Question

我正在使用“用户功能指数指南”中的脚本来计算功能多样性指数。但是，该脚本充满了错误和问题。我已经设法让脚本运行但它产生了错误的答案。

它使用三个矩阵，如下所示：

丰度：

        S1 S2 S3 S4 S5 S6 S7 S8 S9
Palm    6  3  0  1  0 16  0  2  3
Forest  2  0  2  1  2  1  3  0  2

minsp：

maxsp：

我认为它应该产生两个不同的数字，具体取决于两个网站Palm和Forest。相反，它产生两个完全相同的数字。我看过它，发现问题从这些代码行开始变为蒸汽：

Siteminsp <- abundance*minsp 
Siteminsp <- minsp[apply(Siteminsp,1,function (x) sum(x,na.rm=T))!=0, ,drop = FALSE]

Sitemaxsp <- abundance*maxsp 
Sitemaxsp <- maxsp[apply(Sitemaxsp,1,function (x) sum(x,na.rm=T))!=0, ,drop = FALSE]

而不是abundance*minsp产生一个新的矩阵，其中每个物种的最小值乘以每个地点的物种丰度。该函数将第一物种丰度乘以第一物种最小值，然后将第一物种丰度乘以第二个最小值，然后将第二物种丰度乘以第三个值，依此类推。

我是否想念这里应该发生什么？

有人可以解释一下出了什么问题，因为我完全没有想法。如果它有用，这是整个脚本：

rownames(Abundance1) <- Abundance1[,1]
Abundance <- Abundance1[,-1]
Abundance <- Abundance[,order(colnames(Abundance))]

rownames(min1) <- min1[,1]
minsp <- min1[,-1, drop= FALSE]
minsp <- minsp[order(rownames(min1)), , drop = FALSE]

rownames(max1) <- max1[,1]
maxsp <- max1[,-1, drop=FALSE]
maxsp <- maxsp[order(rownames(max1)), , drop = FALSE]

globalFRI <- function(minsp,maxsp) {
  deltaS <- list()

  for (j in 1:ncol(minsp))
  {
    xx <- cbind(minsp[,j], maxsp[,j])
    xx <- xx[apply(xx,1,function(z) sum(is.finite(z))==2),]
    xx <- xx[order(xx[,1]),]
    z <- c(0, nrow(xx))
    i <- 1
    b <- xx[1, 2]
    while ( i < nrow(xx) )
    {
      if (b < xx[i+1,1]) z <- c(z, i)
      b <- ifelse(b >= xx[i+1,2], b, xx[i+1,2])
      i <- i+1
      if (i==nrow(xx)) break
    }
    group <- factor(rep(1:(length(z)-1), diff(sort(z))))
    deltaS[[j]] <- tapply(xx[,2], group, max) - tapply(xx[,1], group, min)
  }

  globalFRIs <- sapply(deltaS, sum)}

globalFRIAll <- globalFRI(minsp,maxsp)

Abundance2 <- Abundance
Abundance2[Abundance2 != 0] <- 1
abundance <- data.matrix(Abundance)

IndexFRIs <- function(Abundance,minsp,maxsp,globalFRI) {

  Siteminsp <- abundance*minsp 
  Siteminsp <- minsp[apply(Siteminsp,1,function (x) sum(x,na.rm=T))!=0, ,drop = FALSE]

  Sitemaxsp <- abundance*maxsp 
  Sitemaxsp <- maxsp[apply(Sitemaxsp,1,function (x) sum(x,na.rm=T))!=0, ,drop = FALSE]

  deltaS <- list()
  for (j in 1:ncol(Siteminsp))
  {
    xx <- cbind(Siteminsp[,j], Sitemaxsp[,j])
    xx <- xx[apply(xx,1,function(z) sum(is.finite(z))==2),]
    xx <- xx[order(xx[,1]),]
    z <- c(0, nrow(xx))
    i <- 1
    b <- xx[1, 2]
    while ( i < nrow(xx) )
    {
      if (b < xx[i+1,1]) z <- c(z, i)
      b <- ifelse(b >= xx[i+1,2], b, xx[i+1,2])
      i <- i+1
      if (i==nrow(xx)) break
    }
    group <- factor(rep(1:(length(z)-1), diff(sort(z))))
    deltaS[[j]] <- tapply(xx[,2], group, max) - tapply(xx[,1], group, min)
  }
  RI <- sapply(deltaS, sum)
  FRI1 <- RI/globalFRI
  FRIs <-  mean(FRI1)

}

FRIs <- apply(Abundance2,1,IndexFRIs,minsp,maxsp,globalFRIAll)

dput（丰度）：

structure(c(6L, 2L, 3L, 0L, 0L, 2L, 1L, 1L, 0L, 2L, 16L, 1L, 
0L, 3L, 2L, 0L, 3L, 2L), .Dim = c(2L, 9L), .Dimnames = list(c("Palm", 
"Forest"), c("S1", "S2", "S3", "S4", "S5", "S6", "S7", "S8", 
"S9")))

dput（minsp）：

structure(list(min = c(25.3038, 19.575, 60.588, 16.2864, 46.104, 
10.9056, 8.757, 2.1289, 4.173)), .Names = "min", row.names = c("S1", 
"S2", "S3", "S4", "S5", "S6", "S7", "S8", "S9"), class = "data.frame")

dput（maxsp）：

structure(list(max = c(44.7344, 22.6966, 75.1817, 17.8176, 50.7472, 
33.366, 14.3341, 3.3947, 10.251)), .Names = "max", row.names = c("S1", 
"S2", "S3", "S4", "S5", "S6", "S7", "S8", "S9"), class = "data.frame")

我认为丰富* minsp的输出应该是这样的：

      Palm    Forest
S1  151.8228  50.6076
S2  58.725    0
S3  0         121.176
S4  16.2864   16.2864
S5  92.208    0
S6  174.489   10.9056
S7  0         26.271
S8  4.2578    0
S9  8.346     12.519

Answer 1

您实际上并不是在寻找矩阵乘法，而是将两个向量相乘：

> apply(abundance, 1, `*`, unlist(minsp))
#       Palm   Forest
#S1 151.8228  50.6076
#S2  58.7250   0.0000
#S3   0.0000 121.1760
#S4  16.2864  16.2864
#S5   0.0000  92.2080
#S6 174.4896  10.9056
#S7   0.0000  26.2710
#S8   4.2578   0.0000
#S9  12.5190   8.3460

甚至更好：

t(abundance) * unlist(minsp)

一个矩阵问题的乘法

1 个答案: