Question

我只能使用struct.pack模块将小数转换为二进制单精度IEEE754，或者使用numpy.frombuffer

执行相反的操作（float16或float32）

是否可以使用Numpy？

将小数转换为二进制半精度浮点数

我需要打印转化结果，因此如果我输入"117.0"，则应打印"0101011101010000"

Answer 1

如果我输入＆＃34; 117.0＆＃34;，则应打印＆＃34; 0101011101010000＆＃34;

>>> import numpy as np
>>> bin(np.float16(117.0).view('H'))[2:].zfill(16)
'0101011101010000'

Answer 2

float16建议的Mark Dickinson方法必须后跟# http://ruby-doc.org/stdlib-2.0.0/libdoc/open-uri/rdoc/OpenURI.html require 'open-uri' # https://github.com/flori/json require 'json' # http://stackoverflow.com/questions/9008847/what-is-difference-between-p-and-pp require 'pp' buffer = open('http://api.cryptocoincharts.info/tradingPair/eth_btc').read # api_key = "FtHwuH8w1RDjQpOr0y0gF3AWm8sRsRzncK3hHh9" result = JSON.parse(buffer) puts result["markets"] # result.each do |user| # puts "#{user['id']}\t#{user['name']}\t#{user['email']}" # puts "Registered: #{user['created_at']}\n\n" # end # my_hash = JSON.parse('{"hello": "goodbye"}') # puts my_hash["hello"] => "goodbye"方法才能获得所需的二进制表示形式：

tostring()

Python中的十进制到二进制半精度IEEE 754

2 个答案: