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我正在尝试以表格格式显示我的数据，但我不知道如何重新排列我的数据以以正确的格式显示它。我习惯于为绘图争吵数据，但在准备表格时我发现自己有点迷茫。这似乎是非常基本的事情，但我无法在这里找到关于我做错了什么的解释。

我有 3 列数据，Type、Year和n。现在格式化的数据会生成一个如下所示的表：

Type    Year    n
Type C  1   5596
Type D  1   1119
Type E  1   116
Type A  1   402
Type F  1   1614
Type B  1   105
Type C  2   26339
Type D  2   14130
Type E  2   98
Type A  2   3176
Type F  2   3071
Type B  2   88

我想要做的是Type作为行名，Year作为列名，并n像这样填充表格内容：

         1      2        
Type A   402    3176   
Type B   105    88
Type C   26339  5596
Type D   1119   14130
Type E   116    98
Type F   1614   3071

从这一点开始，错误可能已经在上游发生。使用完整的原始数据集，我通过执行以下操作得到了这个输出：

exampletable <- df %>%
  group_by(Year) %>%
  count(Type) %>%
  select(Type, Year, n)

这是dput()输出

structure(list(Type = c("Type C", "Type D", "Type E", "Type A", 
"Type F", "Type B", "Type C", "Type D", "Type E", "Type A", "Type F", 
"Type B", "Type C", "Type D", "Type E", "Type A", "Type F", "Type B", 
"Type C", "Type D", "Type E", "Type A", "Type F", "Type B", "Type C", 
"Type D", "Type E"), Year = c(1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 
2, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 5, 5, 5), n = c(5596, 
1119, 116, 402, 1614, 105, 26339, 14130, 98, 3176, 3071, 88, 
40958, 17578, 104, 3904, 3170, 102, 33145, 23800, 93, 1264, 7084, 
1262, 34642, 24911, 504)), class = c("spec_tbl_df", "tbl_df", 
"tbl", "data.frame"), row.names = c(NA, -27L), spec = structure(list(
    cols = list(Type = structure(list(), class = c("collector_character", 
    "collector")), Year = structure(list(), class = c("collector_double", 
    "collector")), n = structure(list(), class = c("collector_double", 
    "collector"))), default = structure(list(), class = c("collector_guess", 
    "collector")), skip = 1), class = "col_spec"))
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2 回答 2

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您可以获取宽格式的数据并将Type列更改为行名。

tidyr::pivot_wider(df, names_from = Year, values_from = n) %>%
   tibble::column_to_rownames('Type')

#          1     2     3     4     5
#Type C 5596 26339 40958 33145 34642
#Type D 1119 14130 17578 23800 24911
#Type E  116    98   104    93   504
#Type A  402  3176  3904  1264    NA
#Type F 1614  3071  3170  7084    NA
#Type B  105    88   102  1262    NA
于 2020-05-13T06:29:12.333 回答
0

您可以使用tidyr包来获得更广泛的格式并使用tibble包来将列转换为行名

dataset <- read.csv(file_location)
dataset <- tidyr::pivot_wider(dataset, names_from = Year, values_from = n)

tibble::column_to_rownames(dataset, var = 'Type')
       1     2
Type C 5596 26339
Type D 1119 14130
Type E  116    98
Type A  402  3176
Type F 1614  3071
Type B  105    88
于 2020-05-13T07:15:22.530 回答