Question

我有一个带有两个参数的函数，一个链接和一个State name缩写。该函数从远程站点获取电子表格让我成为我想要的状态数据的数据框。

library(lubridate)
library(tidyr)
library(gdata)
library(dplyr)
options(stringsAsFactors = FALSE)

nsw_link <-  'http://www.abs.gov.au/ausstats/abs@archive.nsf/log?openagent&3101051.xls&3101.0&Time%20Series%20Spreadsheet&3F92BFA30BC29940CA257F1D001427C3&0&Jun%202015&17.12.2015&Latest'
qld_link <- "http://www.abs.gov.au/ausstats/abs@archive.nsf/log?openagent&3101053-.xls&3101.0&Time%20Series%20Spreadsheet&2927EBD7E6856BABCA257F1D0014283D&0&Jun%202015&17.12.2015&Latest"
vic_link <- "http://www.abs.gov.au/ausstats/abs@archive.nsf/log?openagent&3101052.xls&3101.0&Time%20Series%20Spreadsheet&E3B2958632AB29ECCA257F1D001427FB&0&Jun%202015&17.12.2015&Latest"
nt_link <- "http://www.abs.gov.au/ausstats/abs@archive.nsf/log?openagent&3101057.xls&3101.0&Time%20Series%20Spreadsheet&CCB60AB638D60938CA257F1D0014291C&0&Jun%202015&17.12.2015&Latest"
tas_link <- "http://www.abs.gov.au/ausstats/abs@archive.nsf/log?openagent&3101056.xls&3101.0&Time%20Series%20Spreadsheet&8CA5625A306A4805CA257F1D001428D7&0&Jun%202015&17.12.2015&Latest"
act_link <- "http://www.abs.gov.au/ausstats/abs@archive.nsf/log?openagent&3101058.xls&3101.0&Time%20Series%20Spreadsheet&4234206BA89A82F6CA257F1D00142959&0&Jun%202015&17.12.2015&Latest"
sa_link <- "http://www.abs.gov.au/ausstats/abs@archive.nsf/log?openagent&3101054.xls&3101.0&Time%20Series%20Spreadsheet&A342BFFB06A62F4FCA257F1D0014286D&0&Jun%202015&17.12.2015&Latest"
wa_link <- "http://www.abs.gov.au/ausstats/abs@archive.nsf/log?openagent&3101055.xls&3101.0&Time%20Series%20Spreadsheet&DC6208699BE4D2FFCA257F1D0014289E&0&Jun%202015&17.12.2015&Latest"

get_ERP_data <- function(link, state){

    # Get the xls file and slice only the columns needed for male and female ERPs
    xls_data <- read.xls(link, sheet = 'Data1')
    xls_data <- tbl_df(xls_data)
    xls_data <- xls_data[, 1:203]

    names(xls_data) <- gsub(pattern = "Estimated.Resident.Population....", '',  names(xls_data))
    names(xls_data) <- gsub('[.]+', '', names(xls_data))
    names(xls_data) <- gsub('100andover', '101', names(xls_data))
    names(xls_data) <- gsub("(\\d+)$", ".\\1", names(xls_data))


    xls_data <- xls_data[28:54, ]
    names(xls_data)[1] <- 'Year'


    xls_data$Year <- paste('01', xls_data$Year, sep = '-')
    xls_data$Year <- dmy(xls_data$Year)
    xls_data$Year <- year(xls_data$Year)


    # make a long version of the ERP data
    xls_data_long <- xls_data %>% gather(Sex_Age, Population, Male.0:Female.101)

    # Make two new column, sex and age, from the sex_age column
    xls_data_sep_log <- xls_data_long %>% separate(Sex_Age, c('Sex', 'Age'))

    # Recode observations from 101 to 'over 100'
    xls_data_sep_log$Age[grep(xls_data_sep_log$Age, pattern = '101')] <- 'Over 100'
    xls_data_sep_log$State <- state


    df_name <- paste(state, 'data', sep = '_')

    assign(df_name,xls_data_sep_log,envir = .GlobalEnv)


}


link_list <- list(act_link, nsw_link, nt_link, qld_link, sa_link, tas_link, vic_link, wa_link)
states <- c('ACT', 'NSW', 'NT', 'QLD', 'SA', 'TAS', 'VIC', 'WA')

例如，如果我想要昆士兰估计的常住人口数据，我会运行：

get_ERP_data(qld_link, 'QLD')

这使我成为我需要的数据的纵向数据框架。

我想扩展这个，以便我可以将一个向量或状态列表作为参数传递给链接列表，并为每个链接返回一个数据框参数中的论点？我并不特别关心在函数输出中绑定结果数据帧。

与上面的例子类似，我如何让我的函数运行如下：

get_ERP_data(as.list(qld_link, nsw_link), c('QLD', 'NSW'))

我尝试使用sapply和do.call的变体但是没有走得太远。有没有办法使用一些相当于pythons * args / ** kwargs 将列表作为参数传递给我的函数？

任何帮助都将不胜感激。

将可互换项目传递给R中的函数参数

0 个答案: