library(Rmpfr)
mynumber <- new("mpfr", .Data = list(new("mpfr1", prec = 50L, exp = c(1045L,
0L), sign = 1L, d = c(151748608L, -358118319L)), new("mpfr1",
prec = 50L, exp = c(20L, 0L), sign = 1L, d = c(-1114947584L,
-1905679017L)), new("mpfr1", prec = 50L, exp = c(-55L, -1L
), sign = 1L, d = c(-1449918464L, -906197701L)), new("mpfr1",
prec = 50L, exp = c(221L, 0L), sign = 1L, d = c(819707904L,
-1329031570L))))
mynumber
是其中包含4个数字的类mpfr
对象。我想以0列表示cbind
mynumber
,即
> cbind(rep(0, 4), mynumber)
mynumber
[1,] 0 ?
[2,] 0 ?
[3,] 0 ?
[4,] 0 ?
这给了我????在第二列中,因此我尝试将mynumber
更改为类numeric
首先
mydata <- cbind(rep(0, 4), sapply(mynumber, asNumeric))
> mydata
[,1] [,2]
[1,] 0 Inf
[2,] 0 5.833223e+05
[3,] 0 2.189941e-17
[4,] 0 2.327185e+66
但是,由于mynumber
中的第一个数字确实很大,因此使用asNumeric将其改为Inf
。
编辑:我的最终目标是运行:
mydata <- cbind(rep(0, 4), sapply(mynumber, asNumeric))
> mydata/rowSums(mydata)
[,1] [,2]
[1,] 0 NaN
[2,] 0 1
[3,] 0 1
[4,] 0 1
并且没有打印出NaN。
一种选择是用list
包装,然后创建一个tibble/data.frame
对象,因为cbind
转换为matrix
,并且matrix
只能容纳一个类
library(tibble)
tibble(col1 = 0, col2 = list(mynumber))
# A tibble: 1 x 2
# col1 col2
# <dbl> <list>
#1 0 <mpfr>
即使是cbind
和character
类的[numeric
也会为所有列返回character
,当要绑定的vector
属于不同类时,这不是一个好选择
cbind(letters[1:4], 1:4)
在装入程序包后通过检查methods
中的cbind
methods('cbind')
#[1] cbind,ANY-method cbind,Mnumber-method cbind.bigq* cbind.bigz*
#[5] cbind.data.frame cbind.grouped_df* cbind.ts*
因此,如果它对cbind
使用了正确的Mnumber
方法,则不应给出Inf
cbind(rep(0,4), mynumber)
#'mpfrMatrix' of dim(.) = (4, 2) of precision 50 .. 53 bits
# [,1] [,2]
#[1,] 0. 3.4556867084990952e+314
#[2,] 0. 583322.33392099757
#[3,] 0. 2.1899410233914937e-17
#[4,] 0. 2.3271850367397449e+66
或利用价值的循环利用
cbind(0, mynumber)
#'mpfrMatrix' of dim(.) = (4, 2) of precision 50 .. 53 bits
# [,1] [,2]
#[1,] 0. 3.4556867084990952e+314
#[2,] 0. 583322.33392099757
#[3,] 0. 2.1899410233914937e-17
#[4,] 0. 2.3271850367397449e+66
此外,如果我们检查掩盖的功能,则在加载程序包时会说
以下对象从'package:base'中被屏蔽:
cbind,pmax,pmin,rbind
通过使用cbind
中的base
,可以复制?
。对于OP,cbind
来自base
base::cbind(0, mynumber)
# mynumber
#[1,] 0 ?
#[2,] 0 ?
#[3,] 0 ?
#[4,] 0 ?
如果cbind
中的Rmpfr
被屏蔽,则使用::
mydata <- Rmpfr::cbind(0, mynumber)
mydata
#'mpfrMatrix' of dim(.) = (4, 2) of precision 50 .. 53 bits
# [,1] [,2]
#[1,] 0. 3.4556867084990952e+314
#[2,] 0. 583322.33392099757
#[3,] 0. 2.1899410233914937e-17
#[4,] 0. 2.3271850367397449e+66
mydata/rowSums(mydata)
#'mpfrMatrix' of dim(.) = (4, 2) of precision 53 bits
# [,1] [,2]
#[1,] 0. 1.0000000000000000
#[2,] 0. 1.0000000000000000
#[3,] 0. 1.0000000000000000
#[4,] 0. 1.0000000000000000
我想知道您创建的对象是否存在问题。当我更改它以确保每个exp
字段都只有一个值时,我得到了所需的结果:
library(Rmpfr)
x <- new("mpfr", .Data = list(new("mpfr1", prec = 50L, exp = 1045L,
sign = 1L, d = c(151748608L, -358118319L)), new("mpfr1",
prec = 50L, exp = 20L, sign = 1L, d = c(-1114947584L, -1905679017L
)), new("mpfr1", prec = 50L, exp = -55L, sign = 1L, d = c(-1449918464L,
-906197701L)), new("mpfr1", prec = 50L, exp = 221L, sign = 1L,
d = c(819707904L, -1329031570L))))
cbind(c(0, 0, 0, 0), x)
#> 'mpfrMatrix' of dim(.) = (4, 2) of precision 50 .. 53 bits
#> [,1] [,2]
#> [1,] 0. 3.4556867084990952e+314
#> [2,] 0. 583322.33392099757
#> [3,] 0. 2.1899410233914937e-17
#> [4,] 0. 2.3271850367397449e+66
由reprex package(v0.3.0)在2020-02-29创建
zeros <- mpfr(c(0,0,0,0), precBits = getPrec(mynumber))
M <- cbind(mynumber, zeros)
M %*% t(M) # just to try