Question

说，您有以下数据：

data <- tibble::tribble(~Countries, ~States,   ~Continents,
                        "Country 1",      1L, "continent 1",
                        "Country 1",      2L, "continent 1",
                        "Country 1",      3L, "continent 1",
                        "Country 1",      4L, "continent 1",
                        "Country 2",      1L, "continent 1",
                        "Country 2",      2L, "continent 1",
                        "Country 2",      3L, "continent 1",
                        "Country 2",      4L, "continent 1",
                        "Country 3",      1L, "continent 1",
                        "Country 3",      2L, "continent 1",
                        "Country 3",      3L, "continent 1",
                        "Country 3",      4L, "continent 1",
                        "Country 1",      1L, "continent 2",
                        "Country 1",      2L, "continent 2",
                        "Country 1",      3L, "continent 2",
                        "Country 1",      4L, "continent 2",
                        "Country 2",      1L, "continent 2",
                        "Country 2",      2L, "continent 2",
                        "Country 2",      3L, "continent 2",
                        "Country 2",      4L, "continent 2",
                        "Country 3",      1L, "continent 2",
                        "Country 3",      2L, "continent 2",
                        "Country 3",      3L, "continent 2",
                        "Country 3",      4L, "continent 2")

此数据可能具有许多格式不同的粒度不同的变量。我想了解数据的结构，因此可以说，在上述数据中，最高级别的数据是具有2个值的大陆，下一个粒度级别是具有3个值的县，而最低级别是具有4个值的州。

一种理解这一点的粗略方法可能是将变量数最少的变量保留在左侧（即大陆），将变量数最多的变量保留在数据集的右侧。
更容易弄清混乱数据的方法是创建某种树形图，并在顶部，大洲，此处和底部状态（在这里，如树叶）中看到最少的粒度数据/节点。

首先，我们可以使用一些技巧，例如，在唯一值数目相同的情况下，如果出现平局，则在第一个/顶部显示两个或多个变量中的一个。

如果很难做到第二，那么我们如何至少做到第一呢？ ...可以通过评估任何通用的混乱数据中每个变量的不同值，然后对变量进行排序！任何其他带有R代码的方法都将非常有帮助。

第一点的解决方案如下：

data <- tibble::tribble( ~Continents,  ~Countries,   ~States,
                         "continent 1", "Country 1",   1L,
                         "continent 1", "Country 1",   2L,
                         "continent 1", "Country 1",   3L,
                         "continent 1", "Country 1",   4L,
                         "continent 1", "Country 2",   1L,
                         "continent 1", "Country 2",   2L,
                         "continent 1", "Country 2",   3L,
                         "continent 1", "Country 2",   4L,
                         "continent 1", "Country 3",   1L,
                         "continent 1", "Country 3",   2L,
                         "continent 1", "Country 3",   3L,
                         "continent 1", "Country 3",   4L,
                         "continent 2", "Country 1",   1L,
                         "continent 2", "Country 1",   2L,
                         "continent 2", "Country 1",   3L,
                         "continent 2", "Country 1",   4L,
                         "continent 2", "Country 2",   1L,
                         "continent 2", "Country 2",   2L,
                         "continent 2", "Country 2",   3L,
                         "continent 2", "Country 2",   4L,
                         "continent 2", "Country 3",   1L,
                         "continent 2", "Country 3",   2L,
                         "continent 2", "Country 3",   3L,
                         "continent 2", "Country 3",   4L)

Answer 1

如果我答对了，下面的代码将回答您的问题：

data[order(sapply(data, function(x) length(unique(x))))] # returns the data in the desired order

# simple function for plotting the 'tree'. 
plotTree <- function(lengths, names, space = 0.3){
  L    <- lengths[O <- order(lengths)]
  N    <- names[O]
  XMax <- max(L)
  YMax <- (length(L))
  plot(NULL, xlim = c(-XMax, XMax), ylim = c(-YMax, YMax), axes = F, xlab = "", ylab = "")
  for (i in 1:length(L)){
    rect(-L[i], YMax - 1 - i *  (space + 1), L[i], YMax - i *  (space + 1), col = i)
    text(0, YMax - 1/2 - i * (space + 1), N[i], col = if (i == 1) "white" else "black")
  }
}

# usage
plotTree(sapply(data, function(x) length(unique(x))), names(data), space = 0.3)

确定整个数据的结构

1 个答案: