如何将二进制小数转换为 R 中的十进制小数?
How to convert a binary fraction number into decimal fraction number in R?
我需要编写一个函数,将二进制小数转换为 R 中的十进制小数。例如f(0.001) # 0.125
我做了什么:我在R包中搜索了相关函数:
DescTools::BinToDec(0.001) # NA
DescTools::BinToDec("0.001") # NA
base::strtoi(0.001, base=2) # NA
base::strtoi("0.001", base=2) # NA
base::packBits(intToBits(0.001), "integer") # 0
base::packBits(intToBits("0.001"), "integer") # 0
compositions::unbinary(0.001) # 0.001
compositions::unbinary("0.001") # NA
我在 SOF 中搜索,发现以下内容:
base2decimal <- function(base_number, base = 2) {
split_base <- strsplit(as.character(base_number), split = "")
return(sapply(split_base, function(x) sum(as.numeric(x) * base^(rev(seq_along(x) - 1)))))}
base2decimal(0.001) # NA
base2decimal("0.001") # NA
0.001 是:
(0 * 2^(-1)) + (0 * 2^(-2)) + (1 * 2^(-3)) # 0.125
(0 * 1/2) + (0 * (1/2)^2) + (1 * (1/2)^3) # 0.125
(0 * 0.5) + (0 * (0.5)^2) + (1 * 0.5^3) # 0.125
所以,内积之和 (0,0,1) * (0.5^1, 0.5^2, 0.5^3)
之类的东西似乎可以解决问题,我不知道在一般情况下该怎么做。
javascript 案例:
口齿不清案例:
library(cwhmisc) # int, frac
from2to10 <- function(n) {
SignOfNumber <- ""
if (n < 0) {
n <- abs(n)
SignOfNumber <- "-"}
nWhole <- int(n)
nWhole <- as.character(nWhole)
nFraction <- frac(n)
nFraction <- as.character(nFraction)
DecimalWhole <- sapply(strsplit(nWhole, split=""), function(x) sum(as.numeric(x) * 2^(rev(seq_along(x) - 1))))
if (nFraction == 0) {
DecimalFraction <- ""
paste0(SignOfNumber, DecimalWhole)
} else { # Find decimal fraction part
part3 <- function(x, y, z) { eval(parse(text=(paste(x, y, z,sep="")))) }
y <- as.numeric(strsplit(substr(part3("\"",n,"\""), which(strsplit(part3("\"",n,"\""), "")[[1]]==".") + 1, nchar(part3("\"",n,"\""))),"")[[1]])
DecimalFraction <- sum(y * (0.5^(1:length(y))))
paste0(SignOfNumber, DecimalWhole + DecimalFraction)
}
}
from2to10(0.001) # "0.125"
as.numeric(from2to10(0.001)) # 0.125
您可以扩展您在问题中发布的 solution 以包括从小数点分隔符位置开始的两个负幂,如下所示:
base2decimal <- function(base_number, base = 2) {
base_number = paste(as.character(base_number), ".", sep = "")
return (mapply(function (val, sep) {
val = val[-sep];
if (val[[1]] == "-") {
sign = -1
powmax = sep[[1]] - 3
val = val[-1]
} else {
sign = 1
powmax = sep[[1]] - 2
};
sign * sum(as.numeric(val) * (base ^ seq(powmax, by = -1, length = length(val))))},
strsplit(base_number, NULL), gregexpr("\.", base_number)))
}
此代码也适用于其他小于(或等于)10 的碱基:
base2decimal(c('0.101', '.101', 0.101, 1101.001, 1101, '-0.101', '-.101', -0.101, -1101.001, -1101))
#[1] 0.625 0.625 0.625 13.125 13.000 -0.625 -0.625 -0.625 -13.125
#[10] -13.000
base2decimal(1110.111)
# 14.875
base2decimal(256.3, 8)
# [1] 174.375
我需要编写一个函数,将二进制小数转换为 R 中的十进制小数。例如f(0.001) # 0.125
我做了什么:我在R包中搜索了相关函数:
DescTools::BinToDec(0.001) # NA
DescTools::BinToDec("0.001") # NA
base::strtoi(0.001, base=2) # NA
base::strtoi("0.001", base=2) # NA
base::packBits(intToBits(0.001), "integer") # 0
base::packBits(intToBits("0.001"), "integer") # 0
compositions::unbinary(0.001) # 0.001
compositions::unbinary("0.001") # NA
我在 SOF 中搜索,发现以下内容:
base2decimal <- function(base_number, base = 2) {
split_base <- strsplit(as.character(base_number), split = "")
return(sapply(split_base, function(x) sum(as.numeric(x) * base^(rev(seq_along(x) - 1)))))}
base2decimal(0.001) # NA
base2decimal("0.001") # NA
0.001 是:
(0 * 2^(-1)) + (0 * 2^(-2)) + (1 * 2^(-3)) # 0.125
(0 * 1/2) + (0 * (1/2)^2) + (1 * (1/2)^3) # 0.125
(0 * 0.5) + (0 * (0.5)^2) + (1 * 0.5^3) # 0.125
所以,内积之和 (0,0,1) * (0.5^1, 0.5^2, 0.5^3)
之类的东西似乎可以解决问题,我不知道在一般情况下该怎么做。
javascript 案例:
library(cwhmisc) # int, frac
from2to10 <- function(n) {
SignOfNumber <- ""
if (n < 0) {
n <- abs(n)
SignOfNumber <- "-"}
nWhole <- int(n)
nWhole <- as.character(nWhole)
nFraction <- frac(n)
nFraction <- as.character(nFraction)
DecimalWhole <- sapply(strsplit(nWhole, split=""), function(x) sum(as.numeric(x) * 2^(rev(seq_along(x) - 1))))
if (nFraction == 0) {
DecimalFraction <- ""
paste0(SignOfNumber, DecimalWhole)
} else { # Find decimal fraction part
part3 <- function(x, y, z) { eval(parse(text=(paste(x, y, z,sep="")))) }
y <- as.numeric(strsplit(substr(part3("\"",n,"\""), which(strsplit(part3("\"",n,"\""), "")[[1]]==".") + 1, nchar(part3("\"",n,"\""))),"")[[1]])
DecimalFraction <- sum(y * (0.5^(1:length(y))))
paste0(SignOfNumber, DecimalWhole + DecimalFraction)
}
}
from2to10(0.001) # "0.125"
as.numeric(from2to10(0.001)) # 0.125
您可以扩展您在问题中发布的 solution 以包括从小数点分隔符位置开始的两个负幂,如下所示:
base2decimal <- function(base_number, base = 2) {
base_number = paste(as.character(base_number), ".", sep = "")
return (mapply(function (val, sep) {
val = val[-sep];
if (val[[1]] == "-") {
sign = -1
powmax = sep[[1]] - 3
val = val[-1]
} else {
sign = 1
powmax = sep[[1]] - 2
};
sign * sum(as.numeric(val) * (base ^ seq(powmax, by = -1, length = length(val))))},
strsplit(base_number, NULL), gregexpr("\.", base_number)))
}
此代码也适用于其他小于(或等于)10 的碱基:
base2decimal(c('0.101', '.101', 0.101, 1101.001, 1101, '-0.101', '-.101', -0.101, -1101.001, -1101))
#[1] 0.625 0.625 0.625 13.125 13.000 -0.625 -0.625 -0.625 -13.125
#[10] -13.000
base2decimal(1110.111)
# 14.875
base2decimal(256.3, 8)
# [1] 174.375