带有 rowSums 函数的 for 循环答案

【问题标题】：A for loop with rowSums function带有 rowSums 函数的 for 循环
【发布时间】：2017-09-20 15:58:50
【问题描述】：

我是 R 的初学者，我编写了一个双循环来计算 chi2 值，以便在 6610 个术语和 10 个类中选择特征。这是我的 for 循环：

library(raster)
  #for x^2 [n,r] = term n, class r. n starts from col #7 and r starts from col #6617

  chi2vals <- matrix(0:0,6610,10)
  chi2avgs <- vector("numeric",6610L)

  for(r in 1:10){
    for(n in 1:6610){

      A = sum(data1.sub.added[,6+n]==1 & data1.sub.added[,6616+r]==1)
      M = sum(data1.sub.added[,6+n]==1)
      P = sum(data1.sub.added[,6616+r]==1)
      N = nrow(data1.sub.added)
      E = ((A*N)-(M*P))**2
      F = (N-P)*(N-M)
      chi2vals[n,r] = (N/(P*M))*(E/F) # for term n
    }

    Prcj = sum(data1.sub.added[,6616+r]==1)/sum(data1.sub.added[,6616:6626]==1) #probability of class c_r
    pchi <- Prcj * chi2vals
    chi2avgs[n] = rowSums(pchi)[n]
  }

代码正确计算直到pchi <- Prcj * chi2vals 行的所有内容。结果是一个很好的 p*chi2 值矩阵：

> head(pchi)
             [,1]        [,2]         [,3]      [,4]       [,5]         [,6]       [,7]        [,8]         [,9]
[1,] 128.36551442 0.239308113  0.683517530 1.5038665  0.6145058 3.656857e-01  1.3311564   2.6977448  0.410702803
[2,]   0.06632758 0.067970859  0.019178551 0.2900692  1.5300639 4.430705e-08  0.2599859   0.6362953  0.098745147
[3,]   1.85641330 1.411925435  3.590747764 7.3018416 38.8044465 4.102248e-01  6.4118078  13.0164994  1.709506238
[4,]   0.11063892 0.005039029  0.244964758 0.1622654  0.1156411 8.274468e+00  0.2564959   0.0577651  0.242946022
[5,]   0.04788648 0.049072885  0.001420669 0.2094211  1.7200152 2.045923e-01  0.1877019   0.1468187  0.005493183
[6,]   5.39946188 6.899336618 60.735646913 7.4351538 10.7005784 9.946261e+00 35.8868899 178.7112406 11.382740754
           [,10]
[1,]  0.26436516
[2,]  0.14414444
[3,]  0.90292073
[4,]  0.01168997
[5,]  0.06641298
[6,] 19.68599142

但最终的 chi2avgs 值大多是零：

> head(chi2avgs)
[1] 0.000000   0.000000   0.000000   0.000000   2.638835   0.000000

但是，除了循环之外，我将 n 替换为任意数字时，最后一行效果很好：

  chi2avgs[1] = rowSums(pchi)[1]
  chi2avgs[2] = rowSums(pchi)[2]
  chi2avgs[3] = rowSums(pchi)[3]
  chi2avgs[4] = rowSums(pchi)[4]
  chi2avgs[5] = rowSums(pchi)[5]

> head(chi2avgs)
[1] 136.476367   3.112781  75.416334   9.481914   2.638835   0.000000

我想知道是什么导致了这个问题。你知道我该如何解决它吗？

【问题讨论】：

标签： r variables for-loop

【解决方案1】：

你可以直接尝试不带[n]的rowsums

chi2avgs = rowSums(pchi)

【讨论】：