通过创建函数，代码错误来修改R列

Question

我创建了这些行（函数）来修改数据框架的特定列，我想使用此函数将其用于不同的列和数据框架，但是该函数不起作用，我收到了错误代码消息。

change.date <-  function(df_date,col_nb,first.year, second.year){
  df_date$col_nb <- gsub(first.year, second.year,  df_date$col_nb)
  df_date$col_nb <- as.Date(df_date$col_nb)
  df_date$col_nb <-  as.numeric(df_date$col_nb)
    
}

change.date(df_2020,df_2020[1], "2020","2020")

Error in $<-.data.frame`(*tmp*`, "col_nb", value = character(0)):
replacement table has 0 rows, replaced table has 7265

我可复制的数据是：

df_2020 <- dput(test_qst)
structure(list(Date = structure(c(1588809600, 1588809600, 1588809600, 
1588809600, 1588809600, 1588809600, 1588809600, 1588809600, 1588809600, 
1588809600, 1588809600, 1588809600, 1588809600, 1588809600), class = c("POSIXct", 
"POSIXt"), tzone = "UTC"), Depth = c(1.72, 3.07, 3.65, 4.58, 
5.39, 6.31, 7.27, 8.57, 9.73, 10.78, 11.71, 12.81, 13.79, 14.96
), salinity = c(34.7299999999999, 34.79, 34.76, 34.78, 34.77, 
34.79, 34.76, 34.71, 34.78, 34.78, 34.7999999999999, 34.86, 34.7999999999999, 
34.83)), class = c("tbl_df", "tbl", "data.frame"), row.names = c(NA, 
-14L))

Answer 1

您可以尝试

change.date <-  function(df_date,col_nb,first.year, second.year){
  df_date[[col_nb]] <- gsub(first.year, second.year,  df_date[[col_nb]])
  
   
  df_date[[col_nb]] <- as.Date(df_date[[col_nb]])
  df_date[[col_nb]] <-  as.numeric(df_date[[col_nb]])
  df_date
}

change.date(df_2020, "Date", "2020","2020")

    Date Depth salinity
   <dbl> <dbl>    <dbl>
 1 18389  1.72     34.7
 2 18389  3.07     34.8
 3 18389  3.65     34.8
 4 18389  4.58     34.8
 5 18389  5.39     34.8
 6 18389  6.31     34.8
 7 18389  7.27     34.8
 8 18389  8.57     34.7
 9 18389  9.73     34.8
10 18389 10.8      34.8
11 18389 11.7      34.8
12 18389 12.8      34.9
13 18389 13.8      34.8
14 18389 15.0      34.8

Answer 2

使用GSUB时，您可能会发现一个问题是您丢失了日期。除非您需要数字时间尺度，否则最好保留绘图和分析日期。

使用dplyr，这会提取年份，更改它们，然后再次创建日期（即使它们是同年）：

library(dplyr)

change.date <-  function(df_date, col_nb = "Date", first.year, second.year) {

  col_nb <- which(colnames(df_date) %in% col_nb)      
  
  df_date %>% 
    mutate(year = lubridate::year(.[[col_nb]])) %>% 
    mutate(year = ifelse(year == first.year, second.year, year)) %>% 
    mutate(Date = lubridate::make_date(year, lubridate::month(.[[col_nb]]), lubridate::day(.[[col_nb]]))) %>% 
    select(-year)
}

change.date(df_2020, "Date", 2020, 2020)

# A tibble: 14 x 3

   Date       Depth salinity
   <date>     <dbl>    <dbl>
 1 2020-05-07  1.72     34.7
 2 2020-05-07  3.07     34.8
 3 2020-05-07  3.65     34.8
 4 2020-05-07  4.58     34.8
 5 2020-05-07  5.39     34.8
 6 2020-05-07  6.31     34.8
 7 2020-05-07  7.27     34.8
 8 2020-05-07  8.57     34.7
 9 2020-05-07  9.73     34.8
10 2020-05-07 10.8      34.8
11 2020-05-07 11.7      34.8
12 2020-05-07 12.8      34.9
13 2020-05-07 13.8      34.8
14 2020-05-07 15.0      34.8

如果您确实想要数字日期，请使用此而不是第二行：

mutate(Date = as.numeric(lubridate::make_date(year, lubridate::month(.[[col_nb]]), lubridate::day(.[[col_nb]])))) %>% 

对您的功能的评论是在情况下保持一致。骆驼盒，蛇案例或较少的dot情况都是可以接受的，但是使用组合使跟踪变量更困难，例如df_date versus first.year。