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试图在 ggplot2 邮件列表中回答某人的问题，但我无法弄清楚： https ://groups.google.com/forum/#!topic/ggplot2/YgCqQX8JbPM

OP 希望将不同的起始参数应用于 nls 模型的数据子集。我的想法是他应该阅读 dplyr 和 purrr，但在尝试了几个小时后，我遇到了障碍。不确定这是一个错误还是我缺乏 purrr 的经验。

library(tidyverse)

# input dataset
df <- data.frame(Group = c(rep("A", 7), rep("B", 7), rep("C", 7)),
                 Time = c(rep(c(1:7), 3)),
                 Result = c(100, 96.9, 85.1, 62.0, 30.7, 15.2, 9.6, 
                            10.2, 14.8, 32.26, 45.85, 56.25, 70.1, 100,
                            100, 55.61, 3.26, -4.77, -7.21, -3.2, -5.6))

# nest the datasets for computing models
df_p <-
df %>%
group_by(Group) %>%
nest

# add model parameters as rows/columns
df_p$starta = c(-3, 4,-3)
df_p$startb = c(85, 85, 85)
df_p$startc = c(4, 4, 4)
df_p$startd = c(10,10,10)

# compute models using nls
df_p %>%
mutate(model2 = map(data, ~nls(Result ~ a+(b-a)/(1+(Time/c)^d), data = ., start = c(a = starta, b = startb, c = startc, d = startd)))
        )

#Error in mutate_impl(.data, dots) : 
#  parameters without starting value in 'data': a, b, d

感觉与这个错误有关，但这已经修复了一段时间...... https://github.com/hadley/dplyr/issues/1447

据我所知，它正在寻找嵌套 tibble 范围内的变量，但我希望它在 mutate 调用的范围内。我不知道有没有办法解决这个问题。

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2 回答 2

6

示例数据很棘手，因为 B 组基本上有相反的时间。为此找到好的初始值不是我的问题。所以我为 B 组制作了新数据。以下是如何设置数据框以便nls()在map2().


library(tidyverse)

df <- data.frame(Group = c(rep("A", 7), rep("B", 7), rep("C", 7)),
                 Time = c(rep(c(1:7), 3)),
                 Result = c(100, 96.9, 85.1, 62.0, 30.7, 15.2, 9.6, 
                            ## I replaced these values!!
                            ## Group B initial values are NOT MY PROBLEM
                            105, 90, 82, 55, 40, 23, 7, 
                            100, 55.61, 3.26, -4.77, -7.21, -3.2, -5.6))

## ggplot(df, aes(x = Time, y = Result, group = Group)) + geom_line()

df_p <-
  df %>%
  group_by(Group) %>%
  nest() %>% 
  ## init vals are all the same, but this shows how to make them different
  mutate(start = list(
    list(a = -3, b = 85, c = 4, d = 10),
    list(a = -3, b = 85, c = 4, d = 10),
    list(a = -3, b = 85, c = 4, d = 10)
  )

)

df_p %>%
  mutate(model2 = map2(data, start,
                       ~ nls(Result ~ a+(b-a)/(1+(Time/c)^d),
                             data = .x, start = .y)))
#> # A tibble: 3 × 4
#>    Group             data      start    model2
#>   <fctr>           <list>     <list>    <list>
#> 1      A <tibble [7 × 2]> <list [4]> <S3: nls>
#> 2      B <tibble [7 × 2]> <list [4]> <S3: nls>
#> 3      C <tibble [7 × 2]> <list [4]> <S3: nls>
于 2016-12-29T04:40:57.937 回答
2

无法找到一组参数来生成您设置的模型，但我认为就设置模型拟合过程而言，这是您可以做的；starta, startb .. etc基本上，您可以将所有参数以及Result和列包装到数据中Time，然后您可以使用 访问参数.$，注意在这种情况下，您需要该unique函数选择一个值，因为该值在取消嵌套时已被广播。使用简单的模型公式，a + b*Time它会在model2列中生成模型，您可以按照此路线调整传递给的初始参数nls以适应您指定的更复杂的公式：

library(tidyverse)

df_p %>% unnest %>% group_by(Group) %>% nest %>%
         mutate(model2 = map(data, ~nls(Result ~ a + b*Time, data = ., 
                                        start = c(a = unique(.$starta), 
                                                  b = unique(.$startb))
                                       )
                             )
               )

# A tibble: 3 × 3
#   Group             data    model2
#  <fctr>           <list>    <list>
#1      A <tibble [7 × 6]> <S3: nls>
#2      B <tibble [7 × 6]> <S3: nls>
#3      C <tibble [7 × 6]> <S3: nls>
于 2016-12-29T03:06:37.177 回答