使用Nas值最大化R中的函数

Question

我想在代码中最大化funcToOpt。

数据描述：

1. wb和X1是T x N矩阵，Nt是长度为N的向量
2. wb的每一行的总和为1
3. X1是横截面标准化的（平均值= 1 sd = 0）

标准化意味着theta*X1的横截面平均值为零，这意味着rowSums(wb + (theta1*X1) / Nt)始终为1。 当我使用没有NAs值的X1矩阵时，这种情况发生在我身上，是否有人知道在使用带有NA的矩阵时如何将wi的行和等于1？ /强>

最大化代码：

X1 <- t(scale(t((X1))))    # X1 STANDARDIZATION
N <- dim(X1)[2]
T <- dim(X1)[1]

M <- wb + r + X1                  # Nt, counting the NAs values
Nt <- rowSums(!is.na(M)) 

funcToOpt <- function (theta1)
{
  inner <- rowSums((wb + (theta1*X1) / Nt) * r,na.rm = TRUE)  

  return (sum(inner, na.rm=TRUE) / T)
}

max<-optimize(funcToOpt, c(-50, 50), maximum = TRUE)

theta1=max[[1]]                                 

wi=wb+(theta1*X1) / Nt     
rowSums(wi,na.rm=T)

没有NA的MATRIX 5X5

r=matrix(c(0.1,0.3,0.4,0.3,0.4,0.5,0.3,0.3,0.2,0.1,0.1,0.1,0.1,0.1,0.2,0.9,0.3,0.2,0.1,0.2,0.2,0.4,0.4,0.5,0.1),ncol = 5)
wb=matrix(c(0.3,0.3,-0.4,-0.3,0.4,-0.5,-0.3,0.3,0.2,0.1,0.1,0.1,0.1,0.1,0.2,0.9,0.3,0.6,0.5,0.2,0.2,0.6,0.4,0.5,0.1),ncol = 5)
X1=matrix(c(1:20,0,4,5,0,3),ncol = 5)

MATRIX 5x5，NA

r=matrix(c(0.1,0.3,0.4,0.3,0.4,0.5,0.3,0.3,0.2,0.1,0.1,0.1,0.1,0.1,0.2,0.9,0.3,0.2,0.1,0.2,0.2,0.4,0.4,0.5,0.1),ncol = 5)
wb=matrix(c(0.3,0.3,-0.4,-0.3,0.4,-0.5,-0.3,0.3,0.2,0.1,0.1,0.1,0.1,0.1,0.2,0.9,0.3,0.6,0.5,0.2,0.2,0.6,0.4,0.5,0.1),ncol = 5)
X1=matrix(c(1:20,NA,4,5,NA,3),ncol = 5)