在 R 中使用 pmap 函数答案

【问题标题】：using pmap function in R在 R 中使用 pmap 函数
【发布时间】：2021-08-13 12:43:06
【问题描述】：

我第一次尝试使用 pmap，但在分配参数时遇到了困难。这是我的测试数据集：

  overall <- data.table(dependant = rep(c("SPS", "DEPENDANT", "EMP"), 3),
                        exposure = rnorm(9, 0, 1), 
                        age = c(1,2,3,1,2,3,3,1,2), 
                        gender = rep(c("F", "F", "M"), 3))

我最初是在做这样的事情：

  # spouse
  SPS <- overall[dependant == "SPS", .(exposure = sum(exposure)), by = c("dependant", "age", "gender")]
  sumExposureSPS <- sum(SPS$exposure)
  SPSnormalized <- SPS[, exposure := exposure/sumExposureSPS][, .(age, gender, exposure)]

  
  # dependant  
  DEPENDENT <- overall[dependant == "DEPENDENT", .(exposure = sum(exposure)), by = c("dependant", "age", "gender")]
  sumExposureDEPENDENT <- sum(DEPENDENT$exposure)
  DEPENDENTnormalized <- DEPENDENT[, exposure := exposure/sumExposureDEPENDENT][, .(age, gender, exposure)]


  # employee
  EMP <- overall[dependant == "EMP", .(exposure = sum(exposure)), by = c("dependant", "age", "gender")]
  sumExposureEMP <- sum(EMP$exposure)
  EMPnormalized <- EMP[, exposure := exposure/sumExposureEMP][, .(age, gender, exposure)]

但这是非常重复的，实际上只是名称不同，执行的操作总是相同的。因此我写了一个函数：

  calculateSubset <- function(overall, 
                              dependantCode){
    
    subset <- overall[dependant == dependantCode, .(exposure = sum(exposure)), by = c("dependant", "age", "gender")]
    sumExposureSubset <- sum(subset$exposure)
    subsetNormalized <- subset[, exposure := exposure/sumExposureSubset][, .(age, gender, exposure)]
    
    return(subset)
  }

所以我把它简化为：

  SPSnormalized <- calculateSubset(overall = overall, 
                                   dependantCode = "SPS")

  DEPENDENTnormalized <- calculateSubset(overall = overall, 
                                   dependantCode = "DEPENDENT")

  EMPnormalized <- calculateSubset(overall = overall, 
                                   dependantCode = "EMP")

但是，这仍然是重复的。我似乎有一些人使用pmap 来完全摆脱重复代码的例子。

如何在 pmap 中传递参数，以便最终获得所需的输出？

【问题讨论】：

标签： r purrr pmap

【解决方案1】：

为简单起见，交换函数calculateSubset 中的参数。默认情况下，map 系列会沿列表进行迭代，以作为函数的第一个参数传递。

calculateSubset <- function( dependantCode, df = overall){
  
  subset <- df[dependant == dependantCode, .(exposure = sum(exposure)), by = c("dependant", "age", "gender")]
  sumExposureSubset <- sum(subset$exposure)
  subsetNormalized <- subset[, exposure := exposure/sumExposureSubset][, .(age, gender, exposure)]
  
  return(subset)
}

c("SPS", "DEPENDANT", "EMP") %>% map(calculateSubset)
# Note that the above map() call is equivalent but more concise than this pmap() call: list(c("SPS", "DEPENDENT", "EMP")) %>% pmap(calculateSubset)
[[1]]
   dependant age gender exposure
1:       SPS   1      F 0.522064
2:       SPS   3      F 0.477936

[[2]]
   dependant age gender   exposure
1: DEPENDANT   2      F -0.3019417
2: DEPENDANT   1      F  1.3019417

[[3]]
   dependant age gender  exposure
1:       EMP   3      M 0.8140009
2:       EMP   2      M 0.1859991

【讨论】：

【解决方案2】：

您应该在group_by 中包含dependant 而不是子集和获取sum -

library(dplyr)

overall %>%
  group_by(dependant, age, gender) %>%
  summarise(exposure = sum(exposure), .groups = 'drop') %>%
  mutate(exposure = prop.table(exposure)) %>%
  ungroup

# dependant gender   age exposure
#  <chr>     <chr>  <dbl>    <dbl>
#1 DEPENDANT F          1   0.971 
#2 DEPENDANT F          2   0.0292
#3 EMP       M          2   0.158 
#4 EMP       M          3   0.842 
#5 SPS       F          1   1.39  
#6 SPS       F          3  -0.388

在summarise 之后，数据按dependant 和age 分组，因此使用prop.table 将给出该组中的比率。

【讨论】：