订购一份“混合"的向量(带有字母的数字) [英] Order a "mixed" vector (numbers with letters)

问题描述

如何订购

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)

并对行进行排序(先按V2，然后按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" 

要对数据框进行排序，请改用mixedorder

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无法处理多个向量.通过将所有字符向量转换为具有混合排序水平的元素 并将所有向量传递给标准order函数，我制作了一个可以解决此问题的函数.

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，在特定的情况下，由于V2是数字，因此multi.mixedorder获得的结果与标准order相同.

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等效于c(19:10). c的意思是 concat ，即从多个空头中提取一个长向量，但是在您的情况下，您只有一个向量(19:10)，因此无需连接任何东西.但是，在V1的情况下，您有10个长度为1的向量，因此您需要像以前一样连接.
• 您需要stringsAsFactors=FALSE不能将V1和V3转换为(错误排序)因子(默认设置).

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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