard indexing conventions so that for example ranges of columns can be specified easily, or single columns can be dropped (see the examples). The ‘drop’ argument is passed on to the indexing method for matrices and data frames: note that the default for matrices is different from that for indexing. Value: An object similar to ‘x’ contain just the selected elements (for a vector), rows and columns (for a matrix or data frame), and so on. Author(s): Peter Dalgaard and Brian Ripley See Also: ‘[’, ‘transform’ Examples: subset(airquality, Temp 80, select = c(Ozone, Temp)) subset(airquality, Day == 1, select = -Temp) subset(airquality, select = Ozone:Wind) with(airquality, subset(Ozone, Temp 80)) ## sometimes requiring a logical 'subset' argument is a nuisance nm - rownames(state.x77) start_with_M - nm %in% grep(^M, nm, value=TRUE) subset(state.x77, start_with_M, Illiteracy:Murder) Rich Shepard wrote: While reading ?subset I'm referred to learn about dropvalues() as a following operation. Yet, when I issue ?dropvalue I see, No documentation for '?dropvalues' in specified packages and libraries:. How do I identify the library/package that contains a specific function such as, in this case, dropvalues()? Rich __ 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. -- View this message in context: http://r.789695.n4.nabble.com/Identifying-Package-for-Function-tp3835227p3835252.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 labels to x,y points
#This should work (again, without your data ??) colon-read.table(c:\\alon.txt,header=T,row.names=1) row.names(colon) -paste(g,c(1:nrow(colon)),sep=) with(colon[1:20,], plot(norm1, norm2, type='n',xlab='x norm1 sample',ylab='y norm2 sample',main='Norm1 vs Norm2 - 20 genes')) with(colon, text(norm1, norm2, label = row.names(colon[1:20,])) baumeist wrote: Hi, I am new to R. I have a matrix that I have assigned to the object “colon”. colon-read.table(c:\\alon.txt,header=T,row.names=1) attach(colon) names(colon) The dimenstions are 2000 62. Each of the 62 columns (titled norm1, norm2, norm3, etc) has 2000 different numbers (‘continuous’ values) within it. I have also assigned a name for each of the 2000 rows of the dataframe with a prefix (i.e. g1 …. g2000) using the code (not sure if I did this right): colon-paste(g,c(1:nrow(colon)),sep=) I have plotted the first 20 values from two of the columns(samples). x-c(norm1[1:20]) y-c(norm2[1:20]) plot(x,y,type='n',xlab='x norm1 sample',ylab='y norm2 sample',main='Norm1 vs Norm2 - 20 genes') points(x,y,pch=15,col='blue') Now I wish to assign labels to each point (above each point (i.e. pos=3) in the plot with “g1 to g20 corresponding to each row but I am having trouble with this step. I have tried: text(x,y, label = row.names(colon[1:20])) but nothing happens. Any suggestions? Thanks in advance MAB -- View this message in context: http://r.789695.n4.nabble.com/adding-labels-to-x-y-points-tp3828461p3830363.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 labels to x,y points
I don't have access to your alon.txt file (see ?dput for future posts), but... I'm pretty sure info you want isn't in row.names(colon[1:2]) it should just be text(x,y, label = colon[1:20]) ?? HTH baumeist wrote: Hi, I am new to R. I have a matrix that I have assigned to the object “colon”. colon-read.table(c:\\alon.txt,header=T,row.names=1) attach(colon) names(colon) The dimenstions are 2000 62. Each of the 62 columns (titled norm1, norm2, norm3, etc) has 2000 different numbers (‘continuous’ values) within it. I have also assigned a name for each of the 2000 rows of the dataframe with a prefix (i.e. g1 …. g2000) using the code (not sure if I did this right): colon-paste(g,c(1:nrow(colon)),sep=) I have plotted the first 20 values from two of the columns(samples). x-c(norm1[1:20]) y-c(norm2[1:20]) plot(x,y,type='n',xlab='x norm1 sample',ylab='y norm2 sample',main='Norm1 vs Norm2 - 20 genes') points(x,y,pch=15,col='blue') Now I wish to assign labels to each point (above each point (i.e. pos=3) in the plot with “g1 to g20 corresponding to each row but I am having trouble with this step. I have tried: text(x,y, label = row.names(colon[1:20])) but nothing happens. Any suggestions? Thanks in advance MAB -- View this message in context: http://r.789695.n4.nabble.com/adding-labels-to-x-y-points-tp3828461p3828545.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 labels to x,y points
Actually, you appear to have re-assigned your object “colon” (from c:\\alon.txt) with a character vector of intended row.names. so use row.names(colon) -paste(g,c(1:nrow(colon)),sep=) B77S wrote: I don't have access to your alon.txt file (see ?dput for future posts), but... I'm pretty sure the info you want is not in row.names(colon[1:2]) should just be using: text(x,y, label = colon[1:20,]) ?? HTH baumeist wrote: Hi, I am new to R. I have a matrix that I have assigned to the object “colon”. colon-read.table(c:\\alon.txt,header=T,row.names=1) attach(colon) names(colon) The dimenstions are 2000 62. Each of the 62 columns (titled norm1, norm2, norm3, etc) has 2000 different numbers (‘continuous’ values) within it. I have also assigned a name for each of the 2000 rows of the dataframe with a prefix (i.e. g1 …. g2000) using the code (not sure if I did this right): colon-paste(g,c(1:nrow(colon)),sep=) I have plotted the first 20 values from two of the columns(samples). x-c(norm1[1:20]) y-c(norm2[1:20]) plot(x,y,type='n',xlab='x norm1 sample',ylab='y norm2 sample',main='Norm1 vs Norm2 - 20 genes') points(x,y,pch=15,col='blue') Now I wish to assign labels to each point (above each point (i.e. pos=3) in the plot with “g1 to g20 corresponding to each row but I am having trouble with this step. I have tried: text(x,y, label = row.names(colon[1:20])) but nothing happens. Any suggestions? Thanks in advance MAB -- View this message in context: http://r.789695.n4.nabble.com/adding-labels-to-x-y-points-tp3828461p3828665.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] graph bugs using R on MAC
Have you been shown how to save a graph as a JPEG or PNG? try this: png(myGraph.png) plot(your_data) dev.off() A png will appear in your working directory, which can be imported into the Word document. You can do the same with a JPEG see ?jpeg or ?png and ?dev.off HTH bonnieyuan wrote: This is has been bugging me for a long time. Nobody around me seems to have this problem. I hope someone on the forum could help me. When I generate a R graph and want to bring the image into Word. I cannot copy and paste it like many of my classmates can do. The Select in Edit menu have all options grayed out. So what I ended up doing is to save it as pdf file and bring it into Word. But if I use Identify in the Plot statement, there will be values labels on the graph when you click on the observation. For some reason, these labels don't show up in the pdf file. So the last resort I have is prinscreen, which always have irrelevant things showing. I've tried updating R/OS and all softwares on my MAC, but nothing seems to do anything. I have OX 10.6.8. Does this problem sound familiar to anyone? Thanks in advance! Bonnie Yuan __ 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. -- View this message in context: http://r.789695.n4.nabble.com/graph-bugs-using-R-on-MAC-tp3822460p3822480.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] AIC
This isn't a question about R; more appropriate for stackexchange. Here is one string that might interest you: http://stats.stackexchange.com/questions/4997/can-aic-compare-across-different-types-of-model Tania Sav wrote: Hello, I'm using AIC() to choose a better model. I have 3 options: 1) y=ax+b 2) y=ax^2 +bx + c 3) y=a(1-exp(-x/b)) + c Is it ok to use AIC() for determine wich is the better equation to use for fit to my data? Thanks Tania Tania Sav wrote: Hello, I'm using AIC() to choose a better model. I have 3 options: 1) y=ax+b 2) y=ax^2 +bx + c 3) y=a(1-exp(-x/b)) + c Is it ok to use AIC() for determine wich is the better equation to use for fit to my data? Thanks Tania -- View this message in context: http://r.789695.n4.nabble.com/AIC-tp3820732p3822647.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] onet.permutation()
** I am not a statistician, but... The default arguments are: onet.permutation(x, nsim=2000, plotit=TRUE) you are making nsim=4 (this parameter is not the size of each sample, it is the number of samples taken) You are building the distribution for a statistical test through random sampling, so you want to have MANY MANY random samples, not just 4. song_gpqg wrote: I saw the manual of this function but not sure what to do. I have a array contain 7 numbers and want to choose 4 to do permutation test. But using this function with parameters as onet.permutation(scores,4), it returns 0. Instead, with no parameter, onet.permutation(), it returns something but every time it's different. Please tell me how am I supposed to do it. Thanks a lot -- View this message in context: http://r.789695.n4.nabble.com/onet-permutation-tp3820372p3820671.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] Mystified - comparing chron times
I'm sure Sarah's solution works (and she knows more about R than myself), but
I ran into a similar problem and used:
as.character(start.time)==as.character(expected_start.time)
good luck regardless.
-BS
Sarah Goslee wrote:
Sounds like a case for FAQ 7.31, or, yet another machine precision issue.
Try all.equal() instead of ==
Sarah
On Fri, Sep 16, 2011 at 7:36 AM, mebstyne lt;mebst...@me.comgt; wrote:
I have two local variables: startTime and expectedStartTime. Both are
chron
related objects.
When I look at the class for the objects I can see they are of class
times.
When I print them to the console, they both read: 09:30:00
When I print them as.numeric(), they both read: 0.3958333
When I try and compare them: (as.numeric(startTime) ==
as.numeric(expectedStartTime)) it returns FALSE.
I'm mystified. I would expect them to be true.
Perhaps a key to the riddle is how the two objects were created.
startTime was created by reading a text field from a socket, converting
it
into a chron object using chron(x, %m/$d/%Y %H:%M:%S), then finally I
created a time out of the chron by doing a quick butchering of the
integer
portion of the numeric: (times(as.numeric(x) - as.integer(x))
expectedStartTime was created by the command times('09:30:00')
Any suggestions? Tips? Alternative approaches? I've pulled too many
hairs
triaging this.
All hands welcomed on this little challenge.
Big picture goal of what I'm doing: I have a list of chron objects with
both
dates and times portion filled out and I'm trying to determine if the
time
is a specific time (specific to the minute).
Thanks!
-Michael
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] installation of BiodiversityR package
My guess is that anyone willing to help will want more information. What version of R do you have(?), for example. BiodiversityR Depends on R version ≥ 2.13.1, and vegan ≥ 1.17-12 SG wrote: I have installed R on windows 7 machine. In the install packages I can't find BiodiversityR package. I was wondering if I can get some help with the installation process. Thanks. -- View this message in context: http://r.789695.n4.nabble.com/installation-of-BiodiversityR-package-tp3816807p3816823.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] R functions
You'll never figure it out if you don't play around with your data. Assuming you have been able to import the data, a good place to start is to look at what tools you have available.Check out this: http://cran.r-project.org/doc/contrib/Short-refcard.pdf check out things like ?which ?max ?unique You have 2 choices, hope that someone hands you a solution, or fiddle around until you become proficient. Good luck. sujitha wrote: Hi group, I am trying to right a code to do the following This is how the test file looks like: Chr start end sample1 sample2 chr2 9896633 9896683 0 0 chr2 9896639 9896690 0 0 chr2 14314039 14314098 0 -0.35 chr2 14404467 14404502 0 -0.35 chr2 14421718 14421777 -0.43 -0.35 chr2 16031710 16031769 -0.43 -0.35 chr2 16036178 16036237 -0.43 -0.35 chr2 16048665 16048724 -0.43 -0.35 chr2 37491676 37491735 0 0 chr2 37702947 37703009 0 0 Now I want to summarize the values like Sample Chr Start End Values Probes 1 chr2 9896633 14404502 0 4 1 chr2 14421718 16048724 -0.43 4 1 chr2 37491676 37703001 0 2 2 chr2 9896633 9896690 0 2 2 chr2 14314039 16048724 -0.35 6 2 chr2 37491676 37703009 0 2 Here the start for the first line would be the least value until values are similiar (4) then the end would be highest value. The values is the unique value among the common values. Can I get some ideas or suggestions to perform this because I am new to hard core program in R? waiting for your suggestions, Thanks, suji -- View this message in context: http://r.789695.n4.nabble.com/R-functions-tp3816748p3816844.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] R functions
If your data is named 'test_file' then use dput(test_file) You can copy and paste the results here so people can more easily try and help you. see ?dput sujitha wrote: Hi group, I am trying to right a code to do the following This is how the test file looks like: Chr start end sample1 sample2 chr2 9896633 9896683 0 0 chr2 9896639 9896690 0 0 chr2 14314039 14314098 0 -0.35 chr2 14404467 14404502 0 -0.35 chr2 14421718 14421777 -0.43 -0.35 chr2 16031710 16031769 -0.43 -0.35 chr2 16036178 16036237 -0.43 -0.35 chr2 16048665 16048724 -0.43 -0.35 chr2 37491676 37491735 0 0 chr2 37702947 37703009 0 0 Now I want to summarize the values like Sample Chr Start End Values Probes 1 chr2 9896633 14404502 0 4 1 chr2 14421718 16048724 -0.43 4 1 chr2 37491676 37703001 0 2 2 chr2 9896633 9896690 0 2 2 chr2 14314039 16048724 -0.35 6 2 chr2 37491676 37703009 0 2 Here the start for the first line would be the least value until values are similiar (4) then the end would be highest value. The values is the unique value among the common values. Can I get some ideas or suggestions to perform this because I am new to hard core program in R? waiting for your suggestions, Thanks, suji -- View this message in context: http://r.789695.n4.nabble.com/R-functions-tp3816748p3817055.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] R functions
Suji, # Here is your data (test). test - structure(list(Chr = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = chr2, class = factor), start = c(9896633L, 9896639L, 14314039L, 14404467L, 14421718L, 16031710L, 16036178L, 16048665L, 37491676L, 37702947L), end = c(9896683L, 9896690L, 14314098L, 14404502L, 14421777L, 16031769L, 16036237L, 16048724L, 37491735L, 37703009L), sample1 = c(0, 0, 0, 0, -0.43, -0.43, -0.43, -0.43, 0, 0), sample2 = c(0, 0, -0.35, -0.35, -0.35, -0.35, -0.35, -0.35, 0, 0)), .Names = c(Chr, start, end, sample1, sample2), class = data.frame, row.names = c(NA, -10L)) # Here is where you will likely want to start (but there are many ways to skin a cat). test2 - data.frame(c(rle(test[,4])[[2]], rle(test[,5])[[2]]), c(rle(test[,4])[[1]], rle(test[,5])[[1]])) names(test2) - c(Values, Probes) test2 Values Probes 1 0.00 4 2 -0.43 4 3 0.00 2 4 0.00 2 5 -0.35 6 6 0.00 2 # Obviously this is not exactly what you wanted, only the last 2 columns. Obviously, the tricky part remains... but I hope this helps. ## B77S wrote: If your data is named 'test_file' then use dput(test_file) You can copy and paste the results here so people can more easily try and help you. see ?dput sujitha wrote: Hi group, I am trying to right a code to do the following This is how the test file looks like: Chr start end sample1 sample2 chr2 9896633 9896683 0 0 chr2 9896639 9896690 0 0 chr2 14314039 14314098 0 -0.35 chr2 14404467 14404502 0 -0.35 chr2 14421718 14421777 -0.43 -0.35 chr2 16031710 16031769 -0.43 -0.35 chr2 16036178 16036237 -0.43 -0.35 chr2 16048665 16048724 -0.43 -0.35 chr2 37491676 37491735 0 0 chr2 37702947 37703009 0 0 Now I want to summarize the values like Sample Chr Start End Values Probes 1 chr2 9896633 14404502 0 4 1 chr2 14421718 16048724 -0.43 4 1 chr2 37491676 37703001 0 2 2 chr2 9896633 9896690 0 2 2 chr2 14314039 16048724 -0.35 6 2 chr2 37491676 37703009 0 2 Here the start for the first line would be the least value until values are similiar (4) then the end would be highest value. The values is the unique value among the common values. Can I get some ideas or suggestions to perform this because I am new to hard core program in R? waiting for your suggestions, Thanks, suji -- View this message in context: http://r.789695.n4.nabble.com/R-functions-tp3816748p3817431.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] reshaping data
I have the following data (see RawData using dput below) How do I get it in the following 3 column format (CO2 measurements are the elements of the original data frame). I'm sure the package reshape is where I should look, but I haven't figured out how. Thanks ahead of time Month Year CO2 J 1958 F 1958 M 1958315.71 A 1958317.45 M.1 1958317.5 J.1 1958 J.2 1958315.86 A.1 1958314.93 S 1958313.19 O 1958 N 1958313.34 D 1958314.67 J 1959315.58 F 1959316.47 # here is the data RawData - structure(list(Year = c(1958, 1959, 1960, 1961, 1962, 1963, 1964, 1965, 1966, 1967, 1968, 1969, 1970, 1971, 1972, 1973, 1974, 1975, 1976, 1977, 1978, 1979, 1980, 1981, 1982, 1983, 1984, 1985, 1986, 1987, 1988, 1989, 1990, 1991, 1992, 1993, 1994, 1995, 1996, 1997, 1998, 1999, 2000, 2001, 2002, 2003, 2004), J = c(NA, 315.58, 316.43, 316.89, 317.94, 318.74, 319.57, 319.44, 320.62, 322.33, 322.57, 324, 325.06, 326.17, 326.77, 328.54, 329.35, 330.4, 331.74, 332.92, 334.97, 336.23, 338.01, 339.23, 340.75, 341.37, 343.7, 344.97, 346.29, 348.02, 350.43, 352.76, 353.66, 354.72, 355.98, 356.7, 358.36, 359.96, 362.05, 363.18, 365.32, 368.15, 369.14, 370.28, 372.43, 374.68, 376.79), F = c(NA, 316.47, 316.97, 317.7, 318.56, 319.08, NA, 320.44, 321.59, 322.5, 323.15, 324.42, 325.98, 326.68, 327.63, 329.56, 330.71, 331.41, 332.56, 333.42, 335.39, 336.76, 338.36, 340.47, 341.61, 342.52, 344.51, 346, 346.96, 348.47, 351.72, 353.07, 354.7, 355.75, 356.72, 357.16, 358.91, 361, 363.25, 364, 366.15, 368.87, 369.46, 371.5, 373.09, 375.63, 377.37), M = c(315.71, 316.65, 317.58, 318.54, 319.69, 319.86, NA, 320.89, 322.39, 323.04, 323.89, 325.64, 326.93, 327.18, 327.75, 330.3, 331.48, 332.04, 333.5, 334.7, 336.64, 337.96, 340.08, 341.38, 342.7, 343.1, 345.28, 347.43, 347.86, 349.42, 352.22, 353.68, 355.39, 357.16, 357.81, 358.38, 359.97, 361.64, 364.03, 364.57, 367.31, 369.59, 370.52, 372.12, 373.52, 376.11, 378.41 ), A = c(317.45, 317.71, 319.03, 319.48, 320.58, 321.39, NA, 322.13, 323.7, 324.42, 325.02, 326.66, 328.13, 327.78, 329.72, 331.5, 332.65, 333.31, 334.58, 336.07, 337.76, 338.89, 340.77, 342.51, 343.56, 344.94, 347.08, 348.35, 349.55, 350.99, 353.59, 355.42, 356.2, 358.6, 359.15, 359.46, 361.26, 363.45, 364.72, 366.35, 368.61, 371.14, 371.66, 372.87, 374.86, 377.65, 380.52 ), M.1 = c(317.5, 318.29, 320.03, 320.58, 321.01, 322.24, 322.23, 322.16, 324.07, 325, 325.57, 327.38, 328.07, 328.92, 330.07, 332.48, 333.09, 333.96, 334.87, 336.74, 338.01, 339.47, 341.46, 342.91, 344.13, 345.75, 347.43, 348.93, 350.21, 351.84, 354.22, 355.67, 357.16, 359.34, 359.66, 360.28, 361.68, 363.79, 365.41, 366.79, 369.29, 371, 371.82, 374.02, 375.55, 378.35, 380.63), J.1 = c(NA, 318.16, 319.59, 319.78, 320.61, 321.47, 321.89, 321.87, 323.75, 324.09, 325.36, 326.7, 327.66, 328.57, 329.09, 332.07, 332.25, 333.59, 334.34, 336.27, 337.89, 339.29, 341.17, 342.25, 343.35, 345.32, 346.79, 348.25, 349.54, 351.25, 353.79, 355.13, 356.22, 358.24, 359.25, 359.6, 360.95, 363.26, 364.97, 365.62, 368.87, 370.35, 371.7, 373.3, 375.4, 378.13, 379.57 ), J.2 = c(315.86, 316.55, 318.18, 318.58, 319.61, 319.74, 320.44, 321.21, 322.4, 322.55, 324.14, 325.89, 326.35, 327.37, 328.05, 330.87, 331.18, 331.91, 333.05, 334.93, 336.54, 337.73, 339.56, 340.49, 342.06, 343.99, 345.4, 346.56, 347.94, 349.52, 352.39, 353.9, 354.82, 356.17, 357.03, 357.57, 359.55, 361.9, 363.65, 364.47, 367.64, 369.27, 370.12, 371.62, 374.02, 376.62, 377.79), A.1 = c(314.93, 314.8, 315.91, 316.79, 317.4, 317.77, 318.7, 318.87, 320.37, 320.92, 322.11, 323.67, 324.69, 325.43, 326.32, 329.31, 329.4, 330.06, 330.94, 332.75, 334.68, 336.09, 337.6, 338.43, 339.82, 342.39, 343.28, 344.69, 345.91, 348.1, 350.44, 351.67, 352.91, 354.03, 355, 355.52, 357.49, 359.46, 361.49, 362.51, 365.77, 366.94, 368.12, 369.55, 371.49, 374.5, 375.86), S = c(313.19, 313.84, 314.16, 314.99, 316.26, 316.21, 316.7, 317.81, 318.64, 319.26, 320.33, 322.38, 323.1, 323.36, 324.84, 327.51, 327.44, 328.56, 329.3, 331.58, 332.76, 333.91, 335.88, 336.69, 337.97, 339.86, 341.07, 343.09, 344.86, 346.44, 348.72, 349.8, 350.96, 352.16, 353.01, 353.7, 355.84, 358.06, 359.46, 360.19, 363.9, 364.63, 366.62, 367.96, 370.71, 372.99, 374.06), O = c(NA, 313.34, 313.83, 315.31, 315.42, 315.99, 316.87, 317.3, 318.1, 319.39, 320.25, 321.78, 323.07, 323.56, 325.2, 327.18, 327.37, 328.34, 328.94, 331.16, 332.54, 333.86, 336.01, 336.85, 337.86, 339.99, 341.35, 342.8, 344.17, 346.36, 348.88, 349.99, 351.18, 352.21, 353.31, 353.98, 355.99, 357.75, 359.6, 360.77, 364.23, 365.12, 366.73, 368.09, 370.24, 373, 374.24), N = c(313.34, 314.81, 315, 316.1, 316.69, 317.07, 317.68, 318.87, 319.79, 320.72, 321.32, 322.85, 324.01, 324.8, 326.5,
Re: [R] reshaping data
The terminology (melt, cast, recast) just isn't intuitive to me; but I understand how to use melt now. Thanks! Justin Haynes wrote: look at the melt function in reshape, specifically ?melt.data.frame require(reshape) Raw.melt-melt(RawData,id.vars='Year',variable_name='Month') there is an additional feature in the melt function for handling na values. names(Raw.melt)[3]-'CO2' head(Raw.melt) Year MonthCO2 1 1958 J NA 2 1959 J 315.58 3 1960 J 316.43 4 1961 J 316.89 5 1962 J 317.94 6 1963 J 318.74 you can order your data.frame if you'd like Raw.melt-Raw.melt[order(Raw.melt$Year,Raw.melt$Month),] head(Raw.melt) Year MonthCO2 1 1958 J NA 48 1958 F NA 95 1958 M 315.71 142 1958 A 317.45 189 1958 M.1 317.50 236 1958 J.1 NA On Wed, Sep 7, 2011 at 7:35 AM, B77S lt;bps0...@auburn.edugt; wrote: I have the following data (see RawData using dput below) How do I get it in the following 3 column format (CO2 measurements are the elements of the original data frame). I'm sure the package reshape is where I should look, but I haven't figured out how. Thanks ahead of time Month Year CO2 J 1958 F 1958 M 1958315.71 A 1958317.45 M.1 1958317.5 J.1 1958 J.2 1958315.86 A.1 1958314.93 S 1958313.19 O 1958 N 1958313.34 D 1958314.67 J 1959315.58 F 1959316.47 # here is the data RawData - structure(list(Year = c(1958, 1959, 1960, 1961, 1962, 1963, 1964, 1965, 1966, 1967, 1968, 1969, 1970, 1971, 1972, 1973, 1974, 1975, 1976, 1977, 1978, 1979, 1980, 1981, 1982, 1983, 1984, 1985, 1986, 1987, 1988, 1989, 1990, 1991, 1992, 1993, 1994, 1995, 1996, 1997, 1998, 1999, 2000, 2001, 2002, 2003, 2004), J = c(NA, 315.58, 316.43, 316.89, 317.94, 318.74, 319.57, 319.44, 320.62, 322.33, 322.57, 324, 325.06, 326.17, 326.77, 328.54, 329.35, 330.4, 331.74, 332.92, 334.97, 336.23, 338.01, 339.23, 340.75, 341.37, 343.7, 344.97, 346.29, 348.02, 350.43, 352.76, 353.66, 354.72, 355.98, 356.7, 358.36, 359.96, 362.05, 363.18, 365.32, 368.15, 369.14, 370.28, 372.43, 374.68, 376.79), F = c(NA, 316.47, 316.97, 317.7, 318.56, 319.08, NA, 320.44, 321.59, 322.5, 323.15, 324.42, 325.98, 326.68, 327.63, 329.56, 330.71, 331.41, 332.56, 333.42, 335.39, 336.76, 338.36, 340.47, 341.61, 342.52, 344.51, 346, 346.96, 348.47, 351.72, 353.07, 354.7, 355.75, 356.72, 357.16, 358.91, 361, 363.25, 364, 366.15, 368.87, 369.46, 371.5, 373.09, 375.63, 377.37), M = c(315.71, 316.65, 317.58, 318.54, 319.69, 319.86, NA, 320.89, 322.39, 323.04, 323.89, 325.64, 326.93, 327.18, 327.75, 330.3, 331.48, 332.04, 333.5, 334.7, 336.64, 337.96, 340.08, 341.38, 342.7, 343.1, 345.28, 347.43, 347.86, 349.42, 352.22, 353.68, 355.39, 357.16, 357.81, 358.38, 359.97, 361.64, 364.03, 364.57, 367.31, 369.59, 370.52, 372.12, 373.52, 376.11, 378.41 ), A = c(317.45, 317.71, 319.03, 319.48, 320.58, 321.39, NA, 322.13, 323.7, 324.42, 325.02, 326.66, 328.13, 327.78, 329.72, 331.5, 332.65, 333.31, 334.58, 336.07, 337.76, 338.89, 340.77, 342.51, 343.56, 344.94, 347.08, 348.35, 349.55, 350.99, 353.59, 355.42, 356.2, 358.6, 359.15, 359.46, 361.26, 363.45, 364.72, 366.35, 368.61, 371.14, 371.66, 372.87, 374.86, 377.65, 380.52 ), M.1 = c(317.5, 318.29, 320.03, 320.58, 321.01, 322.24, 322.23, 322.16, 324.07, 325, 325.57, 327.38, 328.07, 328.92, 330.07, 332.48, 333.09, 333.96, 334.87, 336.74, 338.01, 339.47, 341.46, 342.91, 344.13, 345.75, 347.43, 348.93, 350.21, 351.84, 354.22, 355.67, 357.16, 359.34, 359.66, 360.28, 361.68, 363.79, 365.41, 366.79, 369.29, 371, 371.82, 374.02, 375.55, 378.35, 380.63), J.1 = c(NA, 318.16, 319.59, 319.78, 320.61, 321.47, 321.89, 321.87, 323.75, 324.09, 325.36, 326.7, 327.66, 328.57, 329.09, 332.07, 332.25, 333.59, 334.34, 336.27, 337.89, 339.29, 341.17, 342.25, 343.35, 345.32, 346.79, 348.25, 349.54, 351.25, 353.79, 355.13, 356.22, 358.24, 359.25, 359.6, 360.95, 363.26, 364.97, 365.62, 368.87, 370.35, 371.7, 373.3, 375.4, 378.13, 379.57 ), J.2 = c(315.86, 316.55, 318.18, 318.58, 319.61, 319.74, 320.44, 321.21, 322.4, 322.55, 324.14, 325.89, 326.35, 327.37, 328.05, 330.87, 331.18, 331.91, 333.05, 334.93, 336.54, 337.73, 339.56, 340.49, 342.06, 343.99, 345.4, 346.56, 347.94, 349.52, 352.39, 353.9, 354.82, 356.17, 357.03, 357.57, 359.55, 361.9, 363.65, 364.47, 367.64, 369.27, 370.12, 371.62, 374.02, 376.62, 377.79), A.1 = c(314.93, 314.8, 315.91, 316.79, 317.4, 317.77, 318.7, 318.87, 320.37, 320.92, 322.11, 323.67, 324.69, 325.43, 326.32, 329.31, 329.4, 330.06, 330.94, 332.75, 334.68, 336.09, 337.6, 338.43, 339.82, 342.39, 343.28, 344.69, 345.91, 348.1, 350.44, 351.67, 352.91, 354.03, 355, 355.52, 357.49, 359.46, 361.49, 362.51, 365.77, 366.94, 368.12, 369.55, 371.49, 374.5, 375.86), S = c(313.19, 313.84, 314.16
Re: [R] Alternatives to integrate?
package caTools see ?trapz . wrote: Hi all, is there any alternative to the function integrate? Any comments are welcome. Thanks in advance. __ 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. -- View this message in context: http://r.789695.n4.nabble.com/Alternatives-to-integrate-tp3783624p3783645.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] R Help finding Mean
see ?mean
Then avoid other peoples code.
cenae27 wrote:
bob-read.csv('shi.csv', header=T)
newmean-matrix(0, test, dim(bob)[2]-6);a-0; for (i in
c(4,8:(dim(bob)[2])))
{a-a+1;newmean[,a]-tapply(bob[,i], bob$Exam, mean)}
colnames(newmean)-colnames(bob)[c(4,8:(dim(bob)[2]))]
Could anyone please help me what does the above code does ... I want to
find mean ... but would like to know what exactly is the above code doing.
Thanks for your help.
Cenae
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Re: [R] Howto convert Linear Regression data to text
If I understand you correctly, see ?paste and the following to extract the values you require: summary(res)[[4]][1] summary(res)[[4]][2] summary(res)[[8]] HTH ashz wrote: Dear all, How can I covert lm data to text in the form of y=ax+b, r2 and how do I calculate R-squared(r2)? Thanks. Code: x=18:29 y=c(7.1,7,7.7,8.2,8.8,9.7,9.9,7.1,7.2,8.8,8.7,8.5) res=lm(y~x) -- View this message in context: http://r.789695.n4.nabble.com/Howto-convert-Linear-Regression-data-to-text-tp3766230p3767009.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] Importing data from MS EXCEL (.xls) to R XXXX
I agree with Ken.. if you can, save it as a CSV file. But if you have a
bunch of these, then it isn't very efficient. I use read.xlsx() from the
package xlsx.
I notice that you are using the full path.. have you tried changing
directories?... I find it is best to compartmentalize my work and (with a
few exceptions) work within a folder.
good luck.
Dan Abner wrote:
Hello everyone,
What is the simplest, most RELIABLE way to import data from MS EXCEL
(.xls)
format to R? In the past I have used the read.xls() function from the
xlsReadWrite package, however, I have been wrestling with it all afternoon
long with no success. I continue to receive the following error message:
{widge-read.xls(F:\\Classes\\Z1.Data\\stat.3010\\WidgeOne.xls,
+ colNames=TRUE,sheet=1)}
Error in .Call(ReadXls, file, colNames, sheet, type, from, rowNames, :
Incorrect number of arguments (11), expecting 10 for 'ReadXls'
Any insight/suggestions/assistance is appreciated.
Thank you,
Dan
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Re: [R] Ordinary Least Products regression in R
If this is the same as geometric mean regression (aka: line of organic correlation-- Kruskal 1953), I ended up writing my own function (although one may exist and I didn't see it). Bill Hyman wrote: Dear all, Does any one know if any R package or function can do Ordinary Least Products regression? Many thanks! Bill __ 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. -- View this message in context: http://r.789695.n4.nabble.com/Ordinary-Least-Products-regression-in-R-tp3697635p3698343.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] Odd behaviour of as.POSIXct
day doesn't exist?
That would be the 1st problem.
Johannes Egner wrote:
Dear all,
how come the first loop in the below fails, but the second performs as
expected?
days - as.Date( c(2000-01-01, 2000-01-02) )
for(day in days)
{
as.POSIXct(day)
}
for( n in 1:length(days) )
{
show(as.POSIXct(days[n]))
}
Many thanks, Jo
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[R] installation of package 'mapproj' had non-zero exit status
## Hello.. I have asked a similar question, but this is not fixed as before. ## I am running the following using Ubuntu OS: R version 2.13.1 (2011-07-08) Copyright (C) 2011 The R Foundation for Statistical Computing ISBN 3-900051-07-0 Platform: x86_64-pc-linux-gnu (64-bit) ## when I do this: install.packages(mapproj, dependencies=T) ## I get this: Installing package(s) into â/home/brad/R/x86_64-pc-linux-gnu-library/2.13â (as âlibâ is unspecified) also installing the dependency âmapsâ trying URL 'http://cran.case.edu/src/contrib/maps_2.1-6.tar.gz' Content type 'application/x-gzip' length 1371854 bytes (1.3 Mb) opened URL == downloaded 1.3 Mb trying URL 'http://cran.case.edu/src/contrib/mapproj_1.1-8.3.tar.gz' Content type 'application/x-gzip' length 23955 bytes (23 Kb) opened URL == downloaded 23 Kb * installing *source* package âmapsâ ... ** libs ** arch - gcc -std=gnu99 -O3 -pipe -gGmake.c -o Gmake Gmake.c: In function âget_lhâ: Gmake.c:111: warning: cast from pointer to integer of different size Gmake.c:113: warning: cast from pointer to integer of different size Gmake.c: In function âmainâ: Gmake.c:211: warning: cast from pointer to integer of different size Gmake.c:214: warning: cast from pointer to integer of different size Gmake.c:217: warning: cast from pointer to integer of different size Gmake.c:219: warning: cast from pointer to integer of different size Gmake.c:221: warning: cast from pointer to integer of different size Gmake.c:224: warning: cast from pointer to integer of different size Gmake.c:227: warning: cast from pointer to integer of different size gcc -std=gnu99 -O3 -pipe -gLmake.c -o Lmake Lmake.c: In function âmainâ: Lmake.c:223: warning: cast from pointer to integer of different size Lmake.c:228: warning: cast from pointer to integer of different size Lmake.c:230: warning: cast from pointer to integer of different size Lmake.c:232: warning: cast from pointer to integer of different size Lmake.c:235: warning: cast from pointer to integer of different size Converting world to world2 f convert.awk world.line world2.line /bin/bash: f: command not found make: [world2.line] Error 127 (ignored) make county.L state.L usa.L nz.L world.L world2.L italy.L france.L make[1]: Entering directory `/tmp/RtmpssTER5/R.INSTALL21eb6525/maps/src' ./Lmake 0 s b county.line county.linestats ../inst/mapdata/county.L ./Lmake 0 s b state.line state.linestats ../inst/mapdata/state.L ./Lmake 0 s b usa.line usa.linestats ../inst/mapdata/usa.L ./Lmake 0 s b nz.line nz.linestats ../inst/mapdata/nz.L ./Lmake 0 s b world.line world.linestats ../inst/mapdata/world.L ./Lmake 0 s b world2.line world2.linestats ../inst/mapdata/world2.L Cannot read left and right at line 1 make[1]: *** [world2.L] Error 1 make[1]: Leaving directory `/tmp/RtmpssTER5/R.INSTALL21eb6525/maps/src' make: *** [ldata] Error 2 ERROR: compilation failed for package âmapsâ * removing â/home/brad/R/x86_64-pc-linux-gnu-library/2.13/mapsâ ERROR: dependency âmapsâ is not available for package âmapprojâ * removing â/home/brad/R/x86_64-pc-linux-gnu-library/2.13/mapprojâ The downloaded packages are in â/tmp/RtmpwXL9El/downloaded_packagesâ Warning messages: 1: In install.packages(mapproj, dependencies = T) : installation of package 'maps' had non-zero exit status 2: In install.packages(mapproj, dependencies = T) : installation of package 'mapproj' had non-zero exit status ## Any idea as to why? this also happens when I try to install the 'maps' package -- View this message in context: http://r.789695.n4.nabble.com/installation-of-package-mapproj-had-non-zero-exit-status-tp3662940p3662940.html Sent from the R help mailing list archive at Nabble.com. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Confusing piece of R code
Have you looked at the manual for any of these?
?get
?sprintf
?assign
#and so on...
?
Bazman76 wrote:
m0-epxression((4*theta1*theta2-theta3^2)/(2*x*theta3^2)-0.5*theta1*x)
params-all.vars(m0) this reads all the
params from m0 so theta1,2 and 3 correct?
params-params[-which(params==x)] checks which params are
multiplied by x?
np-length(params)
for(i in 1:6){
esp-get(sprintf(m%d,i-1))what does get do? sprinf
formats strings? so what is it doinf here?
assign(sprintf(m%d,i),D(esp,x)) what doeas assign so what
in sprintf doing and what does D do?
}
really really confused?
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[R] identifying a 'run' in a vector
Hi, How can I discern which elements in x (see below) are in 'order', but more specifically.. only the 1st 'ordered run'? I would like for it to return elements 1:8... there may be ordered values after 1:8, but those are not of interest. x - c(1, 2, 3, 4, 5, 6, 7, 8, 20, 21, 22, 45) Thanks for any suggestions. -- View this message in context: http://r.789695.n4.nabble.com/identifying-a-run-in-a-vector-tp3650295p3650295.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] identifying a 'run' in a vector
well.. the following works, but if you have another idea I am still interested. 1:(which(diff(x)!=1)[1]) B77S wrote: Hi, How can I discern which elements in x (see below) are in 'order', but more specifically.. only the 1st 'ordered run'? I would like for it to return elements 1:8... there may be ordered values after 1:8, but those are not of interest. x - c(1, 2, 3, 4, 5, 6, 7, 8, 20, 21, 22, 45) Thanks for any suggestions. -- View this message in context: http://r.789695.n4.nabble.com/identifying-a-run-in-a-vector-tp3650295p3650318.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] identifying a 'run' in a vector
Yes Gabor, you definition (a sequence of numbers which each increase by 1 over the prior number) is what I meant. Sorry it that was not clear and I thank you and Joshua for your time and your explanation. This should work fine. Gabor Grothendieck wrote: On Wed, Jul 6, 2011 at 7:32 PM, B77S lt;bps0...@auburn.edugt; wrote: Hi, How can I discern which elements in x (see below) are in 'order', but more specifically.. only the 1st 'ordered run'? I would like for it to return elements 1:8... there may be ordered values after 1:8, but those are not of interest. x - c(1, 2, 3, 4, 5, 6, 7, 8, 20, 21, 22, 45) Since the definition of an ordered run is not given we assume that it is a sequence of numbers which each increase by 1 over the prior number. If that is not it then you will need to clarify the problem definition. First calculate a logical vector which is TRUE at each position which starts a new run. Note that the first position in x always starts a new run even if that run is a singleton so it can be set to TRUE. The remaining elements can be computed using diff as shown. The resulting logical vector is the argument to cumsum below. Taking the cumulative sum of this logical vector gives a vector the same length as x but with each element of the 1st run replaced with 1, each element of the 2nd run replaced with 2 and so on. Finally, since we only want the 1st run we pick out those positions of x where the cumsum equals 1. x[cumsum(c(TRUE, diff(x) != 1)) == 1] [1] 1 2 3 4 5 6 7 8 -- Statistics Software Consulting GKX Group, GKX Associates Inc. tel: 1-877-GKX-GROUP email: ggrothendieck at gmail.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. -- View this message in context: http://r.789695.n4.nabble.com/identifying-a-run-in-a-vector-tp3650295p3650509.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] Linking 2 columns in 2 databases and applying a function
1st of all Dr Utz, thanks for your recent pubs on regional differences between the piedmont and coastal plain streams. and to add to Daniels post (giving your binary yes/no): df-merge(x,y,all.x=T,all.y=F) df[exceed] - ifelse(df$Qdf$Threshold_Q , 1, 0) ## now look at df df Daniel Malter wrote: For example, you can merge the two data frames and do a direct comparison: df-merge(x,y,all.x=T,all.y=F) df df$Qdf$Threshold_Q HTH, Daniel Ryan Utz-2 wrote: Hi all, I have two datasets, one that represents a long-term time series and one that represents summary data for the time series. It looks something like this: x-data.frame(Year=c(2001,2001,2001,2001,2001,2001,2002,2002,2002,2002,2002,2002), Month=c(1,1,1,2,2,2),Q=c(5,5,5,6,6,6,3,3,3,4,4,5)) y-data.frame(Year=c(2001,2001,2002,2002),Month=c(1,2,1,2),Threshold_Q=c(5,5,4,4)) What I'd like to do is link the Year and Month fields in both dataframes then determine if Q exceeds Q_Threshold (by noting it with something like 1 or 0 in a new field in the dataframe x). If I were doing this in the more-familiar-to-me Matlab, I'd just write a pair of nested for-loops. But as we know, this won't fly in R. I've tried reading the help pages and seeking for solutions on the net, with no luck (I'm relatively new to R and the help pages are still a bit opaque to me). It seems like the functions apply or lapply are key, but I can't make sense of their syntax. Any advice/help?!? Many thanks, Ryan -- Ryan Utz, Ph.D. Aquatic Ecologist/STREON Scientist National Ecological Observatory Network Home/Cell: (724) 272-7769 Work: (720) 746-4844 ext. 2488 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- View this message in context: http://r.789695.n4.nabble.com/Linking-2-columns-in-2-databases-and-applying-a-function-tp3617710p3617861.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] automatically generate the output name of my for loops
?paste
something like...
paste (group, i, sep=_)
jiliguala wrote:
Hello R users,
I am new to R and am having difficulty with the output name of my for
loops.
here is the problem:
for (i in c(1:100))
{
the name of the groups - which(k1$cluster==i)
}
how can it automatically generate the name for 100 cluster(just like
group_1, group_2...)? what should i put in the bold letter place?
really thank you for helping me
Daniel
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Re: [R] automatically generate the output name of my for loops
?paste
something like...
paste (group, i, sep=_)
jiliguala wrote:
Hello R users,
I am new to R and am having difficulty with the output name of my for
loops.
here is the problem:
for (i in c(1:100))
{
the name of the groups - which(k1$cluster==i)
}
how can it automatically generate the name for 100 cluster(just like
group_1, group_2...)? what should i put in the bold letter place?
really thank you for helping me
Daniel
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[R] scatter plot: multiple Y variables and error bars
#Hi all, #Using the example data that follows, can someone please show me how to get a scatterplot of points with #error bars in the Y direction. something like this works for one Y: xYplot(Cbind(y1, l1, u1) ~x1, data=y) #but this: xYplot(Cbind(y1, l1, u1) + Cbind(y2, l2, u2)~x1, data=y) # doesn't give me what I would have expected, which is both sets of points to have their respective error # bars. Any examples would be greatly appreciated, and I am not partial to xYplot, so please share #anything you like. y1 - c(1, 1.2, 0.9, 1, 1.2) u1 - c(1.3, 1.4, 1.3, 1.2, 1.4) l1 - c(0.8, 0.9, 0.85, 0.8, 0.9) x1 - c(1:5) y2 - c(1.2, 1.4, 1.2, 1.4, 1.5) u2 - c(1.5, 1.8, 1.6, 1.6, 1.7) l2 - c(1.1, 1.3, 1.0, 1.2, 1.4) y - data.frame(y1,u1,l1,x1) ## thanks ahead of time! -- View this message in context: http://r.789695.n4.nabble.com/scatter-plot-multiple-Y-variables-and-error-bars-tp3531563p3531563.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] Clearing Console; of weeks of codes!
If you don't want your history, why not just do this. q() Save workspace image? [y/n/c]: n ?? Am I missing something? 1Rnwb wrote: Thanks, I thought that removing the list would take care of it. the question is I do not see a .Rhistory file in my current working directory, so where it is stored. it is not visible in C:\Program files\R either. Serarching the C;\ and D:\ drives shows some old .Rhistory files but not the recent ones. Thanks -- View this message in context: http://r.789695.n4.nabble.com/Clearing-Console-of-weeks-of-codes-tp3447506p3449876.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] extract element from list by rownames
This probably is not ideal, but this works on a list of mine.. ## so you can see the structure of my list str(srMT) List of 4 $ mode : chr discrete $ ks.stat : chr mean $ observed :List of 4 ..$ filter: num [1:13, 1:4] 0.213 0.207 0.144 0.311 0.24 ... .. ..- attr(*, dimnames)=List of 2 .. .. ..$ ID_2: chr [1:13] BC_SP1 BU1_SP1 BU2_SP1 MK_SP1 ... .. .. ..$ : chr [1:4] vel1 vel2 vel3 vel4 ..$ chi2 : num [1:3] 6.93e+02 6.93e+02 1.93e-12 ..$ df: num [1:3] 36 39 3 ..$ p : num [1:3] 0 0 1 $ bootstrap:List of 6 ..$ filter : num [1:13, 1:4, 1:1000] 0.245 0.254 0.247 0.244 0.25 ... # LIST just keeps going and going and going # as an example (srMT[[3]][1]) $filter ID_2 vel1 vel2 vel3 vel4 BC_SP1 0.2127431 0.1923574 0.2839013 0.3109983 BU1_SP1 0.2072090 0.3094771 0.2219164 0.2613975 BU2_SP1 0.1437481 0.3270975 0.2401788 0.2889757 MK_SP1 0.3114887 0.2503257 0.2177326 0.2204530 MO_SP1 0.2395539 0.2079802 0.3208321 0.2316338 MU1_SP1 0.2336137 0.1923791 0.2386081 0.3353991 MU2_SP1 0.3607257 0.2466847 0.1501613 0.2424284 MU3_SP1 0.3583597 0.2499569 0.1652739 0.2264095 RB_SP1 0.2912842 0.2742916 0.1768310 0.2575933 SB1_SP1 0.2108506 0.2075117 0.3399161 0.2417215 SB2_SP1 0.2910056 0.2306834 0.2507012 0.2276098 SB4_SP1 0.2958040 0.2358855 0.2433465 0.2249641 SC_SP1 0.2509970 0.2129797 0.3071544 0.2288689 ## some rownames b - c(MK_SP1, SB1_SP1, SB4_SP1) ## this works to pull them out based on row names data.frame((srMT[[3]][1]))[b,] filter.vel1 filter.vel2 filter.vel3 filter.vel4 MK_SP10.3114887 0.2503257 0.2177326 0.2204530 SB1_SP1 0.2108506 0.2075117 0.3399161 0.2417215 SB4_SP1 0.2958040 0.2358855 0.2433465 0.2249641 I hope this helps, good luck! Alfredo Alessandrini wrote: Hi, I've a list of list. I want to extract an element by the rownames. I can extract it by: data[[1]][[1]][[4]][1] But I want to exctract it by a command like this: data[[1]][[B0]][[smac]][[cont]][1] It's possible? Thanks, Alfredo str(data) List of 1 $ :List of 4 ..$ :List of 4 .. ..$ : num [1, 1:3] 0.4 0.458 0.5 .. ..$ : num [1:41, 1] 0.4 0.403 0.405 0.407 0.41 ... .. ..$ : num [1:41, 1] 0.000128 0.000328 0.000528 0.000728 0.000992 ... .. ..$ :List of 1 .. .. ..$ : num [1:49, 1] 0 0 -0.00626 0.99832 0 ... .. .. ..- attr(*, dim)= int [1:3] 1 1 1 .. .. ..- attr(*, dimnames)=List of 3 .. .. .. ..$ : chr cont .. .. .. ..$ : NULL .. .. .. ..$ : NULL .. ..- attr(*, dim)= int [1:3] 4 1 1 .. ..- attr(*, dimnames)=List of 3 .. .. ..$ : chr [1:4] begin.mode.end lambda sensi smac .. .. ..$ : NULL .. .. ..$ : NULL ..$ :List of 4 .. ..$ : num [1, 1:3] 0.58 0.67 0.782 .. ..$ : num [1:82, 1] 0.58 0.583 0.585 0.588 0.59 ... .. ..$ : num [1:82, 1] 0.0017 0.00168 0.00165 0.00162 0.0016 ... .. ..$ :List of 1 .. .. ..$ : num [1:49, 1] -0.00597 0.71045 -0.05844 0.99187 -0.00807 ... .. .. ..- attr(*, dim)= int [1:3] 1 1 1 .. .. ..- attr(*, dimnames)=List of 3 .. .. .. ..$ : chr cont .. .. .. ..$ : NULL .. .. .. ..$ : NULL .. ..- attr(*, dim)= int [1:3] 4 1 1 .. ..- attr(*, dimnames)=List of 3 .. .. ..$ : chr [1:4] begin.mode.end lambda sensi smac .. .. ..$ : NULL .. .. ..$ : NULL ..$ :List of 4 .. ..$ : num [1, 1:3] 0.7 0.82 1.03 .. ..$ : num [1:133, 1] 0.7 0.703 0.705 0.708 0.71 ... .. ..$ : num [1:133, 1] 0.0007 0.000775 0.00085 0.000925 0.001 ... .. ..$ :List of 1 .. .. ..$ : num [1:49, 1] -0.030584 0.559164 -0.000269 0.995624 -0.006616 ... .. .. ..- attr(*, dim)= int [1:3] 1 1 1 .. .. ..- attr(*, dimnames)=List of 3 .. .. .. ..$ : chr cont .. .. .. ..$ : NULL .. .. .. ..$ : NULL .. ..- attr(*, dim)= int [1:3] 4 1 1 .. ..- attr(*, dimnames)=List of 3 .. .. ..$ : chr [1:4] begin.mode.end lambda sensi smac .. .. ..$ : NULL .. .. ..$ : NULL ..$ :List of 4 .. ..$ : num [1, 1:3] 1.45 1.64 1.8 .. ..$ : num [1:141, 1] 1.45 1.45 1.46 1.46 1.46 ... .. ..$ : num [1:141, 1] 0.001 0.001 0.001 0.001 0.001 ... .. ..$ :List of 1 .. .. ..$ : num [1:49, 1] -0.00547 0.65798 0 0 0 ... .. .. ..- attr(*, dim)= int [1:3] 1 1 1 .. .. ..- attr(*, dimnames)=List of 3 .. .. .. ..$ : chr cont .. .. .. ..$ : NULL .. .. .. ..$ : NULL .. ..- attr(*, dim)= int [1:3] 4 1 1 .. ..- attr(*, dimnames)=List of 3 .. .. ..$ : chr [1:4] begin.mode.end lambda sensi smac .. .. ..$ : NULL .. .. ..$ : NULL ..- attr(*, dim)= int [1:3] 4 1 1 ..- attr(*, dimnames)=List of 3 .. ..$ : chr [1:4] B0 B2 B3 MIR .. ..$ : NULL .. ..$ : NULL __ 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,
Re: [R] extract element from list by rownames
I'm thinking this isn't what you want.. but also: data.frame((srMT[[3]][1]))[b,][2] filter.vel2 MK_SP10.2503257 SB1_SP1 0.2075117 SB4_SP1 0.2358855 B77S wrote: This probably is not ideal, but this works on a list of mine.. ## so you can see the structure of my list str(srMT) List of 4 $ mode : chr discrete $ ks.stat : chr mean $ observed :List of 4 ..$ filter: num [1:13, 1:4] 0.213 0.207 0.144 0.311 0.24 ... .. ..- attr(*, dimnames)=List of 2 .. .. ..$ ID_2: chr [1:13] BC_SP1 BU1_SP1 BU2_SP1 MK_SP1 ... .. .. ..$ : chr [1:4] vel1 vel2 vel3 vel4 ..$ chi2 : num [1:3] 6.93e+02 6.93e+02 1.93e-12 ..$ df: num [1:3] 36 39 3 ..$ p : num [1:3] 0 0 1 $ bootstrap:List of 6 ..$ filter : num [1:13, 1:4, 1:1000] 0.245 0.254 0.247 0.244 0.25 ... # LIST just keeps going and going and going # as an example (srMT[[3]][1]) $filter ID_2 vel1 vel2 vel3 vel4 BC_SP1 0.2127431 0.1923574 0.2839013 0.3109983 BU1_SP1 0.2072090 0.3094771 0.2219164 0.2613975 BU2_SP1 0.1437481 0.3270975 0.2401788 0.2889757 MK_SP1 0.3114887 0.2503257 0.2177326 0.2204530 MO_SP1 0.2395539 0.2079802 0.3208321 0.2316338 MU1_SP1 0.2336137 0.1923791 0.2386081 0.3353991 MU2_SP1 0.3607257 0.2466847 0.1501613 0.2424284 MU3_SP1 0.3583597 0.2499569 0.1652739 0.2264095 RB_SP1 0.2912842 0.2742916 0.1768310 0.2575933 SB1_SP1 0.2108506 0.2075117 0.3399161 0.2417215 SB2_SP1 0.2910056 0.2306834 0.2507012 0.2276098 SB4_SP1 0.2958040 0.2358855 0.2433465 0.2249641 SC_SP1 0.2509970 0.2129797 0.3071544 0.2288689 ## some rownames b - c(MK_SP1, SB1_SP1, SB4_SP1) ## this works to pull them out based on row names data.frame((srMT[[3]][1]))[b,] filter.vel1 filter.vel2 filter.vel3 filter.vel4 MK_SP10.3114887 0.2503257 0.2177326 0.2204530 SB1_SP1 0.2108506 0.2075117 0.3399161 0.2417215 SB4_SP1 0.2958040 0.2358855 0.2433465 0.2249641 I hope this helps, good luck! Alfredo Alessandrini wrote: Hi, I've a list of list. I want to extract an element by the rownames. I can extract it by: data[[1]][[1]][[4]][1] But I want to exctract it by a command like this: data[[1]][[B0]][[smac]][[cont]][1] It's possible? Thanks, Alfredo str(data) List of 1 $ :List of 4 ..$ :List of 4 .. ..$ : num [1, 1:3] 0.4 0.458 0.5 .. ..$ : num [1:41, 1] 0.4 0.403 0.405 0.407 0.41 ... .. ..$ : num [1:41, 1] 0.000128 0.000328 0.000528 0.000728 0.000992 ... .. ..$ :List of 1 .. .. ..$ : num [1:49, 1] 0 0 -0.00626 0.99832 0 ... .. .. ..- attr(*, dim)= int [1:3] 1 1 1 .. .. ..- attr(*, dimnames)=List of 3 .. .. .. ..$ : chr cont .. .. .. ..$ : NULL .. .. .. ..$ : NULL .. ..- attr(*, dim)= int [1:3] 4 1 1 .. ..- attr(*, dimnames)=List of 3 .. .. ..$ : chr [1:4] begin.mode.end lambda sensi smac .. .. ..$ : NULL .. .. ..$ : NULL ..$ :List of 4 .. ..$ : num [1, 1:3] 0.58 0.67 0.782 .. ..$ : num [1:82, 1] 0.58 0.583 0.585 0.588 0.59 ... .. ..$ : num [1:82, 1] 0.0017 0.00168 0.00165 0.00162 0.0016 ... .. ..$ :List of 1 .. .. ..$ : num [1:49, 1] -0.00597 0.71045 -0.05844 0.99187 -0.00807 ... .. .. ..- attr(*, dim)= int [1:3] 1 1 1 .. .. ..- attr(*, dimnames)=List of 3 .. .. .. ..$ : chr cont .. .. .. ..$ : NULL .. .. .. ..$ : NULL .. ..- attr(*, dim)= int [1:3] 4 1 1 .. ..- attr(*, dimnames)=List of 3 .. .. ..$ : chr [1:4] begin.mode.end lambda sensi smac .. .. ..$ : NULL .. .. ..$ : NULL ..$ :List of 4 .. ..$ : num [1, 1:3] 0.7 0.82 1.03 .. ..$ : num [1:133, 1] 0.7 0.703 0.705 0.708 0.71 ... .. ..$ : num [1:133, 1] 0.0007 0.000775 0.00085 0.000925 0.001 ... .. ..$ :List of 1 .. .. ..$ : num [1:49, 1] -0.030584 0.559164 -0.000269 0.995624 -0.006616 ... .. .. ..- attr(*, dim)= int [1:3] 1 1 1 .. .. ..- attr(*, dimnames)=List of 3 .. .. .. ..$ : chr cont .. .. .. ..$ : NULL .. .. .. ..$ : NULL .. ..- attr(*, dim)= int [1:3] 4 1 1 .. ..- attr(*, dimnames)=List of 3 .. .. ..$ : chr [1:4] begin.mode.end lambda sensi smac .. .. ..$ : NULL .. .. ..$ : NULL ..$ :List of 4 .. ..$ : num [1, 1:3] 1.45 1.64 1.8 .. ..$ : num [1:141, 1] 1.45 1.45 1.46 1.46 1.46 ... .. ..$ : num [1:141, 1] 0.001 0.001 0.001 0.001 0.001 ... .. ..$ :List of 1 .. .. ..$ : num [1:49, 1] -0.00547 0.65798 0 0 0 ... .. .. ..- attr(*, dim)= int [1:3] 1 1 1 .. .. ..- attr(*, dimnames)=List of 3 .. .. .. ..$ : chr cont .. .. .. ..$ : NULL .. .. .. ..$ : NULL .. ..- attr(*, dim)= int [1:3] 4 1 1 .. ..- attr(*, dimnames)=List of 3 .. .. ..$ : chr [1:4] begin.mode.end lambda sensi smac .. .. ..$ : NULL .. .. ..$ : NULL ..- attr(*, dim)= int [1:3] 4 1 1 ..- attr(*, dimnames)=List of 3 .. ..$ : chr [1:4] B0 B2 B3 MIR .. ..$ : NULL .. ..$ : NULL
Re: [R] Removing objects and clearing memory
#replace v with whatever rm(list=(ls()[ls()!=v])) -- View this message in context: http://r.789695.n4.nabble.com/Removing-objects-and-clearing-memory-tp3445763p3445865.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] How to create a 3d Surface plot from a CSV file?
step # 1 ?read.csv -- View this message in context: http://r.789695.n4.nabble.com/How-to-create-a-3d-Surface-plot-from-a-CSV-file-tp3437463p3437478.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] How to call data elements
assuming this is from a list mydata[[1]][1] and mydata[[1]][7] ?? Wonjae Lee wrote: Hi, I have a stupid and simple question. Please forgive me. In an example below, please tell me how to call 1947 in mydata. Thank you in advance. Wonjae mydata [[1]] [1] 194783 234.289 235.6 159 107.608 1947 [8] 60.323 mydata[[1],1] error:unexpected ',' in mydata[[1], mydata[1,[1]] error:unexpected '[' in mydata[1,[ -- View this message in context: http://r.789695.n4.nabble.com/How-to-call-data-elements-tp3430859p3430881.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] problem with all/all.equal
ifelse(length(unique(x))==1, All Equal, Not All Equal) -- View this message in context: http://r.789695.n4.nabble.com/problem-with-all-all-equal-tp3431956p3432167.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] R, Ubuntu, package installation with non-zero exit status
FYI: The problem I had was that my modified .Rprofile text file was being read and had lines to load the libraries of packages that were not installed yet on this Ubuntu machine; I guess I copied the .Rprofile from my laptop. If you have similar problems, simply do this in command line: mv .Rprofile Rprofile install.packages to your hearts desire and then mv Rprofile .Rprofile -- View this message in context: http://r.789695.n4.nabble.com/R-Ubuntu-package-installation-with-non-zero-exit-status-tp3305989p3432176.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] unexpected sort order with merge
That is odd, I noticed some weird sorting with merge() a while back too and always am careful with it now. Fortunately, sort=FALSE seems to work the way one would think most of the time. Although, the following results seem weird too! (adding by=date makes it not sort oddly, regardless of sort=TRUE or FALSE) merge(d1, d2, by=date, sort=F) date icpn.x foo icpn.y bar 1 Jan 2000500 1500 10 2 Feb 2000500 2500 20 3 Mar 2000500 3500 30 4 Apr 2000500 4500 40 5 May 2000500 5500 50 6 Jun 2000500 6500 60 merge(d1, d2, by=date, sort=T) date icpn.x foo icpn.y bar 1 Jan 2000500 1500 10 2 Feb 2000500 2500 20 3 Mar 2000500 3500 30 4 Apr 2000500 4500 40 5 May 2000500 5500 50 6 Jun 2000500 6500 60 merge(d1, d2, by=date) date icpn.x foo icpn.y bar 1 Jan 2000500 1500 10 2 Feb 2000500 2500 20 3 Mar 2000500 3500 30 4 Apr 2000500 4500 40 5 May 2000500 5500 50 6 Jun 2000500 6500 60 merge(d1, d2) date icpn foo bar 1 Apr 2000 500 4 40 2 Feb 2000 500 2 20 3 Jan 2000 500 1 10 4 Jun 2000 500 6 60 5 Mar 2000 500 3 30 6 May 2000 500 5 50 -- View this message in context: http://r.789695.n4.nabble.com/unexpected-sort-order-with-merge-tp3431338p3432250.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] R, Ubuntu, package installation with non-zero exit status
All: I have been looking through the string of posts regarding this same issue, but I haven't been able to fix this problem. I am running Ubuntu 10.4 64bit, R version 2.10.1 I cannot install certain packages (e.g. vegetarian) and each time it says basically the same thing (regardless of the package): ... leaving stuff out .. Error in library(vegetarian) : there is no package called 'vegetarian' Execution halted The downloaded packages are in ‘/tmp/RtmphcJ8mn/downloaded_packages’ Warning message: In install.packages(vegetarian) : installation of package 'vegetarian' had non-zero exit status ### I have tried the following: 1) ensured I have 'build-essential' Ubuntu package installed (I do) 2) attempted install.packages as root (sudo R) to no avail -- View this message in context: http://r.789695.n4.nabble.com/R-Ubuntu-package-installation-with-non-zero-exit-status-tp3305989p3305989.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] R, Ubuntu, package installation with non-zero exit status
also, I have 'r-base-dev' installed as well -- View this message in context: http://r.789695.n4.nabble.com/R-Ubuntu-package-installation-with-non-zero-exit-status-tp3305989p3306005.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] Need help merging two dataframes
## i didn't try this, but I would think it would work newAB -data.frame(AB$id, AB$age, AB$sex, AB$area) colnames(newAB)-c(id,age, sex, area) uni.newAB - unique(newAB) t3-merge(t2, uni.newAB, by=id, all=FALSE) -- View this message in context: http://r.789695.n4.nabble.com/Need-help-merging-two-dataframes-tp3297313p3301627.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] Subsampling out of site*abundance matrix
It is a lot prettier than mine too.
Thanks Jari.
rrarefy
function (x, sample)
{
if (length(sample) 1 length(sample) != nrow(x))
stop(length of 'sample' and number of rows of 'x' do not match)
sample - rep(sample, length = nrow(x))
colnames(x) - colnames(x, do.NULL = FALSE)
nm - colnames(x)
for (i in 1:nrow(x)) {
row - sample(rep(nm, times = x[i, ]), sample[i])
row - table(row)
ind - names(row)
x[i, ] - 0
x[i, ind] - row
}
x
}
--
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Re: [R] Subsampling out of site*abundance matrix
Also, I really appreciate you explaining why you used factor. I'm still not
quite sure what set.seed does (i read ?set.seed) or why you chose 123... but
it and the function below work, so that is all that matters. :)
randSub - function(L1, s.size)
{
set.seed(123)
samptbl - apply(L1, 1, function(x) sample(colnames(L1), s.size, prob=x,
replace=TRUE) )
sampdf - as.data.frame(samptbl)
sampdf1 -vector(list)
for(i in 1:nrow(L1))
{
sampdf1[[i]] - factor(sampdf[[i]], levels= colnames(L1))
}
out - t(sapply(sampdf1, table))
}
--
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Re: [R] Subsampling out of site*abundance matrix
So, after thinking about this a bit, I realized that the previous solution
wasn't exactly what I needed. I really needed replacement=F and to be able
to choose any sample size (n.sample) less than or equal to the site (row)
with the lowest total abundance.
Anyway, I think this works. Forgive me if I have misunderstood something
regarding the previous solutions output. I do not pretend to be
intelligent.Cheers!
### start function ###
RAND_L - function(L.matrix, n.sample){
mainout - vector(list)
for(i in 1:nrow(L.matrix)){
## decomposes species (1:ncol(L.matrix)) into a list of counts per each
out- vector(list)
for(j in 1:length(L.matrix[i,])){
out[[j]] - rep(names(L.matrix[i,])[j], L.matrix[i,j])
}
## puts previous loop products (counts) in a row
out2 - vector()
for(k in 1:length(out)){
out2 - append(out2, as.character(unlist(out[k])))
}
out3- sample(out2, n.sample, replace=F)
mainout[[i]] - out3
mainout[[i]] - factor(mainout[[i]], levels= colnames(L.matrix))
}
finalout - t(sapply(mainout, table))
rownames(finalout)-rownames(L.matrix)
return(finalout)
}
### end function ##
RAND_L(abund2, 100)
spA spB spC spD spa spF spG
site1 11 12 18 8 0 24 27
site2 24 24 0 0 27 25 0
site3 0 0 6 38 0 0 56
site4 27 20 0 0 16 37 0
--
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[R] Subsampling out of site*abundance matrix
Hello, How can I randomly sample individuals within a sites from a site (row) X species abundance (column) data frame or matrix? As an example, the matrix abund2 made below. # (sorry, Im a newbie and this is the only way I know to get an example on here) abund1 -c(150, 300, 0, 360, 150, 300, 0, 240, 150, 0, 60, 0, 150, 0, 540, 0, 0, 300, 0, 240, 300, 300, 0, 360, 300, 0, 600, 0) abund2 - matrix(data=abund1, nrow=4, ncol=7) colnames(abund2) - c(spA, spB, spC, spD, spa, spF, spG) rownames(abund2)-c(site1, site2, site3, site4) # abund2 spA spB spC spD spa spF spG site1 150 150 150 150 0 300 300 site2 300 300 0 0 300 300 0 site3 0 0 60 540 0 0 600 site4 360 240 0 0 240 360 0 How can I make a random subsample of 100 individuals from the abundances given for each site? This is probably really easy. Thanks. Bubba -- View this message in context: http://r.789695.n4.nabble.com/Subsampling-out-of-site-abundance-matrix-tp3263148p3263148.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] Subsampling out of site*abundance matrix
I figured there would be an even more straightforward way, but that works David, thanks. There has to be a way to get the output I want/need (see below). I tried to bind or merge the elements of apply(samptbl, 2, table) but with no success. I could probably make a for loop with a merge statement, it would work.. but I'm guessing unnecessary and just plain ugly. ## what I want/need spA spB spC spD spa spF spG site1 813 613 320 28 site2 31 25 0 0 25 19 0 site300 9 510 040 site4 27 19 0 022 32 0 If you know, I'd appreciate it.. thanks again for the help. David Winsemius wrote: On Feb 6, 2011, at 3:25 PM, B77S wrote: Hello, How can I randomly sample individuals within a sites from a site (row) X species abundance (column) data frame or matrix? As an example, the matrix abund2 made below. # (sorry, Im a newbie and this is the only way I know to get an example on here) abund1 -c(150, 300, 0, 360, 150, 300, 0, 240, 150, 0, 60, 0, 150, 0, 540, 0, 0, 300, 0, 240, 300, 300, 0, 360, 300, 0, 600, 0) abund2 - matrix(data=abund1, nrow=4, ncol=7) colnames(abund2) - c(spA, spB, spC, spD, spa, spF, spG) rownames(abund2)-c(site1, site2, site3, site4) Perfect. Best submission of an example by a newbie in weeks. # abund2 spA spB spC spD spa spF spG site1 150 150 150 150 0 300 300 site2 300 300 0 0 300 300 0 site3 0 0 60 540 0 0 600 site4 360 240 0 0 240 360 0 How can I make a random subsample of 100 individuals from the abundances given for each site? samptbl - apply(abund2, 1, function(x) sample(colnames(abund2), 100, prob=x, replace=TRUE) ) samptbl site1 site2 site3 site4 [1,] spG spa spD spF [2,] spF spF spG spB [3,] spF spB spC spA [4,] spD spa spG spA [5,] spF spa spD spa [6,] spA spB spD spF [7,] spA spF spD spA [8,] spG spF spG spa [9,] spF spF spG spa [10,] spG spB spD spA Snipped apply() always transposes the results when called with row margins. The t() function would fix this if it needed to be arranged with rows by site. You could check by further apply-(cation) of table to the columns: apply(samptbl, 2, table) $site1 spA spB spC spD spF spG 8 13 6 13 32 28 $site2 spa spA spB spF 25 31 25 19 $site3 spC spD spG 9 51 40 $site4 spa spA spB spF 22 27 19 32 This is probably really easy. Thanks. Bubba -- View this message in context: http://r.789695.n4.nabble.com/Subsampling-out-of-site-abundance-matrix-tp3263148p3263148.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. David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- View this message in context: http://r.789695.n4.nabble.com/Subsampling-out-of-site-abundance-matrix-tp3263148p3263488.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] Subsampling out of site*abundance matrix
hehe... very true sir; I apologize, that was very straightforward. Thank you for your time. -- View this message in context: http://r.789695.n4.nabble.com/Subsampling-out-of-site-abundance-matrix-tp3263148p3263598.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.
