Question

我在R中有以下输出：

> out

, , 1

             Estimate Std. Error   z value     Pr(>|z|)
(Intercept) -0.8968692  0.1995274 -4.494968 6.958036e-06
X            0.5803202  0.2190155  2.649676 8.056901e-03
Z            0.2535625  0.2232445  1.135806 2.560376e-01
X:Z          0.3375088  0.2659863  1.268895 2.044784e-01

, , 2

              Estimate Std. Error   z value     Pr(>|z|)
(Intercept) -0.8624345  0.1888867 -4.565884 4.973944e-06
X            0.2544430  0.1784808  1.425604 1.539826e-01
Z            0.4704287  0.2387451  1.970423 4.878996e-02
X:Z          0.4813516  0.2239360  2.149505 3.159440e-02

, , 3

              Estimate Std. Error    z value     Pr(>|z|)
(Intercept) -1.0576301  0.1896436 -5.5769362 2.447916e-08
X            0.1515228  0.1917464  0.7902253 4.293962e-01
Z            0.1551839  0.2686790  0.5775810 5.635470e-01
X:Z          0.1673263  0.2872858  0.5824384 5.602714e-01

现在我想在这三个重复中计算相应的估计系数的平均值。也就是说，对于拦截，平均值为（-0.8968692-0.8624345-1.0576301）/ 3 = -0.9389779

所以我试过了：

> out[1,,]
                    [,1]          [,2]          [,3]
Estimate   -8.968692e-01 -8.624345e-01 -1.057630e+00
Std. Error  1.995274e-01  1.888867e-01  1.896436e-01
z value    -4.494968e+00 -4.565884e+00 -5.576936e+00
Pr(>|z|)    6.958036e-06  4.973944e-06  2.447916e-08

mean(out[1,,])
[1] -1.406388

这是不正确的。

然后我尝试了

 mean(out[1,,])$estimates
 Error in mean(out[1, , ])$estimates : 
 $ operator is invalid for atomic vectors

如何计算相应估算值的平均值？

Answer 1

您可以使用apply。这是一个可重复的例子。

set.seed(357)
xy <- array(runif(27), dim = c(3, 3, 3))
xy

, , 1

           [,1]      [,2]      [,3]
[1,] 0.10728121 0.2248621 0.9337365
[2,] 0.05504568 0.6534852 0.7474587
[3,] 0.27987766 0.4634314 0.7783997

, , 2

          [,1]      [,2]      [,3]
[1,] 0.9913409 0.4601817 0.1716801
[2,] 0.6245529 0.2254525 0.3195294
[3,] 0.5245744 0.5898001 0.8953055

, , 3

          [,1]       [,2]      [,3]
[1,] 0.6831485 0.45645455 0.7077218
[2,] 0.4317410 0.36626006 0.6886439
[3,] 0.1809770 0.04334935 0.3717553

mean(c(0.107, 0.991, 0.683))

[1] 0.5936667

apply(xy, MARGIN = 1:2, FUN = mean)

          [,1]      [,2]      [,3]
[1,] 0.5939235 0.3804995 0.6043794
[2,] 0.3704465 0.4150659 0.5852107
[3,] 0.3284764 0.3655270 0.6818202

或者您可以使用rowMeans

  t(apply(xy, 1, FUN=rowMeans))
  #         [,1]      [,2]      [,3]
  #[1,] 0.5939235 0.3804995 0.6043794
  #[2,] 0.3704465 0.4150659 0.5852107
  #[3,] 0.3284764 0.3655270 0.6818202

计算R中三维数据的相应系数的均值

1 个答案: