R自行对矢量进行排序

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")

人们会期望这个向量的顺序是1:length(df.sorted),但似乎并非如此.看起来R内部根据其逻辑对向量进行排序,但是很难以创建它的方式显示它(并且在输出中看到).

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"重置"订单1:length(df.sorted)？这样,矢量的排序和输出将是同步的.

Answer 1

使用包gtools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,它甚至不是nessecary使它成为一个有序因子,只要一个因素就足够了,如果你给出正确的级别顺序:

>     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您将比较运算符用作<和>.这使生活有时更容易.

> 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

这不是与所有词典短片(例如ls在目录中)walker10_foo sorts高于walker1_foo？相同的东西吗？

在我的书中,最简单的方法是使用一致的数字位数,即我会改为binned_walker01_1.grd等等,为一位数字插入0.

Answer 4

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

您可以这样添加0:

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

如果你需要添加00,你可以这样做:

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

...等等.