R:对多个向量(列)进行混合排序 [英] R: mixedsort on multiple vectors (columns)
问题描述
这是对此问题的后续行动被标记为与此重复的内容，但是建议的解决方案没有工作.
This is a follow-up on this question, which was marked as a duplicate to this, but the suggested solution does not work.
我有以下data.frame:
set.seed(1)
mydf <- data.frame(A=paste(sample(LETTERS, 4), sample(1:20, 20), sep=""),
B=paste(sample(1:20, 20), sample(LETTERS, 4), sep=""),
C=sample(LETTERS, 20), D=sample(1:100, 20), value=rnorm(20))
> mydf
A B C D value
1 G5 6N T 9 -0.68875569
2 J18 8T R 87 -0.70749516
3 N19 1A L 34 0.36458196
4 U12 7K Z 82 0.76853292
5 G11 14N J 98 -0.11234621
6 J1 20T F 32 0.88110773
7 N3 17A B 45 0.39810588
8 U14 19K W 83 -0.61202639
9 G9 15N U 80 0.34111969
10 J20 3T I 36 -1.12936310
11 N8 9A K 70 1.43302370
12 U16 16K G 86 1.98039990
13 G6 10N M 39 -0.36722148
14 J7 18T D 62 -1.04413463
15 N13 5A Y 35 0.56971963
16 U4 11K N 28 -0.13505460
17 G17 4N O 64 2.40161776
18 J15 2T C 17 -0.03924000
19 N2 12A P 59 0.68973936
20 U10 13K X 10 0.02800216
我想根据列A至D对其进行排序，但是A和D混合使用，因此需要自然顺序.
I want to order it according to columns A to D, but A and D are mixed, so natural order is required.
我知道我可以应用常规订购，例如:
I know I can apply regular ordering, like:
mydf2 <- mydf[do.call(order, c(mydf[1:4], list(decreasing = FALSE))),]
> mydf2
A B C D value
5 G11 14N J 98 -0.11234621
17 G17 4N O 64 2.40161776
1 G5 6N T 9 -0.68875569
13 G6 10N M 39 -0.36722148
9 G9 15N U 80 0.34111969
6 J1 20T F 32 0.88110773
18 J15 2T C 17 -0.03924000
2 J18 8T R 87 -0.70749516
10 J20 3T I 36 -1.12936310
14 J7 18T D 62 -1.04413463
15 N13 5A Y 35 0.56971963
3 N19 1A L 34 0.36458196
19 N2 12A P 59 0.68973936
7 N3 17A B 45 0.39810588
11 N8 9A K 70 1.43302370
20 U10 13K X 10 0.02800216
4 U12 7K Z 82 0.76853292
8 U14 19K W 83 -0.61202639
12 U16 16K G 86 1.98039990
16 U4 11K N 28 -0.13505460
但这不是我需要的结果.我需要在9之后的10，而不是在1之后的(您可以检查列A来查看它是否与我所需的顺序不符.)
But this is not the result I need. I need 10 after 9, not after 1 (you can check column A to see it is not in the order I need.)
在我最初的问题的评论中，建议使用multi.mixedorder函数.
In the comments of my original question, it was suggested to use the multi.mixedorder function.
但是，正如您在下面看到的，结果与仅使用order的结果相同，但这仍然不是我想要的.
However, as you can see below, the result is identical to the one using just order, which is still not what I want.
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)
))
}
mydf3 <- mydf[do.call(multi.mixedorder, c(mydf[1:4], list(decreasing = FALSE))),]
> mydf3
A B C D value
5 G11 14N J 98 -0.11234621
17 G17 4N O 64 2.40161776
1 G5 6N T 9 -0.68875569
13 G6 10N M 39 -0.36722148
9 G9 15N U 80 0.34111969
6 J1 20T F 32 0.88110773
18 J15 2T C 17 -0.03924000
2 J18 8T R 87 -0.70749516
10 J20 3T I 36 -1.12936310
14 J7 18T D 62 -1.04413463
15 N13 5A Y 35 0.56971963
3 N19 1A L 34 0.36458196
19 N2 12A P 59 0.68973936
7 N3 17A B 45 0.39810588
11 N8 9A K 70 1.43302370
20 U10 13K X 10 0.02800216
4 U12 7K Z 82 0.76853292
8 U14 19K W 83 -0.61202639
12 U16 16K G 86 1.98039990
16 U4 11K N 28 -0.13505460
推荐答案
好了，multi.mixedsort函数需要修复才能处理各种因素:
OK solved it, the multi.mixedsort function needs a fix to be able to deal with factors:
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 {
factor(as.character(l), levels=mixedsort(levels(l)))
}
}),
list(na.last = na.last, decreasing = decreasing)
))
}
否则，将mydf中的所有因子列转换为字符，
Otherwise convert all factor columns in mydf into character, with:
mydf[] <- lapply(mydf, as.character)
但已修正，因此不需要
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