Question

我有六只鹦鹉鸟，"Beefy", "Scoundrel", "Baston", "Mattdamon", "Jesus", and "Hulkhogan"。这些鸟很多，并且在很多不同的地方。我已经决定追踪过去两周内这种情况发生的地点和频率，我试图找出这些小恶魔今天最多的地方。

mydata <- data.frame(Dates = structure(c(16656, 16657, 16658, 16659,
                                         16660, 16661, 16662, 16663,
                                         16664, 16665, 16666, 16667,
                                         16668, 16669 
                                         ), 
                                       class = "Date"), 
                     PooLoc1 = sample(1:40, 7),
                     PooLoc2 = sample(1:10, 7),
                     PooLoc3 = sample(1:10, 7),
                     PooLoc4 = sample(1:30, 7),
                     PooLoc5 = sample(1:20, 7),
                     PooLoc6 = sample(1:70, 7)
)

head(mydata)
     Dates PooLoc1 PooLoc2 PooLoc3 PooLoc4 PooLoc5 PooLoc6
2015-08-09      24       3       9       1      16      45
2015-08-10      39       2       2      12      12       2
2015-08-11      14       7       6       5      19       4
2015-08-12      26       9       8      27       3      64
2015-08-13      20       4       1      15      20      48
2015-08-14       9       1       4       8       8      61

我可以按列日期订购mydata行，以便轻松找到今天的poos：

mydata <- mydata[order(mydata[["Dates"]], decreasing = TRUE), ]

但是，如何按今天的日期获得的值对列进行排序，以便我可以快速查看mydata的左上角并找到问题的答案？你能用一行吗？

Answer 1

我认为这可能是你想要的。我们只是按第一行排序列（第二行到第二行）。

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现在，左上角的值具有今天日期的最高值。

根据评论中的请求，我们可以使用

在一次调用中订购行和列

mydata[, c(1, order(mydata[1, -1], decreasing = TRUE) + 1)]
#        Dates PooLoc6 PooLoc1 PooLoc4 PooLoc5 PooLoc3 PooLoc2
# 6 2015-08-14      61       9       8       8       4       1
# 5 2015-08-13      48      20      15      20       1       4
# 4 2015-08-12      64      26      27       3       8       9
# 3 2015-08-11       4      14       5      19       6       7
# 2 2015-08-10       2      39      12      12       2       2
# 1 2015-08-09      45      24       1      16       9       3

按列中的行值和R中的行中的列值对数据进行排序

1 个答案: