在R中，我想创建一个5x5 0,1,3,5,7矩阵，以便：
0 1 3 5 7
1 0 3 5 7
1 3 0 5 7
1 3 5 0 7
1 3 5 7 0
显然我可以生成起始矩阵：
z <- c(0,1,3,5,7)
matrix(z, ncol=5, nrow=5, byrow = TRUE)
但我不确定如何移动0的位置。我确定我必须使用某种for/in循环，但我真的不知道我到底需要做什么。
答案 0 :(得分：26)
这个怎么样：
m <- 1 - diag(5)
m[m==1] <- rep(c(1,3,5,7), each=5)
m
# [,1] [,2] [,3] [,4] [,5]
# [1,] 0 1 3 5 7
# [2,] 1 0 3 5 7
# [3,] 1 3 0 5 7
# [4,] 1 3 5 0 7
# [5,] 1 3 5 7 0
答案 1 :(得分：10)
或者我们可以这样做：
z <- c(1,3,5,7)
mat <- 1-diag(5)
mat[mat==1] <- z
t(mat)
# [,1] [,2] [,3] [,4] [,5]
# [1,] 0 1 3 5 7
# [2,] 1 0 3 5 7
# [3,] 1 3 0 5 7
# [4,] 1 3 5 0 7
# [5,] 1 3 5 7 0
另一种解决方案也是为了享受combn：
r <- integer(5)
t(combn(5, 1, function(v) {r[v]<-0;r[-v]<-z;r}))
# [,1] [,2] [,3] [,4] [,5]
# [1,] 0 1 3 5 7
# [2,] 1 0 3 5 7
# [3,] 1 3 0 5 7
# [4,] 1 3 5 0 7
# [5,] 1 3 5 7 0
或使用sapply：
v <- integer(5)
t(sapply(seq(5), function(x) {v[x]<-0;v[-x]<-z;v}))
# [,1] [,2] [,3] [,4] [,5]
# [1,] 0 1 3 5 7
# [2,] 1 0 3 5 7
# [3,] 1 3 0 5 7
# [4,] 1 3 5 0 7
# [5,] 1 3 5 7 0
答案 2 :(得分：8)
这是一个解决方案，通过对rep()的几次调用，对c()，seq()和rbind()的几次调用来构建数据向量，以及然后将其包含在对matrix()：
N <- 5L;
matrix(rep(c(0,rbind(seq(1,(N-1)*2,2),0)),rep(c(1,N),len=N*2-1)),N);
## [,1] [,2] [,3] [,4] [,5]
## [1,] 0 1 3 5 7
## [2,] 1 0 3 5 7
## [3,] 1 3 0 5 7
## [4,] 1 3 5 0 7
## [5,] 1 3 5 7 0
另一个想法是，使用两次调用diag()和cumsum()：
N <- 5L;
(1-diag(N))*(cumsum(diag(N)*2)-1);
## [,1] [,2] [,3] [,4] [,5]
## [1,] 0 1 3 5 7
## [2,] 1 0 3 5 7
## [3,] 1 3 0 5 7
## [4,] 1 3 5 0 7
## [5,] 1 3 5 7 0
注意：对于以下基准测试，我在必要时修改了每个人的解决方案，以确保它们在矩阵大小N上进行参数化。在大多数情况下，这只涉及用N替换一些文字，并用c(1,3,5,7)替换seq(1,(N-1)*2,2)的实例。我认为这是公平的。
library(microbenchmark);
josh <- function(N) { m <- 1-diag(N); m[m==1] <- rep(seq(1,(N-1)*2,2),each=N); m; };
marat <- function(N) matrix(rbind(0,col(diag(N))*2-1),nrow=N,ncol=N);
gregor <- function(N) { x = seq(1,(N-1)*2,2); t(mapply(FUN = append, after = c(0, seq_along(x)), MoreArgs = list(x = x, values = 0))); };
barkley <- function(N) { my_vec <- seq(1,(N-1)*2,2); my_val <- 0; my_mat <- matrix(NA, ncol = length(my_vec)+1, nrow = length(my_vec)+1); for (i in 1:nrow(my_mat)) { my_mat[i, i] <- my_val; my_mat[i, -i] <- my_vec; }; my_mat; };
m0h3n <- function(N) { z <- seq(1,(N-1)*2,2); mat=1-diag(N); mat[mat==1]=z; t(mat); };
bgoldst1 <- function(N) matrix(rep(c(0,rbind(seq(1,(N-1)*2,2),0)),rep(c(1,N),len=N*2-1)),N);
bgoldst2 <- function(N) (1-diag(N))*(cumsum(diag(N)*2)-1);
## small-scale: 5x5
N <- 5L;
ex <- josh(N);
identical(ex,marat(N));
## [1] TRUE
identical(ex,gregor(N));
## [1] TRUE
identical(ex,barkley(N));
## [1] TRUE
identical(ex,m0h3n(N));
## [1] TRUE
identical(ex,bgoldst1(N));
## [1] TRUE
identical(ex,bgoldst2(N));
## [1] TRUE
microbenchmark(josh(N),marat(N),gregor(N),barkley(N),m0h3n(N),bgoldst1(N),bgoldst2(N));
## Unit: microseconds
## expr min lq mean median uq max neval
## josh(N) 20.101 21.8110 25.71966 23.0935 24.8045 108.197 100
## marat(N) 5.987 8.1260 9.01131 8.5535 8.9820 24.805 100
## gregor(N) 49.608 51.9605 57.61397 53.8850 61.7965 98.361 100
## barkley(N) 29.081 32.0750 36.33830 33.7855 41.9110 54.740 100
## m0h3n(N) 22.666 24.8040 28.45663 26.0870 28.4400 59.445 100
## bgoldst1(N) 20.528 23.0940 25.49303 23.5220 24.8050 56.879 100
## bgoldst2(N) 3.849 5.1320 5.73551 5.5600 5.9880 16.251 100
## medium-scale: 50x50
N <- 50L;
ex <- josh(N);
identical(ex,marat(N));
## [1] TRUE
identical(ex,gregor(N));
## [1] TRUE
identical(ex,barkley(N));
## [1] TRUE
identical(ex,m0h3n(N));
## [1] TRUE
identical(ex,bgoldst1(N));
## [1] TRUE
identical(ex,bgoldst2(N));
## [1] TRUE
microbenchmark(josh(N),marat(N),gregor(N),barkley(N),m0h3n(N),bgoldst1(N),bgoldst2(N));
## Unit: microseconds
## expr min lq mean median uq max neval
## josh(N) 106.913 110.7630 115.68488 113.1145 116.1080 179.187 100
## marat(N) 62.866 65.4310 78.96237 66.7140 67.9980 1163.215 100
## gregor(N) 195.438 205.2735 233.66129 213.6130 227.9395 1307.334 100
## barkley(N) 184.746 194.5825 227.43905 198.6455 207.1980 1502.771 100
## m0h3n(N) 73.557 76.1230 92.48893 78.6885 81.6820 1176.045 100
## bgoldst1(N) 51.318 54.3125 95.76484 56.4500 60.0855 1732.421 100
## bgoldst2(N) 18.817 21.8110 45.01952 22.6670 23.5220 1118.739 100
## large-scale: 1000x1000
N <- 1e3L;
ex <- josh(N);
identical(ex,marat(N));
## [1] TRUE
identical(ex,gregor(N));
## [1] TRUE
identical(ex,barkley(N));
## [1] TRUE
identical(ex,m0h3n(N));
## [1] TRUE
identical(ex,bgoldst1(N));
## [1] TRUE
identical(ex,bgoldst2(N));
## [1] TRUE
microbenchmark(josh(N),marat(N),gregor(N),barkley(N),m0h3n(N),bgoldst1(N),bgoldst2(N));
## Unit: milliseconds
## expr min lq mean median uq max neval
## josh(N) 40.32035 43.42810 54.46468 45.36386 80.17241 90.69608 100
## marat(N) 41.00074 45.34248 54.74335 47.00904 50.74608 93.85429 100
## gregor(N) 33.65923 37.82393 50.50060 40.24914 75.09810 83.27246 100
## barkley(N) 31.02233 35.42223 43.08745 36.85615 39.81999 85.28585 100
## m0h3n(N) 27.08622 31.00202 38.98395 32.33244 34.33856 90.82652 100
## bgoldst1(N) 12.53962 13.02672 18.31603 14.92314 16.96433 59.87945 100
## bgoldst2(N) 13.23926 16.87965 28.81906 18.92319 54.60009 62.01258 100
## very large scale: 10,000x10,000
N <- 1e4L;
ex <- josh(N);
identical(ex,marat(N));
## [1] TRUE
identical(ex,gregor(N));
## [1] TRUE
identical(ex,barkley(N));
## [1] TRUE
identical(ex,m0h3n(N));
## [1] TRUE
identical(ex,bgoldst1(N));
## [1] TRUE
identical(ex,bgoldst2(N));
## [1] TRUE
microbenchmark(josh(N),marat(N),gregor(N),barkley(N),m0h3n(N),bgoldst1(N),bgoldst2(N));
## Unit: seconds
## expr min lq mean median uq max neval
## josh(N) 3.698714 3.908910 4.067409 4.046770 4.191938 4.608312 100
## marat(N) 6.440882 6.977273 7.272962 7.223293 7.493600 8.471888 100
## gregor(N) 3.546885 3.850812 4.032477 4.022563 4.221085 4.651799 100
## barkley(N) 2.955906 3.162409 3.324033 3.279032 3.446875 4.444848 100
## m0h3n(N) 3.355968 3.667484 3.829618 3.777151 3.973279 4.649226 100
## bgoldst1(N) 1.044510 1.260041 1.363827 1.369945 1.441194 1.819248 100
## bgoldst2(N) 1.144168 1.391711 1.517189 1.519653 1.629994 2.478636 100
答案 3 :(得分：7)
也许不是有史以来最美丽的解决方案，但其简洁性可能很优雅：
my_vec <- c(1,3,5,7)
my_val <- 0
my_mat <- matrix(NA, ncol = length(my_vec)+1, nrow = length(my_vec)+1)
for (i in 1:nrow(my_mat)) {
my_mat[i, i] <- my_val
my_mat[i, -i] <- my_vec
}
my_mat
[,1] [,2] [,3] [,4] [,5]
[1,] 0 1 3 5 7
[2,] 1 0 3 5 7
[3,] 1 3 0 5 7
[4,] 1 3 5 0 7
[5,] 1 3 5 7 0
答案 4 :(得分：6)
您可以使用
n <- 5
matrix(rbind(0,col(diag(n))*2-1),nrow=n,ncol=n)
答案 5 :(得分：6)
有趣的问题！在四处寻找时，我看到append有一个after参数。
x = c(1, 3, 5, 7)
t(mapply(FUN = append, after = c(0, seq_along(x)),
MoreArgs = list(x = x, values = 0)))
# [,1] [,2] [,3] [,4] [,5]
# [1,] 0 1 3 5 7
# [2,] 1 0 3 5 7
# [3,] 1 3 0 5 7
# [4,] 1 3 5 0 7
# [5,] 1 3 5 7 0
答案 6 :(得分：1)
另一个选项，直接构建每一行：
<?php
$text = $article->introtext;
$text = preg_replace('@<div[^>]*class=(["\'])mosimage_caption\\1[^>]*>[^>]* </div>@', '', $text );
$preserv = "";
$text = strip_tags($text, /* exclude */ $preserv );
$text = preg_replace("@<script[^>]*?>.*?</script>@si","",$text);
$text = preg_replace('/{.+?}/','',$text);
$text = preg_replace('/(( )|( ))+/',' ',$text);
$text = preg_replace('/"/',' ',$text);
$text = str_replace(array("\r\n", "\n", "\r"), "", $text);
$text = str_replace("caption","", $text);
$shorttext = substr($text,0,300);
echo '<p>' . $shorttext . '...</p><p><a href="' . JRoute::_(ContentHelperRoute::getArticleRoute($article->slug, $article->catid)) . '">Read more</a></p>';
?>