Question

我有一个带有距离矩阵的列表：

obs <- list(AA=list(A=dist(runif(100)),
                 B=dist(runif(100)),
                 C=dist(runif(100))),
            BB=list(A=dist(runif(100)),
                    B=dist(runif(100)),
                    C=dist(runif(100))))

obs <- lapply(obs, function(x)
  lapply(x, function(x) as.data.frame(as.matrix(x))))

还有一个只有一个等级的人：

distances <- lapply(list(A=rnorm(100),B=rnorm(100),C=rnorm(100)), function(x)
  as.data.frame(as.matrix(dist(x, "euclidean"))))

我想比较obs第二级和distances第一级中的所有矩阵，如果它们的名称匹配（obs[i]$A与distances$A， B与B，C与C；永远不会所有组合！）。尝试在lapply中运行sapply失败后，我进入了for循环，在此期间我存储结果并从中提取一些值：

coef <- pvals <- res <- vector("list", length(names(obs)))
library(vegan)
for(i in (names(obs)){ 
  res[[i]]$A  <- mantel(obs[[i]]$A, distances$A, "spearman", perm=999)
  #tmp <-  res[[i]]$A
  #coef[i]$A <- tmp[i]$statistic
  #pvals[i]$A <- tmp[i]$signic
}

我遍历第一级obs并修复第二级，然后针对obs[[i]]$A和B进行此操作（不要从上方再次粘贴＃行以节省空间）：

    for(i in (names(obs)){ 
      res[[i]]$B  <- mantel(obs[[i]]$B, distances$B, "spearman", perm=999)
      ...
    }
    for(i in (names(obs)){ 
      res[[i]]$C  <- mantel(obs[[i]]$C, distances$C, "spearman", perm=999)
      ...
    }

现在的问题是也要遍历第二级(obs[i]$A,B,C)，同时还要指向distances中的正确矩阵。将上面的三个循环放到一个家长循环中（通过obs [i] [j]）会更好还是有办法使用lapply？谢谢！

Answer 1

如果我正确理解了您的问题，我会做这样的事情：

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遍历列表中的两个变量

1 个答案: