如果您有类别的数一致,那么这将工作:
#recreate your data first:
dat <- data.frame(Name = c("Beans", "Pears", "Eggs"), Category = c("1.12.5",
"5.7.9", "10.6.5"),stringsAsFactors=FALSE)
spl <- strsplit(dat$Category,"\\.")
len <- sapply(spl,length)
dat[paste0("cat",1:max(len))] <- t(sapply(spl,as.numeric))
结果:
dat
Name Category cat1 cat2 cat3
1 Beans 1.12.5 1 12 5
2 Pears 5.7.9 5 7 9
3 Eggs 10.6.5 10 6 5
如果你有分隔值的不同数量,那么这应该占它:
#example unbalanced data
dat <- data.frame(Name = c("Beans", "Pears", "Eggs"), Category = c("1.12.5",
"5.7.9", "10.6.5"),stringsAsFactors=FALSE)
dat$Category[2] <- "5.7"
spl <- strsplit(dat$Category,"\\.")
len <- sapply(spl,length)
spl <- Map(function(x,y) c(x,rep(NA,max(len)-y)), spl, len)
dat[paste0("cat",1:max(len))] <- t(sapply(spl,as.numeric))
结果:
Name Category cat1 cat2 cat3
1 Beans 1.12.5 1 12 5
2 Pears 5.7 5 7 NA
3 Eggs 10.6.5 10 6 5
你期望每一个项目正好有三类? –
你说得对,我应该对此更加确切,有些答案会少于或多于3个类别,最少1个,最多5个! –