Question

我喜欢数据表并使用它来有条件地重命名或添加因子。但是，我似乎无法一次做多个因素。这是一个例子：

a <- rep(c("A", "B", "C", "D"), each=3)
b <- 1:12
df <- data.frame(a,b)
DT <- data.table(df)

现在添加新列＆＃34;新＆＃34;这对于所有＆＃34; A＆＃34;在列＆＃34; a＆＃34;等于＆＃34; z＆＃34;

DT[a=="A", New:="z"]

这很好用。现在，如果我想改变说法＆＃34; A＆＃34;和＆＃34; C＆＃34;等于＆＃34; z＆＃34;：

DT[a==c("A", "C"), New:="z"]

给我有趣的答案：

dput(DT)
structure(list(a = structure(c(1L, 1L, 1L, 2L, 2L, 2L, 3L, 3L, 
3L, 4L, 4L, 4L), .Label = c("A", "B", "C", "D"), class = "factor"), 
    b = 1:12, New = c("z", NA, "z", NA, NA, NA, NA, "z", NA, 
    NA, NA, NA)), .Names = c("a", "b", "New"), row.names = c(NA, 
-12L), class = c("data.table", "data.frame"), .internal.selfref = <pointer: 0x0000000000140788>, index = structure(integer(0), a = integer(0)))

我确定它很简单，我似乎无法在SO上找到它（排队等待！）。感谢

要确认，我想要的输出是：

dput(DT)
structure(list(a = structure(c(1L, 1L, 1L, 2L, 2L, 2L, 3L, 3L, 
3L, 4L, 4L, 4L), .Label = c("A", "B", "C", "D"), class = "factor"), 
    b = 1:12, New = c("z", "z", "z", NA, NA, NA, "z", "z", "z", 
    NA, NA, NA)), .Names = c("a", "b", "New"), row.names = c(NA, 
-12L), class = c("data.table", "data.frame"), .internal.selfref = <pointer: 0x0000000000140788>, index = structure(integer(0), a = integer(0)))

Answer 1

您应该使用%in%代替==，因此您需要：

DT[a %in% c("A", "C"), New:="z"]

给出：

> DT
    a  b New
 1: A  1   z
 2: A  2   z
 3: A  3   z
 4: B  4  NA
 5: B  5  NA
 6: B  6  NA
 7: C  7   z
 8: C  8   z
 9: C  9   z
10: D 10  NA
11: D 11  NA
12: D 12  NA

使用过的数据：

a <- rep(c("A", "B", "C", "D"), each=3)
b <- 1:12
DT <- data.table(a,b)

您可以在数据框中执行以下操作：

df <- data.frame(a,b)
df$New <- NA
df[df$a %in% c("A", "C"), "New"] <- "z"

获得相同的结果。

使用数据表同时重命名多个因子

1 个答案: