sapply 与复合函数的速度比较

发布于 2024-10-10 12:48:00 字数 348 浏览 1 评论 0原文

> system.time(sapply(rnorm(1000000,0,1), function (x) round(x,2)))
   user  system elapsed 
   2.78    0.11    2.89 
> system.time(round(rnorm(1000000,0,1),2))
   user  system elapsed 
   0.29    0.00    0.30

在阅读了 R Tips 问题的答案后，我尝试了这个。我没想到 sapply 会比上述情况下的等效复合函数慢几个数量级。有谁知道为什么会这样？如果我理解正确，sapply 将会矢量化并且接近最佳速度。

原文

> system.time(sapply(rnorm(1000000,0,1), function (x) round(x,2)))
   user  system elapsed 
   2.78    0.11    2.89 
> system.time(round(rnorm(1000000,0,1),2))
   user  system elapsed 
   0.29    0.00    0.30

I was trying this out after reading the answers to the R tips question. I did not expect sapply to be order of magnitude slower than the equivalent composite function in the above case. Does anyone know why this is the case? If i understand correctly sapply will vectorize and be near optimally fast.

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碍人泪离人颜 2024-10-17 12:48:00

sapply 可能是 lapply 的简单包装，没有矢量化。尝试此代码：

system.time(sapply(rnorm(10), function (x) {print(length(x)); round(x,2)}))

并在此处查看实现： https://svn .r-project.org/R/trunk/src/main/apply.c

probably sapply, which is a simple wrapper of lapply, is not vectorized. try this code:

system.time(sapply(rnorm(10), function (x) {print(length(x)); round(x,2)}))

and see the implementation here: https://svn.r-project.org/R/trunk/src/main/apply.c

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人事已非 2024-10-17 12:48:00

这里没有什么可以应用的 - 你只给它一个向量 - 而不是向量列表，并且 sapply 将结果转换为（单列）矩阵。

sapply 正在为您简化结果，但这样做必须生成一个数组。

如果给它一个列表进行比较：

system.time(sapply(list(rnorm(1000000,0,1)), function (x) round(x,2))) 
user  system elapsed 
 0.22    0.00    0.22 

system.time(sapply(rnorm(1000000,0,1), function (x) round(x,2))) 
user  system elapsed 
4.21    0.00    4.21

There's nothing here to sapply to - you only give it a single vector - not a list of vectors, and sapply converts the result to a (single column) matrix.

sapply is simplifying the result for you, but in doing so has to generate an array.

Compare if you give it a list:

system.time(sapply(list(rnorm(1000000,0,1)), function (x) round(x,2))) 
user  system elapsed 
 0.22    0.00    0.22 

system.time(sapply(rnorm(1000000,0,1), function (x) round(x,2))) 
user  system elapsed 
4.21    0.00    4.21

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