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Numpy array is much slower than list

Given two matrices X1 (N,3136) and X2 (M,3136) (where every element in every row is an binary number) i am trying to calculate hamming distance so that each element in X1 is compared to all of the rows from X2, such that result matrix is (N,M).
I have written two function for it (first one with help of numpy and the other one without numpy):
def hamming_distance(X, X_train):
array = np.array([np.sum(np.logical_xor(x, X_train), axis=1) for x in X])
return array
def hamming_distance2(X, X_train):
a = len(X[:,0])
b = len(X_train[:,0])
hamming_distance = np.zeros(shape=(a, b))
for i in range(0, a):
for j in range(0, b):
hamming_distance[i,j] = np.count_nonzero(X[i,:] != X_train[j,:])
return hamming_distance
My problem is that upper function is much slower than lower one where I use two for loops. Is it possible to improve on first function so that I use only one loop?
PS. Sorry for my english, it isn't my first language, although I was trying to do my best!

Numpy only makes your code much faster if you use it to vectorize your work. In your case you can make use of array broadcasting to vectorize your problem: compare your two arrays and create an auxiliary array of shape (N,M,K) which you can sum along its third dimension:
hamming_distance = (X[:,None,:] != X_train).sum(axis=-1)
We inject a singleton dimension into the first array to make it of shape (N,1,K), the second array is implicitly compatible with shape (1,M,K), so the operation can be performed.
In the comments #ayhan noted that this will create a huge auxiliary array for large M and N, which is quite true. This is the price of vectorization: you gain CPU time at the cost of memory. If you have enough memory for the above to work, it will be very fast. If you don't, you have to reduce the scope of your vectorization, and loop in either M or N (or both; this would be your current approach). But this doesn't concern numpy itself, this is about striking a balance between available resources and performance.

What you are doing is very similar to dot product. Consider these two binary arrays:
1 0 1 0 1 1 0 0
0 0 1 1 0 1 0 1
We are trying to find the number of different pairs. If you directly take the dot product, it gives you the number of (1, 1) pairs. However, if you negate one of them, it will count the different ones. For example, a1.dot(1-a2) counts (1, 0) pairs. Since we also need the number of (0, 1) pairs, we will add a2.dot(1-a1) to that. The good thing about dot product is that it is pretty fast. However, you will need to convert your arrays to floats first, as Divakar pointed out.
Here's a demo:
prng = np.random.RandomState(0)
arr1 = prng.binomial(1, 0.3, (1000, 3136))
arr2 = prng.binomial(1, 0.3, (2000, 3136))
res1 = hamming_distance2(arr1, arr2)
arr1 = arr1.astype('float32'); arr2 = arr2.astype('float32')
res2 = (1-arr1).dot(arr2.T) + arr1.dot(1-arr2.T)
np.allclose(res1, res2)
Out: True
And timings:
%timeit hamming_distance(arr1, arr2)
1 loop, best of 3: 13.9 s per loop
%timeit hamming_distance2(arr1, arr2)
1 loop, best of 3: 5.01 s per loop
%timeit (1-arr1).dot(arr2.T) + arr1.dot(1-arr2.T)
10 loops, best of 3: 93.1 ms per loop

Numpy: 2d list min max is slow [duplicate]

This question already has answers here:
Faster alternatives to numpy.argmax/argmin which is slow
(3 answers)
Closed 6 years ago.
I'm completely new to numpy and unable to find a solution.
I have a 2d list of floating point numbers in python like:
list1[0..8][0..2]
Where e.g.:
print(list1[0][0])
> 0.1122233784
Now I want to find min and max values:
b1 = numpy.array(list1)
list1MinX, list1MinY, list1MinZ = b1.min(axis=0)
list1MaxX, list1MaxY, list1MaxZ = b1.max(axis=0)
I need to do this about a million times in a loop.
It works correctly, but it's about 3x slower than my previous native python approach.
(1:15 min[numpy] vs 0:25 min[native])
What am I doing wrong?
I've read that the list conversion could be the problem, but I don't know how to do it better.
EDIT
As request some non-pseudo code, although in my script the list is created in another way.
import numpy
import random
def moonPositionNow():
#assume we read like from a file, line by line
#nextChunk = readNextLine()
#the file is build like this
#x-coord
#y-coord
#z-coord
#x-coord
#...
#but we don't have that data here, so as a **placeholder** we return a random number
nextChunk = random.random()
return nextChunk
for w in range(1000000):
list1 = [[moonPositionNow() for i in range(3)] for j in range(9)]
b1 = numpy.array(list1)
list1MinX, list1MinY, list1MinZ = b1.min(axis=0)
list1MaxX, list1MaxY, list1MaxZ = b1.max(axis=0)
#Print out results
Although the list creation may be a bottle neck here I guaranty in the original code it's not the problem.
EDIT2:
Updated the example code to clarify, I don't need a numpy array of random numbers.

Since your data is available as a Python list it seems reasonable to me that a native implementation (which likely calls some optimized C code) could be faster than converting to numpy first and then calling optimized C code.
You basically loop over your data twice: once for converting the python objects to numpy arrays, and once for computing the maximum or minimum.
The native implementation (I assume it is something like calling min/max on the Python list) only needs to loop over the data once.
Furthermore, it seems that numpy's min/max functions are surprisingly slow: https://stackoverflow.com/a/12200671/3005167

The problem arises because you are passing a python list to a numpy function. The numpy function is significantly faster if you pass a numpy array as the argument.
#Create numpy numbers
nptest = np.random.uniform(size=(10000, 10))
#Create a native python list
listtest = list(nptest)
#Compare performance
%timeit np.min(nptest, axis=0)
%timeit np.min(listtest, axis=0)
Output
1000 loops, best of 3: 394 µs per loop
100 loops, best of 3: 20 ms per loop
EDIT: Added example on how to evaluate a cost function over a grid.
The following evaluates a quadratic cost function over a grid and then takes the minimum along the first axis. In particular, np.meshgrid is your friend.
def cost_function(x, y):
return x ** 2 + y ** 2
x = linspace(-1, 1)
y = linspace(-1, 1)
def eval_python(x, y):
matrix = [cost_function(_x, _y) for _x in x for _y in y]
return np.min(matrix, axis=0)
def eval_numpy(x, y):
xx, yy = np.meshgrid(x, y)
matrix = cost_function(xx, yy)
return np.min(matrix, axis=0)
%timeit eval_python(x, y)
%timeit eval_numpy(x, y)
Output
100 loops, best of 3: 13.9 ms per loop
10000 loops, best of 3: 136 µs per loop
Finally, if you cannot cast your problem in this form, you can preallocated the memory and then fill in each element.
matrix = np.empty((num_x, num_y))
for i in range(num_x):
for j in range(num_y):
matrix[i, j] = cost_function(i, j)

Faster looping with itertools

I have a function
def getSamples():
p = lambda x : mlab.normpdf(x,3,2) + mlab.normpdf(x,-5,1)
q = lambda x : mlab.normpdf(x,5,14)
k=30
goodSamples = []
rightCount = 0
totalCount = 0
while(rightCount < 100000):
z0 = np.random.normal(5, 14)
u0 = np.random.uniform(0,k*q(z0))
if(p(z0) > u0):
goodSamples.append(z0)
rightCount += 1
totalCount += 1
return np.array(goodSamples)
My implementation to generate 100000 samples is taking much long. How can I make it fast with itertools or something similar?

I would say that the secret to making this code faster does not lie in changing the loop syntax. Here are a few points:
np.random.normal has an additional parameter size that lets you get many values at once. I would suggest using an array of say 1E09 elements and then checking your condition on that for how many are good. You can then estimate how likely that is.
To create your uniform samples, why not use sympy for symbolic evaluation of the pdf? (I don't know if this is faster but it could be since you already know the mean and variance.)
Again, for p could you use a symbolic function?

In general, performance problems are caused by doing things the "wrong way". Numpy can be very fast when used as it is designed to be used, that is by exploiting its vector processing where these vectorized operations are handed off to compiled code. Two bad practices that come from other programing languages/approaches are
Loops: Whenever you think you need a loop stop and think. Most of the time you do not and in fact do not even want one. It is much faster both to write and run code without loops.
Memory allocation: Whenever you know the size of an object, preallocate space for it. Growing memory, particularly in Python lists, is very slow compared to the alternatives.
In this case it is easy to get (approximately) two orders of magnitude speedup; the tradeoff is more memory usage.
Below is some representative code, it is not meant to be blindly used. I have not even verified it produces the correct results. It is more or less a direct translation of your routine. It appears you are drawing random numbers from a probability distribution using the rejection method. There may be more efficient algorithms to do this for your probability distribution.
def getSamples2() :
p = lambda x : mlab.normpdf(x,3,2) + mlab.normpdf(x,-5,1)
q = lambda x : mlab.normpdf(x,5,14)
k=30
N = 100000 # Total number of samples we want
Ngood = 0 # Current number of good samples
goodSamples = np.zeros(N) # Storage for the good samples
while Ngood < N : # Unfortunately a loop, ....
z0 = np.random.normal(5, 14, size=N)
u0 = np.random.uniform(size=N)*k*q(z0)
ind, = np.where(p(z0) > u0)
n = min(len(ind), N-Ngood)
goodSamples[Ngood:Ngood+n] = z0[ind[:n]]
Ngood += n
return goodSamples
This generates random numbers in chunks and saves the good ones. I have not tried to optimize the chunk size (here I just use N, the total number we want, in principle this could/should be different and could even be adjusted based on the number we have left to generate). This still uses a loop, unfortunately, but now this will be run "tens" of times instead of 100,000 times. This also uses the where function and array slicing; these are good general tools to be comfortable with.
In one test with %timeit on my machine I found
In [27]: %timeit getSamples() # Original routine
1 loops, best of 3: 49.3 s per loop
In [28]: %timeit getSamples2()
1 loops, best of 3: 505 ms per loop

Here is kinda itertools "magic", but I'm not sure it can help. Probably it's much better for perfomance to prepare an numpy array (using zeros) and fill it without creating python auto-growing list. Here is both itertools and zero-preparations. (Excuse me in advance for untested code)
from itertools import count, ifilter, imap, takewhile
import operator
def getSamples():
p = lambda x : mlab.normpdf(x, 3, 2) + mlab.normpdf(x, -5, 1)
q = lambda x : mlab.normpdf(x, 5, 14)
k = 30
n = 100000
samples_iter = imap(
operator.itemgetter(1),
takewhile(
lambda i, s: i < n,
enumerate(
ifilter(lambda z: p(z) > np.random.uniform(0,k*q(z)),
(np.random.normal(5, 14) for _ in count()))
)))
goodSamples = numpy.zeros(n)
# set values from iterator, probably there is a better way for that
for i, sample in enumerate(samples_iter):
goodSamples[i] = sample
return goodSamples

How can I speed up transition matrix creation in Numpy?

The following is the most basic way I know of to count transitions in a markov chain and use it to populate a transition matrix:
def increment_counts_in_matrix_from_chain(markov_chain, transition_counts_matrix):
for i in xrange(1, len(markov_chain)):
old_state = markov_chain[i - 1]
new_state = markov_chain[i]
transition_counts_matrix[old_state, new_state] += 1
I've tried speeding it up in 3 different ways:
1) Using a sparse matrix one-liner based on this Matlab code:
transition_matrix = full(sparse(markov_chain(1:end-1), markov_chain(2:end), 1))
Which in Numpy/SciPy, looks like this:
def get_sparse_counts_matrix(markov_chain, number_of_states):
return coo_matrix(([1]*(len(markov_chain) - 1), (markov_chain[0:-1], markov_chain[1:])), shape=(number_of_states, number_of_states))
And I've tried a couple more Python tweaks, like using zip():
for old_state, new_state in zip(markov_chain[0:-1], markov_chain[1:]):
transition_counts_matrix[old_state, new_state] += 1
And Queues:
old_and_new_states_holder = Queue(maxsize=2)
old_and_new_states_holder.put(markov_chain[0])
for new_state in markov_chain[1:]:
old_and_new_states_holder.put(new_state)
old_state = old_and_new_states_holder.get()
transition_counts_matrix[old_state, new_state] += 1
But none of these 3 methods sped things up. In fact, everything but the zip() solution was at least 10X slower than my original solution.
Are there any other solutions worth looking into?
Modified solution for building a transition matrix from lots of chains
The best answer to the above question specifically was DSM's. However, for anyone who wants to populate a transition matrix based on a list of millions of markov chains, the quickest way is this:
def fast_increment_transition_counts_from_chain(markov_chain, transition_counts_matrix):
flat_coords = numpy.ravel_multi_index((markov_chain[:-1], markov_chain[1:]), transition_counts_matrix.shape)
transition_counts_matrix.flat += numpy.bincount(flat_coords, minlength=transition_counts_matrix.size)
def get_fake_transitions(markov_chains):
fake_transitions = []
for i in xrange(1,len(markov_chains)):
old_chain = markov_chains[i - 1]
new_chain = markov_chains[i]
end_of_old = old_chain[-1]
beginning_of_new = new_chain[0]
fake_transitions.append((end_of_old, beginning_of_new))
return fake_transitions
def decrement_fake_transitions(fake_transitions, counts_matrix):
for old_state, new_state in fake_transitions:
counts_matrix[old_state, new_state] -= 1
def fast_get_transition_counts_matrix(markov_chains, number_of_states):
"""50% faster than original, but must store 2 additional slice copies of all markov chains in memory at once.
You might need to break up the chains into manageable chunks that don't exceed your memory.
"""
transition_counts_matrix = numpy.zeros([number_of_states, number_of_states])
fake_transitions = get_fake_transitions(markov_chains)
markov_chains = list(itertools.chain(*markov_chains))
fast_increment_transition_counts_from_chain(markov_chains, transition_counts_matrix)
decrement_fake_transitions(fake_transitions, transition_counts_matrix)
return transition_counts_matrix

Just for kicks, and because I've been wanting to try it out, I applied Numba to your problem. In code, that involves just adding a decorator (although I've made a direct call so I could test the jit variants that numba provides here):
import numpy as np
import numba
def increment_counts_in_matrix_from_chain(markov_chain, transition_counts_matrix):
for i in xrange(1, len(markov_chain)):
old_state = markov_chain[i - 1]
new_state = markov_chain[i]
transition_counts_matrix[old_state, new_state] += 1
autojit_func = numba.autojit()(increment_counts_in_matrix_from_chain)
jit_func = numba.jit(argtypes=[numba.int64[:,::1],numba.double[:,::1]])(increment_counts_in_matrix_from_chain)
t = np.random.randint(0,50, 500)
m1 = np.zeros((50,50))
m2 = np.zeros((50,50))
m3 = np.zeros((50,50))
And then timings:
In [10]: %timeit increment_counts_in_matrix_from_chain(t,m1)
100 loops, best of 3: 2.38 ms per loop
In [11]: %timeit autojit_func(t,m2)
10000 loops, best of 3: 67.5 us per loop
In [12]: %timeit jit_func(t,m3)
100000 loops, best of 3: 4.93 us per loop
The autojit method does some guessing based on runtime inputs, and the jit function has types dictated. You have to be a little careful since numba at these early stages doesn't communicate that there was an error with jit if you pass in the wrong type for an input. It will just spit out an incorrect answer.
That said though, getting a 35x and 485x speed-up without any code change and just adding a call to numba (can be also called as a decorator) is pretty impressive in my book. You could probably get similar results using cython, but it would require a bit more boilerplate and writing a setup.py file.
I also like this solution because the code remains readable and you can write it the way you originally thought about implementing the algorithm.

How about something like this, taking advantage of np.bincount? Not super-robust, but functional. [Thanks to #Warren Weckesser for the setup.]
import numpy as np
from collections import Counter
def increment_counts_in_matrix_from_chain(markov_chain, transition_counts_matrix):
for i in xrange(1, len(markov_chain)):
old_state = markov_chain[i - 1]
new_state = markov_chain[i]
transition_counts_matrix[old_state, new_state] += 1
def using_counter(chain, counts_matrix):
counts = Counter(zip(chain[:-1], chain[1:]))
from_, to = zip(*counts.keys())
counts_matrix[from_, to] = counts.values()
def using_bincount(chain, counts_matrix):
flat_coords = np.ravel_multi_index((chain[:-1], chain[1:]), counts_matrix.shape)
counts_matrix.flat = np.bincount(flat_coords, minlength=counts_matrix.size)
def using_bincount_reshape(chain, counts_matrix):
flat_coords = np.ravel_multi_index((chain[:-1], chain[1:]), counts_matrix.shape)
return np.bincount(flat_coords, minlength=counts_matrix.size).reshape(counts_matrix.shape)
which gives:
In [373]: t = np.random.randint(0,50, 500)
In [374]: m1 = np.zeros((50,50))
In [375]: m2 = m1.copy()
In [376]: m3 = m1.copy()
In [377]: timeit increment_counts_in_matrix_from_chain(t, m1)
100 loops, best of 3: 2.79 ms per loop
In [378]: timeit using_counter(t, m2)
1000 loops, best of 3: 924 us per loop
In [379]: timeit using_bincount(t, m3)
10000 loops, best of 3: 57.1 us per loop
[edit]
Avoiding flat (at the cost of not working in-place) can save some time for small matrices:
In [80]: timeit using_bincount_reshape(t, m3)
10000 loops, best of 3: 22.3 us per loop

