使用字符输入在 R 中定义函数

Question

我有一个输入数据集，它可能看起来像这样：

DF=data.frame(
    Variable = c("Test1", "Test2", "Test3"), 
    Distribution = c("Normal", "Exponential","Poisson"),
    Variable = c(2, 3, 4), 
    SD = c(2, NA, NA))

我想使用数据框 DF 中给出的分布来使用随机概率函数（例如 rnorm rexp 和 rbinom）。

那么，如何将文本输入转换为正确的函数？

如果合适，我想使用 Variable 和 SD 列中的相应值作为平均 values/standard 偏差。

Answer 1

类似于：

DF=data.frame(
  Variable = c("Test1", "Test2", "Test3"), 
  Distribution = c("Normal", "Exponential","Poisson"),
  VariablePrm = c(2, 3, 4), 
  SD = c(2, NA, NA),
  stringsAsFactors = FALSE)

# functions-lookup
fun_vec <- c("rnorm", "rexp", "rpois")
names(fun_vec) <- c("Normal", "Exponential", "Poisson")
DF$fun <- fun_vec[DF$Distribution]

# create expr
my_expr <- function(x) {
  txt <- paste0(x[1], "<-", x[5], "(", 10, ", ", x[3], 
                ifelse(is.na(x[4]), "", 
                       paste0(", ", x[4])), ")")
}
want <- apply(DF, 1, function(x) eval(parse(text = my_expr(x))))
colnames(want) <- DF$Variable
want

Answer 2

@r.user.05apr 解决方案有效，但涉及一些此处不需要的表达式解析。通过创建一个函数列表以便稍后使用它们，我们可能会更容易做到这一点。

# generating data:
DF=data.frame(
  Variable = c("Test1", "Test2", "Test3"), 
  Distribution = c("Normal", "Exponential","Poisson"),
  VariablePrm = c(2, 3, 4), 
  SD = c(2, NA, NA),
  stringsAsFactors = FALSE)

# creating function list and selecting this functions by Distribution column
fun_vec <- c(Normal=rnorm, Exponential=rexp, Poisson=rpois)
DF$fun <- fun_vec[DF$Distribution]

# if SD is NA then simply call function only with variablePrm
# else call with sd
# 10 is the number of observations to generate
generate <- function(x) {
  if(is.na(x$SD)){
    x$fun(10, x$VariablePrm)
  }else{
    x$fun(10, x$VariablePrm, x$SD)
  }
}
# if we apply this functions to each row we will get matrix of results
# each column will have 10 rows of generated data for previously selected distribution
apply(DF, 1, generate)

使用字符输入在 R 中定义函数

Use character input to define function in R

r

function

distribution

character