Is there a more efficient way to generate a 10 kBit (10,000 bits) random binary sequence in Python than appending 0s and 1s in a loop?
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510 kBit as in "ten thousand bits"? That wouldn't normally be called a "signal". Are you talking about a waveform? And/or a continuous signal of 10,000 bits per second or something? Please clarify, preferably a lot.unwind– unwind2012-12-17 18:04:19 +00:00Commented Dec 17, 2012 at 18:04
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3 Answers
If you want a random binary sequence then it's probably quickest just to generate a random integer in the appropriate range:
import random
s = random.randint(0, 2**10000 - 1)
After this it really depends on what you want to do with your binary sequence. You can query individual bits using bitwise operations:
s & (1 << x) # is bit x set?
or you could use a library like bitarray or bitstring if you want to make checking, setting slicing etc. easier:
from bitstring import BitArray
b = BitArray(uint=s, length=10000)
p, = b.find('0b000000')
if b[99]:
b[100] = False
...
4 Comments
Piotr Lopusiewicz
Minor nitpick but I think it should be 1 & (s >> x)
Scott Griffiths
@PiotrLopusiewicz: Either way works in that they evaluate to True or False the same, but yours is always 0 or 1 which I guess is a little neater :)
Quant Metropolis
random.randint(0, 2**n-1) does not work for n>63:
OverflowError: Python int too large to convert to C longScott Griffiths
@QuantMetropolis: Works for me, Python 2.7 and 3.5. Are you sure it's this exact bit of code that gives the error? If so exactly which Python version and implementation are you using?
The numpy package has a subpackage 'random' which can produce arrays of random numbers.
http://docs.scipy.org/doc/numpy/reference/routines.random.html
If you want an array of 'n' random bits, you can use
arr = numpy.random.randint(2, size=(n,))
... but depending on what you are doing with them, it may be more efficient to use e.g.
arr = numpy.random.randint(0x10000, size=(n,))
to get an array of 'n' numbers, each with 16 random bits; then
rstring = arr.astype(numpy.uint16).tostring()
turns that into a string of 2*n chars, containing the same random bits.
1 Comment
Uri May
Actually the right way to create an array of 'n' random bits using numpy is: arr = numpy.random.randint(2, size=(n,))
Here is a one liner:
import random
[random.randrange(2) for _ in range(10000)]
