我想有一个函数调用subset,并传递一个subset参数:
df <- data.frame(abc=c("A","A","B","B"),value=1:4)
subset(df,abc=="A")
## works of course:
# abc value
#1 A 1
#2 A 2
mysubset <- function(df,ssubset)
subset(df,ssubset)
mysubset(df,abc=="A")
## Throws an error
# Error in eval(expr, envir, enclos) : object 'abc' not found
mysubset2 <- function(df,ssubset)
subset(df,eval(ssubset))
mysubset2(df,expression(abc=="A"))
## Works, but needs expression
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我尝试过substitute,但无法找到合适的组合.我怎样才能使这个工作?
A5C*_*2T1 12
你也需要eval()和parse()在那里:
mysubset <- function(df, ssubset) {
subset(df, eval(parse(text=ssubset)))
}
mysubset(df, "abc=='A'")
# abc value
# 1 A 1
# 2 A 2
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请注意,您需要嵌套引号,所以来回切换之间",并'在必要时.
根据您的评论,也许这样的事情也是有趣的:
mysubset <- function(df, ...) {
ssubset <- deparse(substitute(...))
subset(df, eval(parse(text = ssubset)))
}
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用法: mysubset(df, abc=='A')
A5C1D2H2I1M1N2O1R2T1 答案有效,但您可以通过简单地使用以下方法跳过整个解析/解析周期:
mysubset <- function(df, p) {
ps <- substitute(p)
subset(df, eval(ps))
}
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