我不能在里面使用switch,mutate因为它返回的是整个向量而不仅仅是行.作为一个黑客,我正在使用:
pick <- function(x, v1, v2, v3, v4) {
ifelse(x == 1, v1,
ifelse(x == 2, v2,
ifelse(x == 3, v3,
ifelse(x == 4, v4, NA))))
}
这在内部工作mutate,现在很好,因为我通常选择4件事,但这可能会改变.你能推荐另一种选择吗？
例如:
library(dplyr)
df.faithful <- tbl_df(faithful)
df.faithful$x <- sample(1:4, 272, rep=TRUE)
df.faithful$y1 <- rnorm(n=272, mean=7, sd=2)
df.faithful$y2 <- rnorm(n=272, mean=5, sd=2)
df.faithful$y3 <- rnorm(n=272, mean=7, sd=1)
df.faithful$y4 <- rnorm(n=272, mean=5, sd=1)
使用pick:
mutate(df.faithful, y = pick(x, y1, y2, y3, y4))
Source: local data frame [272 x 8]
eruptions waiting x y1 y2 y3 y4 y
1 3.600 79 1 8.439092 5.7753006 8.319372 5.078558 8.439092
2 1.800 54 2 13.515956 6.1971512 6.343157 4.962349 6.197151
3 3.333 74 4 7.693941 6.8973365 5.406684 5.425404 5.425404
4 2.283 62 4 12.595852 6.9953995 7.864423 3.730967 3.730967
5 4.533 85 3 11.952922 5.1512987 9.177687 5.511899 9.177687
6 2.883 55 3 7.881350 1.0289711 6.304004 3.554056 6.304004
7 4.700 88 4 8.636709 6.3046198 6.788619 5.748269 5.748269
8 3.600 85 1 8.027371 6.3535056 7.152698 7.034976 8.027371
9 1.950 51 1 5.863370 0.1707758 5.750440 5.058107 5.863370
10 4.350 85 1 7.761653 6.2176610 8.348378 1.861112 7.761653
.. ... ... . ... ... ... ... ...
我们看到如果x == 1,我将y1中的值复制到y中,依此类推.这是我想要做的,但我希望能够做到这一点,无论我有4或400列的列表.
试图使用switch:
mutate(df.faithful, y = switch(x, y1, y2, y3, 4))
Error in switch(c(1L, 2L, 4L, 4L, 3L, 3L, 4L, 1L, 1L, 1L, 4L, 3L, 1L, :
EXPR must be a length 1 vector
试图使用list:
mutate(df.faithful, y = list(y1, y2, y3, y4)[[x]])
Error in list(c(8.43909205142925, 13.5159559591257, 7.69394050059568, :
recursive indexing failed at level 2
试图使用c:
mutate(df.faithful, y = c(y1, y2, y3, y4)[x])
Source: local data frame [272 x 8]
eruptions waiting x y1 y2 y3 y4 y
1 3.600 79 1 8.439092 5.7753006 8.319372 5.078558 8.439092
2 1.800 54 2 13.515956 6.1971512 6.343157 4.962349 13.515956
3 3.333 74 4 7.693941 6.8973365 5.406684 5.425404 12.595852
4 2.283 62 4 12.595852 6.9953995 7.864423 3.730967 12.595852
5 4.533 85 3 11.952922 5.1512987 9.177687 5.511899 7.693941
6 2.883 55 3 7.881350 1.0289711 6.304004 3.554056 7.693941
7 4.700 88 4 8.636709 6.3046198 6.788619 5.748269 12.595852
8 3.600 85 1 8.027371 6.3535056 7.152698 7.034976 8.439092
9 1.950 51 1 5.863370 0.1707758 5.750440 5.058107 8.439092
10 4.350 85 1 7.761653 6.2176610 8.348378 1.861112 8.439092
.. ... ... . ... ... ... ... ...
不会产生错误,但行为不符合预期.
小智 26
对OP来说太晚了,但是如果这出现在搜索中......
dplyr v0.5有recode()一个矢量化版本switch(),所以
data_frame(
x = sample(1:4, 10, replace=TRUE),
y1 = rnorm(n=10, mean=7, sd=2),
y2 = rnorm(n=10, mean=5, sd=2),
y3 = rnorm(n=10, mean=7, sd=1),
y4 = rnorm(n=10, mean=5, sd=1)
) %>%
mutate(y = recode(x,y1,y2,y3,y4))
按预期生产:
# A tibble: 10 x 6
x y1 y2 y3 y4 y
<int> <dbl> <dbl> <dbl> <dbl> <dbl>
1 2 6.950106 6.986780 7.826778 6.317968 6.986780
2 1 5.776381 7.706869 7.982543 5.048649 5.776381
3 2 7.315477 2.213855 6.079149 6.070598 2.213855
4 3 7.461220 5.100436 7.085912 4.440829 7.085912
5 3 5.780493 4.562824 8.311047 5.612913 8.311047
6 3 5.373197 7.657016 7.049352 4.470906 7.049352
7 2 6.604175 9.905151 8.359549 6.430572 9.905151
8 3 11.363914 4.721148 7.670825 5.317243 7.670825
9 3 10.123626 7.140874 6.718351 5.508875 6.718351
10 4 5.407502 4.650987 5.845482 4.797659 4.797659
(也适用于命名的args,包括字符和因子x.)
按 的每个值进行操作x。这是data.table版本，我认为可以在以下位置完成类似的操作dplyr：
library(data.table)
dt = data.table(x = c(1,1,2,2), a = 1:4, b = 4:7)
dt[, newcol := switch(as.character(x), '1' = a, '2' = b, NA), by = x]
dt
# x a b newcol
#1: 1 1 4 1
#2: 1 2 5 2
#3: 2 3 6 6
#4: 2 4 7 7
您现在可以将dplyr的函数case_when与mutate().
按照您的示例生成数据：
library(dplyr)
df.faithful <- tbl_df(faithful)
df.faithful$x <- sample(1:4, 272, rep=TRUE)
df.faithful$y1 <- rnorm(n=272, mean=7, sd=2)
df.faithful$y2 <- rnorm(n=272, mean=5, sd=2)
df.faithful$y3 <- rnorm(n=272, mean=7, sd=1)
df.faithful$y4 <- rnorm(n=272, mean=5, sd=1)
现在我们pick()使用case_when以下方法定义一个新函数：
pick2 <- function(x, v1, v2, v3, v4) {
out = case_when(
x == 1 ~ v1,
x == 2 ~ v2,
x == 3 ~ v3,
x == 4 ~ v4
)
return(out)
}
你会发现你可以完美地使用它mutate()：
df.faithful %>%
mutate(y = pick2(x, y1, y2, y3, y4))
输出是：
# A tibble: 272 x 8
eruptions waiting x y1 y2 y3 y4 y
<dbl> <dbl> <int> <dbl> <dbl> <dbl> <dbl> <dbl>
1 3.6 79 3 8.73 7.23 8.89 4.04 8.89
2 1.8 54 3 9.97 4.31 7.06 5.05 7.06
3 3.33 74 1 6.65 7.23 4.46 6.49 6.65
4 2.28 62 1 6.40 4.39 5.41 3.49 6.40
5 4.53 85 4 3.96 8.85 7.43 6.51 6.51
6 2.88 55 4 6.36 8.08 5.82 5.06 5.06
7 4.7 88 1 5.91 6.47 6.43 5.88 5.91
8 3.6 85 1 7.77 4.55 6.56 5.05 7.77
9 1.95 51 4 5.74 6.46 6.95 4.26 4.26
10 4.35 85 1 7.04 1.73 5.71 2.53 7.04
# ... with 262 more rows