为什么不能使用 eval parse 将字符串作为公式求值

Question

R 为什么不能使用 eval parse 将字符串作为公式求值

在这个问题中，我想知道在 R 中是否可以将字符串作为公式求值。该公式应由 data.frame 中两个以上预先存在的列组成，这允许在 data.frame.

的行中使用不同的参数值来计算公式

我有以下例子：

# 1) Loading data
data(mtcars)

# 2) Creates a string formula using two arguments ( ideal to use three or more arguments )
mtcars$formulafx <- sprintf( 'mean( mtcars[mtcars[,2] == %s & mtcars[,10] == %i , 1] )' , mtcars$cyl , mtcars$gear )

 # 3) Incorrect result and really slow when used with only one argument formula
mtcars$resultsfx <- eval( parse( text = mtcars$formulafx ) )

我一直在网上搜索类似的问题。我发现以下内容是相关的： Updating a data.frame column with eval function

我已经使用 as.formula 函数进行了相同的测试，但也没有结果。

我想确定：

这种方法在 R 中根本不可能。
R 中还有其他解决方案。

================================

编辑：

感谢 Sathish 的解决方案。它有效，但我已经在由 101.815 行组成的 data.frame 中测试了这个解决方案，结果如下：

user  system elapsed 
165.46   11.45  177.11

将近 3 分钟。实际上这并不是很好。我会感谢建议。

Answer 1

使用lapply

unlist( lapply( mtcars$formulafx, function( x ) eval( parse( text = x ) ) ) )
# [1] 19.750 19.750 26.925 19.750 15.050 19.750 15.050 26.925 26.925 19.750 19.750 15.050 15.050
# [14] 15.050 15.050 15.050 15.050 26.925 26.925 26.925 21.500 15.050 15.050 15.050 15.050 26.925
# [27] 28.200 28.200 15.400 19.700 15.400 26.925

使用sapply

sapply( mtcars$formulafx, function( x ) eval( parse( text = x ) ) )
# mean( mtcars[mtcars[,2] == 6 & mtcars[,10] == 4 , 1] ) 
# 19.750 
# mean( mtcars[mtcars[,2] == 6 & mtcars[,10] == 4 , 1] ) 
# 19.750 
# mean( mtcars[mtcars[,2] == 4 & mtcars[,10] == 4 , 1] ) 
# 26.925 
# mean( mtcars[mtcars[,2] == 6 & mtcars[,10] == 3 , 1] ) 
# 19.750 
# mean( mtcars[mtcars[,2] == 8 & mtcars[,10] == 3 , 1] ) 
# 15.050 
# mean( mtcars[mtcars[,2] == 6 & mtcars[,10] == 3 , 1] ) 
# 19.750 
# mean( mtcars[mtcars[,2] == 8 & mtcars[,10] == 3 , 1] ) 
# 15.050 
# mean( mtcars[mtcars[,2] == 4 & mtcars[,10] == 4 , 1] ) 
# 26.925 
# mean( mtcars[mtcars[,2] == 4 & mtcars[,10] == 4 , 1] ) 
# 26.925 
# mean( mtcars[mtcars[,2] == 6 & mtcars[,10] == 4 , 1] ) 
# 19.750 
# mean( mtcars[mtcars[,2] == 6 & mtcars[,10] == 4 , 1] ) 
# 19.750 
# mean( mtcars[mtcars[,2] == 8 & mtcars[,10] == 3 , 1] ) 
# 15.050 
# mean( mtcars[mtcars[,2] == 8 & mtcars[,10] == 3 , 1] ) 
# 15.050 
# mean( mtcars[mtcars[,2] == 8 & mtcars[,10] == 3 , 1] ) 
# 15.050 
# mean( mtcars[mtcars[,2] == 8 & mtcars[,10] == 3 , 1] ) 
# 15.050 
# mean( mtcars[mtcars[,2] == 8 & mtcars[,10] == 3 , 1] ) 
# 15.050 
# mean( mtcars[mtcars[,2] == 8 & mtcars[,10] == 3 , 1] ) 
# 15.050 
# mean( mtcars[mtcars[,2] == 4 & mtcars[,10] == 4 , 1] ) 
# 26.925 
# mean( mtcars[mtcars[,2] == 4 & mtcars[,10] == 4 , 1] ) 
# 26.925 
# mean( mtcars[mtcars[,2] == 4 & mtcars[,10] == 4 , 1] ) 
# 26.925 
# mean( mtcars[mtcars[,2] == 4 & mtcars[,10] == 3 , 1] ) 
# 21.500 
# mean( mtcars[mtcars[,2] == 8 & mtcars[,10] == 3 , 1] ) 
# 15.050 
# mean( mtcars[mtcars[,2] == 8 & mtcars[,10] == 3 , 1] ) 
# 15.050 
# mean( mtcars[mtcars[,2] == 8 & mtcars[,10] == 3 , 1] ) 
# 15.050 
# mean( mtcars[mtcars[,2] == 8 & mtcars[,10] == 3 , 1] ) 
# 15.050 
# mean( mtcars[mtcars[,2] == 4 & mtcars[,10] == 4 , 1] ) 
# 26.925 
# mean( mtcars[mtcars[,2] == 4 & mtcars[,10] == 5 , 1] ) 
# 28.200 
# mean( mtcars[mtcars[,2] == 4 & mtcars[,10] == 5 , 1] ) 
# 28.200 
# mean( mtcars[mtcars[,2] == 8 & mtcars[,10] == 5 , 1] ) 
# 15.400 
# mean( mtcars[mtcars[,2] == 6 & mtcars[,10] == 5 , 1] ) 
# 19.700 
# mean( mtcars[mtcars[,2] == 8 & mtcars[,10] == 5 , 1] ) 
# 15.400 
# mean( mtcars[mtcars[,2] == 4 & mtcars[,10] == 4 , 1] ) 
# 26.925

为什么不能使用 eval parse 将字符串作为公式求值

Why eval parse cannot be used to evaluate string as formula

parsing

eval

r

dataframe