我创建了一个函数来根据 id 减去一些数据。在 dplyr 更新之前，该函数一直运行良好。最初，该函数不接受列名称作为函数中的输入。我用过Programming with dplyr调整函数以接受列名称，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如何修复它。
Error in location == "4" :
comparison (1) is possible only for atomic and list types
最佳答案
我认为 ! 的优先级导致了问题。当这种情况发生时，看起来应该使用 UQ 代替 !!。
在这种情况下，函数的第一部分将如下所示
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 可能是实现此目的的最佳方法。另一种选择是以函数的形式使用提取括号。这也绕过了优先级问题。
例如，代码如下
!!variable[location == "4"]
可以重写为
`[`(!!variable, location == "4")
对函数的第一部分进行这些更改，事情看起来像
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")
}
运行也没有错误
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
关于r - dplyr:函数错误消息，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/44531973/