我用system()函数来调用一个java程序，算出结果后要用结果计算正确率，重复100次，但是这个java程序很耗时间，我想先产生100个文件，通过设置wait=F参数来并行运算100次这个java，但是不知道怎么样修改system()函数来进行这100次不同的命令。code如下：

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#accuracy test<br />
setwd("/users/mg/impute/beagle")<br />
rm(list=ls())<br />
ncpus=4<br />
t=50 #total replicates<br />
storage.inf=matrix(NA,t,1)<br />
data <- read.table("my22k1.bgl",sep="\t",head=F,stringsAsFactors=F)<br />
#data=data[1:100,]<br />
#data <- read.table("hapmap.unphased.bgl",sep=" ",head=F,stringsAsFactors=F)<br />
sfInit(parallel=ncpus>1, cpus=ncpus)<br />
print(sprintf('%s cpus are used', sfCpus()))<br />
for(rep in 1:t){<br />
  ppl=data[,3:ncol(data)]#input data</p>
<p>  s=round(ncol(ppl)*nrow(ppl)/100)<br />
  ppl.nn <- expand.grid(1:ncol(ppl),1:nrow(ppl))<br />
  colnames(ppl.nn) <- c('ppl.col', 'ppl.row')</p>
<p>  smp.id <- sample((nrow(ppl)*ncol(ppl)),s)<br />
  smp.nn <- ppl.nn[smp.id,]<br />
  colnames(smp.nn) <- c('smp.col', 'smp.row')<br />
  smp <- numeric(s)<br />
  for (i in 1:s){<br />
    smp[i] <- ppl[smp.nn[i,2], smp.nn[i,1]]<br />
    ppl[smp.nn[i,2], smp.nn[i,1]] = "?"<br />
  }<br />
  smp1=smp<br />
  ppl1=ppl<br />
  write.table(cbind(data[,1:2],ppl),"test.bgl",sep=" ",quote=F,row.name=F,col.name=F)<br />
  #system("java -Xmx1000m -jar beagle.jar unphased=test.bgl phased=hapmap.phased.bgl markers=hapmap.markers missing=? out=test")<br />
  system("java -Xmx2000m -jar beagle.jar unphased=test.bgl phased=my22k1.bgl markers=marker1.txt missing=? out=test")<br />
  ppl <- read.table("test.test.bgl.phased.gz",sep=" ",head=T,stringsAsFactors=F)<br />
  ppl=merge(data[,1:2],ppl,by.x="V2",by.y="id",sort=F)<br />
  ppl=ppl[,4:ncol(ppl)]<br />
  for (i in 1:s){<br />
    smp[i] <- ppl[smp.nn[i,2], smp.nn[i,1]]<br />
    ppl[smp.nn[i,2], smp.nn[i,1]] = "?"<br />
  }<br />
  ar=length(smp[smp==smp1])/s<br />
  storage.inf[rep,1]=ar<br />
}<br />
sfStop()<br />
ar <- apply(storage.inf,2,mean)<br />
sd <- apply(storage.inf,2,sd)<br />
se=sd/sqrt(length(storage.inf))<br />
save(ar,sd,se,file="1.Rdata")<br />

</p>