计算模型效率离开一个被摄体出模式中的R

Question

我有一个数据帧df

structure(list(x = c(49, 50, 51, 52, 53, 54, 55, 56, 1, 2, 3, 
    4, 5, 14, 15, 16, 17, 163, 164, 165, 153, 154, 72, 38, 39, 40, 
    23, 13, 14, 15, 5, 6, 74, 75, 76, 77, 78, 79, 80, 81, 82, 127, 
    128, 129, 130, 131, 132, 71, 72, 73, 74, 75, 76, 2, 3, 4, 5, 
    6, 99, 100, 101, 10, 11, 3, 30, 50, 51, 52, 53, 54, 56, 64, 66, 
    67, 68, 69, 34, 35, 37, 39, 2, 46, 47, 17, 18, 99, 100, 102, 
    103, 84, 85, 86, 87, 88, 67, 70, 72), y = c(2268.14043972082, 
    2147.62290922552, 2269.1387550775, 2247.31983098201, 1903.39138268307, 
    2174.78291538358, 2359.51909126411, 2488.39004804939, 212.851575751527, 
    461.398994384333, 567.150629704352, 781.775113821961, 918.303706148872, 
    1107.37695799186, 1160.80594193377, 1412.61328924168, 1689.48879626486, 
    685.154353165934, 574.088067465695, 650.30821636616, 494.185166497016, 
    436.312162090908, 641.231373456365, 494.374217984441, 201.745910386788, 
    486.030122926459, 483.045965372879, 265.693628564011, 285.116345260642, 
    291.023782284086, 229.606221692753, 230.952761338012, 1089.06303295676, 
    1255.88808925333, 1087.75402177912, 1068.248897182, 1212.17254891642, 
    884.222588171535, 938.887718005513, 863.582247020735, 1065.91969416523, 
    790.570635510725, 834.500908313203, 710.755041345197, 814.002362551197, 
    726.814950022846, 828.559687148314, 611.564698476112, 603.238720579422, 
    524.322001078981, 565.296378873638, 532.431853589369, 597.174114277044, 
    260.737164468854, 306.72700499362, 283.410379620422, 366.813913489692, 
    387.570173754128, 606.075737104722, 686.408686154056, 705.914347674276, 
    388.602676983443, 477.858510450125, 128.198042456082, 535.519377609133, 
    1893.38468471169, 1819.83262739703, 1827.31409981102, 1640.5816780664, 
    1689.0365549922, 2112.67917439342, 1028.8780498564, 1098.54431357711, 
    1265.26965941035, 1129.58344809909, 820.922447928053, 749.343583476846, 
    779.678206156474, 646.575242339517, 733.953282899613, 461.156280127354, 
    1184.81825619942, 1281.2920902365, 906.813018662913, 798.186995701282, 
    831.365377249207, 764.519073183124, 672.076289062505, 669.879217186302, 
    1265.48484068702, 1193.29000986667, 1156.81486114406, 1199.7373066445, 
    1116.24029749935, 1341.47673353751, 1401.44881976186, 1640.27575962036 
    ), ID = 1:97), .Names = c("x", "y", "ID"), row.names = c(NA, 
    -97L), class = "data.frame") 

我现在要基于一个交叉验证来计算一个非线性模型的模型效率离开一个ID出。我实现了这一行代码。

library(stats) 
library (hydroGOF) 

id <- unique(df$ID) 
for (i in id){ 
    fit1 <- try(nls(y~A*x^3+B*x^2+C*x+D, data = df[df$ID != i,], start = list(A=0.02, B=-0.6, C= 50, D=200)), silent=TRUE) 
    Out <- if (inherits(fit1, "nls")) NSE(sim = predict(fit1, newdatadata=df[df$ID==i,]), obs = df$y, na.rm=T) 
    } 

不过，我有这样的错误消息：

Error in valindex.default(sim, obs) : 
    Invalid argument: 'length(sim) != length(obs)' !! (96!=97) !! 

有人能帮助我吗？

Answer 1

你有一些小的失误和一个大的逻辑错在上面的代码，我下面的地址：

的所有代码首先应该是这样的：

library(stats) 
library (hydroGOF) 

id <- unique(df$ID) 
for (i in id){ 
    fit1 <- try(nls(y~A*x^3+B*x^2+C*x+D, data = df[df$ID != i,], start = list(A=0.02, B=-0.6, C= 50, D=200)), silent=TRUE) 
    Out <- if (inherits(fit1, "nls")) NSE(sim = predict(fit1, newdata=df[df$ID==i,]), obs = df$y[df$ID==i], na.rm=T) 
} 

在你的问题中， NSE函数您设置参数为newdatadata=df[df$ID==i,])而不是newdata=df[df$ID==i,])即有一个额外的data在那里导致运行函数（您拼写错误的参数:)）时，麻烦。此外，obs参数的长度不正确，因为将使用整个y列，但您只需要df$y[df$ID==i]，它的长度为1（以便与现在长度为1的预测相匹配）。

现在纠正后上面的代码将运行。

但是，一旦你运行它，你会发现它会产生一个警告，说你不能在'sum((obs - mean(obs))^2)=0' => it is not possible to compute 'NSE'时使用NSE。在你的情况，因为你只有一个obs计算'sum((obs - mean(obs))^2)=0'将始终为零。

所以，你不能使用这项技术与NSE，因为它会失败的定义（因为你试图计算一个观察NSE）。您可能应该收集所有的一次性预测，将它们存储在一个变量中，然后对df$y使用该变量的NSE。那可行。

我的意思是以下（含留一出CV完成）：

Out <- c() 
id <- unique(df$ID) 
for (i in id){ 
    fit1 <- try(nls(y~A*x^3+B*x^2+C*x+D, data = df[df$ID != i,], start = list(A=0.02, B=-0.6, C= 50, D=200)), silent=TRUE) 
    Out[i] <- if (inherits(fit1, "nls")) sim = predict(fit1, newdata=df[df$ID==i,]) 
} 

这将现在的工作：

> NSE(Out, df$y) 
[1] 0.3440862