我想传递arrange(){dplyr}一个变量名的向量来排序.通常我只需输入我想要的变量,但我正在尝试创建一个函数,其中排序变量可以作为函数参数输入.
df <- structure(list(var1 = c(1L, 2L, 2L, 3L, 1L, 1L, 3L, 2L, 4L, 4L
), var2 = structure(c(10L, 1L, 8L, 3L, 5L, 4L, 7L, 9L, 2L, 6L
), .Label = c("b", "c", "f", "h", "i", "o", "s", "t", "w", "x"
), class = "factor"), var3 = c(7L, 5L, 5L, 8L, 5L, 8L, 6L, 7L,
5L, 8L), var4 = structure(c(8L, 5L, 1L, 4L, 7L, 4L, 3L, 6L, 9L,
2L), .Label = c("b", "c", "d", "e", "f", "h", "i", …使用旧select_()函数,我可以将命名向量传递给select并立即更改位置和列名:
my_data <- data_frame(foo = 0:10, bar = 10:20, meh = 20:30)
my_newnames <- c("newbar" = "bar", "newfoo" = "foo")
move_stuff <- function(df, newnames) {
select_(df, .dots = newnames)
}
move_stuff(my_data, newnames = my_newnames) )
# this is the desired output
# A tibble: 4 x 2
newbar newfoo
<int> <int>
1 10 0
2 11 1
3 12 2
4 13 3
我尝试使用quosures和拼接做类似的事情 - 选择列效果很好,但是矢量的名称(因此同时重命名列)似乎被忽略了.以下两个都返回数据框,列中包含名称bar和foo,但不是newbar和newfoo:
move_stuff2 <- function(df, newnames) …我已经阅读了几个关于dplyr编程的指南,我仍然对如何解决使用非标准评估(NSE)评估构造/连接字符串的问题感到困惑.我意识到有更好的方法来解决这个例子,而不是使用NSE,但想要学习如何.
t <- tibble( x_01 = c(1, 2, 3), x_02 = c(4, 5, 6))
i <- 1
这是我想要的结果,但是想要mutate()构造变量:
t %>% mutate(d_01 = x_01 * 2)
#> A tibble: 3 x 3
#> x_01 x_02 d_01
#> <dbl> <dbl> <dbl>
#> 1 1.00 4.00 2.00
#> 2 2.00 5.00 4.00
#> 3 3.00 6.00 6.00
这是我第一次尝试使用字符串:
new <- sprintf("d_%02d", i)
var <- sprintf("x_%02d", i)
t %>% mutate(new = var * 2)
#> Error in mutate_impl(.data, dots) : …