在 R purr 中迭代 reading/mutating 个 csv 文件

Question

我在 R 中有一个 csv 文件文件夹，需要根据文件名中的信息在列中循环、清理和创建。我正在尝试使用 purr，这就是我到目前为止所做的。

# get file names
files_names <- list.files("data/", recursive = TRUE, full.names = TRUE) 

# inspect
files_names 

[1] "data/BOC_All_ATMImage_(Aug 2020).txt" "data/BOC_All_ATMImage_(Aug 2021).txt" "data/BOC_All_ATMImage_(Feb 2021).txt"
[4] "data/BOC_All_ATMImage_(May 2021).txt" "data/BOC_All_ATMImage_(Nov 2020).txt" "data/BOC_All_ATMImage_(Nov 2021).txt"

# extract month/year inside brackets and convert to snakecase
# this will be used later to create column names

names_data <- files_names %>% 
  str_extract(., "(?<=\().*?(?=\))") %>% 
  str_to_lower() %>%
  str_replace(., " ", "_")

column_names

[1] "aug_2020" "aug_2021" "feb_2021" "may_2021" "nov_2020" "nov_2021"

现在遍历 csvs，读取每个 csv，进行一些数据清理并创建列


mc_data <-
  map(files_names,
         ~ read_csv(.x, guess_max = 50000) %>%
        janitor::clean_names() %>% 
           mutate(month_year = str_extract(.x, "(?<=\().*?(?=\))"),
                  date_dmy = paste0(day, "-", month_year),
                  date = dmy(date_dmy),
                  fsa = str_sub(postal_code, start = 1, end=3),
                  ?? = 1) %>% 
         select(-date_dmy),
         .id = "group"
  )

我需要再改变一列，该列必须根据提取的 names_data 命名。我目前在上面的假代码中将其作为 ?? 。 names_data 遵循与文件路径相同的顺序，因此我们的想法是在一个循环中完成并在清理后保存每个数据。

Answer 1

我们可以使用胶水语法 map2。也许：

mc_data <-
    map2(files_names, column_names,
        ~ read_csv(.x, guess_max = 50000) %>%
            janitor::clean_names() %>% 
            mutate(month_year = str_extract(.x, "(?<=\().*?(?=\))"),
                   date_dmy = paste0(day, "-", month_year),
                   date = dmy(date_dmy),
                   fsa = str_sub(postal_code, start = 1, end=3),
                   '{.y}' := 1) %>% 
            select(-date_dmy),
        .id = "group"
    )

在 R purr 中迭代 reading/mutating 个 csv 文件

iterate reading/mutating csv files in R purr

r

dplyr

purrr

tidyverse