Question

我正在尝试使用phytools软件包在系统发育图上绘制特征数据。我敢肯定这应该很简单，但是我收到一条无用的错误消息，而且我不知道该怎么办。

这是我的代码，包括数据下载。

<!DOCTYPE html>
<html>
  <head>
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  <body>
    <button onclick="getLocation()">Try It</button>
    <p id="result"></p>
    <script>

      var result = document.getElementById("result");

      function getLocation() {
        if (navigator.geolocation) {
          navigator.geolocation.watchPosition(showPosition, null, { enableHighAccuracy: true });
        } else { 
          x.innerHTML = "Geolocation is not supported by this browser.";
        }
      }

      function showPosition(position) {
        result.innerHTML = "latitude: " + position.coords.latitude + 
          "<br>longitude: " + position.coords.longitude +
          "<br>position accuracy: " + position.coords.accuracy + " [m]" +
          "<br>altitude: " + position.coords.altitude + " [m]" +
          "<br>altitude accuracy: " + position.coords.altitudeAccuracy + " [m]" +
          "<br>heading: " + position.coords.heading + " [degrees]" +
          "<br>speed: " + position.coords.speed + " [m/s]";
      }

    </script>
  </body>
</html>

我尝试绘制树和完整树的较小子集，但在两种情况下都出现错误：

# General library(dplyr) # Phylogenetic libraries. library(caper) library(phytools) #+ data_read p <- read.table(file = 'http://esapubs.org/archive/ecol/E090/184/PanTHERIA_1-0_WR05_Aug2008.txt', header = TRUE, sep = "\t", na.strings = c("-999", "-999.00")) ## Some data cleaning # Remove NAs in response and response where litter size is less than one (doesn't make sense). p <- p %>% filter(!is.na(X15.1_LitterSize)) %>% filter(X15.1_LitterSize >= 1) %>% mutate(y = log1p(X15.1_LitterSize)) %>% dplyr::select(-X15.1_LitterSize, -References, -X24.1_TeatNumber) ## Get phylogeny data. ### read in phylogeny data. # Read in trees tree <- read.nexus('https://onlinelibrary.wiley.com/action/downloadSupplement?doi=10.1111%2Fj.1461-0248.2009.01307.x&file=ELE_1307_sm_SA1.tre') # Select best supported tree tree <- tree[[1]] tree$tip.label <- gsub('_', ' ', tree$tip.label) # Check if species are available. mean(p$MSW05_Binomial %in% tree$tip.label) in_phylo <- p$MSW05_Binomial %in% tree$tip.label # Remove data that is not in the phylogeny. p <- p %>% filter(in_phylo) # Try just vulpes. unneededTips <- tree$tip.label[!grepl('Vulpes', tree$tip.label) | !(tree$tip.label %in% p$MSW05_Binomial)] # Prune tree down to only needed tips. pruneTree <- drop.tip(tree, unneededTips) dotTree(pruneTree, p$y[grepl('Vulpes', p$MSW05_Binomial)]) # Try all species unneededTips <- tree$tip.label[!(tree$tip.label %in% p$MSW05_Binomial)] # Prune tree down to only needed tips. pruneTree <- drop.tip(tree, unneededTips) dotTree(pruneTree, p$y)

Answer 1

对于dotTree和phytools中类似功能（例如contMap）的

，您的特征值必须是一个命名向量，其名称与树中的提示相对应。在您的示例中，您需要确保p$y是一个命名向量（{{1}应该是!is.null(names(p$y))）：

TRUE

您可以对较大的树应用相同的过程。建议您使用## Prune down the non Vulpes tips vulpes_tree <- drop.tip(tree, tree$tip.label[-grep("Vulpes", tree$tip.label)]) ## Naming the variables in p$y all_vulpes <- grepl('Vulpes', p$MSW05_Binomial) traits_to_plot <- p$y[all_vulpes] names(traits_to_plot) <- p$MSW05_Binomial[all_vulpes] ## Plotting the Vulpes and the traits dotTree(vulpes_tree, traits_to_plot)包中的函数cleand.data来匹配树和数据集：

dispRity

用phytool绘制系统发育特征

1 个答案: