Question

X3            X4            X5            X6            X7            X8            X9
-63.2929      -64.9519      -65.9586      -65.2485      -63.3387      -56.749       -51.7419
-68.7098      -70.2028      -71.329       -70.5252      -68.7843      -63.0799      -58.7491
58.3993       -60.0864      -60.7925      -59.6933      -57.2357      49.8444       44.1056
-57.8334      59.6063       60.3079       -59.1852      -56.6365      -49.2706      -43.5538
-37.6082      -39.6347      -39.6647      38.4396       34.9498       -26.6984      -21.9614
37.2886       39.0565       -38.8722      -37.5427      -33.969       25.9074       21.7712
-61.972       -63.0513      63.5082       -62.2337      59.7711       -54.2189      -51.6613
-60.5783      -61.6254      -61.9812      -60.6696      58.066        -52.4656      -49.9653

从上面的矩阵我希望以列方式找到具有最大和最小单元格值的单元格，然后逐行地计算每两个邻居的累积和。例如;对于第1列，最小值为-68.70（标识单元格）并添加（-70.2028和-71.329）等等，最小值和最大值相同。我该怎么做？

data=read.table('test.csv', header=TRUE, sep=',')

matrix=data.matrix(data)

Answer 1

这是你在找什么？（警告：使用错误的编码实践回答hack-y）

data = read.table(text = "X3            X4            X5            X6            X7            X8            X9
-63.2929      -64.9519      -65.9586      -65.2485      -63.3387      -56.749       -51.7419
-68.7098      -70.2028      -71.329       -70.5252      -68.7843      -63.0799      -58.7491
58.3993       -60.0864      -60.7925      -59.6933      -57.2357      49.8444       44.1056
-57.8334      59.6063       60.3079       -59.1852      -56.6365      -49.2706      -43.5538
-37.6082      -39.6347      -39.6647      38.4396       34.9498       -26.6984      -21.9614
37.2886       39.0565       -38.8722      -37.5427      -33.969       25.9074       21.7712
-61.972       -63.0513      63.5082       -62.2337      59.7711       -54.2189      -51.6613
-60.5783      -61.6254      -61.9812      -60.6696      58.066        -52.4656      -49.9653", header = T)

sapply(1:(ncol(data)-2), function(x) {c(sum(data[c(which.min(data[,x])),x:(x+2)]),
                                        sum(data[c(which.max(data[,x])),x:(x+2)]))})

          [,1]     [,2]      [,3]      [,4]      [,5]
[1,] -210.2416 -212.057 -210.6385 -202.3894 -190.6133
[2,]  -62.4796   60.729   61.0456   46.6910  -46.1091

第一行最小，第二行最大

Answer 2

minsum <- c()
maxsum <- c()

temp <- matrix(sample(c(-1,1),replace=T)*70*runif(70), ncol=7)
for(i in 1:(ncol(temp)-2)){
   tmin <- which.min(temp[,i])
   tmax <- which.max(temp[,i])

   minsum <- c(minsum, temp[tmin,i+1] +  temp[tmin,i + 2] )
   maxsum <- c(maxsum, temp[tmax,i+1] +  temp[tmax,i + 2] )
  }

怎么做累积总和明智？

2 个答案: