Question

此帖子是从更早的帖子中解决的。我的新问题是如何正确使用pmap。我传递了列表参数和函数。

我想通过：

day = 1

month = 11

year = 2015

为了更进一步，我想传递在这里获得的值：

library(lubridate)
seq.Date(from = as.Date("2015-01-01"), to = as.Date("2015-02-01"), by = "day") %>% 
  tibble::enframe(., name = "seq", value = "date") %>% 
  mutate(year = year(date),
         month = month(date),
         day = day(date))

  # A tibble: 32 x 5
     seq date        year month   day
   <int> <date>     <dbl> <dbl> <int>
 1     1 2015-01-01  2015     1     1
 2     2 2015-01-02  2015     1     2
 3     3 2015-01-03  2015     1     3
 4     4 2015-01-04  2015     1     4
 5     5 2015-01-05  2015     1     5
 6     6 2015-01-06  2015     1     6
 7     7 2015-01-07  2015     1     7
 8     8 2015-01-08  2015     1     8
 9     9 2015-01-09  2015     1     9
10    10 2015-01-10  2015     1    10

功能：

Temps <- function(day, month, year){
  month = stringr::str_pad(month, width = 2, pad = 0)
  day = stringr::str_pad(day, width = 2, pad = 0)
  url <- paste("https://www.timeanddate.com/weather/spain/madrid/historic?hd=", year, month, day, sep = "")
  temps_obtained <- url %>% 
    html_session() %>% 
    read_html() %>% 
    html_table(fill = TRUE) %>% 
    .[[2]] %>% 
    setNames(.[1,]) %>% 
    as_tibble(., .name_repair = "universal") %>% 
    dplyr::slice(., -1) %>% 
    dplyr::slice(., -n())
  return(temps_obtained)
}


purrr::pmap(list(1, 11, 2015), .f = Temps())

Answer 1

如果我们使用的是pmap，请确保将list参数命名为与函数的参数匹配

purrr::pmap(list(day = 1, month = 11, year = 2015), Temps)

或者，如果我们传递一个未命名的list，则用..提取每个组件

purrr::pmap(list(1, 11, 2015),  ~ Temps(day = ..1, month = ..2, year = ..3))

映射多个参数-使用pmap_dfr

1 个答案: