3

题目已经解释了这个问题。在我的问题中，向量很长，大约 1,500。我尝试的一种方法是生成矩阵如下，

在此处输入图像描述

粗略地说，这个矩阵rbind是三个对角矩阵diag(1, 3)，diag(1,2)和diag(1,1)。但是这些矩阵有不同的列数。因此rbind不适用于此处。有没有什么有效的方法来解决这个问题。

4

5 回答 5

7 
vec <- 1:4
n <- length(vec)
as.vector(rev(setNames(vec, n:1)[as.character(sequence(1:n))]))
# [1] 1 2 3 4 2 3 4 3 4 4

这里有几个技巧；as.vector是不必要的，它只是省略了向量名称。

Tyler <- function() do.call(rbind, mapply(compile, rows, nums, lst))
Julius <- function() as.vector(rev(setNames(vec, n:1)[as.character(sequence(1:n))]))

# Vector of length 3
# Unit: microseconds
#      expr     min      lq  median      uq      max neval
#   Tyler() 144.183 148.383 151.649 155.382 2241.617  1000
#  Julius()  73.724  76.058  80.724  82.590  276.236  1000

# Vector of length 1500
# Unit: seconds
#      expr    min       lq   median       uq      max neval
#  Julius() 1.2181 1.270544 1.469416 1.506019 1.518471    10
# (list of 1500 diagonal matrices took too much memory, couldn't compare)

编辑。

JuliusTwo <- function() rev(vec[n + 1 - sequence(1:n)])

vec <- 1:3
n <- length(vec)
microbenchmark(Julius(), JuliusTwo(), times = 1000)
# Unit: microseconds
#         expr    min     lq median     uq      max neval
#     Julius() 72.326 75.125 76.525 78.392  259.905  1000
#  JuliusTwo() 49.461 51.794 53.194 54.595 1950.450  1000

vec <- 1:1500
n <- length(vec)
microbenchmark(Julius(), JuliusTwo(), Henrik(x2), times = 10)
# Unit: milliseconds
#         expr       min        lq   median        uq       max neval
#     Julius() 1497.9588 1499.9438 1547.660 1582.0843 1590.2048    10
#  JuliusTwo()  157.0313  157.9193  177.682  200.7433  214.9415    10
#   Henrik(x2) 4639.1891 6157.247 7178.9953 7350.8146 7640.8685    10

Matthew <- function() {m <- matrix(rep(vec, n), n);m[lower.tri(m, diag=TRUE)]}
microbenchmark(JuliusTwo(), Matthew(), Arun(vec), times = 100)
# Unit: milliseconds
#         expr       min        lq    median        uq       max neval
#  JuliusTwo() 113.25630 121.69106 126.16566 150.42730 237.51304   100
#    Matthew() 119.59806 126.87538 152.28000 157.42816 415.27231   100
#    Arun(vec)  32.93695  37.78204  40.99725  43.19757  98.69791   100
于 2013-10-27T19:14:30.557 回答
3

你也可以很简单地用lower.tri.

vec <- 1:4
m <- matrix(rep(vec, length(vec)), length(vec))
m[lower.tri(m, diag=TRUE)]
# [1] 1 2 3 4 2 3 4 3 4 4

这比@Julius 的聪明回答要慢一点。请参阅他的基准。

于 2013-10-27T20:26:19.480 回答
2

我不知道效率，但这是我使用的解决方案：

lst <- list( 
    diag(1, 3),
    diag(1,2) ,
    diag(1,1)
)

cols <- sapply(lst, ncol)
mcol <- max(cols)
rows <- sapply(lst, nrow)
nums <- (mcol - cols)*rows


compile <- function(x, y, z) {
    if (y == 0) return(z)
    cbind(matrix(rep(0, y), nrow = x), z)
}

do.call(rbind, mapply(compile, rows, nums, lst))

#'      [,1] [,2] [,3]
#' [1,]    1    0    0
#' [2,]    0    1    0
#' [3,]    0    0    1
#' [4,]    0    1    0
#' [5,]    0    0    1
#' [6,]    0    0    1
于 2013-10-27T18:44:06.343 回答
2

已经有几个不错的解决方案。尽管如此，我还是尝试了另一种zoo选择：

library(zoo) 

# the vector
x1 <- c("a1", "a2", "a3")
n <- length(x1)

# convert to zoo object
x2 <- zoo(x1)

# lag the vector with a vector of lags
x3 <- lag(x2, k = seq(from = 0, by = 1, length.out = n))

# convert back to vector
na.omit(as.vector(x3))
# [1] "a1" "a2" "a3" "a2" "a3" "a3"
于 2013-10-27T19:29:22.880 回答
1

迭代解决方案

seq_generator=function(vec) if (length(vec)-1>0) c(vec,seq_generator(vec[-1])) else tail(vec,1)
seq_generator(1:4)
于 2013-10-27T19:21:37.713 回答