Doubling binary digits
How to double a number of binary digits in an integer? For example, if bin(x)="1001" then bin(y) must be "11000011". Is there any smart and fast algorithm ?
UPDATE: Here is an elegant solution:
''.join([''.join开发者_如何学运维(i) for i in zip(X,X)])
where X is bin(int_x)[2:]
However, I am interested in a more faster way and for the integers of any size. Maybe an arithmetical transformation should help.
Here's one way that should be reasonably fast: convert your number to a binary string, then reinterpret the result as being in base 4. Now to make sure that all the '1's are doubled properly, multiply the result by 3.
>>> x = 9
>>> bin(x)
'0b1001'
>>> y = int(bin(x)[2:], 4)*3
>>> bin(y)
'0b11000011'
(Reference http://graphics.stanford.edu/~seander/bithacks.html#Interleave64bitOps):
If your number is below 256, you may use
@magic
def double_digits_holger8(x):
m = (x * 0x0101010101010101 & 0x8040201008040201) * 0x0102040810204081
return ((m >> 49) & 0x5555) | ((m >> 48) & 0xAAAA)
and if it is below 65536,
@more_magic
def double_digits_binmag16(x):
x = (x | x << 8) & 0x00FF00FF
x = (x | x << 4) & 0x0F0F0F0F
x = (x | x << 2) & 0x33333333
x = (x | x << 1) & 0x55555555
return x | x << 1
Comparison with other solutions (the function must take an integer and return an integer for fair comparison):
Method Time per 256 calls
--------------------------------
Do nothing 46.2 usec
Holger8 256 usec
BinMag16 360 usec
Mark 367 usec # http://stackoverflow.com/questions/2928886/doubling-binary-digits/2929198#2929198
Max 720 usec # http://stackoverflow.com/questions/2928886/doubling-binary-digits/2928938#2928938
Peter 1.08 msec # http://stackoverflow.com/questions/2928886/doubling-binary-digits/2928973#2928973
Phiµµ w/o Log 1.11 msec # http://stackoverflow.com/questions/2928886/doubling-binary-digits/2929106#2929106
Jim16 1.26 msec # http://stackoverflow.com/questions/2928886/doubling-binary-digits/2929038#2929038
Elegant 1.66 msec # int(''.join([''.join(i) for i in zip(X,X)]),2)
More Elegant 2.05 msec # int(''.join(chain(*zip(X, X))), 2)
Benchmark source code can be found in http://gist.github.com/417172.
The straightforward solution just using integer arithmetic would be:
def doubledigits(n):
result = 0
power = 1
while n > 0:
if n%2==1:
result += 3*power
power *= 4
n //= 2
return result
any_number - int
str(n) - produces string from int.
str::replace(pattern, replaced_value) - replaces all patterns in string to replaced_value.
int(str) - makes int from string.
n=any_number
result_number = int(str(n).replace("0","00").replace("1","11"))
$ python2.6
Python 2.6.5 (r265:79063, Mar 25 2010, 14:13:28)
>>> def dd(n): return eval("0b" + "".join(d * 2 for d in str(bin(n))[2:]))
...
>>> dd(9)
195
y = 0;
for(i = 15; i >= 0; i--) {
if((1 << i) & x) {
y |= 3;
}
y <<= 2;
}
def doubledigits(x):
from math import log
print (bin(x))
numdigits = x.bit_length()
result = 1 << (numdigits*2)
for i in range(numdigits, -1, -1):
mask = 1 << i
if (x & mask > 0):
rmask = 0b11 << (2*i)
result = result | rmask
return result
should do it.
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