说我有以下简化的数据集:
<=4
我想替换let letter = prefixes[bingoBall/18]
中所有值为2的值。替换值在dt2中给出。可以通过id将两个表连接起来。
如果最终值不等于2,则最终值应保持不变。如果等于2,则最终值应变为dt <- data.table(id = 1:5, val = c(1, 2, 3, 2, 4))
dt2 <- data.table(id = c(2, 4), val = c(2, 3))
。
所需的输出:
dt
我尝试过的方法(它可以工作,但看起来并不优雅):
paste0(dt$val, ".", dt2$val)
问题:如何更好地进行转换?
答案 0 :(得分:3)
您正在寻找更新加入
dt[dt2, on=.(id), val := paste0(x.val, ".", i.val)]
输出:
id val
1: 1 1
2: 2 2.2
3: 3 3
4: 4 2.3
5: 5 4
数据:
#val column needs to be of character type to suppress the warning
dt <- data.table(id = 1:5, val = as.character(c(1, 2, 3, 2, 4)))
dt2 <- data.table(id = c(2, 4), val = c(2, 3))
答案 1 :(得分:2)
library(data.table)
# example data
dt <- data.table(id = 1:5, val = c(1, 2, 3, 2, 4))
dt2 <- data.table(id = c(2, 4), val = c(2, 3))
如果两个数据集都正确排序,则可以使用基数R,如下所示:
dt$val[dt$id %in% dt2$id] = paste0(dt$val[dt$id %in% dt2$id], ".", dt2$val)
dt
# id val
# 1: 1 1
# 2: 2 2.2
# 3: 3 3
# 4: 4 2.3
# 5: 5 4
否则,您可以使用此:
dt_merged = merge(dt, dt2, by="id", all.x=T)[, val:=ifelse(is.na(val.y),
val.x,
paste0(val.x, ".", val.y))]
dt_merged = dt_merged[, c("id","val")]
dt_merged
# id val
# 1: 1 1
# 2: 2 2.2
# 3: 3 3
# 4: 4 2.3
# 5: 5 4