关于矩阵计算的一个问题，请高手指教

<br />
myfun <-<br />
function(filename, dlm = "", what = character(0)){<br />
   # determine ncol<br />
   d <- length(scan(file(filename), what=what, nlines = 1, sep=dlm));</p>
<p>   # scan the file into R as a matrix object<br />
   M <- matrix(scan(file(filename), what=what, sep=dlm), ncol=d, byrow=T);</p>
<p>   firstCol <- M[,1];<br />
   laterCol <- matrix(as.numeric(M[,-1]), ncol=d-1, byrow=F);</p>
<p>   aggregate(laterCol, list(group=firstCol), function(x)mean(x,na.rm=T));<br />
}<br />

例如

<br />
> myfun("c:/mydata.txt")<br />

这里假设数据不多。如果数据很大的话，你最好先手动把第一列组名和后面的数值分开成两个文件，然后需要稍微改一下程序。

回复 第4楼 的 camelbbs：If you find a way to separate the first column of group names from the later columns of numbers, say using Excel or some script editor with "column editing mode", and you saved them in two files: mygroupname.txt and mydata.txt (both should contain the same number of rows), then the modified function goes like this

<br />
myfun1 <-<br />
function(groupName, dataFile, dlm = ""){<br />
   # determine ncol<br />
   d <- length(scan(file(dataFile), what=numeric(0), nlines = 1, sep=dlm));</p>
<p>   # scan the file into R as a matrix object<br />
   M <- matrix(scan(file(dataFile), what=numeric(0), sep=dlm), ncol=d, byrow=T);</p>
<p>   firstCol <- matrix(scan(file(groupName), what=character(0), sep=dlm), ncol=1, byrow=T)</p>
<p>   aggregate(M, list(group=firstCol), function(x)mean(x,na.rm=T));<br />
}<br />

<br />
> myfun1("c:/mygroupname.txt","c:/mydata.txt")<br />

回复 第1楼 的 camelbbs：

关键还是要把向量运算用起来：

<br />
d <- read.table(textConnection(<br />
'sox 7.2 3.8 6.8 9.2 5.6<br />
sox 5.4 2.3 4.6 8.9 9.0<br />
sox 6.7 NA 7.8 9.0 3.1<br />
goo 2.4 6.7 NA 9.0 2.1<br />
goo 2.1 5.6 7.8 9.7 1.2<br />
pkk 2.5 4.3 6.5 4.9 0.2<br />
pkk 2.1 3.4 3.2 NA 4.6<br />
pkk 3.2 5.6 6.7 9.1 2.2'))</p>
<p>sapply(split(d[,-1], d[,1]), colMeans, na.rm = T)<br />

主要是空间效率的考虑。因为第一列是字符型，以后是数值型，所以如果一次性不区分地读，会生成一个大矩阵。然后再提取这个矩阵的数值部分后又会生成一个大矩阵，这样会造成两倍的空间消耗。所以最好开始时把第一列和后面分开读，这样就不会重复生成矩阵在内存里了。