2017-08-28 88 views
0

我想,以适应使用dplyrbroom(最终mclapply)并行使用wrapnls许多非线性配合,但我从nlxb得到一个解析计算错误:如何在dplyr中运行nlxb和wrapnls?

Error in parse(text = joe) (from #11) : <text>:1:6: unexpected input 
1: b1.10% <- 20 

我得到同时使用do这个错误lapply方法。

library(nlmrt) 
library(dplyr) 
library(purrr) 
library(broom) 

data_frame(x = seq(0, 200, 0.1), 
      y = 1.2*exp(-(times - 10)^2/(2*4.2^2)) + 2.4*exp(-(times - 50)^2/(2*3.8^2)) + 5.3*exp(-(times - 80)^2/(2*5.1^2)) + rnorm(length(times), sd = 0.05)) %>% 
    do({ 
    xl <- quantile(.$x, 0.1, na.rm = TRUE) 
    xm <- quantile(.$x, 0.5, na.rm = TRUE) 
    xh <- quantile(.$x, 0.8, na.rm = TRUE) 
    starts <- c(a1 = 5, a2 = 5, a3 = 5, 
       b1 = xl, b2 = xm, b3 = xh, 
       c1 = 5, c2 = 5, c3 = 5) 
    fmla <- y ~ a1*exp(-(x - b1)^2/(2*c1^2)) + a2*exp(-(x - b2)^2/(2*c2^2)) + a3*exp(-(x - b3)^2/(2*c3^2)) 
    df <- data_frame(x = .$x, y = .$y) 
    mod <- wrapnls(fmla, lower = 0, upper = 200, start = starts, data = df) 
    tidy(mod) 
    }) 

有没有办法解决这个问题?

回答

1

问题不在于do方面,它是do内的代码,所以您可以直接调试该部分。该starts载体越来越与位数级联的b#名称:

names(starts) 

## [1] "a1"  "a2"  "a3"  "b1.10%" "b2.50%" "b3.80%" "c1"  "c2"  "c3" 

添加unname到位数计算修复该问题。

data_frame(x = seq(0, 200, 0.1), 
      y = 1.2*exp(-(x - 10)^2/(2*4.2^2)) + 2.4*exp(-(x - 50)^2/(2*3.8^2)) + 5.3*exp(-(x - 80)^2/(2*5.1^2)) + rnorm(length(x), sd = 0.05)) %>% 
    do({ 
    xl <- quantile(.$x, 0.1, na.rm = TRUE) %>% unname() 
    xm <- quantile(.$x, 0.5, na.rm = TRUE) %>% unname() 
    xh <- quantile(.$x, 0.8, na.rm = TRUE) %>% unname() 
    starts <- c(a1 = 5, a2 = 5, a3 = 5, 
       b1 = xl, b2 = xm, b3 = xh, 
       c1 = 5, c2 = 5, c3 = 5) 
    fmla <- y ~ a1*exp(-(x - b1)^2/(2*c1^2)) + a2*exp(-(x - b2)^2/(2*c2^2)) + a3*exp(-(x - b3)^2/(2*c3^2)) 
    df <- data_frame(x = .$x, y = .$y) 
    mod <- wrapnls(fmla, lower = 0, upper = 200, start = starts, data = df) 
    tidy(mod) 
    }) 

## term estimate std.error statistic p.value 
## 1 a1 2.386492 0.007455097 320.1155  0 
## 2 a2 5.296250 0.006437509 822.7174  0 
## 3 a3 1.199384 0.007132559 168.1562  0 
## 4 b1 49.997697 0.013702894 3648.6960  0 
## 5 b2 80.004023 0.007150546 11188.5193  0 
## 6 b3 10.077847 0.028644821 351.8209  0 
## 7 c1 3.798829 0.013702940 277.2273  0 
## 8 c2 5.094727 0.007150573 712.4921  0 
## 9 c3 4.175235 0.028944448 144.2499  0 
+0

这比我想象的要简单。谢谢,我会记住不要在有特殊评估的情况下直接使用'quantile'魔法正在进行:) – wdkrnls