如何求解线性方程组?

本文介绍了如何求解线性方程组?的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

我需要求解线性方程组:

I need to solve a system of linear equations:

aQ + bP = c dQ + eP = f

位置:

a ~ N(100;10) b ~ N(-1;0.1) c ~ N(10;1) d ~ N(10;0.1) e ~ N(100;10) f ~ N(10;0.1)

到目前为止，我已经写过:

So far I have written:

a <- rnorm(100, mean=100, sd=10) b <- rnorm(100, mean=-1, sd=0.1) c <- rnorm(100, mean=10, sd=1) d <- rnorm(100, mean=10, sd=0.1) e <- rnorm(100, mean=100, sd=10) f <- rnorm(100, mean=10, sd=0.1) P <- vector() Q <- vector() for (i in 1:100) { coefs <- matrix(c(a[i],d[i],b[i],e[i]),2,2) ys <- array(c(c,f),2) solve(coefs[i], ys[i]) }

问题是for循环只为我提供了P和Q的一种解决方案，我希望for循环计算100个值集来进行a，b，c，d，e和f并存储向量Q和P上的数据.

The problem is that the for loop is only giving me one solution for P and Q and I would like that the for loop to calculate the 100 set of values do a, b, c, d, e and f and store the data on the vectors Q and P.

推荐答案

您可以尝试使用apply()

# data df = data.frame(a,b,c,d,e,f) # apply function out = t(apply(df, 1, function(x){ coefs = matrix( c(x['a'], x['d'], x['b'], x['e']), 2, 2); ys = array(c(x['c'],x['f']),2); out = solve(coefs, ys); names(out) = c('P','Q'); out}))

或者使用sapply()

out = t(sapply(seq(100), function(i){ coefs = matrix(c(a[i], d[i], b[i] ,e[i]),2,2); ys = array(c(c[i],f[i]),2); out = solve(coefs, ys); names(out) = c('P','Q'); out}))

如果您想使用for循环，这是您可以做的

and if you want to use for loop, here is what you could do

# declare matrix to store the output out = matrix(ncol=2, nrow=100) # populate declared matrix using for loop for(i in 1:100){ coefs = matrix(c(a[i],d[i],b[i],e[i]),2,2) ys = array(c(c[i],f[i]),2) out[i,] = as.vector(solve(coefs, ys)) colnames(out) = c('P','Q') out}