enter image description here我想通过使用for循环，再次从两个矩阵中创建一个10 x 8矩阵，尺寸为10 x 8。
我有矩阵a和e，我想将以下代码的结果保存到矩阵中。但是当我运行代码时，矩阵chi是一个空矩阵，除了最后一列的最后一行。我是R的新人，因此可以提供任何帮助。谢谢。
chi <- matrix(nrow = 10, ncol = 8, byrow = T)
i <- nrow(a)
j <- ncol(a)
k <- nrow(e)
l <- ncol(e)
m <- nrow(chi)
n <- ncol(chi)
for (i in 1:nrow(a)) {
for (j in 1:ncol(a)) {
for (k in 1:nrow(e)) {
for (l in 1:ncol(e))
chi[m, n] <- ((a[i, j] - e[k, l]) ^ 2 / (e[k, l] * (1 - e[k, l])))
}
}
}
答案 0 :(得分：1)
重新考虑使用 any 嵌套for循环，因为所有输入 a 和 e 都是等长对象：
chi <- ((a - e) ^ 2 / (e * (1 - e)))
使用嵌套的for循环方法，每次内部循环遍历都会覆盖尝试的矩阵单元分配，并且仅保存最后一个实例。
为演示，请考虑以下随机矩阵（为可重复性而播种）：
set.seed(1162018)
a <- matrix(runif(800), nrow = 10, ncol = 8)
e <- matrix(runif(800), nrow = 10, ncol = 8)
具有以下输出：
chi2 <- ((a - e) ^ 2 / (e * (1 - e)))
chi2
# [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
# [1,] 1.090516287 5.314965506 0.30221649 4.3078030566 0.08185310 0.33991625 7.475638e-01 7.136321e+01
# [2,] 0.339472596 0.037831564 1.00181544 0.0075194551 0.27228312 20.74823838 2.308509e-04 1.264312e-04
# [3,] 0.001493967 0.009102797 17.76508355 0.0318190760 0.08133848 0.90538852 1.425952e-01 3.600838e-02
# [4,] 25.941857200 2.182678801 0.52170472 0.5485710933 0.57015681 0.09332506 2.631002e-01 4.897862e-01
# [5,] 4.341993499 0.075724451 0.03409925 0.0058830640 0.15290151 0.83227284 2.982630e+02 2.615268e-01
# [6,] 0.327661207 0.058150213 0.17328257 0.3161902785 4.48620227 0.14685330 2.996204e+00 1.888419e+01
# [7,] 0.456397833 1.446942556 0.51597191 0.2051742161 0.20440765 0.58169351 5.345522e+00 1.320896e-03
# [8,] 12.844776005 0.753941152 0.36425134 0.0003481929 0.34011118 2.38649404 1.082046e-01 1.817180e-01
# [9,] 0.042779101 0.119540004 1.41313002 0.1262586599 0.36583013 1.76476721 1.353301e+00 1.670491e-01
# [10,] 4.729182008 5.257386394 0.62181731 0.0000251250 0.32324943 0.08491841 6.627723e+00 2.127289e+00
请注意，chi2的第一个，第二个，最后一个元素一直与您的原始公式一致，就像仅使用单个值一样。 all.equal()证明科学符号之间没有价值差异。
((a[1, 1] - e[1, 1]) ^ 2 / (e[1, 1] * (1 - e[1, 1])))
# [1] 1.090516
((a[1, 2] - e[1, 2]) ^ 2 / (e[1, 2] * (1 - e[1, 2])))
# [1] 1.090516
# ...
((a[10, 8] - e[10, 8]) ^ 2 / (e[10, 8] * (1 - e[10, 8])))
# [1] 2.127289
all.equal(2.127289e+00, 2.127289)
# [1] TRUE
循环处理不正确
但是，将for循环调整为使用chi[i,j]赋值会产生值，但仔细观察并不会不准确地与原始公式对齐：
chi <- matrix(nrow = 10, ncol = 8, byrow = T)
i <- nrow(a)
j <- ncol(a)
k <- nrow(e)
l <- ncol(e)
m <- nrow(chi)
n <- ncol(chi)
for (i in 1:nrow(a)) {
for (j in 1:ncol(a)) {
for (k in 1:nrow(e)) {
for (l in 1:ncol(e))
chi[i,j] <- ((a[i, j] - e[k, l]) ^ 2 / (e[k, l] * (1 - e[k, l])))
}
}
}
chi
# [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
# [1,] 3.409875713 1.91797098 0.983457185 0.72023148 0.96731753 0.047236836 2.20811240 0.6073649
# [2,] 0.011756997 2.96049899 3.614632753 1.30476270 2.49116488 0.074379894 1.01941080 0.3796867
# [3,] 2.061628776 0.03227113 0.691592758 2.58226782 0.17603261 4.377353084 1.07957101 0.9584883
# [4,] 5.477395731 0.07409188 5.287871705 1.86472765 2.02597697 0.078780553 6.20319269 2.6099405
# [5,] 4.342937737 3.57579681 1.016981597 2.83351392 1.11431922 0.083484410 0.08412765 0.5525810
# [6,] 0.008175703 2.63310577 0.005053893 3.69703754 0.05993078 0.004768071 5.92075341 4.2435415
# [7,] 1.051921956 0.31217144 5.624012725 0.90161687 0.43301151 0.156739757 0.72284317 1.2243496
# [8,] 4.941310521 4.85504735 0.021515999 3.66512027 0.08358373 3.603038468 0.38618455 6.1389345
# [9,] 0.559136535 5.08204325 2.999036687 2.72726724 5.99168376 0.319859158 0.59398961 3.6221932
# [10,] 0.001668949 2.97353267 4.703763876 0.04979429 5.31715581 0.053267595 2.09966809 2.1272893
此处，for循环仅返回最后一个实例，因为chi[i,j]在循环过程中被多次覆盖。结果，chi矩阵的所有元素都使用 e 的 last 元素：
((a[1, 1] - e[10, 8]) ^ 2 / (e[10, 8] * (1 - e[10, 8])))
# [1] 3.409876
((a[1, 2] - e[10, 8]) ^ 2 / (e[10, 8] * (1 - e[10, 8])))
# [1] 1.917971
# ...
((a[10, 8] - e[10, 8]) ^ 2 / (e[10, 8] * (1 - e[10, 8])))
# [1] 2.127289
相反，使用chi[k,l]进行循环分配。
for (i in 1:nrow(a)) {
for (j in 1:ncol(a)) {
for (k in 1:nrow(e)) {
for (l in 1:ncol(e))
chi[k,l] <- ((a[i, j] - e[k, l]) ^ 2 / (e[k, l] * (1 - e[k, l])))
}
}
}
chi
# [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
# [1,] 5.649285e-01 5.813300e+00 0.035949545 10.14845208 0.002533313 0.405749651 0.711058301 2.592142e+01
# [2,] 7.481556e+00 4.531135e-05 0.455696004 0.09284383 0.192074706 4.178867177 0.105489574 3.541626e-01
# [3,] 4.953702e-04 6.703029e+00 41.109139456 0.08957573 1.511080005 0.254656165 0.004840752 2.805246e-01
# [4,] 1.152237e+01 2.556255e-02 0.018652264 0.65975403 0.515919955 0.280219679 0.124379946 7.777978e-01
# [5,] 2.126765e+00 5.356927e-01 0.251885418 0.06540162 0.008580900 0.003271672 41.259025738 2.963719e-06
# [6,] 1.401345e-01 1.603721e-02 0.334385097 0.05865054 0.622973490 0.608273911 0.888928067 1.046868e+01
# [7,] 1.018507e-01 1.756129e-01 0.005676374 0.72309875 0.011666290 0.314863595 12.420604213 7.778975e-02
# [8,] 6.082752e+00 1.250805e-01 0.287099891 0.17209992 0.050136187 1.339028574 1.059674334 2.627769e-01
# [9,] 8.005223e-02 9.260464e-02 2.823995704 0.04935770 0.020361815 0.258144647 0.275514317 9.392584e-03
# [10,] 4.952038e-01 3.870331e+00 0.089420009 1.05729955 0.002429084 0.349966871 6.702385325 2.127289e+00
结果，所有矩阵元素都使用 a 的 last 值：
((a[10, 8] - e[1, 1]) ^ 2 / (e[1, 1] * (1 - e[1, 1])))
# [1] 0.5649285
all.equal(5.649285e-01, 0.5649285)
# [1] TRUE
((a[10, 8] - e[1, 2]) ^ 2 / (e[1, 2] * (1 - e[1, 2])))
# [1] 5.8133
all.equal(5.813300e+00, 5.8133)
# [1] TRUE
# ...
((a[10, 8] - e[10, 8]) ^ 2 / (e[10, 8] * (1 - e[10, 8])))
# [1] 2.127289