使用 R：将矩阵转换为向量序列以实现阶函数的可变参数方法

Question

排序函数描述了它在列表中的读取方式

?order

... 
a sequence of numeric, complex, character or logical vectors, all of the same length, or a classed R object.

-----------------------------------------------------

> order
function (..., na.last = TRUE, decreasing = FALSE, method = c("auto", 
    "shell", "radix")) 
{
    z <- list(...)
    decreasing <- as.logical(decreasing)
    if (length(z) == 1L && is.numeric(x <- z[[1L]]) && !is.object(x) && 
        length(x) > 0) {
        if (.Internal(sorted_fpass(x, decreasing, na.last))) 
            return(seq_along(x))
    }

大多数人使用 order 以一种被黑的、非可变的形式：

myData.sorted = myData[ order(-myData[,date.idx],-myData[,(1+date.idx)]), ];

我写了一个函数来使这个表单可变：

        #########################################
        ## how I want it, doesn't work
        #fdf = sdf[order(vecs), ];

        #########################################
        ## non-variadic approach, does work
        fdf = sdf[order( vecs[,1],vecs[,2],vecs[,3] ), ];

所以我有一个矩阵，我想根据它的可变列数来分解它，然后将该矩阵转换为 order 函数可以处理的向量序列。 unlist？也许 as.list?

如何根据列数将矩阵转换为向量序列？

更新

convertDateStringToFormat = function (strvec,format.out="%Y",format.in="%Y-%m-%d %H:%M:%S",numeric=TRUE)
    {
    p.obj = strptime(strvec, format=format.in);
    o.obj = strftime(p.obj, format=format.out);
    
    if(numeric) { as.numeric(o.obj); } else { o.obj; }
    }

library(datasets);
data(iris);
df = iris[1:10,];
df$date.strings = c("3/24/2010 18:33", "9/3/2009 17:28", "10/14/2009 11:40", "7/3/2015 11:16","11/18/2010 1:29","4/23/2011 0:08","10/6/2010 11:13","7/26/2009 13:23","4/9/2008 13:40","8/20/2008 11:32");
df$year = convertDateStringToFormat(df$date.strings,"%Y","%m/%d/%Y %H:%M");
df$week = convertDateStringToFormat(df$date.strings,"%W","%m/%d/%Y %H:%M");
df$day = convertDateStringToFormat(df$date.strings,"%j","%m/%d/%Y %H:%M");
df$date.strings = NULL;

> df
   Sepal.Length Sepal.Width Petal.Length Petal.Width Species year week day
1           5.1         3.5          1.4         0.2  setosa 2010   12  83
2           4.9         3.0          1.4         0.2  setosa 2009   35 246
3           4.7         3.2          1.3         0.2  setosa 2009   41 287
4           4.6         3.1          1.5         0.2  setosa 2015   26 184
5           5.0         3.6          1.4         0.2  setosa 2010   46 322
6           5.4         3.9          1.7         0.4  setosa 2011   16 113
7           4.6         3.4          1.4         0.3  setosa 2010   40 279
8           5.0         3.4          1.5         0.2  setosa 2009   29 207
9           4.4         2.9          1.4         0.2  setosa 2008   14 100
10          4.9         3.1          1.5         0.1  setosa 2008   33 233
>

这里有一个 ... 步骤，但我们得到一个矩阵 vecs 如下所示：

vecs = matrix(
            c(2010,2009,2009,2015,2010,2011,2010,2009,2008,2008,
            -12,-35,-41,-26,-46,-16,-40,-29,-14,-33,
            83,246,287,184,322,113,279,207,100,233),
            
    nrow=10,ncol=3,byrow=F);

> vecs
      [,1] [,2] [,3]
 [1,] 2010  -12   83
 [2,] 2009  -35  246
 [3,] 2009  -41  287
 [4,] 2015  -26  184
 [5,] 2010  -46  322
 [6,] 2011  -16  113
 [7,] 2010  -40  279
 [8,] 2009  -29  207
 [9,] 2008  -14  100
[10,] 2008  -33  233
>

所以我试试这个：vec2 = as.data.frame(vecs); class(vec2) = "list"; 基于另一个 post (alfymbohm) How to convert a matrix to a list of column-vectors in R?

目前，这有效：

df[order( vecs[,1],vecs[,2],vecs[,3] ), ];


   Sepal.Length Sepal.Width Petal.Length Petal.Width Species year week day
10          4.9         3.1          1.5         0.1  setosa 2008   33 233
9           4.4         2.9          1.4         0.2  setosa 2008   14 100
3           4.7         3.2          1.3         0.2  setosa 2009   41 287
2           4.9         3.0          1.4         0.2  setosa 2009   35 246
8           5.0         3.4          1.5         0.2  setosa 2009   29 207
5           5.0         3.6          1.4         0.2  setosa 2010   46 322
7           4.6         3.4          1.4         0.3  setosa 2010   40 279
1           5.1         3.5          1.4         0.2  setosa 2010   12  83
6           5.4         3.9          1.7         0.4  setosa 2011   16 113
4           4.6         3.1          1.5         0.2  setosa 2015   26 184

