将影响多列的函数的输出覆盖到数据框中答案

【问题标题】：Overwriting the output of an function affecting multiple columns to a data frame将影响多列的函数的输出覆盖到数据框中
【发布时间】：2019-03-26 09:01:55
【问题描述】：

我有一个包含许多个人旅行历史的大型数据框（旅行）。在 travel$Travel 1 Destination 列中，我只需要国家/地区数据，但许多条目都输入了地区、城市或其他数据。

我尝试编写一个带有 2 个参数的函数 (city_replace_with_country)：

city_name：已经在 travel$`Travel 1 Destination' 列中的城市名称
country_name: 国家名称

我希望我的函数在数据框中找到这些错误，将它们粘贴到另一个预先存在的列 (travel$regions_visited)，然后用正确的替换 travel$`Travel 1 Destination' 中的城市/地区名称国名。

此代码适用于个别示例，例如。班加罗尔和印度

for (i in 1:nrow(travel) ) {
  if(grepl("bangalore", travel$`Travel 1 Destination`[i], ignore.case = TRUE)){
  travel$regions_visited[i] <- paste(travel$regions_visited[i], "Bangalore", sep = " ")
  travel$`Travel 1 Destination`[i] <- gsub("bangalore", "india", travel$`Travel 1 Destination`[i], ignore.case = TRUE, perl = TRUE)
  }}

理想情况下，我可以使用该函数对数据框进行许多更正

city_replace_with_country <- function(city_name, country_name) {
    for (i in 1:nrow(travel)) {
    if(grepl(city_name, travel$`Travel 1 Destination`[i], ignore.case = TRUE)){
      travel$regions_visited[i] <- paste(travel$regions_visited[i], city_name, sep = " ")
      travel$`Travel 1 Destination`[i] <- gsub(city_name, country_name, travel$`Travel 1 Destination`[i], ignore.case = TRUE, perl = TRUE)
    }}}

当我尝试按如下方式使用此功能时：

city_replace_with_country("bangalore", "india")

函数的输出似乎没有写入/存储到数据帧。如果我尝试：

travel <- city_replace_with_country("bangalore", "india")

或

travel$`Travel 1 Destination` <- city_replace_with_country("bangalore", "india")

它返回一个 NULL 值。

任何有关如何使其工作的建议将不胜感激。非常感谢。

这是一些示例数据。例如，在这种情况下，我想将波哥大更改为哥伦比亚：

travel_example <- structure(list(X1 = c(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 
13, 14, 15, 16, 17, 18, 19, 20, 21), `Travel 1 Dates` = c("03/09/18 to 02/10/18", 
NA, "15/08/18 - 24/08/18", "13/09/2018 to 19/09/2018", "15/07- 14/08", 
"21/9/18-29/9/18", "10/09/18-3/10/18", "28/5/18-31/7/18", "1/9/18-16/9/18", 
NA, NA, NA, "15/07/18-03/10/18", "15/09/18-30/09/18", "18/09/18-05/10/18", 
"7/9/18-23/9/18", "14/8/18-11/9/18", "25/7/18-13/8/18", "24/9/18-30/9/18", 
NA, "18/9/18-2/10/18"), `Travel 1 Destination` = c("Colombia", 
"salvador, Bogata, Honduras", "China,Cambodia", "Lagos, Nigeria", 
"Uganda", "Indonesia", "Kenya", "Dubai, Japan", "Sri Lanka", 
"Jakarta,Indonesia", "South Africa, Mozambique, Ethiopia", NA, 
"Ukraine, Slovakia, India", "Tanzania", "Ghana", "Sri Lanka", 
"Angola", "Tajikistan , Afghanistan", "Morocco", NA, "Tanzania"
), regions_visited = c("character(0)", "character(0)", "character(0)", 
"character(0)", "character(0)", "character(0)", "Nairobi", "character(0)", 
"character(0)", "character(0)", "character(0)", "character(0)", 
"character(0)", "character(0)", "character(0)", "character(0)", 
"Luanda only", "c(\"south\", \"north\", \"very rural\")", "character(0)", 
"character(0)", "character(0)")), class = c("spec_tbl_df", "tbl_df", 
"tbl", "data.frame"), row.names = c(NA, -21L), spec = structure(list(
    cols = list(X1 = structure(list(), class = c("collector_double", 
    "collector")), `Travel 1 Dates` = structure(list(), class = c("collector_character", 
    "collector")), `Travel 1 Destination` = structure(list(), class = c("collector_character", 
    "collector")), regions_visited = structure(list(), class = c("collector_character", 
    "collector"))), default = structure(list(), class = c("collector_guess", 
    "collector")), skip = 1), class = "col_spec"))

【问题讨论】：

请与dput分享数据框样本
谢谢。添加了一个示例。
抱歉，刚刚更正了dput输出（原来用read.csv解析，什么时候应该是read_csv）。

标签： r function

【解决方案1】：

在您的情况下，您可以使用regex 提取逗号后的最后一个单词，然后获取所有国家/地区名称，如下所示：

travel_example$`Travel 1 Destination` <- sub('.*,\\s*', '', travel_example$`Travel 1 Destination`)
travel_example$`Travel 1 Destination`
 [1] "Colombia"    "Honduras"    "Cambodia"    "Nigeria"     "Uganda"     
 [6] "Indonesia"   "Kenya"       "Japan"       "Sri Lanka"   "Indonesia"  
[11] "Ethiopia"    NA            "India"       "Tanzania"    "Ghana"      
[16] "Sri Lanka"   "Angola"      "Afghanistan" "Morocco"     NA           
[21] "Tanzania"

【讨论】：