Here's a faster method. The idea is to count the number of occurrences of each transition, and use the counts in a vectorized update of the matrix. (I'm assuming that the same transition can occur multiple times in markov_chain.) The Counter class from the collections library is used to count the number of occurrences of each transition.
from collections import Counter
def update_matrix(chain, counts_matrix):
counts = Counter(zip(chain[:-1], chain[1:]))
from_, to = zip(*counts.keys())
counts_matrix[from_, to] += counts.values()
Timing example, in ipython:
In [64]: t = np.random.randint(0,50, 500)
In [65]: m1 = zeros((50,50))
In [66]: m2 = zeros((50,50))
In [67]: %timeit increment_counts_in_matrix_from_chain(t, m1)
1000 loops, best of 3: 895 us per loop
In [68]: %timeit update_matrix(t, m2)
1000 loops, best of 3: 504 us per loop
It's faster, but not orders of magnitude faster. For a real speed up, you might consider implementing this in Cython.

Ok, few ideas to tamper with, with some slight improvement (at cost of human undestanding)
Let's start with a random vector of integers between 0 and 9 of length 3000:
L = 3000
N = 10
states = array(randint(N),size=L)
transitions = np.zeros((N,N))
Your method, on my machine, has a timeit performance of 11.4 ms.
The first thing for a little improvement is to avoid to read the data twice, storing it in a temporary variable:
old = states[0]
for i in range(1,len(states)):
new = states[i]
transitions[new,old]+=1
old=new
This gives you a ~10% improvement and drops the time to 10.9 ms.
A more involuted approach uses the strides:
def rolling(a, window):
shape = (a.size - window + 1, window)
strides = (a.itemsize, a.itemsize)
return np.lib.stride_tricks.as_strided(a, shape=shape, strides=strides)
state_2 = rolling(states, 2)
for i in range(len(state_2)):
l,m = state_2[i,0],state_2[i,1]
transitions[m,l]+=1
The strides allow you to read the consecutive numbers of the array tricking the array to think that the rows start in a different way (ok, it's not well described, but if you take some time to read about strides you will get it)
This approach loses performance, going to 12.2 ms, but it is the hallway to trick the system even more. flattening both the transition matrix and the strided array to one dimensional arrays, you can speed up the performance a little more:
transitions = np.zeros(N*N)
state_2 = rolling(states, 2)
state_flat = np.sum(state_2 * array([1,10]),axis=1)
for i in state_flat:
transitions[i]+=1
transitions.reshape((N,N))
This goes down to 7.75 ms. It's not an order of magnitude, but it's a 30% better anyway :)

Fastest Way to generate 1,000,000+ random numbers in python

I am currently writing an app in python that needs to generate large amount of random numbers, FAST. Currently I have a scheme going that uses numpy to generate all of the numbers in a giant batch (about ~500,000 at a time). While this seems to be faster than python's implementation. I still need it to go faster. Any ideas? I'm open to writing it in C and embedding it in the program or doing w/e it takes.
Constraints on the random numbers:
A Set of 7 numbers that can all have different bounds:
eg: [0-X1, 0-X2, 0-X3, 0-X4, 0-X5, 0-X6, 0-X7]
Currently I am generating a list of 7 numbers with random values from [0-1) then multiplying by [X1..X7]
A Set of 13 numbers that all add up to 1
Currently just generating 13 numbers then dividing by their sum
Any ideas? Would pre calculating these numbers and storing them in a file make this faster?
Thanks!

