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R语言学习笔记1-基础篇

R语言学习笔记1-基础篇

作者: RudyHe | 来源:发表于2015-12-03 16:40 被阅读123次
    - Basic Function
        - rnorm(10)
        - mean(abs(rnorm(100)) sd(rnorm(100))
        - source("xxx.R")
        - x<- c(1,2,4) q <- c(x,x,8)
        - y    # print out y
        - q()    # quit from command line
        - seq(1,6,by=3)    # 1 4
        - seq(0,1,length.out=11)    # 0.0,0.1,0.2,...0,9,1.0
        - help(seq), ?seq, example(seq)
    
    - Function
        oddcount <- function(x) {    # oddcount is function name
            k <- 0
            return (k) 
        }
    
    - For statement
        for (n in x) {
            if(n%%2==1) k <- k+1
        }
        for (i in 1:length(x)) {
            if(x[i]%%2==1) k <- k+1
        }
    
    - Data Structure:
        - string: y<-"abc" mode(y)    # "character"
            - u<-paste("abc","def","e")    # abc def e
            - v<-strsplit(u," ")    # split according to blanks 
        - matrix: m<-rbind(c(1,4),c(2,2))    # row bind
            - m[1,2]    # 4
            - m[1,] # row 1
            - m[,2] # column 2
        - list: x<-list(u=2,v="abc")
            - x$u # 2
        - hn <- hist(Nile) # class
            - print(hn)
            - str(hn) # structure
            - attr(,"class") # "histogram"
        - d <- data.frame(list(kids=c("Jack","Jill"),ages=c(12,10)))
            - d, d$age
    
    - Regression Example
        - examsquiz<-read.table("ExamsQuiz.txt",header=FALSE)
        - class(examsquiz) # "data.frame"
        - head(examsquiz) # less
        - lma<-lm(examsquiz[,2] ~ examsquiz[,1])    # using col 1 to fit col 2
        - lma<-lm(examsquiz$V2 ~ examsquiz$V1)    # same as above
        - attribute(lma), summary(lma)    # like print, str
        - lma, lma$coef
        - lmb<-lm(examsquiz[,2] ~ examsquiz[,1] + examsquiz[,3])    # + is not concate symbol
    

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