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r - 确定整个数据的结构

问题描述

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")

该数据可能有许多不同格式的变量，具有不同的粒度级别。我想了解数据的结构，以便我可以说数据的最高级别是具有 2 个值的大陆，下一级粒度是具有 3 个值的县，最低级别是具有 4 个值的州。

  1. 理解这一点的一种粗略方法可能是将具有最少不同值的变量保留在左侧，即大陆，而将具有最多不同值的变量（即数据集右侧的州）保留。

  2. 了解这些杂乱数据的更简单方法是创建某种树形图，并在顶部、大陆、此处查看最细粒度的数据，在底部状态（此处）查看最细粒度的数据，作为叶子/节点。

作为第一个切入点，我们可以使用技巧，例如在唯一值的数量相同的情况下，在第一个/顶部显示两个或多个变量中的任何一个。

如果做第二个很困难，我们怎么能至少做第一个？...可能是通过评估任何通用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 代码的方法都会非常有帮助。

第一点的解决方案如下所示：

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)

标签： rdata.tabletidyversejanitor

解决方案

如果我猜对了，下面的代码会回答您的问题：

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)

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