You can speed things up a bit from what mtrw posted above just by doing what you initially described (generating a bunch of random numbers and multiplying and dividing accordingly)...
Also, you probably already know this, but be sure to do the operations in-place (*=, /=, +=, etc) when working with large-ish numpy arrays. It makes a huge difference in memory usage with large arrays, and will give a considerable speed increase, too.
In [53]: def rand_row_doubles(row_limits, num):
....: ncols = len(row_limits)
....: x = np.random.random((num, ncols))
....: x *= row_limits
....: return x
....:
In [59]: %timeit rand_row_doubles(np.arange(7) + 1, 1000000)
10 loops, best of 3: 187 ms per loop
As compared to:
In [66]: %timeit ManyRandDoubles(np.arange(7) + 1, 1000000)
1 loops, best of 3: 222 ms per loop
It's not a huge difference, but if you're really worried about speed, it's something.
Just to show that it's correct:
In [68]: x.max(0)
Out[68]:
array([ 0.99999991, 1.99999971, 2.99999737, 3.99999569, 4.99999836,
5.99999114, 6.99999738])
In [69]: x.min(0)
Out[69]:
array([ 4.02099599e-07, 4.41729377e-07, 4.33480302e-08,
7.43497138e-06, 1.28446819e-05, 4.27614385e-07,
1.34106753e-05])
Likewise, for your "rows sum to one" part...
In [70]: def rand_rows_sum_to_one(nrows, ncols):
....: x = np.random.random((ncols, nrows))
....: y = x.sum(axis=0)
....: x /= y
....: return x.T
....:
In [71]: %timeit rand_rows_sum_to_one(1000000, 13)
1 loops, best of 3: 455 ms per loop
In [72]: x = rand_rows_sum_to_one(1000000, 13)
In [73]: x.sum(axis=1)
Out[73]: array([ 1., 1., 1., ..., 1., 1., 1.])
Honestly, even if you re-implement things in C, I'm not sure you'll be able to beat numpy by much on this one... I could be very wrong, though!

EDIT Created functions that return the full set of numbers, not just one row at a time.
EDIT 2 Make the functions more pythonic (and faster), add solution for second question
For the first set of numbers, you might consider numpy.random.randint or numpy.random.uniform, which take low and high parameters. Generating an array of 7 x 1,000,000 numbers in a specified range seems to take < 0.7 second on my 2 GHz machine:
def LimitedRandInts(XLim, N):
rowlen = (1,N)
return [np.random.randint(low=0,high=lim,size=rowlen) for lim in XLim]
def LimitedRandDoubles(XLim, N):
rowlen = (1,N)
return [np.random.uniform(low=0,high=lim,size=rowlen) for lim in XLim]
>>> import numpy as np
>>> N = 1000000 #number of randoms in each range
>>> xLim = [x*500 for x in range(1,8)] #convenient limit generation
>>> fLim = [x/7.0 for x in range(1,8)]
>>> aa = LimitedRandInts(xLim, N)
>>> ff = LimitedRandDoubles(fLim, N)
This returns integers in [0,xLim-1] or floats in [0,fLim). The integer version took ~0.3 seconds, the double ~0.66, on my 2 GHz single-core machine.
For the second set, I used #Joe Kingston's suggestion.
def SumToOneRands(NumToSum, N):
aa = np.random.uniform(low=0,high=1.0,size=(NumToSum,N)) #13 rows by 1000000 columns, for instance
s = np.reciprocal(aa.sum(0))
aa *= s
return aa.T #get back to column major order, so aa[k] is the kth set of 13 numbers
>>> ll = SumToOneRands(13, N)
This takes ~1.6 seconds.
In all cases, result[k] gives you the kth set of data.

Try r = 1664525*r + 1013904223
from "an even quicker generator"
in "Numerical Recipes in C" 2nd edition, Press et al., isbn 0521431085, p. 284.
np.random is certainly "more random"; see
Linear congruential generator .
In python, use np.uint32 like this:
python -mtimeit -s '
import numpy as np
r = 1
r = np.array([r], np.uint32)[0] # 316 py -> 16 us np
# python longs can be arbitrarily long, so slow
' '
r = r*1664525 + 1013904223 # NR2 p. 284
'
To generate big blocks at a time:
# initialize --
np.random.seed( ... )
R = np.random.randint( 0, np.iinfo( np.uint32 ).max, size, dtype=np.uint32 )
...
R *= 1664525
R += 1013904223

Making your code run in parallel certainly couldn't hurt. Try adapting it for SMP with Parallel Python

As others have already pointed out, numpy is a very good start, fast and easy to use.
If you need random numbers on a massive scale, consider eas-ecb or rc4. Both can be parallelised, you should reach performance in several GB/s.
achievable numbers posted here

If you have access to multiple cores, the computations can be done in parallel with dask.array:
import dask.array as da
x = da.random.random(size=(rows, cols)).compute()
# .compute is not necessary here, because calculations
# can continue in a lazy form and .compute is used
# on the final result

import random
for i in range(1000000):
print(random.randint(1, 1000000))
Here's a code in Python that you can use to generate one million random numbers, one per line!

Just a quick example of numpy in action:
data = numpy.random.rand(1000000)
No need for loop, you can pass in how many numbers you want to generate.