similar to: How to find points of intersection

Hi, Sorry for the very short explanation about the problem of intersection. I have a wave function monitored from the heart beat in a particular interval of times.?Apart?fom that, there is?a line with positive slope (e.g: y = x+2) which lies across the wave and intersect on a number of points. My problem is i have no exact equation for?such a?complex harmonic wave produced by the heart

Dear All, Let: dp: depth of the?river tp: temperature with respect to depth We can?have a simple scatter plot, between depth as y-axis and temperature as x-axis,?by using a?plot function as shown?below. #####################? dp <- c(1,4,3,2,5,7,9,8,9,2) tp <- 1:10 plot(tp,dp, type= 'l') ##################### ? Could someone advice me on?the way to?plot the same pair of

Dear All, I was looking for the color index in?image function,?such as from topo.colors(n) and etc. but still never found it. For instance, from the help menu. ########################################### # Volcano data visualized as matrix. Need to transpose and flip # matrix horizontally. image(t(volcano)[ncol(volcano):1,]) # A prettier display of the volcano x <- 10*(1:nrow(volcano)) y

Dear All, Let a, b and c are?three matrices with same no. of column but different no. of row. a <- matrix(1, 1, 2) b <- matrix(2, 2, 2) c <- matrix(3, 3, 2) Could someone help me to combine these matrices together as a single matrix? Thank you Fir

Dear All, I've been?searching for?appropriate codes to compute the rate of change and the curvature?of ?nonparametric regression model whish was denoted by a smooth function?but?unfortunately?don't manage to?do?it. I presume that such characteristics from a smooth curve can be determined by the first and second derivative operators. The following are the example of fitting a

Dear All, I have a series of data in which the first column consist of a combination of date and time, for instance 17 April 2008 at 4.01pm, such data is recorded as: 4/17/2008 16:01 I'd like to seperate it into four different columns which consist of Day, Month,Year and Time, respectively. Could someone please advice me on this mater? Thank you, Fir

Dear All, Let? 499 piece-wise lines were buit up by 500 pair of observations, via R code below. x <- 1:500 y <- rnorm(500) plot(x, y, type = 'b') I was trying to compute all the slopes?for the ?lines which were connected between two adjacent points. For instance,?slopes?of lines between first and second points, second and third points, and so on, until?between??points 499 and 500,

Dear All, Let have 10 pair of observations, as shown below. ###################### x <- 1:10 y <- c(1,3,2,4,5,10,13,15,19,22) plot(x,y) ###################### Two fitted? models, with ranges of [1,5] and [5,10],?can be easily fitted separately by lm function as shown below: ####################### mod1 <- lm(y[1:5] ~ x[1:5]) mod2 <- lm(y[5:10] ~ x[5:10]) #######################

Dear All, Could anyone?give me a hand?to suggest few packages in R to running Kalman prediction and filtration ? Thanks Fir

Dear All, I'm sorry if my question does not suit with this R group. I have recently installed?R software with version 2.9.2, but i found the program took almost 1 minute as soon as it was opened, before it can be used. However,?the previous version 2.9.1 only take few seconds?after the menu bar was clicked. This circumstance has caused me to wait for couple of minutes as several R windows

Dear All, ? Let: dp: depth of the river tp: temperature with respect to depth These pair of observations are in 3 different groups i.e: Obs. 1,3,5,7?from the first group Obs. 2,4 and?10?from second group Obs 6,8 and 9?from third group. We can have a simple scatter plot, between depth as y-axis and temperature as x-axis, with each pairs are denoted by a red dot, by using a plot function shown

Hi,   I have an original data frame with 8 columns of variables, which are stored in 'data1' frame.   data1 <- read.csv("E:\\PHD GLASGOW UNIVERSITY\\Data\\R\\Colin\\Cailness21.csv") attach(data1) names(data1)   [1] "Date"       "d"          "m"          "y"          "Time"      [6] "Depth"      "Temp"      

Hi, I tried to use tapply function to find the mean of the data?in each group as the following command, but the result are NA, as there are several missing values in each group. tapply(data,group,mean) Could someone please advice me the way to ?ignore the missing data in order for the fucntion to run successfully? Thanks Fir

Hi, I want find all roots for the following polynomial : a <- c(-0.07, 0.17); b <- c(1, -4); cc <- matrix(c(0.24, 0.00, -0.08, -0.31), 2); d <- matrix(c(0, 0, -0.13, -0.37), 2); e <- matrix(c(0.2, 0, -0.06, -0.34), 2) A1 <- diag(2) + a %*% t(b) + cc; A2 <- -cc + d; A3 <- -d + e; A4 <- -e fn <- function(z) { y <- diag(2) - A1*z - A2*z^2 - A3*z^3 - A4*z^4

Dear All, Could?someone please suggest me the way to calculate the optimal threshold in POT method via?any available ?packages in R? Thanks, Fir

I'd like to plot a point at the intersection of these two curves. Thanks x <- seq(.2, .3, by = .01) f <- function(x){ x*cos(x)-2*x**2+3*x-1 } plot(x,f(x), type = "l") abline(h = 0)

Dear List, I know this is not the first post on this topic, but I need basic help I guess. Assuming the simple case of two lines with one intercept, how can I make R calculate this intercept, NOT using locator(). par (xaxs="i", yaxs="i") plot( 1, bty="n" ,xlim=c(0,300) , ylim=c(0,300) , xlab="X", ylab="Y") curve(100-0.5*x, -50,250, add=T,

Hi, The script below is my current coding in order to produce a contour plot of temperature across altitude and time. In my case,?time,altitude and temperature?are represented by x, y and z variables. ############################################## Brazilan.Pallete <- colorRampPalette(c("blue","green","yellow","red")) image.plot(x, y, z, col =

Hi, At present, i have two distinct and real values for the coefficient, which is  required in AR(2) model. Based on my revision, for distinct and real values of the coefficients in AR(2) model, the correlation structure separated by lag h can be computed by p(h) = a*z1^(-h) + b*z2^(h), where p(h) is the autocorrelation separated by lag h, a and b can be determined by initial values, z1 and z2

Dear All,   I am working on a time series of hourly river flow measurements from 2000 - 2003  and have been trying to compute the extremum index from the original series as well as a new series from  sampling with replacement . The extremum Index produced from the original data series looks  fine but sadly, a strange value of index is given from the sampling. The flow measurements are in range of