search for: momadou

...matrix with 3 columns (Date, MeanArea and SdArea). I want to draw a figure showing the variable MeanArea in terms of the Date. But instead to use the variable SdArea as bar error, I want to use ?polygon error?. I use this code but the output does not seem good. Poly<read.table("C:\\Users\\Momadou\\Documents\\AreaMonthly.csv",sep=";",dec=",",header=TRUE) Poly y <-MeanArea x <-SdArea z <-Date matplot(x,cbind(y,z),type="n") polygon(c(x,rev(x)),c(y,rev(z)),col="grey50") Thank you in advance -- View this message in context: http:/...

HI, Sorry Carl, I received your message in my spam folder. Sarah proposed me a good example of code. Thank you Momadou  ________________________________ De : Carl Witthoft [via R] <ml-node+s789695n4188375h96@n4.nabble.com> Envoyé le : Mardi 13 Décembre 2011 3h34 Objet : Re: Polygon Please read the posting guide and provide a (small) reproducible example of your data. The statement "...output do...

Hi, I created a figure with R and I want to save it in .pdf. I used this code: > pdf("res.pca.pdf",width=10,height=8) > library(FactoMineR) > res.pca<-PCA(acp) > res.pca > dev.off() When I go in my folder, I find an empty file ( 0 Ko). Do you know where is the problem. Thank you in advance -- View this message in context:

Hi, I try to build a loop difficultly. I have in a folder called Matrices several files (.csv) called Mat2002273, Mat2002274 to Mat2002361. I want to calculate for each file the mean of the column called Pixelvalues. I try this code but as result, I have this message: Mat2002273 not found

Hi, Instead to put colour in my histogram, I want to put symbols like lines, dots etc. Do you know the function that does it? Thank you in advance -- View this message in context: http://r.789695.n4.nabble.com/Not-colour-but-symbols-tp4490030p4490030.html Sent from the R help mailing list archive at Nabble.com.

HI, This code is part of a code I used to do a linear regression: >points(var1~var2,data=Regress,pch=21,bg="grey") >reg11<-lm(var1~var2,data=Regress) >abline(lm(var1~var2,data=Regress),lty=2,lwd=2,col="grey") >legend("topleft",legend= >c("NDII from composite", >"y= 0.0007x - 0.1156",expression(paste(r^2 ==

Hi everybody, I try to do an exponential model with decreasing. In my data, there is no data missing but there negative values. I adapted this following code whose the origin code comes from in this forum: Regress<-read.table("C:\\Users\\Regression.csv",sep=";",dec=",",header=TRUE) Regress f <- function(x,a,b) {a * exp(b * x)} x<-Regress$EWT

Hi I want to overlay my raster and its boundary which is a shapefile. When I used thise code separately, all is ok: # Open raster >Image<-read.table("C:\\Users\\Documents\\Nouveau\\Frequence.txt",sep="",dec=",",header=TRUE) >testo<-rasterFromXYZ(Image) >plot(testo) >testo2 <- aggregate(testo,fact=10, fun=mean) >plot(testo2) # open

Hi, I did a linear regression with 5 explanatory variables. Now, to see the contribution of each variable, I use varpart from vegan library. But varpart don?t accepts my 5 explanatory variables, it accept only 4. 1- How must I do to use my 5 explanatory variables? 2- Is it the sum of variance fraction of each variable must be equal to 1. Thanks for your help. Komine -- View this

Hi, After performing a multiple linear regression, I am looking for an R package that can calculate the fraction [a] a partitioning of variation. This fraction measures the proportion of variance of y explained by the explanatory variable x1 (for example) when other variables (x2, x3 ...) are held constant. Thank you in advance for your help -- View this message in context:

Hi everybody, I try to do an exponential model with decreasing. In my data, there is no data missing but there negative values. I adapted this following code whose the origin code comes from in this forum: Regress<-read.table("C:\\Users\\Regression.csv",sep=";",dec=",",header=TRUE) Regress f <- function(x,a,b) {a * exp(b * x)} x<-Regress$EWT

Hi, I used GAM function with this code: library(mgcv) modgam=gam(Z~X+Y,data=table) pred=predict(modgam,type="response") vis.gam(modgam,view=c("X","Y"),plot.type="contour",zlim=c(1,16)) My problem is on my figure, the predicted values begin by 0 to 16 with an interval equal 2: 1- I want that the predicted values (Z) begin by 1 until 16 with an interval

Hi, I used simple linear regression with the R software and EXCEL on the same data. Although , I find the same R2=0.84, I find different estimated values (intercept and slope). For the R software (slope =0.0009, Intercept = -0.1478), for EXCEL (slope =927.7, Intercept = 154,41). When I use the estimated values from the R software, the results seem bad, however the results of Exel seem correct.

Hi all R users I did a logistic regression with my binary variable Y (0/1) and 2 explanatory variables. Now I try to draw my nomogram with predictive value. I visited the help of R but I have problem to understand well the example. When I use glm fonction, I have a problem, thus I use lrm. My code is: modele<-lrm(Y~L+P,data=donnee) fun<- function(x) plogis(x-modele$coef[1]+modele$coef[2])