purrr::pmap 与其他默认输入 [英] purrr::pmap with other default inputs

问题描述

我想知道如果我有超过 3 个输入作为参数映射到具有其他默认输入的函数，如何使用 pmap() 函数.

I am wondering how to use pmap() function if I have more than 3 inputs as parameters to map into a function with other default inputs.

这是一个可重现的例子:

Here is a reproducible example:

a=c(5, 100, 900)
b=c(1, 2, 3)
ablist=list(mean=a,sd=b)
pmap(ablist, ~rnorm( mean=a , sd=b , n = 9))

输出:

 [[1]]
 [1]   5.734723  99.883171 895.962561   5.346905  98.723191 903.373177   4.172267  96.424440 897.437970

 [[2]]
 [1]   4.427977  98.348139 899.287248   4.404674  99.178516 900.983974   3.836353 101.520355 899.992332

 [[3]]
 [1]   4.961772  95.927525 899.096313   4.444354 101.694591 904.172462   6.231246  97.773325 897.611838

但是正如您所看到的，输出并没有按照向量的顺序映射 mean 和 sd.

But as you can see, the output is not mapping the mean and sd in the order of vectors.

我想要 [[1]] 和 rnorm(mean=5,sd=1,n=9) 等等.出于好奇，我想知道 pmap() 为这个演示做了什么.

I want to have [[1]] with rnorm(mean=5,sd=1,n=9) and so on. Out of curiosity, I am wondering what pmap() is doing for this demo.

顺便说一下，我知道在这个例子中，我可以轻松地使用 map2() 而没有任何麻烦，但在我的实际代码中，我有 10 个输入，所以我需要使用 pmap().

By the way, I know in this example, I can easily use map2() without any hassle but in my real code, I have 10 inputs so I need to use pmap().

提前感谢您的回复！

推荐答案

当您使用 pmap 时，您可以使用 ..1、 引用您的参数>..2 等.这应该给你你想要的:

When you are using pmap, you can refer to your arguments with ..1, ..2, etc. This should give you what you want:

pmap(ablist, ~rnorm(mean = ..1, sd = ..2, n = 9))

或者，您可以提供一个包含所有参数的命名列表.这也适用:

Alternatively you can supply a named list with all of the arguments included. This works as well:

abclist = list(
  mean = c(5, 100, 900),
  sd = c(1, 2, 3), 
  n = rep(9, 3)
)

pmap(abclist, rnorm)

您的代码仅运行 rnorm(mean = c(5, 100, 900), sd = c(1, 2, 3), n = 9) 3 次并将其存储在列表.

Your code is just running rnorm(mean = c(5, 100, 900), sd = c(1, 2, 3), n = 9) 3 times and storing it in a list.

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