R 数据表:从数值中减去水平意味着
我想知道如何从R 数据表:从数值中减去水平意味着,r,dataframe,data.table,R,Dataframe,Data.table,我想知道如何从data.table中的值中减去levels means。我的MWE如下所示 set.seed(12345) A <- rep(x=paste0("A", 1:2), each=6) B <- rep(x=paste0("B", 1:3), each=2, times=2) Rep <- rep(x=1:2, times=3) Y <- rnorm(n=12, mean = 50, sd = 5) library(data.table) dt <-
data.table
中的值中减去levels means。我的MWE如下所示
set.seed(12345)
A <- rep(x=paste0("A", 1:2), each=6)
B <- rep(x=paste0("B", 1:3), each=2, times=2)
Rep <- rep(x=1:2, times=3)
Y <- rnorm(n=12, mean = 50, sd = 5)
library(data.table)
dt <- data.table(A, B, Rep, Y)
dt[, j=mean(Y), by=.(A, B)]
dt[, j=mean(Y), by=.(A)]
dt[, j=mean(Y), by=.(A, B)] - dt[, j=mean(Y), by=.(A)]
Error in Ops.data.frame(dt[, j = mean(Y), by = .(A, B)], dt[, j = mean(Y), :
‘-’ only defined for equally-sized data frames
得到错误的原因是两个数据表的维度不同。另一方面,您可以使用
数据进行链式转换。table
:
dt[, j := mean(Y), .(A, B)][, j := j - mean(Y), .(A)]
dt
A B Rep Y j
1: A1 B1 1 52.92764 3.6373822
2: A1 B1 2 53.54733 3.6373822
3: A1 B2 1 49.45348 -1.0071061
4: A1 B2 2 47.73251 -1.0071061
5: A1 B3 1 53.02944 -2.6302761
6: A1 B3 2 40.91022 -2.6302761
7: A2 B1 1 53.15049 0.1752053
8: A2 B1 2 48.61908 0.1752053
9: A2 B2 1 48.57920 -3.7182851
10: A2 B2 2 45.40339 -3.7182851
11: A2 B3 1 49.41876 3.5430798
12: A2 B3 2 59.08656 3.5430798
对于更新,您可以执行以下操作:
dt[, j := mean(Y), .(A, B)][, j := j - mean(Y), .(A)][, j := j - mean(Y), .(B)][, j := j + mean(Y)]
dt
A B Rep Y j
1: A1 B1 1 52.92764 1.731088
2: A1 B1 2 53.54733 1.731088
3: A1 B2 1 49.45348 1.355590
4: A1 B2 2 47.73251 1.355590
5: A1 B3 1 53.02944 -3.086678
6: A1 B3 2 40.91022 -3.086678
7: A2 B1 1 53.15049 -1.731088
8: A2 B1 2 48.61908 -1.731088
9: A2 B2 1 48.57920 -1.355590
10: A2 B2 2 45.40339 -1.355590
11: A2 B3 1 49.41876 3.086678
12: A2 B3 2 59.08656 3.086678
谢谢@Psidom的回答。想知道如何做这样复杂的计算
dt[,j=mean(Y),by=(A,B)]-dt[,j=mean(Y),by=(A)]-dt[,j=mean(Y),by=(B)]+dt[,j=mean(Y)]
。同样地,谢谢你,如果你把所有的业务都联系在一起,你应该能够得到你想要的。例如dt[,j:=mean(Y),(A,B)][,j:=j-mean(Y),(A)][,j:=j-mean(Y),(B)][,j:=j+mean(Y)]
。注意:OP使用j
作为[.data.table
的命名参数,但实际上您正在创建一个名为j
的列,这是一些人(像我一样)所做的可能会觉得困惑。“弗兰克。我看到了你关心的问题。然而,一个命名为“代码”> GROPBY 的结果将导致不同的维数,这使得减法变得困难。这就是为什么我认为创建一个新的向量更直接的方式。是的,我赞成并认为这是正确的方法,我只是想找到另一个名字。,这就是我的意思。
dt[, j := mean(Y), .(A, B)][, j := j - mean(Y), .(A)][, j := j - mean(Y), .(B)][, j := j + mean(Y)]
dt
A B Rep Y j
1: A1 B1 1 52.92764 1.731088
2: A1 B1 2 53.54733 1.731088
3: A1 B2 1 49.45348 1.355590
4: A1 B2 2 47.73251 1.355590
5: A1 B3 1 53.02944 -3.086678
6: A1 B3 2 40.91022 -3.086678
7: A2 B1 1 53.15049 -1.731088
8: A2 B1 2 48.61908 -1.731088
9: A2 B2 1 48.57920 -1.355590
10: A2 B2 2 45.40339 -1.355590
11: A2 B3 1 49.41876 3.086678
12: A2 B3 2 59.08656 3.086678