为什么 eval parse 不能用于将字符串评估为公式

Question

R 为什么 eval parse 不能用于计算字符串作为公式

在这个问题中，我想知道在 R 中是否可以将字符串评估为公式。 该公式应由 data.frame 中的两个以上预先存在的列组成， 这允许使用跨 data.frame 行的不同参数值来评估公式。

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 ) )   

我一直在互联网上搜索类似的问题。 我发现以下是相关的： 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。它可以工作，但我已经在包含 101.815 行的 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

差不多 3 分钟。在实践中，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