[R 언어 데이터 처리] R 데이터 처리의 팁

1. 주요 내용 창설 새로운 변수 수정 데이터 수정 변수 이름 처리 부족값 데이터 정렬 데이터 병합 데이터 선별 샘플
2. R 언어 코드

rm(list=ls())
gc()
manager 1,2,3,4,5)
date "10/24/08", "10/28/08", "10/1/08", "10/12/08", "5/1/09")
country "US", "US", "UK", "UK", "UK")
gender "M", "F", "F", "M", "F")
age 32,45,25,39,99)
q1 5,3,3,3,2)
q2 4,5,5,3,2)
q3 5,2,5,4,1)
q4 5,5,5,NA,2)
q5 5,5,2,NA,1)
leadership data.frame(manager, date ,country, gender, age, q1,q2,q3,q4,q5, stringsAsFactors=F)


############     
leadership 5)

############    
leadership$age[leadership$age==99] NA

leadership$agecat2 NA

leadership 75] "Elder"
  agecat2[age>=55 & age<=75] "Middle Aged"
  agecat2[age<55] "Young"}
)

#####################     #################
library(plyr)
leadership "managerID", date="testDate"))



##################     ################
(leadership 6:10],na.rm=T)))
options(digits=3)###      



###################    ###############

leadership[order(age),]
leadership[order(gender,age),]
leadership[order(gender,-age),]


####################    ###############

manager 1,2,3,4,5)
q6 4,3,4,2,1)
leadership2 data.frame(manager,q6)
leadership2
merge(leadership,leadership2,by.x ="managerID",by.y = "manager")


######################    #############
leadership[,c(6:10)]

leadership[c("q1","q2","q3","q4","q5")]

myvars "q",1:5,sep="")
(newdata leadership[gender=='M' & age>30,]
subset(leadership, age>=35 | age<24, select=gender:q4)


#####################  ########################
leadership[sample(1:nrow(leadership),3,replace=F),]