创建一个三角形矩阵 [英] creating a triangular matrix

问题描述

必须有一种优雅的方法来做到这一点，但我不知道是这样的:

There must an elegant way to do this but I can't figure out so:

列是从1到0右移的概率

Columns are probabilities from 1 to 0 going right

行是从0到1下降的概率

Rows are probabilities from 0 to 1 going down

此模糊代码产生所需的结果(但我想使用比此更大的矩阵来完成此操作):

This kludgy code produces see the desired result (but I want to do it with a much larger matrix than this):

# Vector entries are rowname - colname, if >= 0
#
rb0 <-  c(NA,NA,NA,NA,NA,NA,NA,NA,NA,NA, 0)
rb1 <-  c(NA,NA,NA,NA,NA,NA,NA,NA,NA, 0,.1)
rb2 <-  c(NA,NA,NA,NA,NA,NA,NA,NA, 0,.1,.2)
rb3 <-  c(NA,NA,NA,NA,NA,NA,NA, 0,.1,.2,.3)
rb4 <-  c(NA,NA,NA,NA,NA,NA, 0,.1,.2,.3,.4)
rb5 <-  c(NA,NA,NA,NA,NA, 0,.1,.2,.3,.4,.5)
rb6 <-  c(NA,NA,NA,NA, 0,.1,.2,.3,.4,.5,.6)
rb7 <-  c(NA,NA,NA, 0,.1,.2,.3,.4,.5,.6,.7)
rb8 <-  c(NA,NA, 0,.1,.2,.3,.4,.5,.6,.7,.8)
rb9 <-  c(NA, 0,.1,.2,.3,.4,.5,.6,.7,.8,.9)
rb10 <- c( 0,.1,.2,.3,.4,.5,.6,.7,.8,.9,1 )
indbias <- rbind(rb0,rb1,rb2,rb3,rb4,rb5,rb6,rb7,rb8,rb9,rb10)
colnames(indbias) <- seq(1,0,by=-.1)
rownames(indbias) <- seq(0,1,by=.1)
indbias

谢谢！

推荐答案

 mat <- matrix(NA, 10,10)
 mat[row(mat)+col(mat) >=11] <- (row(mat)+col(mat) -11)[row(mat)+col(mat)>=11]/10
 mat
      [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
 [1,]   NA   NA   NA   NA   NA   NA   NA   NA   NA   0.0
 [2,]   NA   NA   NA   NA   NA   NA   NA   NA  0.0   0.1
 [3,]   NA   NA   NA   NA   NA   NA   NA  0.0  0.1   0.2
 [4,]   NA   NA   NA   NA   NA   NA  0.0  0.1  0.2   0.3
 [5,]   NA   NA   NA   NA   NA  0.0  0.1  0.2  0.3   0.4
 [6,]   NA   NA   NA   NA  0.0  0.1  0.2  0.3  0.4   0.5
 [7,]   NA   NA   NA  0.0  0.1  0.2  0.3  0.4  0.5   0.6
 [8,]   NA   NA  0.0  0.1  0.2  0.3  0.4  0.5  0.6   0.7
 [9,]   NA  0.0  0.1  0.2  0.3  0.4  0.5  0.6  0.7   0.8
[10,]    0  0.1  0.2  0.3  0.4  0.5  0.6  0.7  0.8   0.9

我认为这比plyr解决方案要快得多，而且我碰巧认为它更容易理解.基本上，它为右下角三角形"中的条目设置一个测试，然后将该测试"矩阵bu 10的结果相除.您可以使用以下代码查看测试矩阵:

I think this will be much faster than a plyr solution and I happen to think it is easier to comprehend. It basically sets up a test for the entries that are in the lower right hand "triangle" and then divides the results of that "test" matrix bu 10. You can look at the test matrix with this code:

row(mat)+col(mat) -11

我认为有可能使矩阵一次如sebastian-c所示，然后进行一次测试以进行NA设置可能会更快(对row和col的调用次数为三分之一)但它似乎只有速度的三分之一.看起来这两个seq调用花费的时间比多余的要多:

I thought it possible that making the matrix once as sebastian-c illustrated and then doing a single test to do the NA setting might be faster ( with one third the number of calls to row and col) but it appears to be only one third as fast. It looks like the two seq calls take more time than the extra :

mat <- round(outer(seq(-0.5, 0.5, 0.1), seq(-0.5, 0.5, 0.1), `+`), 1)
is.na(mat) <- row(mat)+col(mat) <= 11
mat

我确实基于鲜为人知的embed函数找到了另一种解决方案:

I did find another solution based on the little known embed function:

mat <- embed(seq(-1,1, by=0.1), 11 )[,11:1]
is.na(mat) <- row(mat)+col(mat) <= 11

尽管它比新解决方案快50％，但仍然比原始解决方案慢.

Although it is 50% faster than the new solution, it is still slower than the original.

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