而我想要的工作失败了。我用vec2来区分

vec2 = as.data.frame(vecs); class(vec2) = "list";
df[order(vec2), ];

它（order 函数）抛出以下错误：

Error in order(vec2) : unimplemented type 'list' in 'orderVector1'

我认为您的方法是我在其他地方找到的按名单演员的想法。

理想情况下，我想要一个函数，例如

vec2 = castMatrixToSequenceOfLists(vecs);

哪里

    
castMatrixToSequenceOfLists = function(mat)
    {
    list_length = ncol(mat);
    out_list = vector("list", list_length);
    for(i in 1:list_length)
        {
        out_list[[i]] = mat[,i]; # double brackets [[1]]
        }
    out_list;
    }

没有成功！抛出相同的错误（order 函数）：

vec2 = castMatrixToSequenceOfLists(vecs);
df[order(vec2), ];


Error in order(vec2) : unimplemented type 'list' in 'orderVector1'

同样，可变参数当前不起作用，因为根据 order.

的手册，矩阵不是“向量序列”

如何根据列数将矩阵转换为向量序列，以便 order 函数接受它？

解决方案

mat_order <- function(x) do.call(order, split(x, (seq(x) - 1) %/% nrow(x)))

> df[mat_order(vecs),]
   Sepal.Length Sepal.Width Petal.Length Petal.Width Species year week day
10          4.9         3.1          1.5         0.1  setosa 2008   33 233
9           4.4         2.9          1.4         0.2  setosa 2008   14 100
3           4.7         3.2          1.3         0.2  setosa 2009   41 287
2           4.9         3.0          1.4         0.2  setosa 2009   35 246
8           5.0         3.4          1.5         0.2  setosa 2009   29 207
5           5.0         3.6          1.4         0.2  setosa 2010   46 322
7           4.6         3.4          1.4         0.3  setosa 2010   40 279
1           5.1         3.5          1.4         0.2  setosa 2010   12  83
6           5.4         3.9          1.7         0.4  setosa 2011   16 113
4           4.6         3.1          1.5         0.2  setosa 2015   26 184

这在可变参数形式中按预期工作。

Answer 1

这行得通吗：

x <- matrix(rnorm(100), ncol=10)
lapply(1:ncol(x), function(i)x[,i])
# [[1]]
# [1]  0.48517941 -0.17305691 -0.77043863  0.60336573 -1.45311257  0.79958015  1.13640966  0.02676497  0.29389045
# [10] -0.01102340
# 
# [[2]]
# [1] -0.54202918 -0.31705192 -0.54335095  0.95893715  1.50479417  0.30277200  0.89060424  1.04398275 -0.05292274
# [10] -1.08171141
# 
# [[3]]
# [1] -0.4263822 -0.7633086 -0.0920494 -0.8624237  0.4733904  1.1280913 -1.3591717 -2.0045355 -0.9451451  0.5850331
# 
# [[4]]
# [1]  0.43011274 -0.31818318 -0.82670988 -1.41186748 -0.11159258  0.97936154 -0.96050860 -0.05459925 -0.64583762
# [10] -1.05754833
# 
# [[5]]
# [1]  0.03352171 -1.41914682 -0.65342097 -0.65543412 -0.64277411  0.20129441  0.79787560  0.74036594  0.85009985
# [10]  0.57234638
# 
# [[6]]
# [1]  1.53409626 -0.09687169  0.03232748 -0.29846023 -1.68693869 -0.35000084 -0.01507354  0.67449541  0.32737139
# [10] -0.25879175
# 
# [[7]]
# [1] -0.03431753 -0.73440722  1.60681714  0.05675589 -0.91227635 -0.82333341  1.24233167 -0.67889010  0.15424119
# [10]  0.11909912
# 
# [[8]]
# [1] -0.31600385  1.05633518  1.39758192  0.46613354 -1.56959308  0.01917428 -0.45930649 -0.90180761  0.14538694
# [10]  0.19565070
# 
# [[9]]
# [1]  0.24165283  1.14789319 -0.01238587 -0.20014950  0.73042111  0.47187272  2.63819369 -0.81273739 -1.83783324
# [10]  0.59991982
# 
# [[10]]
# [1] -1.0260512 -2.1172737  1.3514048  0.7677437 -0.9399838 -1.0775248  1.2656769 -0.5748148 -1.8108845  0.1093450

Answer 2

如果您想将矩阵的列传递给 order 就像调用 order(mat[,1], mat[,2], mat[,3]) 等一样，那么这一行函数可以实现：

mat_order <- function(x) do.call(order, split(x, (seq(x) - 1) %/% nrow(x)))

它首先使用一些模块化数学将矩阵列 split 转换为向量列表，然后对结果使用 do.call(order, ...)，其效果是传递每个列表元素（即每个向量）作为变量。

使用 R：将矩阵转换为向量序列以实现阶函数的可变参数方法

Using R: casting a matrix as a sequence of vectors to implement a variadic approach for order function

r

casting

variadic-functions

variadic

更新

解决方案