如何为每行用numpy random.choice创建2d数组？

Question

我正在尝试创建一个numpy随机选择的2d数组（由六列和许多行组成），每行的唯一值介于1到50之间，不是数组的全部

np.sort(np.random.choice(np.arange(1,50),size=(100,6),replace=False))

但这会引起错误。

ValueError: Cannot take a larger sample than population when 'replace=False'

有没有可能用一个衬套做到这一点

编辑

好的，我得到了答案。

这些是jupyter％time cellmagic的结果

#@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

我在@Paul Panzer的解决方案上更改了 np.empty和np.random.randint的dtypes ，因为它在我的电脑上不起作用。

3.6.0 |Anaconda custom (64-bit)| (default, Dec 23 2016, 11:57:41) [MSC v.1900 64 bit (AMD64)]

最快的是

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

谢谢你们。

Answer 1

这是一种有建设性的方法，首先绘制（50个选择），然后绘制（49个选择），等等。对于大集合，这是很有竞争力的（表中的pp）：

# 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

代码包括基准测试。算法有点复杂，因为映射到自由点很麻烦：

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)