1

我尝试制作函数图或我的前 20 个基因。这就是为什么我创建了一个数据框列表，这些数据框存在或不同的列包含值和名称。数据框中的这些列之一是基因列。我的代码将前 20 个基因变成了函数图。但是现在我在一些存在少于 20 个基因的数据框中遇到了问题。这会导致我的代码中止。

因为我希望每页最多 5 个函数图，所以我不能只定义一个计数器。

谢谢你的反馈。

我的列表或数据框 listGroups 示例

group1_2: 'data.frame': 68 obs. of 7 variables:
    ..$ p_val: num [1:68] 1.15 1.43 ...
    ..$ score: num [1:68] 15.5 27.14 ...
    ..$ gene: Factor w/ 68 levels "BRA1", "NED",...: 41 52 ...
group2_3: 'data.frame': 3 obs. of 7 variables:
    ..$ p_val: num [1:3] 1.15 1.43 ...
    ..$ score: num [1:3] 15.5 27.14 ...
    ..$ gene: Factor w/ 3 levels "BCL12", "DEF1",...: 41 52 ...

代码

groupNames <- c("cluster1_2","cluster2_3","cluster3_4","cluster4_5","cluster5_6")
for (i in 1:length(listGroups)) {
  Grouplist <- listGroups[[i]]
  genesList <- Grouplist['gene']
  lengths(geneList)
  print(groupNames[i])
  # Make Featureplots for top20 DE genes per cluster_group
  pdf(file=paste0(sampleFolder,"/Featureplots_cluster_",groupNames[i],"_",sampleName,".pdf"))
  print(FeaturePlot(object = seuratObj, features = c(as.character(genesList[1:5,]))))
  print(FeaturePlot(object = seuratObj, features = c(as.character(genesList[6:10,]))))
  print(FeaturePlot(object = seuratObj, features = c(as.character(genesList[11:15,]))))
  print(FeaturePlot(object = seuratObj, features = c(as.character(genesList[16:20,]))))
  dev.off()
}
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3 回答 3

2

对于每个基因列表，您可以使用您选择的这样的基因制作一个图（图会看起来更好，更大的尺寸，如在您的 PDF 中）：

  • 使用combine=FALSE并限制要绘制的特征数量，rownames(pbmc_small)[1 : min(20, nrow(pbmc_small))]以避免错误
  • 然后导出单个图的列表（允许主题化）并使用cowplot::plot_grid
  • 而不是在函数 ( ) 内绘图plot(out)，您可以导出为 pdf（可能将文件名作为第二个参数传递给函数）。
library(Seurat)
genelist <- list(
    l1 = sample(rownames(pbmc_small), 23),
    l2 = sample(rownames(pbmc_small), 14),
    l3 = sample(rownames(pbmc_small), 4))

plotFeatures <- function(x){
    p <- FeaturePlot(object = pbmc_small, 
        features = x[1 : min(20, length(x))], 
        combine = FALSE, label.size = 2)
    out <- cowplot::plot_grid(plotlist = p, ncol = 5, nrow = 4)
    plot(out)
}
lapply(genelist, plotFeatures)
于 2020-02-06T15:27:53.777 回答
1

未经测试，这样的东西应该可以工作。我们不是为每 5 个基因调用 print 5 次，而是根据基因的数量在循环中调用它n次。如果我们有 10 个基因， forloop将打印两次，如果有 20 个则我们调用 print 4 次，依此类推：

groupNames <- c("cluster1_2","cluster2_3","cluster3_4","cluster4_5","cluster5_6")

for (i in 1:length(listGroups)) {
  Grouplist <- listGroups[[i]]
  genesList <- Grouplist['gene']
  #lengths(geneList)
  print(groupNames[i])
  # Make Featureplots for top20 DE genes per cluster_group

  # make chunks of 5 each. 
  myChunks <- split(genesList, ceiling(seq_along(genesList)/5))

  pdf(file=paste0(sampleFolder,"/Featureplots_cluster_",groupNames[i],"_",sampleName,".pdf"))

  # loop through genes plotting 5 genes each time.
  for(x %in% seq(myChunks) ){
    print(FeaturePlot(object = seuratObj, features = myChunks[[ x ]]))
  }

  dev.off()
}
于 2020-02-06T14:34:11.993 回答
0

感谢zx8754和user12728748的输入。我为我的问题找到了两种解决方案。

        for (i in 1:length(listGroups)) {
      Grouplist <- listGroups[[1]]
      genesList <- Grouplist['gene']
      print(groupNames[1])

    ## Solution 1
    # Here all genes are printed. I didn't find a way yet to limited to 20

      # make chunks of 5 each. 
      myChunks <- split(genesList,ceiling(seq(lengths(genesList))/5))
      # Make Featureplots for top20 DE genes per cluster_group
      pdf(file=paste0(sampleFolderAggr,"results/Featureplots_",groupNames[i],"_",sampleNameAggr,".pdf"))
      # loop through genes plotting 5 genes each time.
      for(x in 1:min(5, length(myChunks) ){
        # Create a list of 5 genes
        my5Genes <- as.list(myChunks[[x]])

        print(FeaturePlot(object = seuratObj, features = c(as.character(my5Genes$gene))))
      }
      dev.off()

    ## Solution 2

    pdf(file=paste0(sampleFolderAggr,"results/Featureplots_",groupNames[i],"_",sampleNameAggr,".pdf"))

    plotFeatures <- function(x){
        p <- FeaturePlot(object = seuratObj, features = c(as.character(x[1: min(20, lengths(x)),])), combine = FALSE, label.size = 2)
        out <- cowplot::plot_grid(plotlist = p, ncol = 5, nrow = 4)
        # Make Featureplots for top20 DE genes per cluster_group
        plot(out)
        }
      lapply(genelist, plotFeatures)
      dev.off()
    }
于 2020-02-06T15:49:14.797 回答