在Python中有任何罐装的Python方法将整数(或长)转换为二进制字符串吗?

谷歌上有无数的dec2bin()函数…但我希望我可以使用内置函数/库。


当前回答

备选方案总结:

n=42
assert  "-101010" == format(-n, 'b')
assert  "-101010" == "{0:b}".format(-n)
assert  "-101010" == (lambda x: x >= 0 and str(bin(x))[2:] or "-" + str(bin(x))[3:])(-n)
assert "0b101010" == bin(n)
assert   "101010" == bin(n)[2:]   # But this won't work for negative numbers.

贡献者包括John Fouhy, Tung Nguyen, mVChr, Martin Thoma。还有马丁·彼得斯。

其他回答

下面是一个使用divmod构造二进制列表的(调试)程序:

程序

while True:
    indecimal_str = input('Enter positive(decimal) integer: ')
    if indecimal_str == '':
        raise SystemExit
    indecimal_save = int(indecimal_str)
    if indecimal_save < 1:
        print('Rejecting input, try again')
        print()
        continue
    indecimal = int(indecimal_str)
    exbin = []
    print(indecimal, '<->', exbin)
    while True:
        if indecimal == 0:
            print('Conversion:', indecimal_save, '=', "".join(exbin))
            print()
            break
        indecimal, r = divmod(indecimal, 2)
        if r == 0:
            exbin.insert(0, '0')
        else:
            exbin.insert(0, '1')
        print(indecimal, '<->', exbin)

输出

Enter positive(decimal) integer: 8
8 <-> []
4 <-> ['0']
2 <-> ['0', '0']
1 <-> ['0', '0', '0']
0 <-> ['1', '0', '0', '0']
Conversion: 8 = 1000

Enter positive(decimal) integer: 63
63 <-> []
31 <-> ['1']
15 <-> ['1', '1']
7 <-> ['1', '1', '1']
3 <-> ['1', '1', '1', '1']
1 <-> ['1', '1', '1', '1', '1']
0 <-> ['1', '1', '1', '1', '1', '1']
Conversion: 63 = 111111

Enter positive(decimal) integer: 409
409 <-> []
204 <-> ['1']
102 <-> ['0', '1']
51 <-> ['0', '0', '1']
25 <-> ['1', '0', '0', '1']
12 <-> ['1', '1', '0', '0', '1']
6 <-> ['0', '1', '1', '0', '0', '1']
3 <-> ['0', '0', '1', '1', '0', '0', '1']
1 <-> ['1', '0', '0', '1', '1', '0', '0', '1']
0 <-> ['1', '1', '0', '0', '1', '1', '0', '0', '1']
Conversion: 409 = 110011001

有点类似的解决方案

def to_bin(dec):
    flag = True
    bin_str = ''
    while flag:
        remainder = dec % 2
        quotient = dec / 2
        if quotient == 0:
            flag = False
        bin_str += str(remainder)
        dec = quotient
    bin_str = bin_str[::-1] # reverse the string
    return bin_str 

使用lambda的一行代码:

>>> binary = lambda n: '' if n==0 else binary(n/2) + str(n%2)

测试:

>>> binary(5)
'101'

编辑:

但是接下来:(

t1 = time()
for i in range(1000000):
     binary(i)
t2 = time()
print(t2 - t1)
# 6.57236599922

在比较中

t1 = time()
for i in range(1000000):
    '{0:b}'.format(i)
t2 = time()
print(t2 - t1)
# 0.68017411232

Python 3.6增加了一种新的字符串格式化方法,称为格式化字符串字面量或“f-strings”。 例子:

name = 'Bob'
number = 42
f"Hello, {name}, your number is {number:>08b}"

输出将是“你好,Bob,您的号码是00001010!”

关于这个问题的讨论可以在这里找到-在这里

计算二进制数:

print("Binary is {0:>08b}".format(16))

计算一个数的十六进制小数:

print("Hexa Decimal is {0:>0x}".format(15))

计算所有的二进制直到16::

for i in range(17):
   print("{0:>2}: binary is {0:>08b}".format(i))

要计算十六进制小数,直到17

 for i in range(17):
    print("{0:>2}: Hexa Decimal is {0:>0x}".format(i))
##as 2 digit is enogh for hexa decimal representation of a number