对于想了解如何为每行用numpyrandom.choice创建2d数组？的读者，本文将提供新的信息，我们将详细介绍创建2行4列的数组arr，并且为您提供关于Numpyrandom.choice仅适用于

对于想了解如何为每行用numpy random.choice创建2d数组？的读者，本文将提供新的信息，我们将详细介绍创建2行4列的数组arr，并且为您提供关于Numpy random.choice 仅适用于小循环，但不适用于更大的循环、numpy.random.choice、numpy.random.rand()/randn()/randint()/normal()/choice()/RandomState()、numpy.random.random & numpy.ndarray.astype & numpy.arange的有价值信息。

• 如何为每行用numpy random.choice创建2d数组？（创建2行4列的数组arr）
• Numpy random.choice 仅适用于小循环，但不适用于更大的循环
• numpy.random.choice
• numpy.random.rand()/randn()/randint()/normal()/choice()/RandomState()
• numpy.random.random & numpy.ndarray.astype & numpy.arange

如何为每行用numpy random.choice创建2d数组？（创建2行4列的数组arr）

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'' solutionnp.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 solutionnp.random.rand(1_000_000, 50).argpartition(6,axis=1)[:,:6]+1Wall time: 1.36 s#@CoryKramer''s solutionnp.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'' solutionpp(1_000_000)Wall time: 557 ms

谢谢你们。

答案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 npimport typesfrom timeit import timeitdef 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 + 1def 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]+1def 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)

Numpy random.choice 仅适用于小循环，但不适用于更大的循环

def simulate_game(team1,rating1,team2,rating2):
    if team2 is ''bye'':
        return team1
    else :
        matchup_arr = [team1,team2]
        pteam1 = 1/(1+10**((rating2-rating1)/250))
        pteam2 = 1- pteam1
        prob_arr = [pteam1,pteam2]
        winner = np.random.choice(matchup_arr,size=1,p=prob_arr)
    return winner

np.random.rand(2) #生成两个[0,1)之间的数

[0.6555729  0.76240372]

np.random.rand(2，2) #生成2行*2列的矩阵

[[0.58360206 0.91619225]
 [0.78203671 0.06754087]]

np.random.randn(2) #生成两个[0,1)之间的浮点数

[-0.03609815  0.25591596]

np.random.randn(2，2) #生成2行*2列的矩阵

[[-0.3021947  -0.37880516]
 [-1.84794341 -1.27301629]]

np.random.randint(2)#1个0~2的整数

1

np.random.randint(1,2)#1个1~2的整数

1

np.random.randint(1,5,size=3) #3个1~5的整数

[3 1 2]

numpy.random.normal(loc=0.0, scale=1, size=(2, 3))

array([[ 2.36637254, -0.28686707, -0.5227531 ],
       [ 0.35879566,  0.70202319, -0.84655717]])

[[-0.15040995 -0.43780718 -0.22292445]
 [-0.89388124 -0.39465164  0.24113838]]

np.random.choice([''a'',''b'',''c'',''d'',''e''])

c

np.random.choice(5, 6)#6个小于5的元素

[2 3 3 3 1 2]

np.random.choice(5, 3, p=[0.1, 0, 0.3, 0.6, 0])#p:每个条目出现的概率。如果没有，假设样本在A中的所有条目都具有均匀分布。

[0 3 2]

np.random.RandomState(0).randint(8)#产生随机整数

4

n1 = numpy.random.RandomState(0).random_sample()
n2 = numpy.random.RandomState(0).random_sample(size=(2,3))

0.548813503927 
[[ 0.5488135   0.71518937  0.60276338]
 [ 0.54488318  0.4236548   0.64589411]]

xb = np.random.random((nb, d)).astype(''float32'') #创建一个二维随机数矩阵（nb行d列）

xb[:, 0] += np.arange(nb) / 1000. #将矩阵第一列的每个数加上一个值