Question

如何将字符串作为32位浮点数的二进制IEEE 754表示形式？

示例

1.00 - ＆gt; '00111111100000000000000000000000'

Answer 1

您可以使用struct包执行此操作：

import struct
def binary(num):
    return ''.join(bin(ord(c)).replace('0b', '').rjust(8, '0') for c in struct.pack('!f', num))

将其打包为网络字节排序的浮点数，然后将每个结果字节转换为8位二进制表示形式并将它们连接起来：

>>> binary(1)
'00111111100000000000000000000000'

修改：有人要求扩大解释。我将使用中间变量扩展它来评论每一步。

def binary(num): # Struct can provide us with the float packed into bytes. The '!' ensures that # it's in network byte order (big-endian) and the 'f' says that it should be # packed as a float. Alternatively, for double-precision, you could use 'd'. packed = struct.pack('!f', num) print 'Packed: %s' % repr(packed) # For each character in the returned string, we'll turn it into its corresponding # integer code point # # [62, 163, 215, 10] = [ord(c) for c in '>\xa3\xd7\n'] integers = [ord(c) for c in packed] print 'Integers: %s' % integers # For each integer, we'll convert it to its binary representation. binaries = [bin(i) for i in integers] print 'Binaries: %s' % binaries # Now strip off the '0b' from each of these stripped_binaries = [s.replace('0b', '') for s in binaries] print 'Stripped: %s' % stripped_binaries # Pad each byte's binary representation's with 0's to make sure it has all 8 bits: # # ['00111110', '10100011', '11010111', '00001010'] padded = [s.rjust(8, '0') for s in stripped_binaries] print 'Padded: %s' % padded # At this point, we have each of the bytes for the network byte ordered float # in an array as binary strings. Now we just concatenate them to get the total # representation of the float: return ''.join(padded)

以及一些例子的结果：

>>> binary(1) Packed: '?\x80\x00\x00' Integers: [63, 128, 0, 0] Binaries: ['0b111111', '0b10000000', '0b0', '0b0'] Stripped: ['111111', '10000000', '0', '0'] Padded: ['00111111', '10000000', '00000000', '00000000'] '00111111100000000000000000000000' >>> binary(0.32) Packed: '>\xa3\xd7\n' Integers: [62, 163, 215, 10] Binaries: ['0b111110', '0b10100011', '0b11010111', '0b1010'] Stripped: ['111110', '10100011', '11010111', '1010'] Padded: ['00111110', '10100011', '11010111', '00001010'] '00111110101000111101011100001010'

Answer 2

这是一个丑陋的......

>>> import struct
>>> bin(struct.unpack('!i',struct.pack('!f',1.0))[0])
'0b111111100000000000000000000000'

基本上，我只使用struct模块将float转换为int ...

使用ctypes：

的情况稍好一些

>>> import ctypes
>>> bin(ctypes.c_uint.from_buffer(ctypes.c_float(1.0)).value)
'0b111111100000000000000000000000'

基本上，我构造了一个float并使用相同的内存位置，但我将其标记为c_uint。 c_uint的值是一个python整数，您可以使用内置的bin函数。

Answer 3

使用bitstring模块找到另一个解决方案。

import bitstring
f1 = bitstring.BitArray(float=1.0, length=32)
print f1.bin

输出：

00111111100000000000000000000000

Answer 4

通过将其分成两部分可以更清晰地处理这个问题。

第一种是将float转换为具有等效位模式的int：

def float32_bit_pattern(value):
    return sum(ord(b) << 8*i for i,b in enumerate(struct.pack('f', value)))

接下来将int转换为字符串：

def int_to_binary(value, bits):
    return bin(value).replace('0b', '').rjust(bits, '0')

现在合并它们：

>>> int_to_binary(float32_bit_pattern(1.0), 32)
'00111111100000000000000000000000'

Answer 5

为了完整起见，您可以使用numpy实现此目的：

f = 1.00
int32bits = np.asarray(f, dtype=np.float32).view(np.int32).item()  # item() optional

然后，您可以使用填充，使用b格式说明符

打印此内容

print('{:032b}'.format(int32bits))

Answer 6

Dan的答案（带有Python3彩色版本）的小猪尾巴：

import struct

BLUE = "\033[1;34m"
CYAN = "\033[1;36m"
GREEN = "\033[0;32m"
RESET = "\033[0;0m"


def binary(num):
    return [bin(c).replace('0b', '').rjust(8, '0') for c in struct.pack('!f', num)]


def binary_str(num):
    bits = ''.join(binary(num))
    return ''.join([BLUE, bits[:1], GREEN, bits[1:10], CYAN, bits[10:], RESET])


def binary_str_fp16(num):
    bits = ''.join(binary(num))
    return ''.join([BLUE, bits[:1], GREEN, bits[1:10][-5:], CYAN, bits[10:][:11], RESET])

x = 0.7
print(x, "as fp32:", binary_str(0.7), "as fp16 is sort of:", binary_str_fp16(0.7))

Answer 7

在浏览了很多类似的问题后，我写了一些希望能做到我想要的东西。

f = 1.00
negative = False
if f < 0:
    f = f*-1
    negative = True

s = struct.pack('>f', f)
p = struct.unpack('>l', s)[0]
hex_data =  hex(p)

scale = 16
num_of_bits = 32
binrep = bin(int(hex_data, scale))[2:].zfill(num_of_bits)
if negative:
    binrep = '1' + binrep[1:]

binrep就是结果。每个部分都将被解释。

f = 1.00
negative = False
if f < 0:
    f = f*-1
    negative = True

如果为负数，则将数字转换为正数，并将变量负数设置为false。这样做的原因是正和负二进制表示之间的差异只是在第一位，这是比用负数做整个过程时弄清楚出错的简单方法。

s = struct.pack('>f', f)                          #'?\x80\x00\x00'
p = struct.unpack('>l', s)[0]                     #1065353216
hex_data =  hex(p)                                #'0x3f800000'

s是二进制f的十六进制表示。然而，它不是我需要的漂亮形式。这就是p进来的地方。它是十六进制的int表示。然后另一个转换得到一个漂亮的十六进制。

scale = 16
num_of_bits = 32
binrep = bin(int(hex_data, scale))[2:].zfill(num_of_bits)
if negative:
    binrep = '1' + binrep[1:]

scale是十六进制的基数16。 num_of_bits为32，因为float为32位，稍后使用它来填充其他位置，使用0来得到32.从this question获取binrep的代码。如果数字是负数，只需更改第一位。

我知道这很难看，但我没有找到一个好方法而且我需要它很快。欢迎提出意见。

Answer 8

这比被问到的要多一点，但这是我在找到这个条目时所需要的。此代码将给出IEEE 754 32位浮点数的尾数，基数和符号。

import ctypes
def binRep(num):
    binNum = bin(ctypes.c_uint.from_buffer(ctypes.c_float(num)).value)[2:]
    print("bits: " + binNum.rjust(32,"0"))
    mantissa = "1" + binNum[-23:]
    print("sig (bin): " + mantissa.rjust(24))
    mantInt = int(mantissa,2)/2**23
    print("sig (float): " + str(mantInt))
    base = int(binNum[-31:-23],2)-127
    print("base:" + str(base))
    sign = 1-2*("1"==binNum[-32:-31].rjust(1,"0"))
    print("sign:" + str(sign))
    print("recreate:" + str(sign*mantInt*(2**base)))

binRep(-0.75)

输出：

bits: 10111111010000000000000000000000
sig (bin): 110000000000000000000000
sig (float): 1.5
base:-1
sign:-1
recreate:-0.75

Answer 9

对于IEEE 754 binary64，使用这两个简单函数（ Python> = 3.6 ），您可以轻松地将浮点数转换为二进制，反之亦然。

import struct

def bin2float(b):
    ''' Convert binary string to a float.

    Attributes:
        :b: Binary string to transform.
    '''
    h = int(b, 2).to_bytes(8, byteorder="big")
    return struct.unpack('>d', h)[0]


def float2bin(f):
    ''' Convert float to 64-bit binary string.

    Attributes:
        :f: Float number to transform.
    '''
    [d] = struct.unpack(">Q", struct.pack(">d", f))
    return f'{d:064b}'

例如：

print(float2bin(1.618033988749894))
print(float2bin(3.14159265359))
print(float2bin(5.125))
print(float2bin(13.80))

print(bin2float('0011111111111001111000110111011110011011100101111111010010100100'))
print(bin2float('0100000000001001001000011111101101010100010001000010111011101010'))
print(bin2float('0100000000010100100000000000000000000000000000000000000000000000'))
print(bin2float('0100000000101011100110011001100110011001100110011001100110011010'))

输出为：

0011111111111001111000110111011110011011100101111111010010100100
0100000000001001001000011111101101010100010001000010111011101010
0100000000010100100000000000000000000000000000000000000000000000
0100000000101011100110011001100110011001100110011001100110011010
1.618033988749894
3.14159265359
5.125
13.8

我希望你喜欢它，它对我来说很完美。

Answer 10

在我看来，您可以使用.format来最简单地表示位：

我的代码看起来像：

file:///D:/wamp/www/Otagtransit/index.php

输出：

def fto32b(flt):
# is given a 32 bit float value and converts it to a binary string
if isinstance(flt,float):
    # THE FOLLOWING IS AN EXPANDED REPRESENTATION OF THE ONE LINE RETURN
            #   packed = struct.pack('!f',flt) <- get the hex representation in (!)Big Endian format of a (f) Float
            #   integers = []
            #   for c in packed:
            #       integers.append(ord(c))    <- change each entry into an int
            #   binaries = []
            #   for i in integers:
            #       binaries.append("{0:08b}".format(i)) <- get the 8bit binary representation of each int (00100101)
            #   binarystring = ''.join(binaries) <- join all the bytes together
            #   return binarystring
    return ''.join(["{0:08b}".format(i) for i in [ord(c) for c in struct.pack('!f',flt)]])
return None

Answer 11

其中一些答案不像Python 3那样有效，或者没有给出负浮点数的正确表示。我发现以下内容对我有用（虽然这给出了我需要的64位表示）

def float_to_binary_string(f):
    def int_to_8bit_binary_string(n):
        stg=bin(n).replace('0b','')
        fillstg = '0'*(8-len(stg))
        return fillstg+stg
    return ''.join( int_to_8bit_binary_string(int(b)) for b in struct.pack('>d',f) )

Answer 12

将浮点数转换为0..1

def float_bin(n, places = 3): 
    if (n < 0 or n > 1):
        return "ERROR, n must be in 0..1"

    answer = "0."
    while n > 0:
        if len(answer) > places: 
            return answer

        b = n * 2
        if b > 1:
            answer += '1'
            n = b - 1
        else:
            answer += '0'
            n = b

    return answer

Answer 13

我做了一个非常简单的。请检查一下。如果您认为有任何错误，请告诉我。这对我来说很好用。

sds=float(input("Enter the number : "))
sf=float("0."+(str(sds).split(".")[-1]))
aa=[]

while len(aa)<15:
    dd=round(sf*2,5)
    if dd-1>0:
        
        aa.append(1)
        sf=dd-1
        
    else:
        
        sf=round(dd,5)
        aa.append(0)
    
des=aa[:-1]
print("\n")
AA=([str(i) for i in des])

print("So the Binary Of : %s>>>"%sds,bin(int(str(sds).split(".")[0])).replace("0b",'')+"."+"".join(AA))

或者在整数的情况下使用 bin(integer).replace("0b",'')

Python中float的二进制表示（位不是十六进制）

13 个答案: