您可以使用以下
struct软件包：
import structdef binary(num): return ''.join(bin(ord(c)).replace('0b', '').rjust(8, '0') for c in struct.pack('!f', num))它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其串联起来：
>>> binary(1)'00111111100000000000000000000000'
编辑 ：有人要求扩大解释。我将使用中间变量对此进行扩展，以注释每个步骤。
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 pre point # # [62, 163, 215, 10] = [ord(c) for c in '>xa3xd7n'] 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)以及一些示例的结果：
>>> binary(1)Packed: '?x80x00x00'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: '>xa3xd7n'Integers: [62, 163, 215, 10]Binaries: ['0b111110', '0b10100011', '0b11010111', '0b1010']Stripped: ['111110', '10100011', '11010111', '1010']Padded: ['00111110', '10100011', '11010111', '00001010']'00111110101000111101011100001010'
