如何使用 R (tidyverse) 中的函数迭代地创建多个小标题？答案

【问题标题】：How can I iteratively create multiple tibbles using a function in R (tidyverse)?如何使用 R (tidyverse) 中的函数迭代地创建多个小标题？
【发布时间】：2019-12-17 01:39:12
【问题描述】：

我必须创建 10 个具有相同变量的小标题（数据只是不同），这导致我复制以下代码


dpd <- list_sheets[[1]] %>%
    filter(LASTUSER %in% users ) %>% 
    mutate(year = as.numeric(format(DATEMODIFI, format = "%Y")),
           month = as.numeric(format(DATEMODIFI, format = "%m")),
           day = as.numeric(format(DATEMODIFI, format = "%d"))) %>% 
    select(-DATEMODIFI) %>% 
    filter(year == 2019),

  fuse <- list_sheets[[2]]%>%
    filter(LASTUSER %in% users ) %>% 
    mutate(year = as.numeric(format(DATEMODIFI, format = "%Y")),
           month = as.numeric(format(DATEMODIFI, format = "%m")),
           day = as.numeric(format(DATEMODIFI, format = "%d"))) %>% 
    select(-DATEMODIFI) %>% 
    filter(year == 2019),

  ohprimary <- list_sheets[[3]]%>%
    filter(LASTUSER %in% users ) %>% 
    mutate(year = as.numeric(format(DATEMODIFI, format = "%Y")),
           month = as.numeric(format(DATEMODIFI, format = "%m")),
           day = as.numeric(format(DATEMODIFI, format = "%d"))) %>% 
    select(-DATEMODIFI) %>% 
    filter(year == 2019),

  ohsecondary <- list_sheets[[4]]%>%
    filter(LASTUSER %in% users ) %>% 
    mutate(year = as.numeric(format(DATEMODIFI, format = "%Y")),
           month = as.numeric(format(DATEMODIFI, format = "%m")),
           day = as.numeric(format(DATEMODIFI, format = "%d"))) %>% 
    select(-DATEMODIFI) %>% 
    filter(year == 2019),

  poles <- list_sheets[[5]]%>%
    filter(LASTUSER %in% users ) %>% 
    mutate(year = as.numeric(format(DATEMODIFI, format = "%Y")),
           month = as.numeric(format(DATEMODIFI, format = "%m")),
           day = as.numeric(format(DATEMODIFI, format = "%d"))) %>% 
    select(-DATEMODIFI) %>% 
    filter(year == 2019),

  pv <- list_sheets[[6]]%>%
    filter(LASTUSER %in% users ) %>% 
    mutate(year = as.numeric(format(DATEMODIFI, format = "%Y")),
           month = as.numeric(format(DATEMODIFI, format = "%m")),
           day = as.numeric(format(DATEMODIFI, format = "%d"))) %>% 
    select(-DATEMODIFI) %>% 
    filter(year == 2019),

  switch <- list_sheets[[7]]%>%
    filter(LASTUSER %in% users ) %>% 
    mutate(year = as.numeric(format(DATEMODIFI, format = "%Y")),
           month = as.numeric(format(DATEMODIFI, format = "%m")),
           day = as.numeric(format(DATEMODIFI, format = "%d")))%>% 
    select(-DATEMODIFI) %>% 
    filter(year == 2019),

  transformers <- list_sheets[[8]]%>%
    filter(LASTUSER %in% users ) %>% 
    mutate(year = as.numeric(format(DATEMODIFI, format = "%Y")),
           month = as.numeric(format(DATEMODIFI, format = "%m")),
           day = as.numeric(format(DATEMODIFI, format = "%d")))%>% 
    select(-DATEMODIFI) %>% 
    filter(year == 2019),

  ugprimary <- list_sheets[[9]]%>%
    filter(LASTUSER %in% users ) %>% 
    mutate(year = as.numeric(format(DATEMODIFI, format = "%Y")),
           month = as.numeric(format(DATEMODIFI, format = "%m")),
           day = as.numeric(format(DATEMODIFI, format = "%d")))%>% 
    select(-DATEMODIFI) %>% 
    filter(year == 2019),

  ugsecondary <- list_sheets[[10]]%>%
    filter(LASTUSER %in% users ) %>% 
    mutate(year = as.numeric(format(DATEMODIFI, format = "%Y")),
           month = as.numeric(format(DATEMODIFI, format = "%m")),
           day = as.numeric(format(DATEMODIFI, format = "%d")))%>% 
    select(-DATEMODIFI) %>% 
    filter(year == 2019)

所以我创建了以下函数，并尝试像这样迭代地创建它们：

createTibble <- function(i){
  list_sheets[[i]]%>%
    filter(LASTUSER %in% users ) %>% 
    mutate(year = as.numeric(format(DATEMODIFI, format = "%Y")),
           month = as.numeric(format(DATEMODIFI, format = "%m")),
           day = as.numeric(format(DATEMODIFI, format = "%d")))%>% 
    select(-DATEMODIFI) %>% 
    filter(year == 2019)


}

features <- c('dpd', 'fuse')

for (feature in seq_along(features)){
  tibble <- sym(features[feature]) # returns the proper symbol e.g. 'dpd' -> dpd 
  tibble <- createTibble(feature) # This just gives me a table called tibble

}

这不会给我两个 tibble（dpd 和 fuse）而是我只会得到一个名为 tibble 的 tibble

我做错了什么？我喜欢 R，但为什么要做这样的事情会很烦人

编辑：

features <- c('dpd', 'fuse')
output <- vector('list', length(features))
for (feature in seq_along(features)){

  output[[feature]] <- createTibble(feature)
  name <- sym(features[feature])
  print(name)
  name <- output[[feature]]

}

【问题讨论】：

您需要创建一个列表来存储输出，即out <- vector('list', length(features))，在for循环内，out[[feature]] <- createTibble(feature)tibble是一个函数名，请使用不同的名称作为对象名
我没有看到任何编辑，same exact result 是什么意思（是来自for 循环的那个吗？）请展示一个可重复的小示例
如果你已经分配了output[[feature]] <- createTibble(feature)并且你的createTibble有效，那么它应该返回output中的两个元素，因为你已经创建了output <- vector('list', length(features))，检查output如果你正在检查@987654335 @，事实并非如此。此外，您可能需要在for 循环之后将“输出”的名称指定为names(output) <- features，即不需要name <- sym(features[feature])
我不建议在全局环境中创建多个对象
我发布了一个包含您想要的要求的解决方案

标签： r tidyverse

【解决方案1】：

为了创建对象的list，用一个空的list进行初始化，并根据for循环的每次迭代中函数的输出为其赋值

features <- c('dpd', 'fuse')
output <- setNames(vector('list', length(features)), features)
for (feature in seq_along(features)){

   output[[feature]] <- createTibble(feature)
}

不建议在全局环境中有多个对象，但如果需要，可以选择assign

for (feature in seq_along(features)){
    assign(features[feature], value = createTibble(feature))
 }

现在，我们检查一下

ls()

对于创建的对象

或使用tidyverse

library(purrr)
map(seq_along(features), ~createTibble(.x)) %>%
   set_names(features) %>%
   list2env(.GlobalEnv)

【讨论】：