[英]R: Calculate the column medians by grouping the ID's
繼續上一篇文章 ,現在我想按ID分組(僅適用於第3列),並計算列的中位數(Point_B),然后將列中的每個值(Point_B)減去其中位數。 NA仍應返回。
注意:我希望將ID分組僅應用於Point_B列,而不應用於Point_A,因為我想計算整個Point_A列的中位數,然后將其與Point_A中的值相減。
例如
ID <- c("A","A","A","B","B","B","C","C","C")
Point_A <- c(1,2,NA,1,2,3,1,2,NA)
Point_B <- c(1,2,3,NA,NA,1,1,1,3)
df <- data.frame(ID,Point_A ,Point_B)
+----+---------+---------+
| ID | Point_A | Point_B |
+----+---------+---------+
| A | 1 | 1 |
| A | 2 | 2 |
| A | NA | 3 |
| B | 1 | NA |
| B | 2 | NA |
| B | 3 | 1 |
| C | 1 | 1 |
| C | 2 | 1 |
| C | NA | 3 |
+----+---------+---------+
我以前的帖子中提供的解決方案無需計算ID就可以計算中位數。 這里是
library(dplyr)
df %>%
mutate_each(funs(median=.-median(., na.rm=TRUE)), -ID)
期望的輸出
+----+---------+---------+
| ID | Point_A | Point_B |
+----+---------+---------+
| A | -1 | -1 |
| A | 0 | 0 |
| A | NA | 1 |
| B | -1 | NA |
| B | 0 | NA |
| B | 1 | 0 |
| C | -1 | 0 |
| C | 0 | 0 |
| C | NA | 2 |
+----+---------+---------+
如何通過ID分組獲得Column3中的值?
我猜你想要一個group_by
(遵循@docendodiscimus的建議):
demed <- function(x) x-median(x,na.rm=TRUE)
df %>%
mutate_each(funs(demed),Point_A) %>%
group_by(ID) %>%
mutate_each(funs(demed),Point_B)
給予
ID Point_A Point_B
1 A -1 -1
2 A 0 0
3 A NA 1
4 B -1 NA
5 B 0 NA
6 B 1 0
7 C -1 0
8 C 0 0
9 C NA 2
我更喜歡類似的data.table
代碼。 它的語法要求多次寫入變量名,但括號要少得多:
require(data.table)
DT <- data.table(df)
DT[,Point_A:=demed(Point_A)
][,Point_B:=demed(Point_B)
,by=ID]
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