R矩阵应用功能

Question

我有以下示例矩阵x。

x <- data.frame(c1=c(1,2,3,2,1,3),
                    c2=c(4,5,6,2,3,4),
                    c3=c(7,8,9,7,1,6),
                    c4=c(4,0,9,1,5,0),
                    c5=c(3,8,0,7,3,6),
                    c6=c(2,8,5,0,5,7),
                    row.names = c("r1","r2","r3","r4","r5","r6"))

我需要将函数f应用于每列，其中cMin是列最小值，cMax是列最大值向量。

cMax <- colMaxs(mat)
cMin <- colMins(mat)

我正在尝试使用Apply函数apply(mat,2,f) ，如下所示，但收到警告，结果也不正确。

f <- function(x) (x - cMin[])/(cMax - cMin)

警告：警告消息：

1: In x - cMin[] :
  longer object length is not a multiple of shorter object length
2: In (x - cMin[])/(cMax - cMin) :
  longer object length is not a multiple of shorter object length
3: In x - cMin[] :
  longer object length is not a multiple of shorter object length
4: In (x - cMin[])/(cMax - cMin) :
  longer object length is not a multiple of shorter object length

有人可以解释如何使用由向量（cMin或cMax）组成的apply函数吗？

Answer 1

从矩阵中减去向量时，由于矩阵的存储机制和回收规则，向量按列对齐。 因此，您可以转置matrix ，使用cMin ， cMax进行计算，然后将其转回：

t((t(mat) - cMin)/(cMax - cMin))

#    c1   c2    c3        c4    c5    c6
#r1 0.0 0.50 0.750 0.4444444 0.375 0.250
#r2 0.5 0.75 0.875 0.0000000 1.000 1.000
#r3 1.0 1.00 1.000 1.0000000 0.000 0.625
#r4 0.5 0.00 0.750 0.1111111 0.875 0.000
#r5 0.0 0.25 0.000 0.5555556 0.375 0.625
#r6 1.0 0.50 0.625 0.0000000 0.750 0.875

Answer 2

library(magrittr)
x <- data.frame(c1=c(1,2,3,2,1,3),
                c2=c(4,5,6,2,3,4),
                c3=c(7,8,9,7,1,6),
                c4=c(4,0,9,1,5,0),
                c5=c(3,8,0,7,3,6),
                c6=c(2,8,5,0,5,7),
                row.names = c("r1","r2","r3","r4","r5","r6"))

cMin <- apply(x, MARGIN = 2, FUN = min)
cMax <- apply(x, MARGIN = 2, FUN = max)

sweep(x, MARGIN = 2, STATS = cMin, FUN = "-") %>%
  sweep(., MARGIN = 2, STATS = (cMax - cMin), FUN = "/")

    c1   c2    c3        c4    c5    c6
r1 0.0 0.50 0.750 0.4444444 0.375 0.250
r2 0.5 0.75 0.875 0.0000000 1.000 1.000
r3 1.0 1.00 1.000 1.0000000 0.000 0.625
r4 0.5 0.00 0.750 0.1111111 0.875 0.000
r5 0.0 0.25 0.000 0.5555556 0.375 0.625
r6 1.0 0.50 0.625 0.0000000 0.750 0.875

Answer 3

从解决方案中我看到，其目标是使每一列线性地缩放到0到1的范围，最小值映射到0，最大值映射到1。

cMin一行，而无需计算cMin和cMax

apply(x, 2, 
      function(each_col) (each_col - min(each_col))/diff(range(each_col)))

# c1   c2    c3        c4    c5    c6
# r1 0.0 0.50 0.750 0.4444444 0.375 0.250
# r2 0.5 0.75 0.875 0.0000000 1.000 1.000
# r3 1.0 1.00 1.000 1.0000000 0.000 0.625
# r4 0.5 0.00 0.750 0.1111111 0.875 0.000
# r5 0.0 0.25 0.000 0.5555556 0.375 0.625
# r6 1.0 0.50 0.625 0.0000000 0.750 0.875

Answer 4

我们可以复制“ cMin”和“ cMax”并进行计算

(mat - cMin[col(mat)])/(cMax[col(mat)] - cMin[col(mat)])
#    c1   c2    c3        c4    c5    c6
#r1 0.0 0.50 0.750 0.4444444 0.375 0.250
#r2 0.5 0.75 0.875 0.0000000 1.000 1.000
#r3 1.0 1.00 1.000 1.0000000 0.000 0.625
#r4 0.5 0.00 0.750 0.1111111 0.875 0.000
#r5 0.0 0.25 0.000 0.5555556 0.375 0.625
#r6 1.0 0.50 0.625 0.0000000 0.750 0.875