Question

我目前有2个阵列对应于坐标（X，Y），第三个数组对应于2d空间中此时的值。它被编码为不是矩阵，因为它是一个相当稀疏的矩阵（并非每个点都有一个值）。现在我想重建矩阵，以便用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：

 var productlist = new Bloodhound({
 datumTokenizer: Bloodhound.tokenizers.whitespace,
 queryTokenizer: Bloodhound.tokenizers.whitespace,

local: [
  { "product": "GSPEN", "description": "Pen" },
  { "product": "GSTS", "description": "Tissue" },
  ]
});

$('#custom-templates .typeahead').typeahead(null, {
name: 'productlist',
display: 'product ',
source: local,
templates: {
empty: [
  '<div class="empty-message">',
    'unable to find any product that match the current query',
  '</div>'
].join('\n'),
suggestion: Handlebars.compile('<div><strong>{{product}}</strong> – {{description}}</div>')
}
});

我的意思是，它并不可怕，但我担心大阵列。是否有一个函数将第1和第2个输入分别作为第1和第2个坐标，第3个输入作为这些坐标的值？或者会有类似的东西吗？

Answer 1

对于你正在做的事情，你可以直接使用列表（不带）循环。示例 -

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

时间测试 -

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

当处理大型数组（并且Z很大）时，向量化方法会更快。

如果Z很小，那么使用for循环方法会更快。

用python中的坐标数组填充矩阵

1 个答案: