search for: midpoint

Hello list: I am using "cut" and "table" to obtain a frequency table from a numeric sample vector.  The idea is to calculate mean and standard deviation on grouped data.  However, I can't extract the midpoints of the class intervals, which seem to be strings treated as factors.  How do i extract the midpoint? Thanks, jose loreto [[alternative HTML version deleted]]

...uency counts using the following: Max.brks <- pretty(range(Max.Spread$Distance), 100) Max.f<-hist(x=Max.Spread$Distance, breaks=Max.brks,plot=FALSE ) Max.cnt<-as.data.frame(cbind(sim,Max.f$mids,Max.f$counts)) colnames(Max.cnt)<-c("Simulation","MidPoint","Count") then I fit this to a gamma distribution function: modl<- vglm (Count ~ MidPoint ,gamma2 ,data =subset(Max.cnt,select=(simulation,MidPoint,Count),trace=TRUE,crit="c") print(coef(modl2,matrix=TRUE)) print(summary(modl2)) This produces the output...

May I raise the issue of the behaviour of image()? In image() dim(z) is supposed to be c(length(x), length(y)). This suggests that the user is supplying the midpoints of the rectangles. For all but the outer rectangles this is true. The outer rectangles, however, are only displayed in half width. So it is not possible to show correctly an image plot of equally-spaced abscissa. I would hazard a guess that these kinds of plots make up the bulk of calls to ima...

...own (the heights of the agglomeration), but how can I get the x-values? > mydata <- c(1,2,3,4,5) > hc <- hclust(dist(mydata), method="average") > hc$height # these are the y-coordinates [1] 1.0 1.0 1.5 2.5 I experimented with the dendrogram object because it gives the midpoints: > dend <- as.dendrogram(hc) > attributes(dend[[1]])$midpoint [1] 0.5 Perhaps I could loop over all possible branches and get the midpoints if I knew how to convert them to x-values. The manual says that midpoint is the "numeric horizontal distance of the node from the left bor...

#If I have two lists as follows a1<- array(1:6, dim=c(2,3)) a2<- array(7:12, dim=c(2,3)) l1<- list(a1,a2) a3<- array(1:4, dim=c(2,2)) a4<- array(5:8, dim=c(2,2)) l2<- list(a3,a4) #how can I create a new list with the mean across all arrays within the list, so all components are included? As an example for [[1]]; cbind((l1[[1]][,1]+l2[[1]][,1])/2,

Dear R users, I need to find the coordinates for the point (midpoint) located half way between two pairs of coordinates (lon1,lat1 and lon2,lat2) assuming a straight line between them. What would be the best way? I tried to find an answer in the help archives but without success. I would greatly appreciate any help. df<- data.frame(lon1=c(-4.568,-4.3980), lat1=c...

Hello r-help, I have a lengthy vector of data (with values anywhere from 1-200), and another index vector of 'groups' representing values 0-2, 3-5, 6-8, ... of length 67. The index vector has the structure (1, 4, 7, ... , 196, 199), where each value is the midpoint of each respective group. I'm trying to convert the data vector such that values falling into each group are changed to the 'number' of that group. That is, the position of the midpoint of that group in the index vector. For example, 4 or 5 would become a 2; 6 would become 3; 9 or 10...

Hello: I'm having quite a bit of difficulty adjusting the x-labels using bargraphCI. I've tried using text and srt=45 to rotate the labels or mtext for 2 lines to break up the labels. However, using either method, I cannot line up the labels with the midpoints of the bars (they line up with some sort of tick mark that is off the midpoint of the bars). Any suggestions would be greatly appreciated. Regards Pedro L. Vera, Ph.D. University of South Florida, Dept of Surgery [[alternative HTML version deleted]]

Starting with the head of a 499 element matrix whose column names are now the labels trom a cut() operation, I needed to get to a vector of midpoints to serve as the basis for plotting a calibration curve ( exp(linear predictor) vs. : > dput(head(dimnames(mtcal)[2][[1]])) # was starting point testvec <- c("(-8.616,-3.084]", "(-3.084,-2.876]", "(-2.876,-2.756]", "(-2.756,-2.668]", "(-2....

...nd Ihaka (2000), "Lexical Scope and Statistical Computing," JCGS, 9, 491-508, is very interesting and helpful in understanding how lexical scope is about. However, I got some questions that I just can't figure out. First all, allow me to copy the two functions given by the authors: midpoint <- function(f, a, b, n=100) { h <- (b - a)/n (b - a)*mean(sapply(seq(a + h/2, b - h/2, length=n), f)) } bv.integrate <- function(f, a, b, n=100, rule=midpoint) { g <- function(y) { fx <- function(x) f(x, y) rule(fx, a[2], b[2], n) } rule(g, a[1],...

