我正在寻找一种方法，可以有条件地只向函数传递一个参数(三个选项之一(。根据选择，我想简单地在数据集中创建一个变量。假设我们有以下数据集：
set.seed(10)
test <- data.frame(time_stamp = sample(seq(as.Date('1999/01/01'), as.Date('2012/01/01'), by="day"), 12))
test
# time_stamp
# 1 2000-05-05
# 2 2009-03-09
# 3 2008-04-24
# 4 2011-03-22
# 5 2003-05-27
# 6 2003-01-01
# 7 2008-10-22
# 8 2003-10-13
# 9 2011-02-26
# 10 2008-08-27
# 11 2011-12-30
# 12 2001-07-18
当我运行我的函数时，我想要的输出如下：
test_fun(type = "halfs")
#or more simply
test_fun(halfs)
# time_stamp half_var
# 1 2000-05-05 H1 2000
# 2 2009-03-09 H1 2009
# 3 2008-04-24 H1 2008
# 4 2011-03-22 H1 2011
# 5 2003-05-27 H1 2003
# 6 2003-01-01 H1 2003
# 7 2008-10-22 H2 2008
# 8 2003-10-13 H2 2003
# 9 2011-02-26 H1 2011
# 10 2008-08-27 H2 2008
# 11 2011-12-30 H2 2011
# 12 2001-07-18 H2 2001
根据所选的参数，我在管道中运行了一个if语句，我认为如果我像这里提到的那样在条件语句周围加上{}，我可以做到这一点，但我无法理解。功能如下：
test_fun <- function(type = c("halfs", "quarts", "other")) {
test %>% {
if (type == "halfs") {
mutate(half_var = ifelse(month(time_stamp) <= 6, paste('H1', year(time_stamp)), paste('H2', year(time_stamp))))
} else if (type == "quarts") {
mutate(quarts_var = case_when(month(time_stamp) <= 3 ~ paste('q1', year(time_stamp)),
month(time_stamp) > 3 & month(time_stamp) <= 6 ~ paste('q2', year(time_stamp)),
month(time_stamp) > 6 & month(time_stamp) <= 9 ~ paste('q3', year(time_stamp)),
month(time_stamp) > 9 ~ paste('q4', year(time_stamp))))
} else (type == "other") {
mutate(other = ifelse(month(time_stamp) <= 6, paste('H1', year(time_stamp)), paste('H2', year(time_stamp))))
}
}
}
我收到了一个关于意外括号的错误，但我认为问题在于管道内的条件if(所有括号都关闭(。
另一种方法可能是使用此处建议的可选参数test_fun <- function(halfs, quarts = NULL, other = NULL))，但这种方式表明必须提供halfs，但事实并非如此。真的，我想要test_fun <- function(halfs = NULL, quarts = NULL, other = NULL))或test_fun <- function(...))之类的东西，但这是做不到的。解决这个问题的方法可能是提供数据作为参数：test_fun <- function(test, halfs = NULL, quarts = NULL, other = NULL))，但我搞不清楚。
任何建议都很好。
语法错误是真实的，必须首先解决。else (type == "other")不是正确的语法。我想你指的是else if (type == "other")。由于您没有if，括号是出乎意料的。
但是，当您通过管道进入代码块时，也需要使用.来放置变量。你在{}内的突变应该使用mutate(., half_var=...)
test_fun <- function(type = c("halfs", "quarts", "other")) {
test %>% {
if (type == "halfs") {
mutate(., half_var = ifelse(month(time_stamp) <= 6, paste('H1', year(time_stamp)), paste('H2', year(time_stamp))))
} else if (type == "quarts") {
mutate(., quarts_var = case_when(month(time_stamp) <= 3 ~ paste('q1', year(time_stamp)),
month(time_stamp) > 3 & month(time_stamp) <= 6 ~ paste('q2', year(time_stamp)),
month(time_stamp) > 6 & month(time_stamp) <= 9 ~ paste('q3', year(time_stamp)),
month(time_stamp) > 9 ~ paste('q4', year(time_stamp))))
} else if (type == "other") {
mutate(., other = ifelse(month(time_stamp) <= 6, paste('H1', year(time_stamp)), paste('H2', year(time_stamp))))
}
}
}
这些计算已经在zoo包中的yearmon和yearqtr中直接可用，因此：
library(zoo)
test %>%
mutate(yearmon = as.yearmon(time_stamp),
yearqtr = as.yearqtr(time_stamp),
yearhalf = paste0(as.integer(yearmon), " H", (cycle(yearmon) > 6) + 1))
给予：
time_stamp yearmon yearqtr yearhalf
1 2005-08-07 Aug 2005 2005 Q3 2005 H2
2 2002-12-27 Dec 2002 2002 Q4 2002 H2
3 2004-07-19 Jul 2004 2004 Q3 2004 H2
4 2008-01-03 Jan 2008 2008 Q1 2008 H1
5 2000-02-08 Feb 2000 2000 Q1 2000 H1
6 2001-12-05 Dec 2001 2001 Q4 2001 H2
7 2002-07-26 Jul 2002 2002 Q3 2002 H2
8 2002-07-15 Jul 2002 2002 Q3 2002 H2
9 2006-12-29 Dec 2006 2006 Q4 2006 H2
10 2004-07-29 Jul 2004 2004 Q3 2004 H2
11 2007-06-16 Jun 2007 2007 Q2 2007 H1
12 2006-05-13 May 2006 2006 Q2 2006 H1
功能
目前还不清楚我们是否真的需要一个功能，但只是为了完成这个：
test_fun <- function(x, type = c("month", "quarter", "half")) {
type <- match.arg(type)
ym <- as.yearmon(x)
if (type == "month") ym
else if (type == "quarter") as.yearqtr(x)
else paste0(as.integer(ym), " H", (cycle(ym) > 6) + 1)
}
library(zoo)
test %>%
mutate(yearmonth = test_fun(time_stamp, "month"),
yearqtr = test_fun(time_stamp, "quarter"),
yearhalf = test_fun(time_stamp, "half"))
带有一个参数的函数
关于这个问题的主题行，它要求一个自变量的函数，我不太确定这是一个好主意，因为它意味着要对使用哪一列进行硬编码，但如果你真的想这样做，我们在下面展示如何做到这一点。实际上，我们提供了第二个参数，以防您改变主意并想要指定time_stamp列，但如果没有指定它，它将适当地默认为在mutate中调用它。
test_fun2 <- function(type = c("month", "quarter", "half"),
x = parent.frame()$.data$time_stamp) {
type <- match.arg(type)
ym <- as.yearmon(x)
if (type == "month") ym
else if (type == "quarter") as.yearqtr(x)
else paste0(as.integer(ym), " H", (cycle(ym) > 6) + 1)
}
library(zoo)
test %>%
mutate(month = test_fun2("month"),
quarter = test_fun2("quarter"),
halfs = test_fun2("half"))
函数，返回三个子集
如果您的意思是希望test_fun3最多返回3列，那么
test_fun3 <- function(x, month = FALSE, quarter = FALSE, half = FALSE) {
ym <- as.yearmon(x)
data <- data.frame(yearmon = ym,
quarter = as.yearqtr(x),
half = paste0(as.integer(ym), " H", (cycle(ym) > 6) + 1))
data[c(month, quarter, half)]
}
test %>%
bind_cols(test_fun3(.$time_stamp, TRUE, TRUE))