[问答] R软件计算矩阵的逆出错，求助！ [推广有奖]

说明这个x是不可逆的
试一下qr(x)$rank只有15

> r1
         [,1]    [,2]    [,3]    [,4]    [,5]    [,6]    [,7]    [,8]    [,9]   [,10]   [,11]   [,12]   [,13]   [,14]   [,15]   [,16]   [,17]   [,18]   [,19]   [,20]
[1,] -0.1200 -0.2230 -0.2611 -0.2881 -0.4679  0.3515 -0.4426 -0.3040 -0.0904 -0.0511 -0.2414 -0.2119 -0.2579  0.1492 -0.4840 -0.2341 -0.1738 -0.2291 -0.3549  0.0675
[2,] -0.1303 -0.5107 -0.3702 -0.1868 -0.2471  0.1682 -0.3117 -0.4058 -0.3027 -0.2676  0.0445 -0.0152 -0.1738  0.1492 -0.3975 -0.3682 -0.0510 -0.2517 -0.4476 -0.2257
[3,]  0.0357 -0.1078 -0.0241 -0.3157 -0.0834 -0.4959 -0.3341 -0.2580 -0.0990 -0.2223 -0.4291 -0.1850 -0.2396 -0.0626 -0.1749  0.0621 -0.1823 -0.1550 -0.0879 -0.0253
[4,] -0.0620 -0.0158 -0.2603 -0.1256 -0.2012 -0.3714 -0.1309 -0.2590  0.0452  0.5421 -0.0709  0.1421 -0.2880 -0.0505 -0.0848 -0.2400  0.0197  0.0293 -0.2741  0.1356
[5,]  0.3994  0.7050  0.7508  0.8544  0.1934  0.5043  1.2747  1.0809  0.3380  0.3644  0.2741  0.0623  0.5270  0.2846  1.0000  0.8712  0.1853  0.0078  0.6307  0.2740
[6,]  0.1909  0.4238  0.1572  0.8140  0.2425  0.8447  0.0629  0.1431  0.2168 -0.0682  0.0430  0.4418  0.0938 -0.0766  0.1114  0.0674 -0.0388  0.3371 -0.0700 -0.0762
[7,]  0.1256 -0.2126  0.3816 -0.1671  0.0598 -0.2253  0.1468 -0.1724  0.0431 -0.1938  0.0936  0.4157 -0.0807  0.2760  0.2482  0.1818  0.1509  0.0199 -0.0700 -0.0643
[8,]  0.4743  0.0652  0.2845  0.1415  0.5141  0.3876  0.1684  0.2867  0.3275  0.3318  0.3438  0.0658  0.8108  0.5456  0.3809  0.0883  0.0198  0.2678 -0.0700  0.4956
[9,] -0.0677 -0.0920  0.0668 -0.0088 -0.1102 -0.1036 -0.1661  0.4003 -0.0256  0.0390  0.0784  0.1332 -0.1172  0.3707 -0.1290 -0.0907  0.1559 -0.0958  0.0822  0.2227
[10,] -0.2991 -0.3846 -0.1060 -0.3606 -0.1345 -0.3899 -0.1846 -0.0416 -0.3333 -0.2131  0.2326  0.1533 -0.0186  0.2008 -0.2836 -0.3023 -0.2520 -0.0873 -0.1673  0.0858
[11,]  0.3476  0.1714  0.2387  0.0104  0.4168  0.2094  0.7257  0.4952  0.1580  0.2906  0.4278  0.1313  0.3497  0.2637  0.2647  0.6443  0.4036  0.4009  0.2500  0.2322
[12,]  0.0294 -0.0796  0.3836  0.0710  0.0087  0.4883  0.2094 -0.1001  0.2786 -0.1269  0.0369 -0.0512  0.0946  0.4297 -0.1347  0.3315  0.1172 -0.1797 -0.0510  0.3237
