【问题标题】:Calculating predicted means (or predicted probabilities) and SE after multiple imputation in R在 R 中进行多重插补后计算预测均值(或预测概率)和 SE
【发布时间】:2017-12-18 08:00:31
【问题描述】:

我想计算预测值和标准误差,但我不能简单地使用 predict(),因为我使用了 15 个多重插补数据集(生成了 Amelia 包)。我在每个数据集上运行回归模型。然后,使用使用鲁宾规则的 Amelia 函数 mi.meld() 将结果组合成一组模型系数和标准误差。

示例数据和代码:

dd<-list()
for (i in 1:15){
dd[[i]] <- data.frame(
  Age=runif(50,20,90),
  Cat=factor(sample(0:4, 50, replace=T)),
  Outcome = sample(0:1, 50, replace=T)
)}

b.out<-NULL
se.out<-NULL
for(i in 1:15) {
  ols.out<-glm(Outcome~Age+factor(Cat), data=dd[[i]],family="binomial")
  b.out <- rbind(b.out, ols.out$coef)
  se.out <- rbind(se.out, coef(summary(ols.out))[,2])}
mod0 <- mi.meld(q = b.out, se = se.out)

> mod0
$q.mi
     (Intercept)         Age factor(Cat)1 factor(Cat)2 factor(Cat)3 factor(Cat)4
[1,]   0.0466825 -0.00577106    0.5291908  -0.09760264    0.4058684    0.3125109

$se.mi
     (Intercept)        Age factor(Cat)1 factor(Cat)2 factor(Cat)3 
factor(Cat)4
[1,]    1.863276 0.02596468     1.604759     1.398322     1.414589     
1.332743

现在是有问题的部分。我想计算以下一组预测值的预测值(在本例中为预测概率)和标准误差:

data.predict <- data.frame(Cat=as.factor(c(0:4)), Age=53.6)
print(data.predict)
  Cat  Age
1   0 53.6
2   1 53.6
3   2 53.6
4   3 53.6
5   4 53.6

如果我在 1 个数据集上拟合了一个模型,我会这样做:

prediction<- predict(mod1, data.predict, type="response",se.fit=T)

但是,我没有模型对象,我只存储了系数.. 现在我研究了两种解决方法,第一种是以这种方式手动计算预测:predict() with arbitrary coefficients in r 但是我不知道如何得到标准错误。我的另一个想法是创建一个假模型对象,就像这个函数创建:https://gist.github.com/MrFlick/ae299d8f3760f02de6bf 并在 predict() 中使用它,但作为标准错误没有使用模型,也没有办法计算预测的标准误差..

有人对如何解决这个问题有建议吗?我试图用示例代码清楚地解释我的问题,但如果我的问题不清楚,请告诉我,以便我提供更多信息。感谢您的帮助!

【问题讨论】:

    标签: r regression predict imputation


    【解决方案1】:

    我确信你在近几年后不需要这个答案,但我刚刚研究了一个类似的问题,我想我会把答案放在这里供后代使用。

    Andrew Heiss 将这个解决方案放在了 gisthub 上 - https://gist.github.com/andrewheiss/a3134085e92c6607db39c5b14e1b879e

    我稍微修改了一下(部分原因是我认为 tidyverse 中“nest”的默认行为可能在他写这篇文章后已经改变了?)

    代码(辛苦工作!)几乎完全来自 Andrew Heiss。这里的注释是我和他的混合。

    这是使用 Amelia Africa 数据集,对于我现实生活中的问题,我有一个不同的数据集(显然)并且前几步做了一些不同的,这一切都很好。

    library(tidyverse)
    library(Amelia)
    library(broom)
    
    # Use the africa dataset from Amelia
    data(africa)
    set.seed(1234)
    imp_amelia <- amelia(x = africa, m = 5, cs = "country", ts = "year", logs = "gdp_pc", p2s = 0) # do the imputations -- for me, it was fine to do this bit in 'mice'
    
    # Gather all the imputed datasets into one data frame and run a model on each
    models_imputed_df <- bind_rows(unclass(imp_amelia$imputations), .id = "m") %>%
      group_by(m) %>%
      nest() %>% 
      mutate(model = data %>% map(~ lm(gdp_pc ~ trade + civlib, data = .)))
    
    # again - for my real life problem the models looked very different to this, and used rms - and this was also totally fine.
    
    models_imputed_df
    #> # A tibble: 5 x 3
    #>   m     data               model   
    #>   <chr> <list>             <list>  
    #> 1 imp1  <tibble [120 × 7]> <S3: lm>
    #> 2 imp2  <tibble [120 × 7]> <S3: lm>
    #> 3 imp3  <tibble [120 × 7]> <S3: lm>
    #> 4 imp4  <tibble [120 × 7]> <S3: lm>
    #> 5 imp5  <tibble [120 × 7]> <S3: lm>
    
    
    # We want to see how GDP per capita varies with changes in civil liberties, so
    # we create a new data frame with values for each of the covariates in the
    # model. We include the full range of civil liberties (from 0 to 1) and the mean
    # of trade.
    
    # ie. this is a 'skelton' data frame of all your variables that you want to make predictions over.
    
    new_data <- data_frame(civlib = seq(0, 1, 0.1), 
                           trade = mean(africa$trade, na.rm = TRUE))
    new_data
    #> # A tibble: 11 x 2
    #>    civlib trade
    #>     <dbl> <dbl>
    #>  1  0.     62.6
    #>  2  0.100  62.6
    #>  3  0.200  62.6
    #>  4  0.300  62.6
    #>  5  0.400  62.6
    #>  6  0.500  62.6
    #>  7  0.600  62.6
    #>  8  0.700  62.6
    #>  9  0.800  62.6
    #> 10  0.900  62.6
    #> 11  1.00   62.6
    
    # write a function to meld predictions
    
    meld_predictions <- function(x) {
      # x is a data frame with m rows and two columns:
      #
      # m  .fitted  .se.fit
      # 1  1.05     0.34
      # 2  1.09     0.28
      # x  ...      ...
    
      # Meld the fitted values using Rubin's rules
      x_melded <- mi.meld(matrix(x$.fitted), matrix(x$.se.fit))
    
      data_frame(.fitted = as.numeric(x_melded$q.mi),
                 .se.fit = as.numeric(x_melded$se.mi))
    }
    
    # We augment/predict using new_data in each of the imputed models, then we group
    # by each of the values of civil liberties (so each value, like 0.1 and 0.2 has
    # 5 values, 1 from each of the imputed models), and then we meld those 5
    # predicted values into a single value with meld_predictions()
    
    predict_melded <- data_frame(models = models_imputed_df$model) %>%
      mutate(m = 1:n(),
             fitted = models %>% map(~ augment(., newdata = new_data))) %>% 
      unnest(fitted) %>% 
      dplyr::select(-models) %>% #### I needed to add this row to make the code work, once you've used the models to get the fit you don't need them in the data object anymore.  I took this line out because it was slowing everything down, then realised the code only works with this line... not sure why?
      group_by(civlib) %>%  
      nest(data=c(m, .fitted, .se.fit)) %>%  # needed to change this here from gisthub to make the nested 'data' have all the imputations in it, not just estimates from one of the imputations.
      mutate(fitted_melded = data %>% map(~ meld_predictions(.))) %>% 
      unnest(fitted_melded) %>% 
      mutate(ymin = .fitted + (qnorm(0.025) * .se.fit),
             ymax = .fitted + (qnorm(0.975) * .se.fit))
    
    
    ## NB. this is still on the link scale -- you'd need to write an extra few lines to exponentiate everything and get your predictions and se's on the response scale
    # Plot!
    ggplot(predict_melded, aes(x = civlib, y = .fitted)) +
      geom_line(color = "blue") +
      geom_ribbon(aes(ymin = ymin, ymax = ymax), alpha = 0.2, fill = "blue")
    

    希望这至少对遇到此问题的其他人有所帮助。

    【讨论】:

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