python之二进制和十进制转换

[belukwill]

二进制转十进制：

bin=str(input('输入二进制数：'))
count = 0
for i in range(0,len(bin)):
    if bin[i] == str(1):
        sum=2**(len(bin)-i-1)
        count=count+sum
print count

十进制转二进制：

number = int(input('输入十进制'))
n=''.join([str((number>>a&0x1))for a in range(7,-1,-1)]);
print n

[bugs2k]

[GCC 4.6.3] on linux2
Type "help", "copyright", "credits" or "license" for more information.
>>> bin(7)
'0b111'
>>> int(0b111)
7
>>> 

[aaseh]

请问，用什么ide？

[belukwill]

我这是算法，你用人家系统函数，没有技术含量

[angel_su]

用python不要跑去造轮子，开发和执行效率都慢。当然你做算法研究就不予置评...

[Rlay_2]

x = bin(1024)
print 'bin:', x

x = oct(1024)
print 'oct:', x

x = hex(1024)
print 'hex:', x

bin: 0b10000000000
oct: 02000
hex: 0x400

[panghuhu250]

In [21]: def bin2dec(s):
    ...:     total = 0
    ...:     for i in s:
    ...:         total = 2*total + (0 if i=='0' else 1)
    ...:     return total

In [22]: def bin2dec2(bin):
    ...:     count = 0
    ...:     for i in range(0,len(bin)):
    ...:         if bin[i] == str(1):
    ...:             sum=2**(len(bin)-i-1)
    ...:             count=count+sum
    ...:     return count

In [23]: a = '101010101010111'

In [24]: int(a, 2)
Out[24]: 21847

In [25]: bin2dec(a)
Out[25]: 21847

In [26]: bin2dec2(a)
Out[26]: 21847

In [27]: %timeit int(a, 2)
10000000 loops, best of 3: 181 ns per loop

In [28]: %timeit bin2dec(a)
1000000 loops, best of 3: 1.36 us per loop

In [29]: %timeit bin2dec2(a)
100000 loops, best of 3: 4.55 us per loop

[huer0625]

int(x[, base]) -> integer

[huaoweiomian]

要的就是简单，没有技术含量，要不然就不用python了。