我想在 python 中創(chuàng)建隨機(jī)變量并使用以下代碼， weights = np.random.random(10)但我想創(chuàng)建隨機(jī)變量，使三分之一的權(quán)重應(yīng)為零。有什么辦法嗎？我也試過(guò)下面的代碼，但這不是我想要的weights = np.random.random(7)
weights.append(0, 0, 0)
3 回答
慕妹3146593
TA貢獻(xiàn)1820條經(jīng)驗(yàn) 獲得超9個(gè)贊
設(shè)置大約 1/3 的重量
這將保證大約三分之一的權(quán)重為 0：
weights = np.random.random(10)/np.random.choice([0,1],10,p=[0.3,0.7])
weights[np.isinf(weights)] = 0
# or
# weights[weights == np.inf] = 0
>>> weights
array([0. , 0.25715864, 0. , 0.80958258, 0.12880619,
0.48781856, 0.52278911, 0.76541417, 0.87736431, 0. ])
它的作用是將您的值的大約 1/3 除以 0，給您inf，然后只需將其替換為inf0
設(shè)置正好 1/3 的權(quán)重
或者，如果您需要它正好是1/3（或者在您的情況下，10 分之 3），您可以將 1/3 的權(quán)重替換為0：
weights = np.random.random(10)
# Replace 3 with however many indices you want changed...
weights[np.random.choice(range(len(weights)),3,replace=False)] = 0
>>> weights
array([0. , 0.36839012, 0. , 0.51468295, 0.45694205,
0.23881473, 0.1223229 , 0.68440171, 0. , 0.15542469])
從權(quán)重中選擇 3 個(gè)隨機(jī)索引并將它們替換為 0
倚天杖
TA貢獻(xiàn)1828條經(jīng)驗(yàn) 獲得超3個(gè)贊
一種簡(jiǎn)單的方法：
>>> import numpy as np
>>>
>>> a = np.clip(np.random.uniform(-0.5, 1, (100,)), 0, np.inf)
>>> a
array([0.39497669, 0.65003362, 0. , 0. , 0. ,
0.75545815, 0.30772786, 0.1805628 , 0. , 0. ,
0. , 0.82527704, 0. , 0.63983682, 0.89283051,
0.25173721, 0.18409163, 0.63631959, 0.59095185, 0. ,
0.85817311, 0. , 0.06769175, 0. , 0.67807471,
0.29805637, 0.03429861, 0.53077809, 0.32317273, 0.52346321,
0.22966515, 0.98175502, 0.54615167, 0. , 0.88853359,
0. , 0.70622272, 0.08106305, 0. , 0.8767082 ,
0.52920044, 0. , 0. , 0.29394736, 0.4097331 ,
0.77977164, 0.62860222, 0. , 0. , 0.14899124,
0.81880283, 0. , 0.1398242 , 0. , 0.50113732,
0. , 0.68872893, 0.15582668, 0. , 0.34789122,
0.18510949, 0.60281713, 0.21097922, 0.77419626, 0.29588479,
0.18890799, 0.9781896 , 0.96220508, 0.52201816, 0.71087763,
0. , 0.43540516, 0.99297503, 0. , 0.69248893,
0.05157044, 0. , 0.75131066, 0. , 0. ,
0.25627591, 0.53367521, 0.58151298, 0.85662171, 0.455367 ,
0. , 0. , 0.21293519, 0.52337335, 0. ,
0.68644488, 0. , 0. , 0.39695189, 0. ,
0.40860821, 0.84549468, 0. , 0.21247807, 0.59054669])
>>> np.count_nonzero(a)
67
它從 [-0.5, 1] 均勻地繪制，然后將低于零的所有內(nèi)容設(shè)置為零。
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