(R) 如何从 slide_index() 的 window 中排除每个当前行的值?
(R) How to exclude each current row's value from slide_index()'s window?
您好,我正在使用 slide_index() 在时间 window 中为数据帧中的每一行捕获值。但是,对于每一行,我还将当前行的值与该行 window 的捕获值进行比较。我使用的部分逻辑要求在每一行的迭代期间将当前行的值从捕获的值中排除。
我认为有两种方法可以解决这个问题:1. 我可以将当前行的值直接传递到我在 slide_index(.f) 中使用的自定义函数逻辑中,或者 2.我可以从为行的滑动 window 捕获的值中排除当前行的值。第一条路线我找不到任何资源,所以我想知道第二条路线是否可行
library(slider)
x <- c(rep(1:16))
i <- as.Date("2019-08-15") + c(0:15)
slide_index(x, i, ~.x, .before = 2, .after = 2)
例如,在 x = 2 的第二次迭代中,从 slide_index() 输出上方的可重现代码将是:[1] 1 2 3 4。但我想要输出到 return only [1] 1 3 4,或者找到一种方法将当前 x 值读入我传递给 slide_index(.f)
的自定义函数
编辑:
group_by
的第二个例子
library(slider)
library(tidyverse)
x <- c(10, 12, 12, 14, 11, 22, 25, 25, 33, 31, 34, 36, 23, 24, 29, 13)
y <- c(10, 12, 12, 14, 11, 22, 25, 25, 33, 31, 34, 36, 23, 24, 29, 13)
group <- c(rep(as.character('A'), 5), rep(as.character('B'), 4), rep(as.character('C'), 1), rep(as.character('D'), 6))
df <- data.frame(group, x, y, dates) %>%
mutate(group = as.factor(group))
df %>%
group_by(group) %>%
do(mutate(., result = slide_index(.x = ., .i = .$dates, ~median(.$y), .before = 2, .after = 2)
%>% unlist()
)
)
我也试过这个但是没用
df %>%
group_by(group) %>%
do(mutate(., result = max(map2(slide_index(.x = seq_along(.), .i = .$dates, .f = ~.$y, .before = 2, .after = 2), seq_along(.), ~.[setdiff(.x, .y)] )
)
)
)
如果有重复,用值序列做索引,然后用map2
删除那个索引并得到相应的值
library(purrr)
library(slider)
i1 <- seq_along(x)
map2(slide_index(i1, i, ~.x, .before = 2, .after = 2),
i1, ~ x[setdiff(.x, .y)])
-输出
[[1]]
[1] 2 2
[[2]]
[1] 1 2 4
[[3]]
[1] 1 2 4 5
[[4]]
[1] 2 2 5 6
[[5]]
[1] 2 4 6 7
[[6]]
[1] 4 5 7 8
[[7]]
[1] 5 6 8 9
[[8]]
[1] 6 7 9 10
[[9]]
[1] 7 8 10 11
[[10]]
[1] 8 9 11 12
[[11]]
[1] 9 10 12 13
[[12]]
[1] 10 11 13 14
[[13]]
[1] 11 12 14 15
[[14]]
[1] 12 13 15 16
[[15]]
[1] 13 14 16
[[16]]
[1] 14 15
更新
如果有分组数据,那么我们做一个group by操作,创建一个list
列
dates <- i
df %>%
group_by(group) %>%
mutate(new = map2(slide_index(row_number(), dates, ~ .x,
.before = 2, .after = 2), row_number(), ~ x[setdiff(.x, .y)])) %>%
ungroup
# A tibble: 16 x 5
# group x y dates new
# <fct> <dbl> <dbl> <date> <list>
# 1 A 10 10 2019-08-15 <dbl [2]>
# 2 A 12 12 2019-08-16 <dbl [3]>
# 3 A 12 12 2019-08-17 <dbl [4]>
# 4 A 14 14 2019-08-18 <dbl [3]>
# 5 A 11 11 2019-08-19 <dbl [2]>
# 6 B 22 22 2019-08-20 <dbl [2]>
# 7 B 25 25 2019-08-21 <dbl [3]>
# 8 B 25 25 2019-08-22 <dbl [3]>
# 9 B 33 33 2019-08-23 <dbl [2]>
#10 C 31 31 2019-08-24 <dbl [0]>
#11 D 34 34 2019-08-25 <dbl [2]>
#12 D 36 36 2019-08-26 <dbl [3]>
#13 D 23 23 2019-08-27 <dbl [4]>
#14 D 24 24 2019-08-28 <dbl [4]>
#15 D 29 29 2019-08-29 <dbl [3]>
#16 D 13 13 2019-08-30 <dbl [2]>
数据
x <- c(c(1, 2, 2),rep(4:16))
i <- as.Date("2019-08-15") + c(0:15)
df <- structure(list(group = structure(c(1L, 1L, 1L, 1L, 1L, 2L, 2L,
2L, 2L, 3L, 4L, 4L, 4L, 4L, 4L, 4L), .Label = c("A", "B", "C",
"D"), class = "factor"), x = c(10, 12, 12, 14, 11, 22, 25, 25,
33, 31, 34, 36, 23, 24, 29, 13), y = c(10, 12, 12, 14, 11, 22,
25, 25, 33, 31, 34, 36, 23, 24, 29, 13), dates = structure(c(18123,
18124, 18125, 18126, 18127, 18128, 18129, 18130, 18131, 18132,
18133, 18134, 18135, 18136, 18137, 18138), class = "Date")),
class = "data.frame", row.names = c(NA,
-16L))
您好,我正在使用 slide_index() 在时间 window 中为数据帧中的每一行捕获值。但是,对于每一行,我还将当前行的值与该行 window 的捕获值进行比较。我使用的部分逻辑要求在每一行的迭代期间将当前行的值从捕获的值中排除。
我认为有两种方法可以解决这个问题:1. 我可以将当前行的值直接传递到我在 slide_index(.f) 中使用的自定义函数逻辑中,或者 2.我可以从为行的滑动 window 捕获的值中排除当前行的值。第一条路线我找不到任何资源,所以我想知道第二条路线是否可行
library(slider)
x <- c(rep(1:16))
i <- as.Date("2019-08-15") + c(0:15)
slide_index(x, i, ~.x, .before = 2, .after = 2)
例如,在 x = 2 的第二次迭代中,从 slide_index() 输出上方的可重现代码将是:[1] 1 2 3 4。但我想要输出到 return only [1] 1 3 4,或者找到一种方法将当前 x 值读入我传递给 slide_index(.f)
的自定义函数编辑: group_by
的第二个例子library(slider)
library(tidyverse)
x <- c(10, 12, 12, 14, 11, 22, 25, 25, 33, 31, 34, 36, 23, 24, 29, 13)
y <- c(10, 12, 12, 14, 11, 22, 25, 25, 33, 31, 34, 36, 23, 24, 29, 13)
group <- c(rep(as.character('A'), 5), rep(as.character('B'), 4), rep(as.character('C'), 1), rep(as.character('D'), 6))
df <- data.frame(group, x, y, dates) %>%
mutate(group = as.factor(group))
df %>%
group_by(group) %>%
do(mutate(., result = slide_index(.x = ., .i = .$dates, ~median(.$y), .before = 2, .after = 2)
%>% unlist()
)
)
我也试过这个但是没用
df %>%
group_by(group) %>%
do(mutate(., result = max(map2(slide_index(.x = seq_along(.), .i = .$dates, .f = ~.$y, .before = 2, .after = 2), seq_along(.), ~.[setdiff(.x, .y)] )
)
)
)
如果有重复,用值序列做索引,然后用map2
删除那个索引并得到相应的值
library(purrr)
library(slider)
i1 <- seq_along(x)
map2(slide_index(i1, i, ~.x, .before = 2, .after = 2),
i1, ~ x[setdiff(.x, .y)])
-输出
[[1]]
[1] 2 2
[[2]]
[1] 1 2 4
[[3]]
[1] 1 2 4 5
[[4]]
[1] 2 2 5 6
[[5]]
[1] 2 4 6 7
[[6]]
[1] 4 5 7 8
[[7]]
[1] 5 6 8 9
[[8]]
[1] 6 7 9 10
[[9]]
[1] 7 8 10 11
[[10]]
[1] 8 9 11 12
[[11]]
[1] 9 10 12 13
[[12]]
[1] 10 11 13 14
[[13]]
[1] 11 12 14 15
[[14]]
[1] 12 13 15 16
[[15]]
[1] 13 14 16
[[16]]
[1] 14 15
更新
如果有分组数据,那么我们做一个group by操作,创建一个list
列
dates <- i
df %>%
group_by(group) %>%
mutate(new = map2(slide_index(row_number(), dates, ~ .x,
.before = 2, .after = 2), row_number(), ~ x[setdiff(.x, .y)])) %>%
ungroup
# A tibble: 16 x 5
# group x y dates new
# <fct> <dbl> <dbl> <date> <list>
# 1 A 10 10 2019-08-15 <dbl [2]>
# 2 A 12 12 2019-08-16 <dbl [3]>
# 3 A 12 12 2019-08-17 <dbl [4]>
# 4 A 14 14 2019-08-18 <dbl [3]>
# 5 A 11 11 2019-08-19 <dbl [2]>
# 6 B 22 22 2019-08-20 <dbl [2]>
# 7 B 25 25 2019-08-21 <dbl [3]>
# 8 B 25 25 2019-08-22 <dbl [3]>
# 9 B 33 33 2019-08-23 <dbl [2]>
#10 C 31 31 2019-08-24 <dbl [0]>
#11 D 34 34 2019-08-25 <dbl [2]>
#12 D 36 36 2019-08-26 <dbl [3]>
#13 D 23 23 2019-08-27 <dbl [4]>
#14 D 24 24 2019-08-28 <dbl [4]>
#15 D 29 29 2019-08-29 <dbl [3]>
#16 D 13 13 2019-08-30 <dbl [2]>
数据
x <- c(c(1, 2, 2),rep(4:16))
i <- as.Date("2019-08-15") + c(0:15)
df <- structure(list(group = structure(c(1L, 1L, 1L, 1L, 1L, 2L, 2L,
2L, 2L, 3L, 4L, 4L, 4L, 4L, 4L, 4L), .Label = c("A", "B", "C",
"D"), class = "factor"), x = c(10, 12, 12, 14, 11, 22, 25, 25,
33, 31, 34, 36, 23, 24, 29, 13), y = c(10, 12, 12, 14, 11, 22,
25, 25, 33, 31, 34, 36, 23, 24, 29, 13), dates = structure(c(18123,
18124, 18125, 18126, 18127, 18128, 18129, 18130, 18131, 18132,
18133, 18134, 18135, 18136, 18137, 18138), class = "Date")),
class = "data.frame", row.names = c(NA,
-16L))