Question

我是python的新手，在这种情况下，我想把我的函数放到2d数组中，所以我可以绘制函数。这是我的三角函数，我用它来模糊逻辑：

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    android:layout_width="wrap_content"
    android:layout_height="wrap_content"
    android:drawableRight="@drawable/your_drawable"/>

我正在尝试使用numpy制作数组， def triangle (z,a,b,c): if (z<=a) | (z>=c): y = 0 elif (a<=z) & (z<=b): y = (z-a) / (b-a) elif (b<=z) & (z<=c): y = (b-z) / (c-b) return y但我无法完成，我试图使用模糊库，但没有任何作用。

Answer 1

a, b, c和z之间看起来np.linspace是常数而a是c。

您可以使用Boolean Indexing，SciPy cookbook/Indexing

a = 1
b = 2
c = 3

def triangle (z, a = a, b = b, c = c):
    y = np.zeros(z.shape)
    y[z <= a] = 0
    y[z >= c] = 0
    first_half = np.logical_and(a < z, z <= b)
    y[first_half] = (z[first_half]-a) / (b-a)
    second_half = np.logical_and(b < z, z < c)
    y[second_half] = (c-z[second_half]) / (c-b)
    return y

z = np.linspace(a, c, num = 51)
y = triangle(z, a, b, c)

q = np.vstack((z, y)) # shape = (2, 50) ... [[z, z, z, ...], [y, y, y, ...]]
q =  q.T # shape = (50, 2) ... [[z, y], [z, y], ....]

当你在比较表达式中使用numpy ndarray时，结果是一个布尔数组：

>>> q = np.linspace(0, 20, num = 50)
>>> print(q)
[  0.           0.40816327   0.81632653   1.2244898    1.63265306
   2.04081633   2.44897959   2.85714286   3.26530612   3.67346939
   4.08163265   4.48979592   4.89795918   5.30612245   5.71428571
   6.12244898   6.53061224   6.93877551   7.34693878   7.75510204
   8.16326531   8.57142857   8.97959184   9.3877551    9.79591837
  10.20408163  10.6122449   11.02040816  11.42857143  11.83673469
  12.24489796  12.65306122  13.06122449  13.46938776  13.87755102
  14.28571429  14.69387755  15.10204082  15.51020408  15.91836735
  16.32653061  16.73469388  17.14285714  17.55102041  17.95918367
  18.36734694  18.7755102   19.18367347  19.59183673  20.        ]
>>> print(q < 5)
[ True  True  True  True  True  True  True  True  True  True  True  True
  True False False False False False False False False False False False
 False False False False False False False False False False False False
 False False False False False False False False False False False False
 False False]
>>> print(q > 15)
[False False False False False False False False False False False False
 False False False False False False False False False False False False
 False False False False False False False False False False False False
 False  True  True  True  True  True  True  True  True  True  True  True
  True  True]
>>> print(np.logical_and(q > 5, q < 15))
[False False False False False False False False False False False False
 False  True  True  True  True  True  True  True  True  True  True  True
  True  True  True  True  True  True  True  True  True  True  True  True
  True False False False False False False False False False False False
 False False]
>>>

您可以使用布尔数组选择符合条件的数组部分：

>>> q[np.logical_and(q > 7, q < 11)]
array([  7.34693878,   7.75510204,   8.16326531,   8.57142857,
         8.97959184,   9.3877551 ,   9.79591837,  10.20408163,  10.6122449 ])
>>>

在赋值语句中使用布尔索引时，右侧仅分配给比较为True的索引：

>>> q[np.logical_and(q > 7, q < 11)] = -1
>>> print(q)
[  0.           0.40816327   0.81632653   1.2244898    1.63265306
   2.04081633   2.44897959   2.85714286   3.26530612   3.67346939
   4.08163265   4.48979592   4.89795918   5.30612245   5.71428571
   6.12244898   6.53061224   6.93877551  -1.          -1.          -1.          -1.
  -1.          -1.          -1.          -1.          -1.          11.02040816
  11.42857143  11.83673469  12.24489796  12.65306122  13.06122449
  13.46938776  13.87755102  14.28571429  14.69387755  15.10204082
  15.51020408  15.91836735  16.32653061  16.73469388  17.14285714
  17.55102041  17.95918367  18.36734694  18.7755102   19.18367347
  19.59183673  20.        ]
>>>

python，将三角函数绘制成2d数组

1 个答案: