我使用minpack.lm包中的 nls.lm 来适应很多非线性模型。
由于初始参数估计时的奇异梯度矩阵，它经常在20次迭代后失败。
问题是当我看到failling之前的迭代（trace = T）我可以看到结果没问题。
可重复的示例：
数据：
df <- structure(list(x1 = c(7L, 5L, 10L, 6L, 9L, 10L, 2L, 4L, 9L, 3L,
11L, 6L, 4L, 0L, 7L, 12L, 9L, 11L, 11L, 0L, 2L, 3L, 5L, 6L, 6L,
9L, 1L, 7L, 7L, 4L, 3L, 13L, 12L, 13L, 5L, 0L, 5L, 6L, 6L, 7L,
5L, 10L, 6L, 10L, 0L, 7L, 9L, 12L, 4L, 5L, 6L, 3L, 4L, 5L, 5L,
0L, 9L, 9L, 1L, 2L, 2L, 13L, 8L, 2L, 5L, 10L, 6L, 11L, 5L, 0L,
4L, 4L, 8L, 9L, 4L, 2L, 12L, 4L, 10L, 7L, 0L, 4L, 4L, 5L, 8L,
8L, 12L, 4L, 6L, 13L, 5L, 12L, 1L, 6L, 4L, 9L, 11L, 11L, 6L,
10L, 10L, 0L, 3L, 1L, 11L, 4L, 3L, 13L, 5L, 4L, 2L, 3L, 11L,
7L, 0L, 9L, 6L, 11L, 6L, 13L, 1L, 5L, 0L, 6L, 4L, 8L, 2L, 3L,
7L, 9L, 12L, 11L, 7L, 4L, 10L, 0L, 6L, 1L, 7L, 2L, 6L, 3L, 1L,
6L, 10L, 12L, 7L, 7L, 6L, 6L, 1L, 7L, 8L, 7L, 7L, 5L, 7L, 10L,
10L, 11L, 7L, 1L, 8L, 3L, 12L, 0L, 11L, 8L, 5L, 0L, 6L, 3L, 2L,
2L, 8L, 9L, 2L, 8L, 2L, 13L, 10L, 2L, 12L, 6L, 13L, 2L, 11L,
1L, 12L, 6L, 7L, 9L, 8L, 10L, 2L, 6L, 0L, 2L, 11L, 2L, 3L, 9L,
12L, 1L, 11L, 11L, 12L, 4L, 6L, 9L, 1L, 4L, 1L, 8L, 8L, 6L, 1L,
9L, 8L, 2L, 10L, 10L, 1L, 2L, 0L, 11L, 6L, 6L, 0L, 4L, 13L, 4L,
8L, 4L, 10L, 9L, 6L, 11L, 8L, 1L, 6L, 5L, 10L, 8L, 10L, 8L, 0L,
3L, 0L, 6L, 7L, 4L, 3L, 7L, 7L, 8L, 6L, 2L, 9L, 5L, 7L, 7L, 0L,
7L, 2L, 5L, 5L, 7L, 5L, 7L, 8L, 6L, 1L, 2L, 6L, 0L, 8L, 10L,
0L, 10L), x2 = c(4L, 6L, 1L, 5L, 4L, 1L, 8L, 9L, 4L, 7L, 2L,
6L, 9L, 11L, 5L, 1L, 3L, 2L, 2L, 12L, 8L, 9L, 6L, 4L, 4L, 2L,
9L, 6L, 6L, 6L, 8L, 0L, 0L, 0L, 8L, 10L, 7L, 7L, 4L, 5L, 5L,
3L, 6L, 3L, 12L, 6L, 1L, 0L, 8L, 6L, 6L, 7L, 8L, 5L, 8L, 11L,
3L, 2L, 12L, 11L, 10L, 0L, 2L, 8L, 8L, 3L, 7L, 2L, 7L, 10L, 7L,
8L, 2L, 4L, 7L, 11L, 1L, 8L, 2L, 5L, 11L, 9L, 7L, 5L, 5L, 3L,
1L, 8L, 4L, 0L, 5L, 0L, 12L, 5L, 9L, 1L, 2L, 0L, 5L, 0L, 2L,
10L, 9L, 10L, 0L, 8L, 10L, 0L, 6L, 8L, 8L, 7L, 1L, 6L, 10L, 1L,
5L, 1L, 6L, 0L, 12L, 7L, 13L, 6L, 9L, 2L, 11L, 10L, 5L, 2L, 0L,
2L, 5L, 6L, 2L, 10L, 4L, 10L, 4L, 9L, 5L, 9L, 11L, 4L, 3L, 1L,
6L, 3L, 7L, 7L, 10L, 3L, 3L, 6L, 3L, 7L, 4L, 1L, 0L, 1L, 4L,
11L, 4L, 10L, 0L, 11L, 0L, 3L, 5L, 11L, 5L, 8L, 10L, 9L, 4L,
3L, 10L, 4L, 10L, 0L, 3L, 9L, 1L, 7L, 0L, 8L, 1L, 11L, 0L, 5L,
4L, 2L, 2L, 0L, 11L, 6L, 13L, 9L, 1L, 9L, 7L, 3L, 1L, 12L, 2L,
2L, 1L, 6L, 4L, 2L, 10L, 6L, 10L, 2L, 3L, 4L, 9L, 2L, 5L, 10L,
0L, 0L, 10L, 9L, 12L, 0L, 7L, 5L, 10L, 6L, 0L, 9L, 4L, 8L, 1L,
3L, 5L, 2L, 4L, 12L, 4L, 5L, 2L, 5L, 0L, 2L, 10L, 8L, 10L, 7L,
3L, 8L, 8L, 6L, 3L, 5L, 6L, 11L, 4L, 5L, 4L, 3L, 10L, 6L, 8L,
6L, 7L, 4L, 8L, 5L, 3L, 7L, 12L, 8L, 4L, 11L, 2L, 3L, 12L, 1L
), x3 = c(1, 1, 1, 1, 3, 1, 0, 3, 3, 0, 3, 2, 3, 1, 2, 3, 2,
3, 3, 2, 0, 2, 1, 0, 0, 1, 0, 3, 3, 0, 1, 3, 2, 3, 3, 0, 2, 3,
0, 2, 0, 3, 2, 3, 2, 3, 0, 2, 2, 1, 2, 0, 2, 0, 3, 1, 2, 1, 3,
3, 2, 3, 0, 0, 3, 3, 3, 3, 2, 0, 1, 2, 0, 3, 1, 3, 3, 2, 2, 2,
1, 3, 1, 0, 3, 1, 3, 2, 0, 3, 0, 2, 3, 1, 3, 0, 3, 1, 1, 0, 2,
0, 2, 1, 1, 2, 3, 3, 1, 2, 0, 0, 2, 3, 0, 0, 1, 2, 2, 3, 3, 2,
3, 2, 3, 0, 3, 3, 2, 1, 2, 3, 2, 0, 2, 0, 0, 1, 1, 1, 1, 2, 2,
0, 3, 3, 3, 0, 3, 3, 1, 0, 1, 3, 0, 2, 1, 1, 0, 2, 1, 2, 2, 3,
2, 1, 1, 1, 0, 1, 1, 1, 2, 1, 2, 2, 2, 2, 2, 3, 3, 1, 3, 3, 3,
0, 2, 2, 2, 1, 1, 1, 0, 0, 3, 2, 3, 1, 2, 1, 0, 2, 3, 3, 3, 3,
3, 0, 0, 1, 1, 0, 1, 0, 1, 0, 0, 1, 3, 2, 0, 0, 1, 1, 2, 1, 3,
1, 0, 0, 3, 3, 2, 2, 1, 2, 1, 3, 2, 3, 0, 0, 2, 3, 0, 0, 0, 1,
0, 3, 0, 2, 1, 3, 0, 3, 2, 3, 3, 0, 1, 0, 0, 3, 0, 1, 2, 1, 3,
2, 1, 3, 3, 0, 0, 1, 0, 3, 2, 1), y = c(0.03688, 0.09105, 0.16246,
0, 0.11024, 0.16246, 0.13467, 0, 0.11024, 0.0807, 0.12726, 0.03934,
0, 0.0826, 0.03688, 0.06931, 0.1378, 0.12726, 0.12726, 0.08815,
0.13467, 0.01314, 0.09105, 0.12077, 0.12077, 0.02821, 0.15134,
0.03604, 0.03604, 0.08729, 0.04035, 0.46088, 0.20987, 0.46088,
0.06672, 0.24121, 0.08948, 0.07867, 0.12077, 0.03688, 0.02276,
0.04535, 0.03934, 0.04535, 0.08815, 0.03604, 0.50771, 0.20987,
0.08569, 0.09105, 0.03934, 0.0807, 0.08569, 0.02276, 0.06672,
0.0826, 0.1378, 0.02821, 0.03943, 0.03589, 0.04813, 0.46088,
0.22346, 0.13467, 0.06672, 0.04535, 0.07867, 0.12726, 0.08948,
0.24121, 0.06983, 0.08569, 0.22346, 0.11024, 0.06983, 0.03589,
0.06931, 0.08569, 0.04589, 0.03688, 0.0826, 0, 0.06983, 0.02276,
0.06238, 0.03192, 0.06931, 0.08569, 0.12077, 0.46088, 0.02276,
0.20987, 0.03943, 0, 0, 0.50771, 0.12726, 0.1628, 0, 0.41776,
0.04589, 0.24121, 0.01314, 0.03027, 0.1628, 0.08569, 0, 0.46088,
0.09105, 0.08569, 0.13467, 0.0807, 0.12912, 0.03604, 0.24121,
0.50771, 0, 0.12912, 0.03934, 0.46088, 0.03943, 0.08948, 0.07103,
0.03934, 0, 0.22346, 0.03589, 0, 0.03688, 0.02821, 0.20987, 0.12726,
0.03688, 0.08729, 0.04589, 0.24121, 0.12077, 0.03027, 0.03688,
0.03673, 0, 0.01314, 0.02957, 0.12077, 0.04535, 0.06931, 0.03604,
0.36883, 0.07867, 0.07867, 0.03027, 0.36883, 0.03192, 0.03604,
0.36883, 0.08948, 0.03688, 0.16246, 0.41776, 0.12912, 0.03688,
0.02957, 0.1255, 0, 0.20987, 0.0826, 0.1628, 0.03192, 0.02276,
0.0826, 0, 0.04035, 0.04813, 0.03673, 0.1255, 0.1378, 0.04813,
0.1255, 0.04813, 0.46088, 0.04535, 0.03673, 0.06931, 0.07867,
0.46088, 0.13467, 0.12912, 0.02957, 0.20987, 0, 0.03688, 0.02821,
0.22346, 0.41776, 0.03589, 0.03934, 0.07103, 0.03673, 0.12912,
0.03673, 0.0807, 0.1378, 0.06931, 0.03943, 0.12726, 0.12726,
0.06931, 0.08729, 0.12077, 0.02821, 0.03027, 0.08729, 0.03027,
0.22346, 0.03192, 0.12077, 0.15134, 0.02821, 0.06238, 0.04813,
0.41776, 0.41776, 0.03027, 0.03673, 0.08815, 0.1628, 0.07867,
0, 0.24121, 0.08729, 0.46088, 0, 0.1255, 0.08569, 0.16246, 0.1378,
0, 0.12726, 0.1255, 0.03943, 0.12077, 0.02276, 0.04589, 0.06238,
0.41776, 0.22346, 0.24121, 0.04035, 0.24121, 0.07867, 0.36883,
0.08569, 0.04035, 0.03604, 0.36883, 0.06238, 0.03934, 0.03589,
0.11024, 0.02276, 0.03688, 0.36883, 0.24121, 0.03604, 0.13467,
0.09105, 0.08948, 0.03688, 0.06672, 0.03688, 0.03192, 0.07867,
0.03943, 0.13467, 0.12077, 0.0826, 0.22346, 0.04535, 0.08815,
0.16246)), .Names = c("x1", "x2", "x3", "y"), row.names = c(995L,
1416L, 281L, 1192L, 1075L, 294L, 1812L, 2235L, 1097L, 1583L,
670L, 1485L, 2199L, 2495L, 1259L, 436L, 803L, 631L, 617L, 2654L,
1813L, 2180L, 1403L, 911L, 927L, 533L, 2024L, 1517L, 1522L, 1356L,
1850L, 222L, 115L, 204L, 1974L, 2292L, 1695L, 1746L, 915L, 1283L,
1128L, 880L, 1467L, 887L, 2665L, 1532L, 267L, 155L, 1933L, 1447L,
1488L, 1609L, 1922L, 1168L, 1965L, 2479L, 813L, 550L, 2707L,
2590L, 2373L, 190L, 504L, 1810L, 2007L, 843L, 1770L, 659L, 1730L,
2246L, 1668L, 1923L, 465L, 1108L, 1663L, 2616L, 409L, 1946L,
589L, 1277L, 2493L, 2210L, 1662L, 1142L, 1331L, 735L, 430L, 1916L,
922L, 208L, 1134L, 127L, 2693L, 1213L, 2236L, 240L, 623L, 108L,
1190L, 9L, 575L, 2268L, 2171L, 2308L, 103L, 1953L, 2409L, 184L,
1437L, 1947L, 1847L, 1570L, 365L, 1550L, 2278L, 270L, 1204L,
384L, 1472L, 205L, 2694L, 1727L, 2800L, 1476L, 2229L, 453L, 2630L,
2426L, 1275L, 523L, 163L, 635L, 1287L, 1349L, 561L, 2261L, 931L,
2339L, 973L, 2113L, 1229L, 2155L, 2554L, 936L, 892L, 433L, 1560L,
697L, 1791L, 1755L, 2351L, 720L, 740L, 1558L, 674L, 1736L, 988L,
321L, 18L, 375L, 959L, 2560L, 1047L, 2429L, 119L, 2468L, 98L,
773L, 1158L, 2520L, 1216L, 1872L, 2364L, 2094L, 1035L, 826L,
2374L, 1028L, 2368L, 176L, 895L, 2090L, 399L, 1789L, 179L, 1800L,
369L, 2568L, 140L, 1207L, 1001L, 518L, 481L, 12L, 2597L, 1474L,
2749L, 2097L, 379L, 2110L, 1615L, 800L, 423L, 2733L, 626L, 662L,
421L, 1363L, 898L, 530L, 2315L, 1365L, 2331L, 468L, 768L, 900L,
2027L, 544L, 1337L, 2376L, 53L, 44L, 2338L, 2075L, 2655L, 78L,
1782L, 1231L, 2291L, 1379L, 212L, 2212L, 1032L, 1929L, 331L,
790L, 1226L, 664L, 1018L, 2735L, 916L, 1157L, 590L, 1343L, 7L,
490L, 2257L, 1853L, 2251L, 1748L, 719L, 1941L, 1885L, 1544L,
725L, 1294L, 1494L, 2601L, 1077L, 1169L, 979L, 709L, 2282L, 1526L,
1797L, 1424L, 1690L, 993L, 1979L, 1268L, 730L, 1739L, 2697L,
1842L, 952L, 2483L, 479L, 864L, 2677L, 283L), class = "data.frame")
起始值
starting_value <- structure(c(0.177698291502873, 0.6, 0.0761564106440883, 0.05,
1.9, 1.1, 0.877181493020499, 1.9), .Names = c("F_initial_x2",
"F_decay_x2", "S_initial_x2", "S_decay_x2", "initial_x1", "decay_x1",
"initial_x3", "decay_x3"))
NLSLM失败
coef(nlsLM(
formula = y ~ (F_initial_x2 * exp(- F_decay_x2 * x2) + S_initial_x2 * exp(- S_decay_x2 * x2)) *
(1 + initial_x1 * exp(- decay_x1 * x1)) *
(1 + initial_x3 * exp(- decay_x3 * x3 )),
data = df,
start = coef(brute_force),
lower = c(0, 0, 0, 0, 0, 0, 0, 0),
control = nls.lm.control(maxiter = 200),
trace = T))
It. 0, RSS = 1.36145, Par. = 0.177698 0.6 0.0761564 0.05 1.9 1.1 0.877181 1.9
It. 1, RSS = 1.25401, Par. = 0.207931 0.581039 0.0769047 0.0577244 2.01947 1.22911 0.772957 5.67978
It. 2, RSS = 1.19703, Par. = 0.188978 0.604515 0.0722749 0.0792141 2.44179 1.1258 0.96305 8.67253
It. 3, RSS = 1.1969, Par. = 0.160885 0.640958 0.0990201 0.145187 3.5853 0.847158 0.961844 13.2183
It. 4, RSS = 1.19057, Par. = 0.142138 0.685678 0.11792 0.167417 4.27977 0.936981 0.959606 13.2644
It. 5, RSS = 1.19008, Par. = 0.124264 0.757088 0.136277 0.188896 4.76578 0.91274 0.955142 21.0167
It. 6, RSS = 1.18989, Par. = 0.118904 0.798296 0.141951 0.194167 4.93099 0.91529 0.952972 38.563
It. 7, RSS = 1.18987, Par. = 0.115771 0.821874 0.145398 0.197773 5.02251 0.914204 0.949906 38.563
It. 8, RSS = 1.18986, Par. = 0.113793 0.837804 0.147573 0.199943 5.07456 0.914192 0.948289 38.563
It. 9, RSS = 1.18986, Par. = 0.112458 0.848666 0.149033 0.201406 5.11024 0.914099 0.947232 38.563
It. 10, RSS = 1.18986, Par. = 0.111538 0.856282 0.150035 0.202411 5.13491 0.914051 0.946546 38.563
It. 11, RSS = 1.18986, Par. = 0.110889 0.861702 0.15074 0.203118 5.15244 0.914013 0.946076 38.563
It. 12, RSS = 1.18986, Par. = 0.110426 0.865606 0.151243 0.203623 5.16501 0.913986 0.945747 38.563
It. 13, RSS = 1.18986, Par. = 0.110092 0.868441 0.151605 0.203986 5.17412 0.913966 0.945512 38.563
It. 14, RSS = 1.18986, Par. = 0.109849 0.87051 0.151868 0.20425 5.18075 0.913952 0.945343 38.563
It. 15, RSS = 1.18985, Par. = 0.109672 0.872029 0.15206 0.204443 5.18561 0.913941 0.94522 38.563
It. 16, RSS = 1.18985, Par. = 0.109542 0.873147 0.152201 0.204585 5.18918 0.913933 0.945131 38.563
It. 17, RSS = 1.18985, Par. = 0.109446 0.873971 0.152305 0.204689 5.19181 0.913927 0.945065 38.563
Error in nlsModel(formula, mf, start, wts) :
singular gradient matrix at initial parameter estimates
问题：
使用在奇异梯度矩阵问题之前找到的最佳参数是有意义的，即在Iteration = 17时发现的那个？
如果有，是否有办法获取？发生错误时，我没有成功保存结果。
我注意到如果我将maxiter的数量减少到17以下的数字，我仍然会出现在新的最后一次迭代中出现的相同错误，这对我来说没有意义
< / LI>例如maxiter = 10
It. 0, RSS = 1.36145, Par. = 0.177698 0.6 0.0761564 0.05 1.9 1.1 0.877181 1.9
It. 1, RSS = 1.25401, Par. = 0.207931 0.581039 0.0769047 0.0577244 2.01947 1.22911 0.772957 5.67978
It. 2, RSS = 1.19703, Par. = 0.188978 0.604515 0.0722749 0.0792141 2.44179 1.1258 0.96305 8.67253
It. 3, RSS = 1.1969, Par. = 0.160885 0.640958 0.0990201 0.145187 3.5853 0.847158 0.961844 13.2183
It. 4, RSS = 1.19057, Par. = 0.142138 0.685678 0.11792 0.167417 4.27977 0.936981 0.959606 13.2644
It. 5, RSS = 1.19008, Par. = 0.124264 0.757088 0.136277 0.188896 4.76578 0.91274 0.955142 21.0167
It. 6, RSS = 1.18989, Par. = 0.118904 0.798296 0.141951 0.194167 4.93099 0.91529 0.952972 38.563
It. 7, RSS = 1.18987, Par. = 0.115771 0.821874 0.145398 0.197773 5.02251 0.914204 0.949906 38.563
It. 8, RSS = 1.18986, Par. = 0.113793 0.837804 0.147573 0.199943 5.07456 0.914192 0.948289 38.563
It. 9, RSS = 1.18986, Par. = 0.112458 0.848666 0.149033 0.201406 5.11024 0.914099 0.947232 38.563
It. 10, RSS = 0.12289, Par. = 0.112458 0.848666 0.149033 0.201406 5.11024 0.914099 0.947232 38.563
Error in nlsModel(formula, mf, start, wts) :
singular gradient matrix at initial parameter estimates
In addition: Warning message:
In nls.lm(par = start, fn = FCT, jac = jac, control = control, lower = lower, :
lmdif: info = -1. Number of iterations has reached `maxiter' == 10.
你看到有什么解释吗？
答案 0 :(得分：4)
问题的根本问题是没有实现融合。这可以通过使用Y = log（X + 1）变换衰减参数然后使用X = exp（Y）-1将其转换回来来解决。这种转变可以有利地改变雅可比。不幸的是，这种转换的应用往往在很大程度上是反复试验。 （另见注1）。
ix <- grep("decay", names(starting_value))
fm <- nlsLM(
formula = y ~ (F_initial_x2 * exp(- log(F_decay_x2+1) * x2) +
S_initial_x2 * exp(- log(S_decay_x2+1) * x2)) *
(1 + initial_x1 * exp(- log(decay_x1+1) * x1)) *
(1 + initial_x3 * exp(- log(decay_x3+1) * x3 )),
data = df,
start = replace(starting_value, ix, exp(starting_value[ix]) - 1),
lower = c(0, 0, 0, 0, 0, 0, 0, 0),
control = nls.lm.control(maxiter = 200),
trace = TRUE)
给出类似的残差平方和但是实现收敛：
> fm
Nonlinear regression model
model: y ~ (F_initial_x2 * exp(-log(F_decay_x2 + 1) * x2) + S_initial_x2 * exp(-log(S_decay_x2 + 1) * x2)) * (1 + initial_x1 * exp(-log(decay_x1 + 1) * x1)) * (1 + initial_x3 * exp(-log(decay_x3 + 1) * x3))
data: df
F_initial_x2 F_decay_x2 S_initial_x2 S_decay_x2 initial_x1 decay_x1
1.092e-01 1.402e+00 1.526e-01 2.275e-01 5.199e+00 1.494e+00
initial_x3 decay_x3
9.449e-01 1.375e+07
residual sum-of-squares: 1.19
Number of iterations to convergence: 38
Achieved convergence tolerance: 1.49e-08
> replace(coef(fm), ix, log(coef(fm)[ix]+1))
F_initial_x2 F_decay_x2 S_initial_x2 S_decay_x2 initial_x1 decay_x1
0.1091735 0.8763253 0.1525997 0.2049852 5.1993194 0.9139096
initial_x3 decay_x3
0.9448779 16.4368001
注1：经过一些实验，我发现在decay_x3上应用转换就足够了。
注2：关于您想要自动注释的注释，注意与lm拟合的三次多项式将更加一致地不会遇到麻烦并且具有较低的残差平方和 - - 1.14 vs. 1.19 - 但是以更多参数为代价--10对8。
# lm poly fit
fm.poly <- lm(y ~ poly(x1, x2, degree = 3), df)
deviance(fm.poly) # residual sum of squares
## [1] 1.141398
length(coef(fm.poly)) # no. of coefficients
## [1] 10
# nlsLM fit transforming decay parameters
deviance(fm)
## [1] 1.189855
length(coef(fm))
## [1] 8
注3：这是另一个模型，它是用二次多项式替换x3部分并在它变为冗余时丢弃F_initial_x2。它还有8个参数，它收敛并且比问题中的模型更好地拟合数据（即具有较低的剩余平方和）。
fm3 <- nlsLM(formula = y ~ (exp(- F_decay_x2 * x2) +
S_initial_x2 * exp(- S_decay_x2 * x2)) *
(1 + initial_x1 * exp(- decay_x1 * x1)) *
cbind(1, poly(x3, degree = 2)) %*% c(p1,p2,p3),
data = df,
start = c(starting_value[-c(1, 7:8)], p1=0, p2=0, p3=0),
lower = c(0, 0, 0, 0, 0, 0, NA, NA),
control = nls.lm.control(maxiter = 200),
