订购“混合"矢量(带字母的数字) [英] Order a "mixed" vector (numbers with letters)

问题描述

如何订购像

c("7","10a","10b","10c","8","9","11c","11b","11a","12") -> alph

在

alph
[1] "7","8","9","10a","10b","10c","11a","11b","11c","12"

并使用它对 data.frame 进行排序，例如

and use it to sort a data.frame, like

V1 <- c("A","A","B","B","C","C","D","D","E","E")
V2 <- 2:1 
V3 <- alph
df <- data.frame(V1,V2,V3)

并排序要获得的行(排序V2然后V3)

and order the row to obtain (order V2 and then V3)

 V1 V2  V3
C  1   9
A  1 10a
B  1 10c
D  1 11b
E  1  12
A  2   7
C  2   8
B  2 10b
E  2 11a
D  2 11c

推荐答案

> library(gtools)
> mixedsort(alph)

[1] "7"   "8"   "9"   "10a" "10b" "10c" "11a" "11b" "11c" "12" 

要对 data.frame 进行排序，请使用 mixedorder 代替

To sort a data.frame you use mixedorder instead

> mydf <- data.frame(alph, USArrests[seq_along(alph),])
> mydf[mixedorder(mydf$alph),]

            alph Murder Assault UrbanPop Rape
Alabama        7   13.2     236       58 21.2
California     8    9.0     276       91 40.6
Colorado       9    7.9     204       78 38.7
Alaska       10a   10.0     263       48 44.5
Arizona      10b    8.1     294       80 31.0
Arkansas     10c    8.8     190       50 19.5
Florida      11a   15.4     335       80 31.9
Delaware     11b    5.9     238       72 15.8
Connecticut  11c    3.3     110       77 11.1
Georgia       12   17.4     211       60 25.8

mixedorder 在多个向量(列)上

显然 mixedorder 不能处理多个向量.我创建了一个函数，通过将所有字符向量转换为具有混合排序级别的因子，并将所有向量传递给标准的 order 函数.

mixedorder on multiple vectors (columns)

Apparently mixedorder cannot handle multiple vectors. I have made a function that circumvents this by converting all character vectors to factors with mixedsorted sorted levels, and pass all vectors on to the standard order function.

multi.mixedorder <- function(..., na.last = TRUE, decreasing = FALSE){
    do.call(order, c(
        lapply(list(...), function(l){
            if(is.character(l)){
                factor(l, levels=mixedsort(unique(l)))
            } else {
                l
            }
        }),
        list(na.last = na.last, decreasing = decreasing)
    ))
}

但是，在您的特定情况下，multi.mixedorder 为您提供与标准 order 相同的结果，因为 V2 是数字.>

However, in your particular case multi.mixedorder gets you the same result as the standard order, since V2 is numeric.

df <- data.frame(
    V1 = c("A","A","B","B","C","C","D","D","E","E"),
    V2 = 19:10,
    V3 = alph,
    stringsAsFactors = FALSE)

df[multi.mixedorder(df$V2, df$V3),]

   V1 V2  V3
10  E 10  12
9   E 11 11a
8   D 12 11b
7   D 13 11c
6   C 14   9
5   C 15   8
4   B 16 10c
3   B 17 10b
2   A 18 10a
1   A 19   7

注意

• 19:10 等价于 c(19:10).c 的意思是 concat，也就是把多个短向量变成一个长向量，但在你的情况下你只有一个向量 (19:10) 所以没有必要连接任何东西.但是，在 V1 的情况下，您有 10 个长度为 1 的向量，因此您需要像之前一样进行连接.
• 您需要 stringsAsFactors=FALSE 才能将 V1 和 V3 转换为(错误排序的)因子(这是默认值).

Notice that

• 19:10 is equivalent to c(19:10). c means concat, that is to make one long vector out of many short, but in you case you only have one vector (19:10) so there's no need to concat anything. However, in the case of V1 you have 10 vectors of length 1, so there you need to concat, as you already do.
• You need stringsAsFactors=FALSE to not convert V1 and V3 to (incorrectly sorted) factors (which is default).

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