Question

我创建了一个函数来根据id减去我的一些数据。该功能正常工作，直到dplyr更新。最初，该函数不接受列名作为函数中的输入。我使用Programming with dplyr来调整函数以接受列名，但是我现在收到一条新的错误消息。

testdf <- structure(list(date = c("2016-04-04", "2016-04-04", "2016-04-04", 
                        "2016-04-04", "2016-04-04", "2016-04-04"), sensorheight = c(1L, 
                                                                                    16L, 1L, 16L, 1L, 16L), farm = c("McDonald", "McDonald", 
                                                                                                                     "McDonald", "McDonald", "McDonald", "McDonald"
                                                                                    ), location = c("4", "4", "5", "5", "Outside", "Outside"), Temp = c(122.8875, 
                                                                                                                                                        117.225, 102.0375, 98.3625, 88.5125, 94.7)), .Names = c("date", 
                                                                                                                                                                                                                "sensorheight", "farm", "location", "Temp"), row.names = c(NA, 
                                                                                                                                                                                                                                                                           6L), class = "data.frame")


DailyInOutDiff <- function (df, variable) {

  DailyInOutDiff04 <- df %>%
    filter(location %in% c(4, 'Outside')) %>% 
    group_by(date, sensorheight, farm) %>%
    arrange(sensorheight, farm, location) %>%
    summarise(Diff = if(n()==1) NA else !!variable[location=="4"] - !!variable[location=='Outside'], 
              location = "4")  %>%
    select(1, 2, 3, 5, 4)

  DailyInOutDiff05 <- df %>%
    filter(location %in% c(5, 'Outside')) %>% 
    group_by(date, sensorheight, farm) %>%
    arrange(sensorheight, farm, location) %>%
    summarise(Diff = if(n()==1) NA else !!variable[location=="5"] - !!variable[location=='Outside'], 
              location = "5")  %>%
    select(1, 2, 3, 5, 4)

  temp.list <- list(DailyInOutDiff04, DailyInOutDiff05)
  final.df = bind_rows(temp.list)
  return(final.df)
}

test <- DailyInOutDiff(testdf, quo(Temp))

我想知道错误消息的含义以及解决方法。

 Error in location == "4" : 
    comparison (1) is possible only for atomic and list types

Answer 1

我认为!的优先顺序会导致问题。如果发生这种情况，则应使用UQ代替!!。

在这种情况下，函数的第一部分看起来像

DailyInOutDiff <- function (df, variable) {

    variable = enquo(variable)

    df %>%
        filter(location %in% c(4, 'Outside')) %>% 
        group_by(date, sensorheight, farm) %>%
        arrange(sensorheight, farm, location) %>%
        summarise(Diff = if(n()==1) NA else UQ(variable)[location == "4"] - 
                    UQ(variable)[location == "Outside"], 
                location = "4")

}

现在运行没有错误。

DailyInOutDiff(testdf, Temp)

        date sensorheight     farm   Diff location
       <chr>        <int>    <chr>  <dbl>    <chr>
1 2016-04-04            1 McDonald 34.375        4
2 2016-04-04           16 McDonald 22.525        4

我认为使用UQ可能是最好的方法。另一种方法是以函数的形式使用提取括号。这也绕过了优先问题。

例如，代码看起来像

!!variable[location == "4"]

可以改写为

`[`(!!variable, location == "4")

对函数的第一部分进行这些更改，看起来像

DailyInOutDiff <- function (df, variable) {

    variable = enquo(variable)

    df %>%
        filter(location %in% c(4, 'Outside')) %>% 
        group_by(date, sensorheight, farm) %>%
        arrange(sensorheight, farm, location) %>%
        summarise(Diff = if(n()==1) NA else `[`(!!variable, location == "4") - 
                    `[`(!!variable, location == "Outside"), 
                location = "4")

}

哪个也运行无错误

DailyInOutDiff(testdf, Temp)

        date sensorheight     farm   Diff location
       <chr>        <int>    <chr>  <dbl>    <chr>
1 2016-04-04            1 McDonald 34.375        4
2 2016-04-04           16 McDonald 22.525        4

dplyr：函数错误消息

1 个答案: