Question

我想生成1000种气体混合物，其中三种成分具有明确的范围。每个混合的总和应该是100.我无法弄清楚如何使用for循环关联组件。

import numpy as np

comp_list = []

c1 = np.arange(80,100, 0.001)

c2 = np.arange(0,14,0.001)

c3 = np.arange(0,4, 0.001)

for i in range(10000):
    comp_sum = c1[i] + c2[i] +c3[i]
    if comp_sum == 100:
        comp_list.append(c1[i], c2[i],c3[i])

print comp_list

Answer 1

正如@AndrasDeak在评论中建议的那样，您只需执行c3 = 100 - c1 - c2即可确保某些结果样本可以满足您的约束条件，然后从结果中获取最多1000个样本，方法：

# I created more than 1000 samples so that I have enough to slice with
In [35]: c1 = np.random.uniform(80, 100, 10000)

In [36]: c2 = np.random.uniform(0, 14, 10000)

In [37]: c3 = 100 - c1 - c2

In [38]: c3
Out[38]: 
array([ 12.68861952,   4.34446942,  -9.74132792, ...,   3.65083356,
        -0.71305583,   9.78624485])

In [39]: masked = np.where((c3 >= 0) & (c3 <= 4))

# only take up to 1000 samples
In [40]: c1 = c1[masked][:1000]

In [41]: c2 = c2[masked][:1000]

In [42]: c3 = c3[masked][:1000]

# sum of the arrays show 100 in all
In [43]: c1 + c2 + c3
Out[43]: 
array([ 100.,  100.,  100.,  100.,  100.,  100.,  100.,  100.,  100.,
        100.,  100.,  100.,  100.,  100.,  100.,  100.,  100.,  100.,
        100.,  100.,  100.,  100.,  100.,  100.,  100.,  100.,  100.,
        100.,  100.,  100.,  100.,  100.,  100.,  100.,  100.,  100.,
        100.,  100.,  100.,  100.,  100.,  100.,  100.,  100.,  100.,
        100.,  100.,  100.,  100.,  100.,  100.,  100.,  100.,  100.,
        100.,  100.,  100.,  100.,  100.,  100.,  100.,  100.,  100.,
        100.,  100.,  100.,  100.,  100.,  100.,  100.,  100.,  100.,
        100.,  100.,  100.,  100.,  100.,  100.,  100.,  100.,  100.,
        100.,  100.,  100.,  100.,  100.,  100.,  100.,  100.,  100.,
        100.,  100.,  100.,  100.,  100.,  100.,  100.,  100.,  100.,

这肯定不是最有效的方法，但对于简单的用例，它可以达到你想要的效果。

Answer 2

您可以使用列表推导使用直接方法：

import numpy as np

c1=np.linspace(80,100,100)

carr=np.array([[c1[i],cc2,100-c1[i]-cc2] for i in range(len(c1)) for cc2 in np.arange(0,min(14,100-c1[i]),1)])

您可以使用

恢复浓度矢量

c1=carr[:,0]
c2=carr[:,1]
c3=carr[:,2]

证明：

In [496]: carr.shape
Out[496]: (944, 3)

In [497]: carr.sum(1)

Out[497]: array([ 100.,  100.,  100.,  100.,  100.,  100.,  100.,  100.,  100.,
        100.,  100.,  100.,  100.,  100.,  100.,  100.,  100.,  100.,
        100.,  100.,  100.,  100.,  100.,  100.,  100.,  100.,  100.,
        100.,  100.,  100.,  100.,  100.,  100.,  100.,  100.,  100.,
        ...

您必须考虑一下您选择的参数以获得大约1000个样本，因为对于每个c1，您将拥有不同数量的c2。然而，这将产生大致均匀的浓度集，你可能只需要小心c=100个案例（我不是）。

使用numpy模拟含有三种成分的1000种气体混合物

2 个答案: