我已经连接了两个数据表,并基于该数据的子集计算均值。当下面的代码不在我编写的函数中时,它可以正常运行,但当我尝试使用函数时,我会遇到这个错误:
Error in `[.data.table`(poll.name, AQ.Date >= Cdate & AQ.Date < Cdate + :
i evaluates to a logical vector length 159 but there are 2797432 rows. Recycling of logical i is no longer allowed as it hides more bugs than is worth the rare convenience. Explicitly use rep(...,length=.N) if you really need to recycle.
我的功能:
myfunc <- function(linked.dat, poll.name) {
linked.dat[,
`:=` (t1.avg = mean(poll.name[AQ.Date >= Cdate & AQ.Date < Cdate + 1], na.rm = TRUE),
t2.avg = mean(poll.name[AQ.Date >= Cdate + 1 & AQ.Date < Cdate + 2], na.rm = TRUE),
t3.avg = mean(poll.name[AQ.Date >= Cdate + 2 & AQ.Date <= Bdate], na.rm = TRUE),
total.avg = mean(poll.name)),
by = ID]
linked.pollname <- linked.dat
return(linked.pollname)
}
因此,将这个函数与示例df一起使用看起来像:
myfunc(df, O3)
一些数据:
df <- structure(list(O3 = c(21.1, 27.3, 23.8, 29.5, 23.8, 27.1, 31.6,
25.8, 31.2, 14, 19.1, 15.5, 15.6, 28.6, 16.9, 27.4, 30.1, 24.4,
21.2, 22.1, 26.1, 19.9), AQ.Date = structure(c(3679, 3681, 3682,
3683, 3680, 3685, 3686, 3687, 3684, 3689, 3673, 3675, 3677, 3678,
3686, 3687, 3688, 3692, 3681, 3693, 3695, 3696), class = "Date"),
ID = c("a", "a", "a", "a", "a", "a", "a", "a", "a", "a",
"a", "a", "b", "b", "b", "b", "b", "b", "b", "b", "b", "b"
), Cdate = structure(c(3673, 3673, 3673, 3673, 3673,
3673, 3673, 3673, 3673, 3673, 3673, 3673, 3677, 3677, 3677,
3677, 3677, 3677, 3677, 3677, 3677, 3677), class = "Date"),
Bdate = structure(c(3690, 3690, 3690, 3690, 3690, 3690,
3690, 3690, 3690, 3690, 3690, 3690, 3696, 3696, 3696, 3696,
3696, 3696, 3696, 3696, 3696, 3696), class = "Date"), Total_weeks = c(2.428571,
2.428571, 2.428571, 2.428571, 2.428571, 2.428571, 2.428571,
2.428571, 2.428571, 2.428571, 2.428571, 2.428571, 2.714286,
2.714286, 2.714286, 2.714286, 2.714286, 2.714286, 2.714286,
2.714286, 2.714286, 2.714286)), row.names = c(NA, -22L), class = "data.frame")
setDT(df)
我不明白这个错误是什么意思。回收指的是什么?为什么它只发生在函数内?如何调整函数以解决错误?
一般回收
回收与不同长度的矢量如何组合成data.frame
(以及其他一些地方(有关。data.frame
(因此data.table
和tbl_df
(的每一列都必须是相同的长度,当某个列的长度不相同时,它被回收。
在大多数(所有?(基R函数中,只要最长的向量是较短向量的偶数倍,就可以无声地循环。例如,
data.frame(x = 1, y = 1:3)
# x y
# 1 1 1
# 2 1 2
# 3 1 3
data.frame(x = 1:2, y = 1:4)
# x y
# 1 1 1
# 2 2 2
# 3 1 3
# 4 2 4
但当提供非偶数组合时,R会出错(通常,但并非所有情况下(:
data.frame(x = 1:3, y = 1:4)
# Error in data.frame(x = 1:3, y = 1:4) :
# arguments imply differing number of rows: 3, 4
我个人的观点是,回收是便利和安全之间的平衡;便利性";我想把一个具有单一不变值的列添加到一个具有多行的帧中,就像上面的第一个例子一样"安全";您可以确定每个函数返回的内容(例如长度(,并且惊讶地发现它们没有被隐藏。
对于后者,考虑一个自定义函数(旨在模拟which.min
(,它可以找到最小值的位置:
myfunc <- function(x) which(x == min(x)) # this is naive, do not use it
用";正常的";数据,它将返回一个值,如
set.seed(42)
myfunc(runif(10))
# [1] 8
然而,也许在处理整数或其他可能发生相等的情况时(在一些罕见的numeric
实例中(,可能会得到不止一个:
myfunc(sample(10, size = 11, replace = TRUE))
# [1] 2 10
因此,如果依赖返回一个值,但它返回两个或多个值,那么。。。你所依赖的东西可能会进行无声的回收,而你对此一无所知。例如,
set.seed(3)
mydat <- data.frame(x = sample(10, size = 12, replace = TRUE))
mydat$y <- myfunc(mydat$x)
mydat
# x y
# 1 5 4
# 2 10 8
# 3 7 4
# 4 4 8
# 5 10 4
# 6 8 8
# 7 8 4
# 8 4 8
# 9 10 4
# 10 7 8
# 11 8 4
# 12 8 8
从我的角度来看,回收只是";"可接受";当它是一个全或一的事情。。。其他任何东西在很多地方都可以正确使用,但在我看来应该非常明确。
tibble
允许全或1,否则会出错:
library(tibble)
tibble(x = 1, y = 1:3)
# # A tibble: 3 x 2
# x y
# <dbl> <int>
# 1 1 1
# 2 1 2
# 3 1 3
tibble(x = 1:2, y = 1:3)
# Error: Tibble columns must have compatible sizes.
# * Size 2: Existing data.
# * Size 3: Column `y`.
# i Only values of size one are recycled.
针对您的问题
您正试图在data.table
构造之外对符号O3
执行非标准求值。我相信你打算在其他条件下采用用户提供的框架列的平均值。
这里有一种方法:传递一个字符串,并在data.table
中使用get(poll.name)
(无论何时需要数据(来获取数据:
myfunc <- function(linked.dat, poll.name) {
linked.dat[,
`:=` (t1.avg = mean(get(poll.name)[AQ.Date >= Cdate & AQ.Date < Cdate + 1], na.rm = TRUE),
t2.avg = mean(get(poll.name)[AQ.Date >= Cdate + 1 & AQ.Date < Cdate + 2], na.rm = TRUE),
t3.avg = mean(get(poll.name)[AQ.Date >= Cdate + 2 & AQ.Date <= Bdate], na.rm = TRUE),
total.avg = mean(get(poll.name))),
by = ID]
linked.pollname <- linked.dat
return(linked.pollname)
}
myfunc(df, "O3")
# O3 AQ.Date ID Cdate Bdate Total_weeks t1.avg t2.avg t3.avg total.avg
# 1: 21.1 1980-01-28 a 1980-01-22 1980-02-08 2.428571 19.1 NaN 24.60909 24.15
# 2: 27.3 1980-01-30 a 1980-01-22 1980-02-08 2.428571 19.1 NaN 24.60909 24.15
# 3: 23.8 1980-01-31 a 1980-01-22 1980-02-08 2.428571 19.1 NaN 24.60909 24.15
# 4: 29.5 1980-02-01 a 1980-01-22 1980-02-08 2.428571 19.1 NaN 24.60909 24.15
# 5: 23.8 1980-01-29 a 1980-01-22 1980-02-08 2.428571 19.1 NaN 24.60909 24.15
# ---
# 18: 24.4 1980-02-10 b 1980-01-26 1980-02-14 2.714286 15.6 28.6 23.51250 23.23
# 19: 21.2 1980-01-30 b 1980-01-26 1980-02-14 2.714286 15.6 28.6 23.51250 23.23
# 20: 22.1 1980-02-11 b 1980-01-26 1980-02-14 2.714286 15.6 28.6 23.51250 23.23
# 21: 26.1 1980-02-13 b 1980-01-26 1980-02-14 2.714286 15.6 28.6 23.51250 23.23
# 22: 19.9 1980-02-14 b 1980-01-26 1980-02-14 2.714286 15.6 28.6 23.51250 23.23