在Python中用坐标数组填充矩阵

Question

我目前有2个对应于坐标（X，Y）的数组，以及第三个对应于2d空间中此点值的数组。它是这样编码的，而不是矩阵，因为它是一个相当稀疏的矩阵（并非每个点都具有一个值）。现在，我想重建矩阵以便使用matplotlib.imshow（）绘制值。

到目前为止，我最简单的方法是执行for循环，如下所示：

X = [1, 1, 3, 5];
Y = [2, 2, 3, 7];
Z = [0.3, -0.5, 1, 1];
matrix = np.zeros([10,10])

for i in range(len(Z)):
    matrix[X[i],Y[i]] = Z[i]

我的意思是，这并不可怕，但我担心会发生很多事情。是否有一个函数将第一和第二输入分别作为第一和第二坐标，而第三输入作为这些坐标的值？还是会有类似的东西？

Answer 1

对于您正在做的事情，您可以直接使用列表（无循环）。范例-

matrix[X,Y] = Z

演示-

In [3]: X = [1, 1, 3, 5];

In [4]: Y = [2, 2, 3, 7];

In [5]: Z = [0.3, -0.5, 1, 1];

In [6]: matrix = np.zeros([10,10])

In [7]: matrix[X,Y] = Z

In [8]: matrix
Out[8]:
array([[ 0. ,  0. ,  0. ,  0. ,  0. ,  0. ,  0. ,  0. ,  0. ,  0. ],
       [ 0. ,  0. , -0.5,  0. ,  0. ,  0. ,  0. ,  0. ,  0. ,  0. ],
       [ 0. ,  0. ,  0. ,  0. ,  0. ,  0. ,  0. ,  0. ,  0. ,  0. ],
       [ 0. ,  0. ,  0. ,  1. ,  0. ,  0. ,  0. ,  0. ,  0. ,  0. ],
       [ 0. ,  0. ,  0. ,  0. ,  0. ,  0. ,  0. ,  0. ,  0. ,  0. ],
       [ 0. ,  0. ,  0. ,  0. ,  0. ,  0. ,  0. ,  1. ,  0. ,  0. ],
       [ 0. ,  0. ,  0. ,  0. ,  0. ,  0. ,  0. ,  0. ,  0. ,  0. ],
       [ 0. ,  0. ,  0. ,  0. ,  0. ,  0. ,  0. ,  0. ,  0. ,  0. ],
       [ 0. ,  0. ,  0. ,  0. ,  0. ,  0. ,  0. ,  0. ,  0. ,  0. ],
       [ 0. ,  0. ,  0. ,  0. ,  0. ,  0. ,  0. ,  0. ,  0. ,  0. ]])

In [9]: matrix1 = np.zeros([10,10])

In [10]: for i in range(len(Z)):
   ....:     matrix1[X[i],Y[i]] = Z[i]

In [13]: (matrix1 == matrix).all()  #Just to show its equal to OP's `for` loop method.
Out[13]: True

计时测试-

In [24]: X = np.arange(1000)

In [25]: Y = np.arange(1000)

In [26]: Z = np.random.rand(1000)

In [27]: %%timeit
   ....: matrix = np.zeros([1000,1000])
   ....: matrix[X,Y] = Z
   ....:
1000 loops, best of 3: 834 µs per loop

In [28]: %%timeit
   ....: matrix1 = np.zeros([1000,1000])
   ....: for i in range(len(Z)):
   ....:     matrix1[X[i],Y[i]] = Z[i]
   ....:
The slowest run took 6.47 times longer than the fastest. This could mean that an intermediate result is being cached
1000 loops, best of 3: 1.43 ms per loop

当处理大数组（并且Z很大）时，矢量化方法会更快。

如果Z小，则使用for循环方法会更快。