当输入为 numeric 和 bigq 时，R 函数 `c( )` 强制转换是什么？

发布于 2025-01-16 16:35:06 字数 735 浏览 4 评论 0原文

考虑这两个对比示例：

> library(Rmpfr)
  > K = 1:5
    > denoms = c(as.bigz(1), as.bigq(rep(1,times = length(K)), K) )
    #Big Rational ('bigq') object of length 6:
    #[1] 1   1   1/2 1/3 1/4 1/5
    > foo = c(1, as.bigq(rep(1,times = length(K)), K) )
    
    # [1] 1 5 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1
    # [39] 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0    

    > bar <- c(as.bigz(1), as.bigq(rep(1,times = length(K)), K) )
    
    #Big Rational ('bigq') object of length 6:
    # [1] 1   1   1/2 1/3 1/4 1/5

我知道方法 c.bigq 可能不完整，导致所需的解释和数据类型强制转换失败。我的问题是：一些 bigq 对象是如何变成 1 和 0 的向量的？输出 foo 是数字。

原文

Consider these two contrasting examples:

> library(Rmpfr)
  > K = 1:5
    > denoms = c(as.bigz(1), as.bigq(rep(1,times = length(K)), K) )
    #Big Rational ('bigq') object of length 6:
    #[1] 1   1   1/2 1/3 1/4 1/5
    > foo = c(1, as.bigq(rep(1,times = length(K)), K) )
    
    # [1] 1 5 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1
    # [39] 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0    

    > bar <- c(as.bigz(1), as.bigq(rep(1,times = length(K)), K) )
    
    #Big Rational ('bigq') object of length 6:
    # [1] 1   1   1/2 1/3 1/4 1/5

I understand that the method c.bigq may be incomplete, leading to failure of the desired interpretation and coercion of data types. My question is: how did some bigq objects get turned into those vectors of 1s and 0s ? The output foo is numeric.

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倚栏听风 2025-01-23 16:35:06

c() 上的调度是根据第一个元素的类完成的，因此 c(1, as.bigq(rep(1,times = length(K)), K) ) 调用默认方法。

第二个参数 as.bigq(rep(1,times = length(K)), K) 具有类 "bigq"，但这无关紧要，因为默认方法会忽略班级。它只是查看类型，并且 typeof(as.bigq(rep(1,times = length(K)), K)) 给出 "raw"，所以它是存储为原始字节。如果您对其调用 unclass() ，您会看到类似的内容，

 [1] 05 00 00 00 01 00 00 00 01 00 00 00 01 00 00 00 01 00 00 00 01 00 00 00 01 00 00 00 01 00 00 00 01 00 00 00 01 00 00
[40] 00 01 00 00 00 01 00 00 00 01 00 00 00 01 00 00 00 01 00 00 00 01 00 00 00
attr(,"denominator")
 [1] 05 00 00 00 01 00 00 00 01 00 00 00 01 00 00 00 01 00 00 00 01 00 00 00 02 00 00 00 01 00 00 00 01 00 00 00 03 00 00
[40] 00 01 00 00 00 01 00 00 00 04 00 00 00 01 00 00 00 01 00 00 00 05 00 00 00

这就是所有这些数字的来源。

另一方面，c(as.bigz(1), as.bigq(rep(1,times = length(K)), K) ) 有一个 bigz object 作为第一个参数，因此它将调用 c.bigz。该方法了解 bigz 和 bigq 对象，并用它们做正确的事情。

The dispatch on c() is done based on the class of the first element, so c(1, as.bigq(rep(1,times = length(K)), K) ) calls the default method.

The second parameter as.bigq(rep(1,times = length(K)), K) has class "bigq", but that's irrelevant, because the default method ignores the class. It just looks at the type, and typeof(as.bigq(rep(1,times = length(K)), K)) gives "raw", so it's stored as raw bytes. If you call unclass() on it, you'll see something like

 [1] 05 00 00 00 01 00 00 00 01 00 00 00 01 00 00 00 01 00 00 00 01 00 00 00 01 00 00 00 01 00 00 00 01 00 00 00 01 00 00
[40] 00 01 00 00 00 01 00 00 00 01 00 00 00 01 00 00 00 01 00 00 00 01 00 00 00
attr(,"denominator")
 [1] 05 00 00 00 01 00 00 00 01 00 00 00 01 00 00 00 01 00 00 00 01 00 00 00 02 00 00 00 01 00 00 00 01 00 00 00 03 00 00
[40] 00 01 00 00 00 01 00 00 00 04 00 00 00 01 00 00 00 01 00 00 00 05 00 00 00

so that's where all those numbers came from.

On the other hand, c(as.bigz(1), as.bigq(rep(1,times = length(K)), K) ) has a bigz object as the first argument, so it will call c.bigz. That method knows about bigz and bigq objects and does the right thing with them.

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