问题:将以2为底的二进制数字字符串转换为int
我只想将以2为底的二进制数字字符串转换为int,如下所示:
>>> '11111111'.fromBinaryToInt()
255
有没有办法在Python中做到这一点?
I’d simply like to convert a base-2 binary number string into an int, something like this:
>>> '11111111'.fromBinaryToInt()
255
Is there a way to do this in Python?
回答 0
您可以使用内置int
函数,然后将输入数字的基数传递给它,即2
二进制数:
>>> int('11111111', 2)
255
这是python2和python3的文档。
You use the built-in int
function, and pass it the base of the input number, i.e. 2
for a binary number:
>>> int('11111111', 2)
255
Here is documentation for python2, and for python3.
回答 1
只需在python交互界面中输入0b11111111:
>>> 0b11111111
255
Just type 0b11111111 in python interactive interface:
>>> 0b11111111
255
回答 2
另一种方法是使用bitstring
模块:
>>> from bitstring import BitArray
>>> b = BitArray(bin='11111111')
>>> b.uint
255
请注意,无符号整数与有符号整数不同:
>>> b.int
-1
该bitstring
模块不是必需的,但它具有许多高性能的方法,可以将输入转换为位或转换为其他形式,以及对其进行操作。
Another way to do this is by using the bitstring
module:
>>> from bitstring import BitArray
>>> b = BitArray(bin='11111111')
>>> b.uint
255
Note that the unsigned integer is different from the signed integer:
>>> b.int
-1
The bitstring
module isn’t a requirement, but it has lots of performant methods for turning input into and from bits into other forms, as well as manipulating them.
回答 3
将int与base一起使用是正确的方法。在发现int也需要基础之前,我曾经这样做过。从本质上讲,这是应用于将二进制转换为十进制的原始方法的列表理解的减少(例如110 = 2 ** 0 * 0 + 2 ** 1 * 1 + 2 ** 2 * 1)
add = lambda x,y : x + y
reduce(add, [int(x) * 2 ** y for x, y in zip(list(binstr), range(len(binstr) - 1, -1, -1))])
Using int with base is the right way to go. I used to do this before I found int takes base also. It is basically a reduce applied on a list comprehension of the primitive way of converting binary to decimal ( e.g. 110 = 2**0 * 0 + 2 ** 1 * 1 + 2 ** 2 * 1)
add = lambda x,y : x + y
reduce(add, [int(x) * 2 ** y for x, y in zip(list(binstr), range(len(binstr) - 1, -1, -1))])
回答 4
如果您想知道幕后发生的事情,那么您就可以开始了。
class Binary():
def __init__(self, binNumber):
self._binNumber = binNumber
self._binNumber = self._binNumber[::-1]
self._binNumber = list(self._binNumber)
self._x = [1]
self._count = 1
self._change = 2
self._amount = 0
print(self._ToNumber(self._binNumber))
def _ToNumber(self, number):
self._number = number
for i in range (1, len (self._number)):
self._total = self._count * self._change
self._count = self._total
self._x.append(self._count)
self._deep = zip(self._number, self._x)
for self._k, self._v in self._deep:
if self._k == '1':
self._amount += self._v
return self._amount
mo = Binary('101111110')
If you wanna know what is happening behind the scene, then here you go.
class Binary():
def __init__(self, binNumber):
self._binNumber = binNumber
self._binNumber = self._binNumber[::-1]
self._binNumber = list(self._binNumber)
self._x = [1]
self._count = 1
self._change = 2
self._amount = 0
print(self._ToNumber(self._binNumber))
def _ToNumber(self, number):
self._number = number
for i in range (1, len (self._number)):
self._total = self._count * self._change
self._count = self._total
self._x.append(self._count)
self._deep = zip(self._number, self._x)
for self._k, self._v in self._deep:
if self._k == '1':
self._amount += self._v
return self._amount
mo = Binary('101111110')
回答 5
递归Python实现:
def int2bin(n):
return int2bin(n >> 1) + [n & 1] if n > 1 else [1]
A recursive Python implementation:
def int2bin(n):
return int2bin(n >> 1) + [n & 1] if n > 1 else [1]
回答 6
如果您使用的是python3.6或更高版本,则可以使用f-string进行转换:
二进制到十进制:
>>> print(f'{0b1011010:#0}')
90
>>> bin_2_decimal = int(f'{0b1011010:#0}')
>>> bin_2_decimal
90
二进制到八进制六进制等
>>> f'{0b1011010:#o}'
'0o132' # octal
>>> f'{0b1011010:#x}'
'0x5a' # hexadecimal
>>> f'{0b1011010:#0}'
'90' # decimal
注意2条以冒号分隔的信息。
这样,您可以通过更改冒号右侧来在{二进制,八进制,十六进制,十进制}之间转换为{二进制,八进制,十六进制,十进制}。
:#b -> converts to binary
:#o -> converts to octal
:#x -> converts to hexadecimal
:#0 -> converts to decimal as above example
尝试将冒号的左侧更改为八进制/十六进制/十进制。
If you are using python3.6 or later you can use f-string to do the
conversion:
Binary to decimal:
>>> print(f'{0b1011010:#0}')
90
>>> bin_2_decimal = int(f'{0b1011010:#0}')
>>> bin_2_decimal
90
binary to octal hexa and etc.
>>> f'{0b1011010:#o}'
'0o132' # octal
>>> f'{0b1011010:#x}'
'0x5a' # hexadecimal
>>> f'{0b1011010:#0}'
'90' # decimal
Pay attention to 2 piece of information separated by colon.
In this way, you can convert between {binary, octal, hexadecimal, decimal} to {binary, octal, hexadecimal, decimal} by changing right side of colon[:]
:#b -> converts to binary
:#o -> converts to octal
:#x -> converts to hexadecimal
:#0 -> converts to decimal as above example
Try changing left side of colon to have octal/hexadecimal/decimal.
回答 7
对于较大的矩阵(10 ** 5行及以上),最好使用矢量化matmult。一杆传入所有行和列。非常快。python中没有循环。我最初将其设计为将许多二进制列(例如MovieLens中10个不同类型列的0/1)转换为每个示例行的单个整数。
def BitsToIntAFast(bits):
m,n = bits.shape
a = 2**np.arange(n)[::-1] # -1 reverses array of powers of 2 of same length as bits
return bits @ a
For large matrix (10**5 rows and up) it is better to use a vectorized matmult. Pass in all rows and cols in one shot. It is extremely fast. There is no looping in python here. I originally designed it for converting many binary columns like 0/1 for like 10 different genre columns in MovieLens into a single integer for each example row.
def BitsToIntAFast(bits):
m,n = bits.shape
a = 2**np.arange(n)[::-1] # -1 reverses array of powers of 2 of same length as bits
return bits @ a