.... Thanks in advance. -Ben ########################### begin code ################################### version dendro.leaf <- function(label) { ans <- list() attr(ans, 'members') <- 1 attr(ans, 'height') <- 0 attr(ans, 'leaf') <- T attr(ans, 'midpoint') <- 0 attr(ans, 'label') <- label attr(ans, 'class') <- 'dendrogram' ans } dendro.merge <- function(d1, d2, height) { ans <- list(d1, d2) members <- attr(d1, 'members') + attr(d2, 'members') attr(ans, 'members') &lt...

I have a data.frame with three columns + the row number Sample MidPoint Count [1,] 1 0.025 1 [2,] 1 0.075 3 [3,] 1 0.125 15 [4,] 1 0.175 12 [5,] 1 0.225 5 [6,] 1 0.275 8 When I export the data to a csv the column names are shifted because the row numbers are being exported as wel...

...ce between line segments. For instance, consider 20 random line segments: x1 <- runif(20) y1 <- runif(20) x2 <- runif(20) y2 <- runif(20) plot(x1, y1, type = "n") segments(x1, y1, x2, y2) Inititally I thought the solution to this problem was to work out the distance between midpoints (it quickly became apparent that this is totally wrong when looking at the plot). So, I thought that perhaps finding the minimum distance between each of the lines endpoints AND their midpoints would be a good proxy for this, so I set up a loop that uses pythagoras to work out these 9 distances an...

I''m using standard barplot (Windows version 1.6.2 of R) to represent a certain weekly metric "v" and I would like to properly overlay on top of it its moving average "mean.8" (window of 8 weeks). I must be doing something wrong since the moving average (using "lines") doesn''t overlay properly, i.e., both x-scales do not match! ...

Anyone know how to get p-values for the t-values from the coefficients produced in vglm? Attached is the code and output ? see comment added to output to show where I need p-values + print(paste("********** Using VGAM function gamma2 **********")) + modl2<- vglm(MidPoint~Count,gamma2,data=modl.subset,trace=TRUE,crit="c") + print(coef(modl2,matrix=TRUE)) + print(summary(modl2)) [1] "********** Using VGAM function gamma2 **********" VGLM linear loop 1 : coefficients = 0.408464609241, 3.255887520104, -0.000220585671 VGLM line...

Hello I have a table describing butterfly range traits. It is composed of three columns as below Species name range size (X) latitude of range midpoint (Y) There are 11 kinds of butterflies. Each has its range size, and the latitude of each range midpoint ranges from 9 to 19. I would like to have the average range size of every degree of latitude. For example, the average range size of latitude degree 10 (10.0-10.99: Butterfly A2, B1, B2) P...

...} and --- a/lib/Support/APInt.cpp +++ b/lib/Support/APInt.cpp @@ -1440,16 +1440,14 @@ APInt APInt::sqrt() const { APInt nextSquare((x_old + 1) * (x_old +1)); if (this->ult(square)) return x_old; - else if (this->ule(nextSquare)) { + if (this->ule(nextSquare)) { APInt midpoint((nextSquare - square).udiv(two)); APInt offset(*this - square); if (offset.ult(midpoint)) return x_old; - else - return x_old + 1; - } else - llvm_unreachable("Error in APInt::sqrt computation"); - return x_old + 1; + return x_old + 1; + } + llvm_unreach...

Hello R-Users, I am new to R and trying my best however I need help with this simple task. I have a dataset, YM1207. X.Symbol Date Time Exchange TickType ReferenceNumber Price Size 12491 3:YMZ7.EC 12/03/2007 08:32:50 EC B 85985770 13379 7 12492 3:YMZ7.EC 12/03/2007 08:32:50 EC A 85985771 13380 4 12493

...))+ theme(axis.ticks = element_blank())+xlab("") + ylab("")+ geom_point(data=Data,aes(x=lon,y=lat,color= value),size=1.5) + scale_colour_gradient2(low = "white", mid = "blue",high = "red", midpoint = 175, guide="colourbar",limits=c(0,350))+ labs(title = "Depth of changes")) . -- Dr Manuel Mendoza Department of Biogeography and Global Change National Museum of Natural History (MNCN) Spanish Scientific Council...

I would like a 3d plot of a matrix such that individual trapezoids that make up the surface are colored according to the z-value of that point (or preferably the midpoint of its four corners, or something similar). MS Excel has something like that. I know that persp can have an nx by ny matrix its "col" argument, I just don't know how to generate that matrix. To calculate the midpoint of the tiles for the matrix z, I could use something like zz <-...