只需使用format函数

format(6, "08b")

一般形式是

format(<the_integer>, "<0><width_of_string><format_specifier>")

def int_to_bin(num, fill):
    bin_result = ''

    def int_to_binary(number):
        nonlocal bin_result
        if number > 1:
            int_to_binary(number // 2)
        bin_result = bin_result + str(number % 2)

    int_to_binary(num)
    return bin_result.zfill(fill)

有点糊弄的方法…

>>> bin8 = lambda x : ''.join(reversed( [str((x >> i) & 1) for i in range(8)] ) )
>>> bin8(6)
'00000110'
>>> bin8(-3)
'11111101'

走老路总是管用的

def intoBinary(number):
binarynumber=""
if (number!=0):
    while (number>=1):
        if (number %2==0):
            binarynumber=binarynumber+"0"
            number=number/2
        else:
            binarynumber=binarynumber+"1"
            number=(number-1)/2

else:
    binarynumber="0"

return "".join(reversed(binarynumber))

. .或者如果你不确定它应该总是8位，你可以把它作为一个参数传递:

>>> '%0*d' % (8, int(bin(6)[2:]))
'00000110'

简单的递归代码:

 def bin(n,number=('')):
   if n==0:
     return(number)
   else:
     number=str(n%2)+number
     n=n//2
     return bin(n,number)