[13,] -0.0253  0.1110  0.0155 -0.0214 -0.0791  0.1648  0.1111 -0.0072  0.2499 -0.0379 -0.1568 -0.1631  0.0134 -0.1829 -0.0126  0.0934  0.0285 -0.0206  0.0331 -0.1417
[14,] -0.1398  0.0395 -0.1457 -0.1257  0.1295  0.0975 -0.1771 -0.2180  0.0503 -0.0668 -0.0425 -0.1546 -0.1643  0.1408 -0.0256 -0.0052  0.0285  0.0062 -0.1232 -0.1260
[15,] -0.0367 -0.0852 -0.1704  0.0500  0.0162 -0.0686 -0.1952 -0.1231 -0.1494  0.1583  0.0128 -0.3477 -0.1828 -0.0902 -0.1434 -0.1793 -0.0113 -0.1874 -0.0912 -0.0052
[16,] -0.1510  0.0753 -0.1546 -0.2406 -0.0964 -0.2859 -0.2372 -0.1972 -0.1351 -0.1422  0.0352  0.1296  0.0305  0.0394 -0.1230 -0.1914  0.0557 -0.1892 -0.3092  0.0161
        [,21]   [,22]   [,23]   [,24]   [,25]   [,26]   [,27]
[1,] -0.1123 -0.2020 -0.4509 -0.1725 -0.2472 -0.1807 -0.3353
[2,] -0.4330 -0.4654 -0.1133 -0.5626 -0.2113 -0.3828 -0.2970
[3,] -0.2752 -0.3090 -0.2071 -0.2634 -0.1333 -0.2556 -0.2964
[4,] -0.0123  0.1157 -0.2815 -0.0315 -0.4043  0.0700 -0.0965
[5,]  0.2837  0.2990  0.8148  0.8103  0.4991  0.4860  1.0174
[6,]  0.5459  0.0350  0.1997  0.8839 -0.0593  0.2809  0.3488
[7,] -0.1827  0.1462 -0.0153 -0.3472 -0.0638  0.0180 -0.1497
[8,]  0.0374  0.5864  0.2344  0.1032  0.4181  1.1109  0.4331
[9,] -0.1212  0.3790 -0.0707 -0.0844  0.0952 -0.1333 -0.1537
[10,] -0.2789  0.0738 -0.4189 -0.3819 -0.0599 -0.2320 -0.3424
[11,]  0.2389  0.6479  0.0956  0.0999  0.2038  0.2265  0.5048
[12,] -0.0214 -0.0562  0.1779  0.0193  0.6779  0.0690  0.2128
[13,]  0.0572 -0.1536 -0.0200 -0.0256 -0.0397  0.0070 -0.0130
[14,] -0.0161  0.0187  0.1092  0.0034 -0.1181 -0.2045  0.0575
[15,] -0.1432  0.0486 -0.1376 -0.1948 -0.2989 -0.2076 -0.1107
[16,]  0.0318 -0.0361 -0.2326  0.2308 -0.2503 -0.2641 -0.2716

大神在吗？
@qoiqpwqr

x=var（r1）  求得r1的方差。   对方差求逆？对吗？

我感兴趣的是你是如何将二楼的数据弄入R里的？有什么好的方法能否分享一下？

土办法是， data1 <- read.table(text=" # 复制1:16行 ")  
data2 <- read.table(text="  剩下的 1:16行" )
data=cbind(data1,data2)

不知道有没有其他好的办法

主要是那个行首的中括号怎么处理，要用regex？

不用管【】
adata2 <-read.table(text="[,21]   [,22]   [,23]   [,24]   [,25]   [,26]   [,27]
                   [1,] -0.1123 -0.2020 -0.4509 -0.1725 -0.2472 -0.1807 -0.3353
                   [2,] -0.4330 -0.4654 -0.1133 -0.5626 -0.2113 -0.3828 -0.2970")

solve(x,tol=1e-21)就能得到一个结果。x的秩最多是16，而你的列数是27个。得到是广义逆。这是要求估计系数的t统计量吗？