How to create 2d array with numpy random.choice for every rows?
I'm trying to create a 2d array (which is a six column and lots of rows) with numpy random choice with unique values between 1 and 50 for every row not all of the array
np.sort(np.random.choice(np.arange(1,50),size=(100,6),replace=False))
But this raises an error.
ValueError: Cannot take a larger sample than population when 'replace=False'
Is it possible to make this with an one liner without a loop
Edit
Okey i get the answer.
These are the results with jupyter %time cellmagic
#@James' solution
np.stack([np.random.choice(np.arange(1,50),size=6,replace=False) for i in range(1_000_000)])
Wall time: 25.1 s
#@Divakar's solution
np.random.rand(1_000_000, 50).argpartition(6,axis=1)[:,:6]+1
Wall time: 1.36 s
#@CoryKramer's solution
np.array([np.random.choice(np.arange(1, 50), size=6, replace=False) for _ in range(1_000_000)])
Wall time: 25.5 s
I changed dtypes of np.empty and np.random.randint on @Paul Panzer's solution because it was not working on my pc.
3.6.0 |Anaconda custom (64-bit)| (default, Dec 23 2016, 11:57:41) [MSC v.1900 64 bit (AMD64)]
Fastest one is
def pp(n):
draw = np.empty((n, 6), dtype=np.int64)
# generating random numbers is expensive, so draw a large one and
# make six out of one
draw[:, 0] = np.random.randint(0, 50*49*48*47*46*45, (n,),dtype=np.uint64)
draw[:, 1:] = np.arange(50, 45, -1)
draw = np.floor_divide.accumulate(draw, axis=-1)
draw[:, :-1] -= draw[:, 1:] * np.arange(50, 45, -1)
# map the shorter ranges (:49, :48, :47) to the non-occupied
# positions; this amounts to incrementing for each number on the
# left that is not larger. the nasty bit: if due to incrementing
# new numbers on the left are "overtaken" then for them we also
# need to increment.
for i in range(1, 6):
coll = np.sum(draw[:, :i] <= draw[:, i, None], axis=-1)
collidx = np.flatnonzero(coll)
if collidx.size == 0:
continue
coll = coll[collidx]
tot = coll
while True:
draw[collidx, i] += coll
coll = np.sum(draw[collidx, :i] <= draw[collidx, i, None], axis=-1)
relidx = np.flatnonzero(coll > tot)
if relidx.size == 0:
break
coll, tot = coll[relidx]-tot[relidx], coll[relidx]
collidx = collidx[relidx]
return draw + 1
#@Paul Panzer' solution
pp(1_000_000)
Wall time: 557 ms
Thank you all.
Solution 1:
Here's a vectorized approach with rand+argsort/argpartition trick from here -
np.random.rand(rows, 50).argpartition(6,axis=1)[:,:6]+1
Sample run -
In [41]: rows = 10
In [42]: np.random.rand(rows, 50).argpartition(6,axis=1)[:,:6]+1
Out[42]:
array([[ 1, 9, 3, 26, 14, 44],
[32, 20, 27, 13, 25, 45],
[40, 12, 47, 16, 10, 29],
[ 6, 36, 32, 16, 18, 4],
[42, 46, 24, 9, 1, 31],
[15, 25, 47, 42, 34, 24],
[ 7, 16, 49, 31, 40, 20],
[28, 17, 47, 36, 8, 44],
[ 7, 42, 14, 4, 17, 35],
[39, 19, 37, 7, 8, 36]])
Just to prove the random-ness -
In [56]: rows = 1000000
In [57]: out = np.random.rand(rows, 50).argpartition(6,axis=1)[:,:6]+1
In [58]: np.bincount(out.ravel())[1:]
Out[58]:
array([120048, 120026, 119942, 119838, 119885, 119669, 119965, 119491,
120280, 120108, 120293, 119399, 119917, 119974, 120195, 119796,
119887, 119505, 120235, 119857, 119499, 120560, 119891, 119693,
120081, 120369, 120011, 119714, 120218, 120581, 120111, 119867,
119791, 120265, 120457, 120048, 119813, 119702, 120266, 120445,
120016, 120190, 119576, 119737, 120153, 120215, 120144, 120196,
120218, 119863])
Timings on one million rows of data -
In [43]: rows = 1000000
In [44]: %timeit np.random.rand(rows, 50).argpartition(6,axis=1)[:,:6]+1
1 loop, best of 3: 1.07 s per loop
Solution 2:
This isn't pure numpy but you could wrap your solution within a list comprehension
>>> rows = 10
>>> cols = 6
>>> np.array([np.random.choice(np.arange(1, 50), size=cols, replace=False) for _ in range(rows)])
array([[ 9, 10, 21, 33, 34, 15],
[48, 46, 36, 7, 37, 45],
[21, 15, 5, 9, 31, 26],
[48, 24, 30, 18, 47, 23],
[22, 31, 19, 32, 3, 33],
[35, 44, 15, 46, 20, 43],
[11, 37, 44, 6, 16, 35],
[42, 49, 41, 28, 12, 19],
[19, 6, 32, 3, 1, 22],
[29, 33, 42, 5, 30, 43]])
Solution 3:
You can create each row by itself and then stack them.
np.stack([np.random.choice(np.arange(1,50),size=6,replace=False) for i in range(100)])
Solution 4:
Here is a constructive approach, draw first (50 choices), second (49 choices) etc. For large sets it's quite competitive (pp in table):
# n = 10
# pp 0.18564210 ms
# Divakar 0.01960790 ms
# James 0.20074140 ms
# CK 0.17823420 ms
# n = 1000
# pp 0.80046050 ms
# Divakar 1.31817130 ms
# James 18.93511460 ms
# CK 20.83670820 ms
# n = 1000000
# pp 655.32905590 ms
# Divakar 1352.44713990 ms
# James 18471.08987370 ms
# CK 18369.79808050 ms
# pp checking plausibility...
# var (exp obs) 208.333333333 208.363840259
# mean (exp obs) 25.5 25.5064865
# Divakar checking plausibility...
# var (exp obs) 208.333333333 208.21113972
# mean (exp obs) 25.5 25.499471
# James checking plausibility...
# var (exp obs) 208.333333333 208.313436938
# mean (exp obs) 25.5 25.4979035
# CK checking plausibility...
# var (exp obs) 208.333333333 208.169585249
# mean (exp obs) 25.5 25.49
Code including benchmarking. Algo is a bit complicated because mapping to free spots is hairy:
import numpy as np
import types
from timeit import timeit
def f_pp(n):
draw = np.empty((n, 6), dtype=int)
# generating random numbers is expensive, so draw a large one and
# make six out of one
draw[:, 0] = np.random.randint(0, 50*49*48*47*46*45, (n,))
draw[:, 1:] = np.arange(50, 45, -1)
draw = np.floor_divide.accumulate(draw, axis=-1)
draw[:, :-1] -= draw[:, 1:] * np.arange(50, 45, -1)
# map the shorter ranges (:49, :48, :47) to the non-occupied
# positions; this amounts to incrementing for each number on the
# left that is not larger. the nasty bit: if due to incrementing
# new numbers on the left are "overtaken" then for them we also
# need to increment.
for i in range(1, 6):
coll = np.sum(draw[:, :i] <= draw[:, i, None], axis=-1)
collidx = np.flatnonzero(coll)
if collidx.size == 0:
continue
coll = coll[collidx]
tot = coll
while True:
draw[collidx, i] += coll
coll = np.sum(draw[collidx, :i] <= draw[collidx, i, None], axis=-1)
relidx = np.flatnonzero(coll > tot)
if relidx.size == 0:
break
coll, tot = coll[relidx]-tot[relidx], coll[relidx]
collidx = collidx[relidx]
return draw + 1
def check_result(draw, name):
print(name[2:], ' checking plausibility...')
import scipy.stats
assert all(len(set(row)) == 6 for row in draw)
assert len(set(draw.ravel())) == 50
print(' var (exp obs)', scipy.stats.uniform(0.5, 50).var(), draw.var())
print(' mean (exp obs)', scipy.stats.uniform(0.5, 50).mean(), draw.mean())
def f_Divakar(n):
return np.random.rand(n, 50).argpartition(6,axis=1)[:,:6]+1
def f_James(n):
return np.stack([np.random.choice(np.arange(1,51),size=6,replace=False) for i in range(n)])
def f_CK(n):
return np.array([np.random.choice(np.arange(1, 51), size=6, replace=False) for _ in range(n)])
for n in (10, 1_000, 1_000_000):
print(f'n = {n}')
for name, func in list(globals().items()):
if not name.startswith('f_') or not isinstance(func, types.FunctionType):
continue
try:
print("{:16s}{:16.8f} ms".format(name[2:], timeit(
'f(n)', globals={'f':func, 'n':n}, number=10)*100))
except:
print("{:16s} apparently failed".format(name[2:]))
if(n >= 10000):
for name, func in list(globals().items()):
if name.startswith('f_') and isinstance(func, types.FunctionType):
check_result(func(n), name)