I'm new to Python. While reading, please mention any other suggestions regarding ways to improve my Python code.
Question: How do I generate a 8xN dimensional array in Python containing random numbers? The constraint is that each column of this array must contain 8 draws without replacement from the integer set [1,8]. More specifically, when N = 10, I want something like this.
[[ 6. 2. 3. 4. 7. 5. 5. 7. 8. 4.]
[ 1. 4. 5. 5. 4. 4. 8. 5. 7. 5.]
[ 7. 3. 8. 8. 3. 8. 7. 3. 6. 7.]
[ 3. 6. 7. 1. 5. 6. 2. 1. 5. 1.]
[ 8. 1. 4. 3. 8. 2. 3. 4. 3. 3.]
[ 5. 8. 1. 7. 1. 3. 6. 8. 1. 6.]
[ 4. 5. 2. 6. 2. 1. 1. 6. 4. 2.]
[ 2. 7. 6. 2. 6. 7. 4. 2. 2. 8.]]
To do this I use the following approach:
import numpy.random
import numpy
def rand_M(N):
M = numpy.zeros(shape = (8, N))
for i in range (0, N):
M[:, i] = numpy.random.choice(8, size = 8, replace = False) + 1
return M
In practice N will be ~1e7. The algorithm above is O(n) in time and it takes roughly .38 secs when N=1e3. The time therefore when N = 1e7 is ~1hr (i.e. 3800 secs). There has to be a much more efficient way.
Timing the function
from timeit import Timer
t = Timer(lambda: rand_M(1000))
print(t.timeit(5))
0.3863314103162543
Create a random array of specified shape and then sort along the axis where you want to keep the limits, thus giving us a vectorized and very efficient solution. This would be based on this smart answer
to MATLAB randomly permuting columns differently
. Here's the implementation -
Sample run -
In [122]: N = 10
In [123]: np.argsort(np.random.rand(8,N),axis=0)+1
Out[123]:
array([[7, 3, 5, 1, 1, 5, 2, 4, 1, 4],
[8, 4, 3, 2, 2, 8, 5, 5, 6, 2],
[1, 2, 4, 6, 5, 4, 4, 3, 4, 7],
[5, 6, 2, 5, 8, 2, 7, 8, 5, 8],
[2, 8, 6, 3, 4, 7, 1, 1, 2, 6],
[6, 7, 7, 8, 6, 6, 3, 2, 7, 3],
[4, 1, 1, 4, 3, 3, 8, 6, 8, 1],
[3, 5, 8, 7, 7, 1, 6, 7, 3, 5]], dtype=int64)
Runtime tests -
In [124]: def sortbased_rand8(N):
...: return np.argsort(np.random.rand(8,N),axis=0)+1
...:
...: def rand_M(N):
...: M = np.zeros(shape = (8, N))
...: for i in range (0, N):
...: M[:, i] = np.random.choice(8, size = 8, replace = False) + 1
...: return M
...:
In [125]: N = 5000
In [126]: %timeit sortbased_rand8(N)
100 loops, best of 3: 1.95 ms per loop
In [127]: %timeit rand_M(N)
1 loops, best of 3: 233 ms per loop
Thus, awaits a 120x
speedup!
How about shuffling, that is to say, permuting?
import random
import numpy
from timeit import Timer
def B_rand_M(N):
a = numpy.arange(1,9)
M = numpy.zeros(shape = (8, N))
for i in range (0, N):
M[:, i] = numpy.random.permutation(a)
return M
# your original implementation
def J_rand_M(N):
M = numpy.zeros(shape = (8, N))
for i in range (0, N):
M[:, i] = numpy.random.choice(8, size = 8, replace = False) + 1
return M
some timings:
def compare(N):
for f in (J_rand_M, B_rand_M):
t = Timer(lambda: f(N)).timeit(6)
print 'time for %s(%s): %.6f' % (f.__name__, N, t)
for i in range(6):
print 'N = 10^%s' % i
compare(10**i)
print
gives
N = 10^0
time for J_rand_M(1): 0.001199
time for B_rand_M(1): 0.000080
N = 10^1
time for J_rand_M(10): 0.001112
time for B_rand_M(10): 0.000335
N = 10^2
time for J_rand_M(100): 0.011118
time for B_rand_M(100): 0.003022
N = 10^3
time for J_rand_M(1000): 0.110887
time for B_rand_M(1000): 0.030528
N = 10^4
time for J_rand_M(10000): 1.100540
time for B_rand_M(10000): 0.304696
N = 10^5
time for J_rand_M(100000): 11.151576
time for B_rand_M(100000): 3.049474
If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
Donate Us With