如何做明智的累计和？

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 

从上面的矩阵我想找到最大和最小单元格列明智的单元格，然后做每两个邻居的明智的累计和。例如;对于第1列，最小值为-68.70（识别单元格）并添加（-70.2028和-71.329）等等，最小值和最大值相同。我会怎么做？

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

matrix=data.matrix(data) 

Answer 1

这是你在找什么？ （警告：使用不良编码做法的hack-y回答）

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min，第二行max。

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]) 
    }