similar to: plot rownames

Displaying 20 results from an estimated 9000 matches similar to: "plot rownames"

2009 Apr 16
3
"reverse truncate" to extract only decimal values
hello there, Is there a way of truncating in the opposite direction so as to retain only the values to the right of the decimal?? i.e. rather than: > trunc(39.5) [1] 39 i would get something like: > revtrunc(39.5) [1] 0.5 I've been searching to no avail but I imagine there is a very simple solution! Tyler -- View this message in context:
2008 May 05
3
merge numerous columns of unequal length
I have numerous objects, each containing continuous data representing the same variable, movement rate, yet each having a different number of rows. e.g. d1<-as.matrix(rnorm(5)) d2<-as.matrix(rnorm(3)) d3<-as.matrix(rnorm(6)) How can I merge these three columns side-by-side in order to create a table regardless of the difference in length? I wish to analyze the output in a spreadsheet
2008 Jun 02
4
NOT-SO-SIMPLE function!
I am trying to set up a function which processes my data according to the following rules: 1. if (x[i]==0) NA 2. if (x[i]>0) log(x[i]/(number of consecutive zeros immediately preceding it +1)) The data this will apply to include a variety of whole numbers not limited to 1 & 0, a number of which may appear consecutively and not separated by zeros. Below is an example with a detailed
2010 Feb 19
3
plot circular histogram
In conducting studies of animal orientation and displacement, I need to produce circular histograms of angles (bearings in radians 0-2pi) where the centre of the circle indicates very few observations for a given bin of angles and outwardly concentric circles indicate greater frequencies of observations for a given bin of angles. I'd like not to have to write the function myself but I
2008 Sep 16
1
analyze subsample of dataframe
Hi there, I'm dealing with a pretty big dataset (~22,000 entries) with numerous entries for every day over a period of several years. I have a column "judy" (for Julian Day) with 0 beginning on Jan. 1st of every new year (I want to compare tendencies between years). However, in order to control for a leap year (2004), I simply need to subtract 1 from every judy value for the year
2009 Mar 28
1
Find inflection points using smooth.spline
Is there any way to identify or infer the inflection points in a smooth spline object? I am doing a comparison of various methods of time-series analysis (polynomial regression, spline smoothing, recursive partitioning) and I am specifically interested in obtaining the julian dates associated with the inflection points inferred by the various models. Tyler e.g.
2008 Jun 19
2
Advanced Filtering problem
http://www.nabble.com/file/p18018170/subdata.csv subdata.csv I've attached 100 rows of a data frame I am working with. I have one factor, id, with 27 levels. There are two columns of reference data, x and y (UTM coordinates), one column "date" in POSIXct format, and one column "diff" in times format (chron package). What I am trying to do is as follows: For each day
2008 May 27
4
help with simple function
I have a matrix of frequency counts from 0-160. x<-as.matrix(c(0,1,0,0,1,0,0,0,1,0,0,0,0,1)) I would like to apply a function creating a new column (x[,2])containing values equal to: a) log(x[m,1]) if x[m,1] > 0; and b) for all x[m,1]= 0, log(next x[m,1] > 0 / count of preceding zero values +1) for example, x[1,2] should equal log(x[2,1]/2) = log(1/2) = -0.6931472 whereas x[3,2] should
2009 Nov 26
2
Export kde object as shapefile
I am trying to estimate home range size using the plug-in method with kernel density estimation in the kernel smoothing (ks) package. Unless there is another way I am not familiar with, in order to calculate spatial area under the space I need to convert my kde () object into a spatial object somehow in order to calculate its spatial area. Could someone demonstrate how this might be done? --
2005 Mar 29
1
improved pairs.formula?
Dear all, I would like to suggest changing the pairs.formula command such that a command like pairs(GNP ~ . - Year - GNP.deflator, longley) would behave in a similar fashion as lm(GNP ~ . - Year - GNP.deflator, longley) i.e., make a pairwise scatterplot of GNP and all other variables in the (longley) dataframe except for Year and GNP.deflator. The above command, with the
2004 Oct 11
3
logistic regression
Hello, I have a problem concerning logistic regressions. When I add a quadratic term to my linear model, I cannot draw the line through my scatterplot anymore, which is no problem without the quadratic term. In this example my binary response variable is "incidence", the explanatory variable is "sun": > model0<-glm(incidence~1,binomial) >
2008 Oct 06
2
Graphics window BUG
I thought this problem would be resolved when I switched to R version 2.7.0 (for Windows), but no - anytime I plot something that produces more than one page of graphics, the graphics window starts by showing the first page, until such time as I hit enter to show me the next page, at which time it speeds through every consecutive page and freezes on the last one. At this point the last page is
2012 Nov 29
1
Coerce rownames to factor for ordering
Hi all, I think this might be an easy one but I cannot think of a way to do this other than what I am currently attempting. I simply want to sort my data frame's rownames by a defined vector so that the plots I generate from that vector are in the desired order Consider the test data below: #Create test data DataToPlot.. <- data.frame("Location1" =
2006 Mar 21
3
ROWNAMES error message
I am getting an error message, which I do not know the source to. I have a matrix SAMPLES that has preexisting rownames that I would like to change. GENE_NAMES contains these rownames. > rownames(SAMPLES) = GENE_NAMES Error in "dimnames<-.data.frame"(`*tmp*`, value = list(list(V1 = c(3843, : invalid 'dimnames' given for data frame > dim(SAMPLES) [1] 12626
2008 Sep 13
2
Symbols on a capscale object plot
Hi, I'm a beginner with R, but I'm getting excellent results with it. Well I've got an capscale object (vegan package) and I want to made a biplot with symbols representing six groups of areas. With the plot.cca function and some par attributes (like 'labels') I was able to substitute the samples names for keyboard symbols ('x','o','#',ect), but it's
2010 Feb 05
1
Strange "rownames"
I find one row in my large dataset. But when I use the "rownames" for the data on the 100,000 row, the result show nothing. I try it by the following example, it still likes that. > tst[rownames(tst)==100000,] [1] x y <0 rows> (or 0-length row.names) ############## > tst=data.frame(x=1:200000,y=200000:1) > tst[rownames(tst)==1,] x y 1 1 200000 >
2007 May 27
1
How to reference or sort rownames in a data frame
As I was working through elementary examples, I was using dataset "plasma" of package "HSAUR". In performing a logistic regression of the data, and making the diagnostic plots (R-2.5.0) data(plasma,package='HSAUR') plasma_1<- glm(ESR ~ fibrinogen * globulin, data=plasma, family=binomial()) layout(matrix(1:4,nrow=2)) plot(plasma_1) I find that data points
2006 Mar 21
1
rownames, colnames, and date and time
I noticed something surprising (in R 2.2.1 on WinXP) According to the documentation, rownames and colnames are character vectors. Assigning a vector of class POSIXct or POSIXlt as rownames or colnames therefore is not strictly according to the rules. In some cases, R performs a reasonable typecast, but in some other cases where the same typecast also would be possible, it does not. Assigning a
2006 Mar 21
1
rownames, colnames, and date and time
I noticed something surprising (in R 2.2.1 on WinXP) According to the documentation, rownames and colnames are character vectors. Assigning a vector of class POSIXct or POSIXlt as rownames or colnames therefore is not strictly according to the rules. In some cases, R performs a reasonable typecast, but in some other cases where the same typecast also would be possible, it does not. Assigning a
2010 Mar 17
2
Troubles on retrieving rownames
Hi guys, I am using the blp() function from RBloomberg package which returns a matrix of prices with the columns corresponding to the security name and the columns to the date. When I have a look at the matrix I can see the rownames (dates) on the left of the prices but when I call the rownames() function it returns me a NULL value. It worked perfectly until I had to reinstall the RBloomberg