trace = TRUE)
，并提供：
> fm3
Nonlinear regression model
model: y ~ (exp(-F_decay_x2 * x2) + S_initial_x2 * exp(-S_decay_x2 * x2)) * (1 + initial_x1 * exp(-decay_x1 * x1)) * cbind(1, poly(x3, degree = 2)) %*% c(p1, p2, p3)
data: df
F_decay_x2 S_initial_x2 S_decay_x2 initial_x1 decay_x1 p1
3.51614 2.60886 0.26304 8.26244 0.81232 0.09031
p2 p3
-0.16968 0.53324
residual sum-of-squares: 1.019
Number of iterations to convergence: 20
Achieved convergence tolerance: 1.49e-08
注意4： nlmrt包中的nlxb收敛而没有做任何特殊的事情。
library(nlmrt)
nlxb(
formula = y ~ (F_initial_x2 * exp(- F_decay_x2 * x2) + S_initial_x2 * exp(- S_decay_x2 * x2)) *
(1 + initial_x1 * exp(- decay_x1 * x1)) *
(1 + initial_x3 * exp(- decay_x3 * x3 )),
data = df,
start = starting_value,
lower = c(0, 0, 0, 0, 0, 0, 0, 0),
control = nls.lm.control(maxiter = 200),
trace = TRUE)
，并提供：
residual sumsquares = 1.1899 on 280 observations
after 31 Jacobian and 33 function evaluations
name coeff SE tstat pval gradient JSingval
F_initial_x2 0.109175 NA NA NA 3.372e-11 15.1
F_decay_x2 0.876313 NA NA NA -5.94e-12 8.083
S_initial_x2 0.152598 NA NA NA 6.55e-11 2.163
S_decay_x2 0.204984 NA NA NA 4.206e-11 0.6181
initial_x1 5.19928 NA NA NA -1.191e-12 0.3601
decay_x1 0.91391 NA NA NA 6.662e-13 0.1315
initial_x3 0.944879 NA NA NA 2.736e-12 0.02247
decay_x3 33.9921 NA NA NA -1.056e-15 2.928e-15
答案 1 :(得分：2)
通常在发生此错误时，问题不是代码而是使用的模型。 singular gradient matrix at the initial parameter estimates可能表示模型没有唯一的唯一解决方案，或者模型对于手头的数据而言是过度指定的。
回答你的问题：
是的，这是有道理的。函数nlsLM首先调用执行迭代的nls.lm。当它到达迭代结束时（由于最佳拟合或因为max.iter），结果将传递给函数nlsModel。该函数对梯度矩阵的QR分解乘以平方权重。并且您的初始渐变矩阵包含仅包含零的列。因此，虽然nls.lm可以进行迭代，但只有在下一步nlsModel才能实际检查和发现梯度矩阵的问题。
有一种方法，但这需要您更改R本身中的选项，特别是error选项。通过将其设置为dump.frames，您可以转储出现错误时存在的所有环境。这些存储在名为last.dump的列表中，您可以使用这些环境查找所需的值。
在这种情况下，参数由函数getPars()返回，该函数位于主力函数nlsModel的环境中：
old.opt <- options(error = dump.frames)
themod <- nlsLM(
formula = y ~ (F_initial_x2 * exp(- F_decay_x2 * x2) +
S_initial_x2 * exp(- S_decay_x2 * x2)) *
(1 + initial_x1 * exp(- decay_x1 * x1)) *
(1 + initial_x3 * exp(- decay_x3 * x3 )),
data = df,
start = starting_value,
lower = c(0, 0, 0, 0, 0, 0, 0, 0),
control = nls.lm.control(maxiter = 200),
trace = TRUE)
thecoefs <- llast.dump[["nlsModel(formula, mf, start, wts)"]]$getPars()
options(old.opt) # reset to the previous value.
