如何通过物种矩阵显示站点中的植物物种生物量? [英] How to display plant species biomass in a site by species matrix?

问题描述

我之前问过如何将两列显示为二进制(存在/不存在)矩阵?".这个问题有两个很好的答案.我现在想更进一步，并按物种列在原始站点上添加第三列，以反映每个样地中每个物种的生物量.

I earlier asked "How to display two columns as binary (presence/absence) matrix?". This question received two excellent answers. I would now like to take this a step further and add a third column to the original site by species columns which reflects the biomass of each species in each plot.

第1栏(地块)指定约200个地块的代码，第2栏(种)指定约1200种的代码，第3栏(生物量)指定干重.每个地块都有> 1个物种，每个物种都可以出现在> 1个地块中.总行数约为2700.

Column 1 (plot) specifies code for ~ 200 plots, column 2 (species) specifies code for ~ 1200 species and Column 3 (biomass) specifies the dryweight. Each plot has > 1 species and each species can occur in > 1 plot. The total number of rows is ~ 2700.

> head(dissim)
    plot species biomass
1 a1f56r  jactom 20.2
2 a1f56r  zinunk 10.3
3 a1f56r  mikcor 0.4
4 a1f56r  rubcle 1.3
5 a1f56r  sphoos 12.4
6 a1f56r nepbis1 8.2

tail(dissim)
           plot species biomass
2707 og100m562r  selcup 4.7
2708 og100m562r  pip139 30.5
2709 og100m562r  stasum 0.1
2710 og100m562r  artani 3.4
2711 og100m562r  annunk 20.7
2712 og100m562r  rubunk 22.6

我想创建一个按物种划分的图，显示每个图中每个物种的生物量(而不是二进制存在/不存在矩阵)，其形式如下:

I would like to create a plot by species matrix that displays the biomass of each species in each plot (rather than a binary presence/absence matrix), something of the form:

    jactom  rubcle  chrodo  uncgla
a1f56r  1.3 0   10.3    0
a1f17r  0   22.3    0   4
a1m5r   3.2 0   3.7 9.7
a1m5r   1   0   0   20.1
a1m17r  5.4 6.9 0   1

非常感谢您提供有关如何增加这种附加级别的复杂性的建议.

Any advice on how to add this additional level of complexity would be very much appreciated.

推荐答案

xtabs和tapply函数返回的表是矩阵:

The xtabs and tapply functions return a table which is a matrix:

# Using MrFlick's example
> xtabs(~a+b,dd)
   b
a   f g h i j
  a 0 1 0 2 3
  b 0 0 2 1 0
  c 0 3 0 0 1
  d 2 2 2 1 1
  e 1 1 2 4 1

# --- the tapply solution is a bit less elegant
> dd$one=1
> with(dd, tapply(one, list(a,b), sum))
   f  g  h  i  j
a NA  1 NA  2  3
b NA NA  2  1 NA
c NA  3 NA NA  1
d  2  2  2  1  1
e  1  1  2  4  1

# If you want to make the NA's become zeros then:

> tbl <- with(dd, tapply(one, list(a,b), sum))
> tbl[is.na(tbl)] <- 0
> tbl
  f g h i j
a 0 1 0 2 3
b 0 0 2 1 0
c 0 3 0 0 1
d 2 2 2 1 1
e 1 1 2 4 1

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