dplyr窗口函数有order_by和with_order

Question

BACKGROUD

dplyr具有窗口功能.当您想要控制窗口函数的顺序时,您可以使用order_by.

数据

mydf <- data.frame(id = c("ana", "bob", "caroline",
                          "bob", "ana", "caroline"),
                   order = as.POSIXct(c("2015-01-01 18:00:00", "2015-01-01 18:05:00",
                                        "2015-01-01 19:20:00", "2015-01-01 09:07:00",
                                        "2015-01-01 08:30:00", "2015-01-01 11:11:00"),
                                        format = "%Y-%m-%d %H:%M:%S"),  
                   value = runif(6, 10, 20),
                   stringsAsFactors = FALSE)

#        id               order    value
#1      ana 2015-01-01 18:00:00 19.00659
#2      bob 2015-01-01 18:05:00 13.64010
#3 caroline 2015-01-01 19:20:00 12.08506
#4      bob 2015-01-01 09:07:00 14.40996
#5      ana 2015-01-01 08:30:00 17.45165
#6 caroline 2015-01-01 11:11:00 14.50865

假设您要使用lag(),您可以执行以下操作.

arrange(mydf, id, order) %>%
group_by(id) %>%
mutate(check = lag(value))

#        id               order    value    check
#1      ana 2015-01-01 08:30:00 17.45165       NA
#2      ana 2015-01-01 18:00:00 19.00659 17.45165
#3      bob 2015-01-01 09:07:00 14.40996       NA
#4      bob 2015-01-01 18:05:00 13.64010 14.40996
#5 caroline 2015-01-01 11:11:00 14.50865       NA
#6 caroline 2015-01-01 19:20:00 12.08506 14.50865

但是,你能避免使用arrange()带order_by().

group_by(mydf, id) %>%
mutate(check = lag(value, order_by = order))

#        id               order    value    check
#1      ana 2015-01-01 18:00:00 19.00659 17.45165
#2      bob 2015-01-01 18:05:00 13.64010 14.40996
#3 caroline 2015-01-01 19:20:00 12.08506 14.50865
#4      bob 2015-01-01 09:07:00 14.40996       NA
#5      ana 2015-01-01 08:30:00 17.45165       NA
#6 caroline 2015-01-01 11:11:00 14.50865       NA

实验

我想对我想要将行号分配给新列的情况应用相同的过程.使用示例数据,您可以执行以下操作.

group_by(mydf, id) %>%
arrange(order) %>%
mutate(num = row_number())

#        id               order    value num
#1      ana 2015-01-01 08:30:00 17.45165   1
#2      ana 2015-01-01 18:00:00 19.00659   2
#3      bob 2015-01-01 09:07:00 14.40996   1
#4      bob 2015-01-01 18:05:00 13.64010   2
#5 caroline 2015-01-01 11:11:00 14.50865   1
#6 caroline 2015-01-01 19:20:00 12.08506   2

我们可以省略排列线吗？看到CRAN手册,我做了以下几点.两次尝试都没有成功.

### Not working
group_by(mydf, id) %>%
mutate(num = row_number(order_by = order))

### Not working
group_by(mydf, id) %>%
mutate(num = order_by(order, row_number()))

我们怎样才能做到这一点？

Answer 1

我不是故意自己回答这个问题.但是,我决定分享我发现的东西,因为我没有看到很多帖子使用order_by,特别是 with_order.我的回答是用with_order()而不是order_by().

group_by(mydf, id) %>%
mutate(num = with_order(order_by = order, fun = row_number, x = order))

#        id               order    value num
#1      ana 2015-01-01 18:00:00 19.00659   2
#2      bob 2015-01-01 18:05:00 13.64010   2
#3 caroline 2015-01-01 19:20:00 12.08506   2
#4      bob 2015-01-01 09:07:00 14.40996   1
#5      ana 2015-01-01 08:30:00 17.45165   1
#6 caroline 2015-01-01 11:11:00 14.50865   1

我想看看这两种方法在速度方面是否会有任何差异.在这种情况下,它们看起来非常相似.

library(microbenchmark)

mydf2 <- data.frame(id = rep(c("ana", "bob", "caroline",
                               "bob", "ana", "caroline"), times = 200000),
                    order = seq(as.POSIXct("2015-03-01 18:00:00", format = "%Y-%m-%d %H:%M:%S"),
                                as.POSIXct("2015-01-01 18:00:00", format = "%Y-%m-%d %H:%M:%S"),
                                length.out = 1200000),
                    value = runif(1200000, 10, 20),
                    stringsAsFactors = FALSE)

jazz1 <- function() {group_by(mydf2, id) %>%
                     arrange(order) %>%
                     mutate(num = row_number())}

jazz2 <- function() {group_by(mydf2, id) %>%
                     mutate(num = with_order(order_by = order, fun = row_number, x = order))}


res <- microbenchmark(jazz1, jazz2, times = 1000000L)
res

#Unit: nanoseconds
#  expr min lq     mean median uq   max neval cld
# jazz1  32 36 47.17647     38 47 12308 1e+06   a
# jazz2  32 36 47.02902     38 47 12402 1e+06   a