请注意，这不是您希望在生产环境中使用或与同事共享的代码。它也不是你问题的解决方案，因为问题是模型，而不是代码。
我做了一个非常简短的测试，看看它是否真的是模型，如果我用它的起始值替换最后一个参数（decay_x3），那么模型就没有问题了。我不知道我们在这里处理什么，所以删除另一个参数可能在现实世界中更有意义，但只是为了证明你的代码是好的：
themod <- nlsLM(
formula = y ~ (F_initial_x2 * exp(- F_decay_x2 * x2) +
S_initial_x2 * exp(- S_decay_x2 * x2)) *
(1 + initial_x1 * exp(- decay_x1 * x1)) *
(1 + initial_x3 * exp(- 1.9* x3 )),
data = df,
start = starting_value[-8],
lower = c(0, 0, 0, 0, 0, 0, 0, 0)[-8],
control = nls.lm.control(maxiter = 200),
trace = TRUE)
在迭代10中退出而没有错误。
编辑： 我一直在深入研究它，基于数据，“额外”解决方案基本上是将X3踢出模型。您只有3个唯一值，参数的初始估计值约为38.所以：
> exp(-38*c(1,2,3)) < .Machine$double.eps
[1] TRUE TRUE TRUE
如果将其与实际Y值进行比较，很明显initial_x3 * exp(- decay_x3 * x3 )对模型没有任何贡献，因为它几乎为0.
如果您在nlsModel中手动计算渐变，则会得到一个不是满秩的矩阵;最后一列只包含0：
theenv <- list2env( c(df, thecoefs))
thederiv <- numericDeriv(form[[3]], names(starting_value), theenv)
thegrad <- attr(thederiv, "gradient")
这就是给你错误的原因。对于您拥有的数据，该模型已被过度指定。
Gabor建议的对数转换可以防止你的最后估计变得如此之大，迫使x3离开模型。由于对数转换，算法不会很容易地跳到这样的极值。为了获得与原始模型相同的估算值，他的decay_x3应该与3.2e16一样高，以指定相同的模型（exp(38)）。因此，对数转换可以保护您免受强制任何变量影响的估计值。
日志转换的另一个副作用是decay_x3值的重要步骤对模型只有中等影响。 Gabor发现的估计已经是一个惊人的1.3e7，但在后转换之后仍然是16 decay_x3的可行值。如果你看一下，这仍然会使模型中的x3成为冗余：
> exp(-16*c(1,2,3))
[1] 1.125352e-07 1.266417e-14 1.425164e-21
但它不会导致导致错误的奇点。
您可以通过设置上限来避免这种情况，例如：
themod <- nlsLM(
formula = y ~ (F_initial_x2 * exp(- F_decay_x2 * x2) +
S_initial_x2 * exp(- S_decay_x2 * x2)) *
(1 + initial_x1 * exp(- decay_x1 * x1)) *
(1 + initial_x3 * exp(- decay_x3 * x3 )),
data = df,
start = starting_value,
lower = c(0, 0, 0, 0, 0, 0, 0, 0),
upper = rep(- log(.Machine$double.eps^0.5),8),
control = nls.lm.control(maxiter = 200),
trace = TRUE)
运行完全正常，给出相同的估算值，并再次得出结论x3是多余的。
所以无论你怎样看待它，x3对y都没有影响，你的模型被过度指定，并且不能很好地适应手头的数据。