在不使用内置 bin 函数的情况下将整数转换为二进制

def trans(x):
    if x == 0: return [0]
    bit = []
    while x:
        bit.append(x % 2)
        x >>= 1
    return bit[::-1]

这样就可以了。如果有内置函数，则滚动自己的函数是没有意义的。

def binary(x):
    return [int(i) for i in bin(x)[2:]]

该函数转换为二进制字符串。剥离，你就准备好了。bin()0b

只是为了好玩 - 作为递归单行的解决方案：

def tobin(x):
    return tobin(x/2) + [x%2] if x > 1 else [x]

请允许我提出以下建议：

def tobin(x,s):
    return [(x>>k)&1 for k in range(0,s)]

这可能是最快的方法，对我来说似乎很清楚。 当性能很重要时，bin way 太慢了。

干杯

# dec2bin.py
# FB - 201012057
import math

def dec2bin(f):
    if f >= 1:
        g = int(math.log(f, 2))
    else:
        g = -1
    h = g + 1
    ig = math.pow(2, g)
    st = ""    
    while f > 0 or ig >= 1: 
        if f < 1:
            if len(st[h:]) >= 10: # 10 fractional digits max
                   break
        if f >= ig:
            st += "1"
            f -= ig
        else:
            st += "0"
        ig /= 2
    st = st[:h] + "." + st[h:]
    return st

# MAIN
while True:
    f = float(raw_input("Enter decimal number >0: "))
    if f <= 0: break
    print "Binary #: ", dec2bin(f)
    print "bin(int(f)): ", bin(int(f)) # for comparison

这是我为大学制作的代码。 单击此处查看代码的 youtube 视频。！https://www.youtube.com/watch?v=SGTZzJ5H-CE

__author__ = 'Derek'
print('Int to binary')
intStr = input('Give me an int: ')
myInt = int(intStr)
binStr = ''
while myInt > 0:
    binStr = str(myInt % 2) + binStr
    myInt //= 2
print('The binary of', intStr, 'is', binStr)
print('\nBinary to int')
binStr = input('Give me a binary string: ')
temp = binStr
newInt = 0
power = 0
while len(temp) > 0:   # While the length of the array if greater than zero keep looping through
    bit = int(temp[-1])   # bit is were you temporally store the converted binary number before adding it to the total
    newInt = newInt + bit * 2 ** power  # newInt is the total,  Each time it loops it adds bit to newInt.
    temp = temp[:-1]  # this moves you to the next item in the string.
    power += 1  # adds one to the power each time.
print("The binary number " + binStr, 'as an integer is', newInt)

填充长度

在大多数情况下，您希望二进制数是特定的长度。例如，您希望 1 是 8 个二进制数字，长度为 [0,0,0,0,0,0,0,1]。我自己用这个：

def convert_to_binary(num, length=8):
    binary_string_list = list(format(num, '0{}b'.format(length)))
    return [int(digit) for digit in binary_string_list]

您可以先使用 format 函数来获取像当前函数一样的二进制字符串。例如，以下代码片段创建一个 8 位的二进制字符串，对应于整数 58。

>>>u = format(58, "08b")
'00111010'

现在迭代字符串以将每个位转换为整数，以获得编码为整数的所需位列表。

>>>[int(d) for d in u]
[0, 0, 1, 1, 1, 0, 1, 0]

不是最有效的，但至少它提供了一种简单的概念方式来理解它......

1） 地板将所有数字重复除以 2，直到达到 1

2）以相反的顺序，创建这个数字数组的位，如果它是偶数，则附加一个0，如果它是奇数，则附加一个1。

下面是它的字面实现：

def intToBin(n):
    nums = [n]
    while n > 1:
        n = n // 2
        nums.append(n)

    bits = []
    for i in nums:
        bits.append(str(0 if i%2 == 0 else 1))
    bits.reverse()
    print ''.join(bits)

下面是一个更好地利用内存的版本：

def intToBin(n):
    bits = []

    bits.append(str(0 if n%2 == 0 else 1))
    while n > 1:
        n = n // 2
        bits.append(str(0 if n%2 == 0 else 1))

    bits.reverse()
    return ''.join(bits)

您可以使用numpy包并获得非常快速的解决方案：

python -m timeit -s "import numpy as np; x=np.array([8], dtype=np.uint8)" "np.unpackbits(x)"
1000000 loops, best of 3: 0.65 usec per loop

python -m timeit "[int(x) for x in list('{0:0b}'.format(8))]"
100000 loops, best of 3: 3.68 usec per loop

unpackbits 仅处理 uint8 类型的输入，但您仍然可以使用 np.view：

python -m timeit -s "import numpy as np; x=np.array([124567], dtype=np.uint64).view(np.uint8)" "np.unpackbits(x)"
1000000 loops, best of 3: 0.697 usec per loop

不是pythonic的方式......但仍然有效：

def get_binary_list_from_decimal(integer, bits):
    '''Return a list of 0's and 1's representing a decimal type integer.

    Keyword arguments:
    integer -- decimal type number.
    bits -- number of bits to represent the integer.

    Usage example:
    #Convert 3 to a binary list
    get_binary_list_from_decimal(3, 4)
    #Return will be [0, 0, 1, 1]
    '''
    #Validate bits parameter.
    if 2**bits <= integer:
        raise ValueError("Error: Number of bits is not sufficient to \
                          represent the integer. Increase bits parameter.")

    #Initialise binary list
    binary_list = []
    remainder = integer
    for i in range(bits-1, -1, -1):
        #If current bit value is less than or equal to the remainder of 
        #the integer then bit value is 1.
        if 2**i <= remainder:
            binary_list.append(1)
            #Subtract the current bit value from the integer.
            remainder = remainder - 2**i
        else:
            binary_list.append(0)

    return binary_list

如何使用它的示例：

get_binary_list_from_decimal(1, 3)
#Return will be [0, 0, 1]

def nToKBit(n, K=64):
   output = [0]*K

   def loop(n, i):
       if n == 0: 
           return output
       output[-i] = n & 1
       return loop(n >> 1, i+1)

   return loop(n, 1)

将十进制转换为二进制取决于您将如何使用 % 和 //

def getbin(num):
    if (num==0):
        k=[0] 
        return k 
    else:
        s = []
        while(num):
            s.append(num%2)
            num=num//2
        return s

只是共享一个处理整数数组的函数：

def to_binary_string(x):
    length = len(bin(max(x))[2:])

    for i in x:
        b = bin(i)[2:].zfill(length)

        yield [int(n) for n in b]

测试：

x1 = to_binary_string([1, 2, 3])
x2 = to_binary_string([1, 2, 3, 4])

print(list(x1)) # [[0, 1], [1, 0], [1, 1]]
print(list(x2)) # [[0, 0, 1], [0, 1, 0], [0, 1, 1], [1, 0, 0]]

将整数转换为具有固定长度的位列表：

[int(x) for x in list('{0:0{width}b}'.format(8, width=5))]

def dectobin(x):
  i,f=str(x).split('.')
  i1=int(i)
  f1=int(f)
  int1=[]
  dec=[]
  count=0
  while i1>0:
    int1.append(i1%2)
    i1=i1//2
  while f1>0 and count<5:
    f1=f1/10**len(f)
    print(f1)
    f2=f1*2
    i3,f3=str(f2).split('.')
    dec.append(i3)
    f1=int(f3)
    count=count+1
  strint=''
  decint=''
  for x in int1:
    strint=strint+str(x)
  for x in dec:
    decint=decint+str(x)
  return(strint+'.'+decint)

print(dectobin(47.234))

#works