我想用另一个表t2替换t1$x3中的一些记录(注意,除x3外,所有其他列都是相同的):
t1 <- data.frame(x1 = c(1,7,3,4,2,6),
x2 = c("I","R","R","I","I","R"),
x3 = c("a","a","a","a","a","a"))
t2 <- data.frame(x1 = c(4,2,6),
x2 = c("I","I","R"),
x3 = c("b","b","b"))
t1
# x1 x2 x3
# 1 1 I a
# 2 7 R a
# 3 3 R a
# 4 4 I a
# 5 2 I a
# 6 6 R a
t2
# x1 x2 x3
# 1 4 I b
# 2 2 I b
# 3 6 R b结果应该是这样的:
data.frame(x1 = c(1,7,3,4,2,6),
x2 = c("I","R","R","I","I","R"),
x3 = c("a","a","a","b","b","b"))
# x1 x2 x3
# 1 1 I a
# 2 7 R a
# 3 3 R a
# 4 4 I b
# 5 2 I b
# 6 6 R b我该怎么做呢?
发布于 2016-05-26 08:14:33
然后左联接更新:
library(dplyr)
left_join(t1, t2, by = c("x1", "x2"), all.x = TRUE) %>%
mutate(x3 = ifelse(is.na(x3.y), as.character(x3.x), as.character(x3.y))) %>%
select(-c(x3.x, x3.y))
# x1 x2 x3
# 1 1 I a
# 2 7 R a
# 3 3 R a
# 4 4 I b
# 5 2 I b
# 6 6 R b使用R基:
# left join
res <- merge(t1, t2, by = c("x1", "x2"), all.x = TRUE)
# update x3
res$x3 <- ifelse(is.na(res$x3.y), as.character(res$x3.x), as.character(res$x3.y))
# subset and reorder
res <- res[match(t1$x1, res$x1), c("x1", "x2", "x3")]发布于 2016-05-26 08:52:41
我们可以试试data.table。它应该非常快,因为我们正在分配(:=)到位。将'data.frame‘转换为'data.table’(setDT(t1)),加入't2‘on 'x1’和'x2',然后将'i.x3‘值赋给'x3’。
library(data.table)
setDT(t1)[t2, x3 := i.x3, on = c("x1", "x2")]
t1
# x1 x2 x3
#1: 1 I a
#2: 7 R a
#3: 3 R a
#4: 4 I b
#5: 2 I b
#6: 6 R b或者我们可以使用match
t1$x3 <- factor(t1$x3, levels = c('a','b'))
t1[match(do.call(paste,t2[-3]), do.call(paste, t1[-3])), 'x3'] <- t2$x3
t1
# x1 x2 x3
#1 1 I a
#2 7 R a
#3 3 R a
#4 4 I b
#5 2 I b
#6 6 R b发布于 2016-05-26 08:48:07
我想这会给你你想要的。
将x3列转换为t1和t2中的字符,然后检查条件
t1$x3 <- as.character(t1$x3)
t2$x3 <- as.character(t2$x3)
t1[t1$x1 %in% t2$x1 & as.character(t1$x2) %in% as.character(t2$x2), "x3"] <- t2$x3
> t1
# x1 x2 x3
#1 1 I a
#2 7 R a
#3 3 R a
#4 4 I b
#5 2 I b
#6 6 R bhttps://stackoverflow.com/questions/37454868
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