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I would like to prepare the data for barplot. But I only have the data frame now. x1=rnorm(10,mean=2) x2=rnorm(20,mean=-1) x3=rnorm(15,mean=3) data=data.frame(x1,x2,x3) If there a way to put data within a specific range? The expected result is as follows: range x1 x2 x3 -10-0 2 5 1 (# points in this

Hi, folks, As seen in the following codes: x1=rlnorm(10) x2=rlnorm(10,mean=2) y=rlnorm(10,mean=10)### Fake dataset linmod=lm(log(y)~log(x1)+log(x2)) After the regression, I would like to know the mean of y. Since log(y) is normal and y is lognormal, I need to know the mean and variance of log(y) first. I tried mean (y) and mean(linmod), but either one is what I want. Any tips? Thanks in

Dear all, Suppose my data frame is as follows: id price distance 1 2 4 1 3 5 ... 2 4 8 2 5 9 ... n 3 7 n 8 9 I would like to calculate the rank-order correlation between price and distance for each id. cor(price,distance,method = "spearman") calculate a correlation for all. Then I tried to use apply(data,list='id',cor(price , distance , method =

Hi, folks, As I understand, Least-squares Estimate (second-moment assumption) and the Method of Maximum Likelihood (full distribtuion assumption) are used for linear regression. I do >?lm, but the help file does not tell me the model employed in R. But in the book 'Introductory Statistics with R', it indicates R estimate the parameters using the method of Least-squares. However it

Hi, folks, Here are the codes: ############## y=1:10 x=c(1:9,1) lin=lm(log(y)~x) ### log(y) is following Normal distribution x=5:14 prediction=predict(lin,newdata=x) ##prediction=predict(lin) ############### 1. The codes do not work, and give the error message: Error in eval(predvars, data, env) : numeric 'envir' arg not of length one. But if I use the code after the pound sign, it

Hi, folks linmod=y~x+z summary(linmod) The summary of linmod shows the standard error of the coefficients. How can we get the sd of y and the robust standard errors in R? Thanks! [[alternative HTML version deleted]]

Hi, folks, runif (n,min,max) is the typical code for generate R.V from uniform dist. But what if we need to fix the mean as 20, and we want the values to be integers only? Thanks [[alternative HTML version deleted]]

Hi, folks, Please first look at the codes: plan_a=c('apple','orange','apple','apple','pear','bread') plan_b=c('bread','bread','orange','bread','bread','yogurt') value=1:6 data=data.frame(plan_a,plan_b,value) library(plyr) library(reshape) mm=melt(data, id=c('plan_a','plan_b'))

Apologize for not being clearer earlier. I would like to ask again. Thank Joris and Markleeds for response. Two examples: 1. Function 'var'. In R, it is the sum of square divided by (n-1) but not by n. (I know this in R class) 2. Function 'lm'. In R, it is the residual sum of square divied by (n-2) not by n, the same as in the least squares estimate. But the assumption following

Hi, folks, Finally Friday~~ Here comes the question: x=c('germany','poor italy','usa','england','poor italy','japan') y=c('Spain','germany','usa','brazil','england','chile') s=1:6 z=3:8 test=data.frame(x,y,s,z) #Now I only concern the countries ('germany','england','brazil').

Hi, folks, I would like to copy the output of summary(lm) and anova (lm) in R to my word file. But the output will be a mess if I just copy after I call summary and anova. ##################### x=rnorm(10) y=rnorm(10,mean=3) lm=lm(y~x) summary(lm) Call: lm(formula = y ~ x) Residuals: Min 1Q Median 3Q Max -1.278567 -0.312017 0.001938 0.297578 1.310113

Hi, folks, I am sorry that I did not state the problem correctly yesterday. Please let me address the problem by the following codes: first=c('u','b','e','k','j','c','u','f','c','e')

I am so frustrated about reading data from this sample csv file. My code is : >test=read.csv(file='test.csv',header=T) warning message: In read.table(file = file, header = header, sep = sep, quote = quote, : incomplete final line found by readTableHeader on 'test.csv' >test [1] ??.?.. <0 rows> (or 0-length row.names) What is the problem here? Thanks

Hi, The format problem is really annoying. How to get rid of it? x [1] 1e+06 And also when I do barplot, x=rnorm(100,mean=1000) barplot(table(cut(x,breaks=c(0,50,100,150,200,250,300,350,400,450,500,1000,2000,5000,10000)))) The data will show as scientific format, when it is larger than 999 (1e+03, 5e+3). I would like all the data in the digit format (1000, 5000). Does any function works

Hi, folks, x=1:10 y=rep(2:6,2) lin=lm(y~x) x=3:12 new=predict(lin,se.fit=T) #se.fit: the standard error of the predicted means, namely, the square root of Var( E[y|x] | x) # How can I generate the variances of the new observations? Namely the square root of var(y|x), ## Which I think should be much larger than the values from se.fit=T. The reason why I need to know the estimations of

Hi, folks, ###### food=c('fruit','fruit','fruit','drink','drink','drink') type=c('apple','apple','orange','water','soda','soda') value=c(2,3,1,5,7,6) data=data.frame(food,type,value) share=c((2+3)/(2+3+1),5/6,1/6,5/(5+7+6),13/18,13/18)

Hi, I need to regress the population on time slot and also days in advance. A sample dataset is like this. I do not know whether I can attach a file in this mailing list or not. If you know how to do it, I will be happy to send you my data file. time slot days in advance US EUR CHINA JAPAN 2 34 56 78 34 6:am-7am 5

Hi, folks, Please let me address the problem by the following codes: first=c('u','b','e','k','j','c','u','f','c','e') second=c('usa','Brazil','England','Korea','Japan','China','usa','France','China','England') third=1:10

Hi, folks, test=matrix(rep(letters[1:3],6),nrow=6,byrow=T) test[2,]=letters[5:7] test[4,]=c('e','f','i') test[1,3]='i' colnames(test)=c('leave','arrive','line') The code will generate a sample dataset. I have thousands rows of data. For the same 'leave' and 'arrive', how can I get the sum of square of percentages of

Hi, folks, Happy work in weekends >_< My question is how to plot the dependent variable against one of the predictors with other predictors as constant. Not for the original data, but after prediction. It means y is the predicted value of the dependent variables. The constane value of the other predictors may be the average or some fixed value. ####### y=1:10 x=10:1 z=2:11