我需要使每行中具有不同的范围，生成的随机整数快速numpy的阵列。
通过我的作品，但速度很慢，当我增加向量号300000的代码：
import numpy as np
import random
population_size = 4
vectors_number = population_size * 3
add_matrix = []
for i in range(0, int(vectors_number/population_size)):
candidates = list(range(population_size*i, population_size*(i+1)))
random_index = random.sample(candidates, 4)
add_matrix.append(random_index)
winning_matrix = np.row_stack(add_matrix)
print(winning_matrix)
每一行都是选择从可变范围4张的随机数。
输出：
[[ 3 0 1 2]
[ 4 6 7 5]
[11 9 8 10]]
最好将通过创建只使用没有循环numpy的是矩阵
下面是遵循量化方法this trick提取独特的随机样本-
ncols = 4
N = int(vectors_number/population_size)
offset = np.arange(N)[:,None]*population_size
winning_matrix = np.random.rand(N,population_size).argsort(1)[:,:ncols] + offset
我们还可以利用np.argpartition更换的最后一步-
r = np.random.rand(N,population_size)
out = r.argpartition(ncols,axis=1)[:,:ncols] + offset
计时 -
In [63]: import numpy as np
...: import random
...:
...: population_size = 64
...: vectors_number = population_size * 300000
In [64]: %%timeit
...: add_matrix = []
...: for i in range(0, int(vectors_number/population_size)):
...: candidates = list(range(population_size*i, population_size*(i+1)))
...: random_index = random.sample(candidates, 4)
...: add_matrix.append(random_index)
...:
...: winning_matrix = np.row_stack(add_matrix)
1 loop, best of 3: 1.82 s per loop
In [65]: %%timeit
...: ncols = 4
...: N = int(vectors_number/population_size)
...: offset = np.arange(N)[:,None]*population_size
...: out = np.random.rand(N,population_size).argsort(1)[:,:ncols] + offset
1 loop, best of 3: 718 ms per loop
In [66]: %%timeit
...: ncols = 4
...: N = int(vectors_number/population_size)
...: offset = np.arange(N)[:,None]*population_size
...: r = np.random.rand(N,population_size)
...: out = r.argpartition(ncols,axis=1)[:,:ncols] + offset
1 loop, best of 3: 428 ms per loop
在你的情况下，循环可以使用压缩map和列表内涵。
winning_matrix = np.vstack ([random.sample (candidate, d2) for candidate in map (lambda i: list(range(population_size*i, population_size*(i+1))), range(0, int(vectors_number/population_size)))])
输出：
array([[ 0, 1, 3, 2],
[ 5, 6, 4, 7],
[11, 10, 9, 8]])
这可细分为
# This is your loop generating the arrays from where you are sampling
range_list = map (lambda i: list(range(population_size*i, population_size*(i+1))), range(0, int(vectors_number/population_size)))
# This does the generation of the matrix, using exactly following your method
winning_matrix = np.vstack ([random.sample (candidate, d2) for candidate in range_list])
在产生具有不同范围的随机整数的情况下（而不是从一个样品），可以按照下面的方法。
如何像这样
# Generating upper and lower bounds for each row.
pair_ranges = product (list (range (1, 5)), list (range (5, 9)))
d2 = 4
np.vstack ([np.random.random_integers (x, y, [1, d2]) for x, y in pair_ranges])
输出：
array([[2, 5, 2, 5],
[5, 6, 2, 4],
[1, 3, 2, 3],
[4, 2, 4, 4],
[2, 6, 4, 6],
[7, 2, 6, 3],
[4, 5, 3, 5],
[4, 6, 3, 6],
[3, 6, 3, 6]])
该行将具有范围之间的随机整数
array([[1, 5],
[1, 6],
[1, 7],
[2, 5],
[2, 6],
[2, 7],
[3, 5],
[3, 6],
[3, 7]])
因为我们只选择4出64的碰撞将是罕见的，因此，我们可以用更换和正确绘制之后。
import numpy as np
def multiperm(y, x, factor=16, remap=False):
draw = np.random.randint(0, factor*x, (y, x))
idx = np.full((y, factor*x), -1, dtype=np.int8 if factor*x < 128 else int)
yi, xi = np.ogrid[:y, :x]
idx[yi, draw] = xi
yd, xd = np.where(idx[yi, draw] != xi)
while yd.size > 0:
ndraw = np.random.randint(0, factor*x, yd.shape)
draw[yd, xd] = ndraw
good = idx[yd, ndraw] == -1
idx[yd[good], ndraw[good]] = xd[good]
good[good] = idx[yd[good], ndraw[good]] == xd[good]
yd, xd = yd[~good], xd[~good]
if remap:
idx = np.zeros((y, factor*x), dtype=np.int8)
idx[yi, draw] = 1
idx[0, 0] -= 1
return idx.ravel().cumsum().reshape(idx.shape)[yi, draw]
else:
return draw + factor*x*yi
from timeit import timeit
print(timeit("multiperm(300_000, 4)", globals=globals(), number=100)*10, 'ms')
# sanity checks
check = multiperm(300_000, 4)
print(np.all(np.arange(300_000) * 64 <= check.T) and np.all(np.arange(1, 300_001) * 64 > check.T))
print(len(set(check.ravel().tolist())) == check.size)
样品运行：
44.83660604993929 ms
True
True
重温与问题，更好地了解，认识仅需要64人口的第一随机4的结果，我来到了这个答案。 还有一个循环，但它是一个循环过少数所需的列，它基本上交换唯一的第一4（入围）列与氨基酸随机其他柱：
import numpy as np
PLAYERS = 64 # per game
GAMES = 300000
FINALISTS = 4 # we only want to know the first four
# every player in every game has a unique id
matrix = np.arange(PLAYERS * GAMES).reshape((GAMES, PLAYERS))
games = np.arange(GAMES)
swaps = np.random.randint(0, PLAYERS, size=(FINALISTS, GAMES))
for i in range(FINALISTS):
# some trickey stuff to create tuples for indexing
dst = tuple(np.vstack([ games, i * np.ones(GAMES, dtype=np.int) ]))
src = tuple(np.vstack([ games, swaps[i] ]))
# do the a swap for location i
matrix[dst], matrix[src] = matrix[src], matrix[dst]
winning_matrix = matrix[:,:FINALISTS]
print(winning_matrix)
