Question

df.sorted <- c("binned_walker1_1.grd", "binned_walker1_2.grd", "binned_walker1_3.grd",
    "binned_walker1_4.grd", "binned_walker1_5.grd", "binned_walker1_6.grd",
    "binned_walker2_1.grd", "binned_walker2_2.grd", "binned_walker3_1.grd",
    "binned_walker3_2.grd", "binned_walker3_3.grd", "binned_walker3_4.grd",
    "binned_walker3_5.grd", "binned_walker4_1.grd", "binned_walker4_2.grd",
    "binned_walker4_3.grd", "binned_walker4_4.grd", "binned_walker4_5.grd",
    "binned_walker5_1.grd", "binned_walker5_2.grd", "binned_walker5_3.grd",
    "binned_walker5_4.grd", "binned_walker5_5.grd", "binned_walker5_6.grd",
    "binned_walker6_1.grd", "binned_walker7_1.grd", "binned_walker7_2.grd",
    "binned_walker7_3.grd", "binned_walker7_4.grd", "binned_walker7_5.grd",
    "binned_walker8_1.grd", "binned_walker8_2.grd", "binned_walker9_1.grd",
    "binned_walker9_2.grd", "binned_walker9_3.grd", "binned_walker9_4.grd",
    "binned_walker10_1.grd", "binned_walker10_2.grd", "binned_walker10_3.grd")

可以预期此向量的顺序为1:length(df.sorted)，但似乎并非如此。看起来R内部根据其逻辑对矢量进行排序，但是很难以它的创建方式显示它（并且在输出中可以看到）。

order(df.sorted)
 [1] 37 38 39  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20 21 22
[26] 23 24 25 26 27 28 29 30 31 32 33 34 35 36

有没有办法将订单“重置”为1:length(df.sorted)？这样，矢量的排序和输出将是同步的。

Answer 1

使用gtools包中的mixedsort（或）mixedorder函数：

require(gtools)
mixedorder(df.sorted)
 [1]  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27
[28] 28 29 30 31 32 33 34 35 36 37 38 39

Answer 2

将其构建为有序因子：

> df.new <- ordered(df.sorted,levels=df.sorted)
> order(df.new)
 [1]  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 ...

编辑：

在@DWins发表评论之后，我想补充说，如果你给出正确的级别顺序，那么它就是一个有序的因素，这个因素就足够了：

>     df.new2 <- factor(df.sorted,levels=df.sorted)
>     order(df.new)
 [1]  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 ...

当您在回归分析中使用这些因素时，差异会很明显，可以区别对待它们。有序因子的优点是它们允许你使用比较运算符＆lt;和＆gt;。这使生活有时变得容易多了。

> df.new2[5] < df.new2[10]
[1] NA
Warning message:
In Ops.factor(df.new[5], df.new[10]) : < not meaningful for factors

> df.new[5] < df.new[10]
[1] TRUE

Answer 3

这与您在ls高于walker10_foo sorts的所有词典短片（例如目录walker1_foo）上获得的不完全相同吗？

在我的书中，最简单的方法是使用一致的数字位数，即我将更改为binned_walker01_1.grd，依此插入0来表示一位数。

Answer 4

回应Dwin对Dirk答案的评论：数据总是在你手上。 “这是R.如果没有，那就没有。” - 西蒙布隆伯格

您可以像这样添加0：

df.sorted <- gsub("(walker)([[:digit:]]{1}_)", "\\10\\2", df.sorted)

如果您需要添加00，请执行以下操作：

df.sorted <- gsub("(walker)([[:digit:]]{1}_)", "\\10\\2", df.sorted)
df.sorted <- gsub("(walker)([[:digit:]]{2}_)", "\\10\\2", df.sorted)

......等等。

R自行对矢量进行排序

4 个答案: