使用dplyr计算变量变化的更简单方法？ [英] Easier way to calculate change in variable using dplyr?
问题描述
我正在尝试找到一种更简单的方法来计算使用 dplyr 的数据帧中的变量（以列表示）的变化。我的玩具数据集是这样的
I am trying to find an easier way to calculate change in a variable (represented by a column) in a data frame using dplyr. My toy data set is something like this
structure(list(CAR = structure(c(1L, 2L, 3L, 4L, 5L, 6L, 1L,
2L, 3L, 4L, 5L, 6L, 1L, 2L, 3L, 4L, 5L, 6L), .Label = c("a",
"b", "c", "d", "e", "f"), class = "factor"), TIME = c(0L, 0L,
0L, 0L, 0L, 0L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L
), VAR = c(20L, 30L, 40L, 50L, 60L, 70L, 30L, 40L, 50L, 60L,
70L, 80L, 40L, 50L, 60L, 70L, 80L, 90L)), .Names = c("CAR", "TIME",
"VAR"), class = "data.frame", row.names = c(NA, -18L))
看起来像
CAR TIME VAR
1 a 0 20
2 b 0 30
3 c 0 40
4 d 0 50
5 e 0 60
6 f 0 70
7 a 1 30
8 b 1 40
9 c 1 50
10 d 1 60
11 e 1 70
12 f 1 80
13 a 2 40
14 b 2 50
15 c 2 60
16 d 2 70
17 e 2 80
18 f 2 90
我正在尝试计算 VAR 在 TIME 等于 0 与其他时间之间，例如 1,2 CAR c $ c>。
I am trying to calculate change in VAR between TIME equal to 0 and other times, e.g., 1,2 for each CAR.
这就是我要做的，首先，这似乎很复杂我在时间得到等于 VAR 的值 0
This is what I do, which seems a very convoluted way, first I get values of VAR at TIME equals to 0
library(dplyr)
X <- local_test %>% filter(TIME == 0) %>% group_by(CAR) %>% mutate(baseline_VAR = VAR)
X 看起来
Source: local data frame [6 x 4]
Groups: CAR
CAR TIME VAR baseline_VAR
1 a 0 20 20
2 b 0 30 30
3 c 0 40 40
4 d 0 50 50
5 e 0 60 60
6 f 0 70 70
然后，我 left_join 与原始数据框 local_tes t
then, I do a left_join with the original data frame local_test
Y <- left_join(local_test, X, by = c("CAR"))
Y 看起来像
CAR TIME.x VAR.x TIME.y VAR.y baseline_VAR
1 a 0 20 0 20 20
2 b 0 30 0 30 30
3 c 0 40 0 40 40
4 d 0 50 0 50 50
5 e 0 60 0 60 60
6 f 0 70 0 70 70
7 a 1 30 0 20 20
8 b 1 40 0 30 30
9 c 1 50 0 40 40
10 d 1 60 0 50 50
11 e 1 70 0 60 60
12 f 1 80 0 70 70
13 a 2 40 0 20 20
14 b 2 50 0 30 30
15 c 2 60 0 40 40
16 d 2 70 0 50 50
17 e 2 80 0 60 60
18 f 2 90 0 70 70
最后，我在 Y 中添加一列，该列计算两个不同变量之间 VAR 的变化 CAR的时间
finally, I add a column in Y, which calculates the change in VAR between two different TIME for CAR
Y %>% group_by(CAR) %>% mutate(change_VAR = VAR.x - baseline_VAR)
最终 Y 看起来
Source: local data frame [18 x 7]
Groups: CAR
CAR TIME.x VAR.x TIME.y VAR.y baseline_VAR change_VAR
1 a 0 20 0 20 20 0
2 b 0 30 0 30 30 0
3 c 0 40 0 40 40 0
4 d 0 50 0 50 50 0
5 e 0 60 0 60 60 0
6 f 0 70 0 70 70 0
7 a 1 30 0 20 20 10
8 b 1 40 0 30 30 10
9 c 1 50 0 40 40 10
10 d 1 60 0 50 50 10
11 e 1 70 0 60 60 10
12 f 1 80 0 70 70 10
13 a 2 40 0 20 20 20
14 b 2 50 0 30 30 20
15 c 2 60 0 40 40 20
16 d 2 70 0 50 50 20
17 e 2 80 0 60 60 20
18 f 2 90 0 70 70 20
这似乎是很多额外的工作，在原始数据框中添加了额外的列。我需要对大数据帧重复执行此操作。有没有一种更简单（单步）的方法来计算 change_VAR ？
This seems like a lot of extra work, with extra columns being added to the original data frame. I need to do this operation for a large data frame repeatedly. Is there an easier (one-step) way to compute change_VAR?
谢谢！
推荐答案
这可以通过将'VAR'与的 min 取差来实现。
This could be done by taking the difference of 'VAR' with the min of 'VAR' grouped by 'CAR'.
local_test %>%
group_by(CAR) %>%
mutate(change_VAR= VAR- min(VAR))
或者如果'VAR的基值'是当'TIME'为0（假设每个组中没有'TIME'的重复项）时，我们将TIME 0的'VAR'子集化并得到差值。
Or if the base value of 'VAR' is when 'TIME' is 0 (assuming there are no duplicates of 'TIME' per each group), we subset the 'VAR' for TIME 0 and get the difference.
local_test %>%
group_by(CAR) %>%
mutate(change_VAR= VAR- VAR[TIME==0])
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