R sort（）data.frame

我有以下数据框

head(stockdatareturnpercent)
                  SPY         DIA        IWM        SMH        OIH        
2001-04-02  8.1985485   7.8349806   7.935566  21.223832  13.975655  
2001-05-01 -0.5621328   1.7198760   2.141846 -10.904936  -4.565291  
2001-06-01 -2.6957979  -3.5838102   2.786250   4.671762 -23.241009 
2001-07-02 -1.0248091  -0.1997433  -5.725078  -3.354391  -9.161594  
2001-08-01 -6.1165559  -5.0276558  -2.461728  -6.218129 -13.956695  
2001-09-04 -8.8900629 -12.2663267 -15.760037 -39.321172 -16.902913 

实际上有更多的股票，但为了说明原因，我不得不把它减下来。每个月我都想知道从最好到最差（或从最差到最好）的表现。我使用了sort（）函数，这就是我想到的

N <- dim(stockdatareturnpercent)[1]  
for (i in 1:N) {  
    s <- sort(stockdatareturnpercent[i,])  
    print(s)  
}  

                 UPS     FDX      XLP      XLU      XLV     DIA      IWM      SPY      XLE      XLB      XLI      OIH      XLK      SMH     MSFT
2001-04-02 0.6481585 0.93135 1.923136 4.712996 7.122751 7.83498 7.935566 8.198549 9.826701 10.13465 10.82522 13.97566 14.98789 21.22383 21.41436
                 SMH       FDX       OIH       XLK        XLE        SPY       XLU      XLP      DIA     MSFT      IWM     UPS      XLV      XLB      XLI
2001-05-01 -10.90494 -5.045544 -4.565291 -4.182041 -0.9492803 -0.5621328 0.6987724 1.457579 1.719876 2.088734 2.141846 3.73587 3.748309 3.774033 4.099748
                 OIH       XLE       XLI     XLU     XLP       XLB      DIA       UPS       SPY       XLV       FDX      XLK     IWM      SMH     MSFT
2001-06-01 -23.24101 -10.02403 -6.594324 -5.8602 -5.0532 -3.955192 -3.58381 -2.814685 -2.695798 -1.177474 0.4987542 1.935544 2.78625 4.671762 5.374764
                MSFT       OIH      XLK       IWM       SMH       XLV       UPS       XLE       SPY        XLU        XLB        XLI        DIA      FDX
2001-07-02 -9.793005 -9.161594 -7.17351 -5.725078 -3.354391 -2.016818 -1.692442 -1.159914 -1.024809 -0.9029407 -0.2723560 -0.2078283 -0.1997433 2.868898
                XLP
2001-07-02 2.998604

---

N使用原始代码将每个排序的行保存在
列表中
：

stockdatareturnpercent <- read.table(textConnection("                  SPY         DIA        IWM        SMH        OIH        
2001-04-02  8.1985485   7.8349806   7.935566  21.223832  13.975655  
2001-05-01 -0.5621328   1.7198760   2.141846 -10.904936  -4.565291  
2001-06-01 -2.6957979  -3.5838102   2.786250   4.671762 -23.241009 
2001-07-02 -1.0248091  -0.1997433  -5.725078  -3.354391  -9.161594  
2001-08-01 -6.1165559  -5.0276558  -2.461728  -6.218129 -13.956695  
2001-09-04 -8.8900629 -12.2663267 -15.760037 -39.321172 -16.902913"))

x <- vector("list", nrow(stockdatareturnpercent))

## use unlist to drop the data.frame structure
for (i in 1:nrow(stockdatareturnpercent)) {  
    x[[i]] <- sort(unlist(stockdatareturnpercent[i,])  )
} 
## use the row names to name each list element
names(x) <- rownames(stockdatareturnpercent)

x
$`2001-04-02`
  DIA       IWM       SPY       OIH       SMH 
7.834981  7.935566  8.198548 13.975655 21.223832 

$`2001-05-01`
    SMH         OIH         SPY         DIA         IWM 
-10.9049360  -4.5652910  -0.5621328   1.7198760   2.1418460 

$`2001-06-01`
   OIH        DIA        SPY        IWM        SMH 
-23.241009  -3.583810  -2.695798   2.786250   4.671762 

$`2001-07-02`
   OIH        IWM        SMH        SPY        DIA 
-9.1615940 -5.7250780 -3.3543910 -1.0248091 -0.1997433 

$`2001-08-01`
   OIH        SMH        SPY        DIA        IWM 
-13.956695  -6.218129  -6.116556  -5.027656  -2.461728 

$`2001-09-04`
   SMH        OIH        IWM        DIA        SPY 
-39.321172 -16.902913 -15.760037 -12.266327  -8.890063 

返回一个矩阵，其中每列都是已排序的行。然后转置：

sortmat <- t(apply(stockdatareturnpercent, 1, sort))

sortmat为此使用
order（）
，因为
sort（）
在使用
*apply时会删除名称：

id <- t(apply(Data,1,order))
lapply(1:nrow(id),function(i)Data[i,id[i,]])

找出在某一时刻哪一个是最好的

如果要使用循环，可以使用列表。正如Joshua所说，你在每个循环中都覆盖了s。初始化列表以首先存储结果。此循环给出的结果与上面代码中的
lappy（）
相同，但没有id矩阵。虽然使用apply还有其他好处，但速度没有提高：

N <- nrow(Data)
s <- vector("list",N)
for (i in 1:N) {
    s[[i]] <- sort(Data[i,])
}

s
只包含最后一行，因为您在每次迭代中都会分配（重写）它。谢谢。虽然这些列似乎已排序，但现在无法判断这些回报属于哪只股票。例如，new data.frame中的第一列表示表现最差的值，但无法将其与股票相关联。最初的data.frame将列dimnames设置为stock ticker，因此我的方法非常愚蠢，因为它使我能够每月查看性能订单。我非常感谢您，因为您的解决方案为我提供了额外的知识和工具。我还要补一些。
sortdf <- as.data.frame(t(apply(stockdatareturnpercent, 1, sort)))

id <- t(apply(Data,1,order))
lapply(1:nrow(id),function(i)Data[i,id[i,]])

matrix(names(Data)[id],ncol=ncol(Data))
     [,1]  [,2]  [,3]  [,4]  [,5] 
[1,] "DIA" "IWM" "SPY" "OIH" "SMH"
[2,] "SMH" "OIH" "SPY" "DIA" "IWM"
[3,] "OIH" "DIA" "SPY" "IWM" "SMH"
[4,] "OIH" "IWM" "SMH" "SPY" "DIA"
[5,] "OIH" "SMH" "SPY" "DIA" "IWM"
[6,] "SMH" "OIH" "IWM" "DIA" "SPY"

N <- nrow(Data)
s <- vector("list",N)
for (i in 1:N) {
    s[[i]] <- sort(Data[i,])
}

zz <- textConnection(" SPY         DIA        IWM        SMH        OIH
  8.1985485   7.8349806   7.935566  21.223832  13.975655
 -0.5621328   1.7198760   2.141846 -10.904936  -4.565291
 -2.6957979  -3.5838102   2.786250   4.671762 -23.241009
 -1.0248091  -0.1997433  -5.725078  -3.354391  -9.161594
 -6.1165559  -5.0276558  -2.461728  -6.218129 -13.956695
 -8.8900629 -12.2663267 -15.760037 -39.321172 -16.902913 ")

Data <- read.table(zz,header=T)
close(zz)