Question

这个问题是我在这里Exponential curve fitting with nls using data.table groups问的上一个问题的基础。

我正在使用nls通过多个组将指数曲线拟合到数据表对象。并非所有数据似乎都适合指数模型，nls有时会引发错误，从而停止其余组的所有进一步计算。

我在下面附加了一个MWE，以尝试使用tryCatch跳过有问题的组，但最终所有我的新列都输出了错误。如何跳过有问题的群组的nls值计算？

## Example data table
DT <- data.table(
     x = c(1,2,3,4,5,6,7,8,
           1,2,3,4,5,6,7,8,
           1,2,3,4,5,6,7,8),
     y = c(15.4,16,16.4,17.7,20,23,27,35,
           25.4,26,26.4,27.7,30,33,37,45,
           27.4,28,28.4,29.7,32,35,39,47),
groups = c(1,1,1,1,1,1,1,1,
           2,2,2,2,2,2,2,2,
           3,3,3,3,3,3,3,3)
)

## Fit exponential curve using starting values a,b,c for each group 
DT[, c("sigma", "a", "b", "c") := {
    c.0 <- min(y) * 0.5
    model.0 <- lm(log(y - c.0) ~ x, data=.SD)
    start <- list(a=exp(coef(model.0)[1]), b=coef(model.0)[2], c=c.0)
    model <- nls(y ~ a * exp(b * x) + c,
        data=.SD, 
        start=start, 
        control=nls.control(maxiter=500))
    c(.(sigma=summary(model)$sigma), as.list(coef(model)))
},
by=.(groups)]

## Modify data table to ruin nls model for group 2
set(DT, i=16L, j="y", value=3)

## Calculation works for group 1 but stops for group 2 and onwards
DT[, c("sigma", "a", "b", "c") := {
    c.0 <- min(y) * 0.5
    model.0 <- lm(log(y - c.0) ~ x, data=.SD)
    start <- list(a=exp(coef(model.0)[1]), b=coef(model.0)[2], c=c.0)
    model <- nls(y ~ a * exp(b * x) + c,
        data=.SD, 
        start=start, 
        control=nls.control(maxiter=500))
    c(.(sigma=summary(model)$sigma), as.list(coef(model)))
},
by=.(groups)]

## My poor attempt at using a tryCatch just gives NA to every column
DT[, c("sigma","a", "b", "c") := {
    c.0 <- min(y) * 0.5
    model.0 <- lm(log(y - c.0) ~ x, data=.SD)
    start <- list(a=exp(coef(model.0)[1]), b=coef(model.0)[2], c=c.0)
    model <- tryCatch(
        {
            nls(y ~ a * exp(b * x) + c,
                data=.SD, 
                start=start, 
                control=nls.control(maxiter=500))
            c(.(sigma=summary(model)$sigma), as.list(coef(model)))
        },
        error=function(err){
            return(NA_real_)
        }
    )
},
by=.(groups)]

Answer 1

无需标记，注释中的时间太长。

类似这样的东西：

DT[, c("sigma", "a", "b", "c") := 
    tryCatch({
        c.0 <- min(y) * 0.5
        model.0 <- lm(log(y - c.0) ~ x, data=.SD)
        start <- list(a=exp(coef(model.0)[1]), b=coef(model.0)[2], c=c.0)
        model <- nls(y ~ a * exp(b * x) + c,
            data=.SD,
            start=start,
            control=nls.control(maxiter=500))
        c(.(sigma=summary(model)$sigma), as.list(coef(model)))
    }, error=function(e) NA_real_),
    by=.(groups)]

输出：

    x    y groups     sigma         a         b        c
 1: 1 15.4      1 0.2986243 0.5265405 0.4565363 14.56728
 2: 2 16.0      1 0.2986243 0.5265405 0.4565363 14.56728
 3: 3 16.4      1 0.2986243 0.5265405 0.4565363 14.56728
 4: 4 17.7      1 0.2986243 0.5265405 0.4565363 14.56728
 5: 5 20.0      1 0.2986243 0.5265405 0.4565363 14.56728
 6: 6 23.0      1 0.2986243 0.5265405 0.4565363 14.56728
 7: 7 27.0      1 0.2986243 0.5265405 0.4565363 14.56728
 8: 8 35.0      1 0.2986243 0.5265405 0.4565363 14.56728
 9: 1 25.4      2        NA        NA        NA       NA
10: 2 26.0      2        NA        NA        NA       NA
11: 3 26.4      2        NA        NA        NA       NA
12: 4 27.7      2        NA        NA        NA       NA
13: 5 30.0      2        NA        NA        NA       NA
14: 6 33.0      2        NA        NA        NA       NA
15: 7 37.0      2        NA        NA        NA       NA
16: 8  3.0      2        NA        NA        NA       NA
17: 1 27.4      3 0.2986243 0.5265401 0.4565364 26.56728
18: 2 28.0      3 0.2986243 0.5265401 0.4565364 26.56728
19: 3 28.4      3 0.2986243 0.5265401 0.4565364 26.56728
20: 4 29.7      3 0.2986243 0.5265401 0.4565364 26.56728
21: 5 32.0      3 0.2986243 0.5265401 0.4565364 26.56728
22: 6 35.0      3 0.2986243 0.5265401 0.4565364 26.56728
23: 7 39.0      3 0.2986243 0.5265401 0.4565364 26.56728
24: 8 47.0      3 0.2986243 0.5265401 0.4565364 26.56728
    x    y groups     sigma         a         b        c

使用nls跳过数据表计算中的错误

1 个答案: