用R中的矢量劃分數據幀或矩陣

Question

這似乎不應該太困難，但是我對此很難接受。比方說，比如我有以下數據幀：

set.seed(99) 
data <- data.frame(Names=rep(c('A','B'),5), 
        First = rnorm(10), 
        Second = rnorm(10), 
        Third = rnorm(10)) 

我要的是用平均A的這個我可以通過計算來劃分整個數據幀：

a.mean < - sapply（數據[$名稱==「A」，2：4]，平均）

但是當我嘗試通過矢量像這樣把整個數據幀我沒有得到正確的價值觀：

normalized.data <- data[2:4]/a.mean 
normalized.data$Names <- data$Names 
sapply(data[normalized.data$Names == 'A', 2:4], mean) 

First  Second  Third 
0.2578018 -0.5864073 0.1156760 

Wheras我想要的是A的歸一化平均數現在等於1.是否有這樣做的方法？

Answer 1

set.seed(99) 
data <- data.frame(Names=rep(c('A','B'),5), 
        First = rnorm(10), 
        Second = rnorm(10), 
        Third = rnorm(10)) 

a.mean <- sapply(data[data$Names == 'A', 2:4], mean) 
data[,2:4] <- sweep(data[,2:4],MARGIN=2,a.mean,"/") 

(norm.mean <- sapply(data[data$Names == 'A', 2:4], mean)) 
## First Second Third 
##  1  1  1 

根據您的應用程序，它可能更容易使Names列是該行的名稱改爲：

data <- data.frame(First = rnorm(10), 
        Second = rnorm(10), 
        Third = rnorm(10), 
        row.names=rep(c('A','B'),5)) 

我也喜歡subset(data,Names=='A')的可讀性（雖然不建議用於編程：看https://github.com/hadley/devtools/wiki/Evaluation ）

Answer 2

set.seed(99) 
data <- data.frame(Names=rep(c('A','B'),5), 
        First = rnorm(10), 
        Second = rnorm(10), 
        Third = rnorm(10)) 

a.mean <- colMeans(data[data$Names == 'A', 2:4]) 


normalized.data <- as.data.frame(t(t(data[,2:4])/a.mean)) 


normalized.data$Names <- data$Names 
colMeans(normalized.data[normalized.data$Names == 'A', 1:3]) 

#First Second Third 
#1  1  1 

Answer 3

哦，別介意：你沒有按照自己認爲的方式劃分。將矩陣除以值的向量不會將每列除以給定的值。

Rgames> foo 
    [,1] [,2] [,3] 
[1,] 5 3 7 
[2,] 5 3 7 
[3,] 5 3 7 
[4,] 5 3 7 
[5,] 5 3 7 
Rgames> foo/c(1,2,3) 
     [,1] [,2]  [,3] 
[1,] 5.000000 1.0 3.500000 
[2,] 2.500000 3.0 2.333333 
[3,] 1.666667 1.5 7.000000 
[4,] 5.000000 1.0 3.500000 
[5,] 2.500000 3.0 2.333333