r - 非标准评估不起作用： h(simpleError(msg, call)) 中的错误

Question

这是我想做的一个工作示例。

# interactive use
subset(iris,Species=="setosa")

# as a function
fn <- function(level,choice) {
  x <- paste0(level,"==","'",choice,"'")
  subset(iris,eval(parse(text=x)))
}
fn("Species","setosa")

如上面的函数所示，我想以编程方式指定“Species”和“setosa”并将它们组合起来形成一个表达式。那行得通。

但是，这似乎不适用于我正在使用的另一个包。

#Biocmanager::install("phyloseq")
library(phyloseq)
data(GlobalPatterns)

# interactive use, this works
subset_taxa(GlobalPatterns,Kingdom=='Archaea')

# as a function, this doesn't work
fn <- function(level,choice) {
  x <- paste0(level,"==","'",choice,"'")
  subset_taxa(GlobalPatterns,eval(parse(text=x)))
}
fn(level="Kingdom",choice="Archaea")


Error in h(simpleError(msg, call)) : 
  error in evaluating the argument 'expr' in selecting a method for function 'eval': object 'x' not found 

该函数的行为是否不同或我做错了什么？

R 4.0.2
phyloseq_1.34.0

Answer 1

phyloseq 包的作者提供的解决方案：

library(phyloseq)
data(GlobalPatterns)

fn <- function(ps, level, choice) {
        x <- paste0(level,"==","'",choice,"'")
        oldTax <- data.frame(tax_table(ps))
        newTax <- subset(oldTax, eval(parse(text=x)))
        tax_table(ps) <- tax_table(as.matrix(newTax))
        return(ps)
}

fn(GlobalPatterns, level="Kingdom",choice="Archaea")