NumPy的：打开散点图成一个二维数组

Question

说我有以下几点：

import pyplot as plt 
import numpy as np 
'''array([[29, 13, 11, 4, 5], #dataMag 
     [19, 16, 25, 9, 10], 
     [16, 22, 14, 18, 26], 
     [ 9, 17, 8, 9, 777]]) 

array([[205, 338, 380, 428, 228], #dataX 
     [199, 546, 430, 95, 374], 
     [418, 85, 260, 236, 241], 
     [308, 481, 133, 136, 83]]) 

array([[ 0.48, 0.83, 0.71, 0.12, 0.], #dataY 
     [ 0.09, 0. , 0.7 , 0.43, 0.54], 
     [ 0.58, 0. , 0.56, 0.18, 0.25], 
     [ 0.96, 0.26, 0.57, 0. , 0.82])''' 

plt.scatter(x=dataX.flat, y=dataY.flat, c=dataMag.flat, vmin=np.min(dataMag), 
      vmax=np.max(dataMag), marker='s', cmap='hot') 
plt.show() 

这给了我下面的结果：

而不是使用三个阵列来获得2D图像，是在Numpy（或Scipy等）中有没有将它们表示为（a，b）2D数组的方法？

Answer 1

编辑我在下面保留我的原始答案，但在相同主题上深入探讨您的上一个问题，请遵循您所遵循的内容。请注意，它不处理重复的值，因此如果将多个值分配给同一位置，则只会保留其中一个值。此外，这会弄乱散点图的规模，所以像我原来的答案可能更适合你的后续工作。但无论如何，下面的代码：

x_, x_idx = np.unique(np.ravel(dataX), return_inverse=True) 
y_, y_idx = np.unique(np.ravel(dataY), return_inverse=True) 
newArray = np.zeros((len(x_), len(y_)), dtype=dataMag.dtype) 
newArray[x_idx, y_idx] = np.ravel(dataMag) 
>>> newArray 
array([[ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 777, 0, 0], 
     [ 22, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], 
     [ 0, 0, 0, 0, 0, 0, 9, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], 
     [ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 8, 0, 0, 0, 0, 0, 0], 
     [ 9, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], 
     [ 0, 19, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], 
     [ 0, 0, 0, 0, 0, 0, 0, 29, 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, 18, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], 
     [ 0, 0, 0, 0, 26, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], 
     [ 0, 0, 0, 0, 0, 0, 0, 0, 0, 14, 0, 0, 0, 0, 0, 0, 0], 
     [ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 9], 
     [ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 13, 0], 
     [ 0, 0, 0, 0, 0, 0, 0, 0, 10, 0, 0, 0, 0, 0, 0, 0, 0], 
     [ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 11, 0, 0, 0], 
     [ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 16, 0, 0, 0, 0, 0], 
     [ 0, 0, 4, 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, 25, 0, 0, 0, 0], 
     [ 0, 0, 0, 0, 0, 17, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], 
     [ 16, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]]) 

原来的答案

如果dataX和dataY其中两个整数数组，实现起来是非常简单的。但是，因为它们似乎不一定是，所以您需要做一些舍入操作，为此您需要首先在每个方向上为阵列选择一个步长，然后可以执行下列操作：

from __future__ import division 

x_step, y_step = 25, 0.10 
x = np.round(dataX/x_step).astype(int) 
y = np.round(dataY/y_step).astype(int) 
x_m, x_M = np.min(x), np.max(x) 
y_m, y_M = np.min(y), np.max(y) 
newArray = np.zeros((x_M - x_m + 1, y_M - y_m + 1), dtype=dataMag.dtype) 
newArray[x - x_m, y - y_m] = dataMag 

>>> newArray 
array([[ 22, 0, 0, 0, 0, 0, 0, 0, 777, 0, 0], 
     [ 0, 0, 0, 0, 9, 0, 0, 0, 0, 0, 0], 
     [ 9, 0, 0, 0, 0, 0, 8, 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, 19, 0, 0, 0, 29, 0, 0, 0, 0, 0], 
     [ 5, 0, 18, 0, 0, 0, 0, 0, 0, 0, 0], 
     [ 0, 0, 26, 0, 0, 0, 14, 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, 9], 
     [ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], 
     [ 0, 0, 0, 0, 0, 0, 0, 0, 13, 0, 0], 
     [ 0, 0, 0, 0, 0, 10, 0, 11, 0, 0, 0], 
     [ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], 
     [ 0, 4, 0, 0, 0, 0, 16, 25, 0, 0, 0], 
     [ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], 
     [ 0, 0, 0, 17, 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], 
     [ 16, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]]) 

当你这样做时，你必须小心谨慎，确保你的舍入步骤足够小，以避免数组中的相同位置没有两个值，因为那样会丢失信息。例如：

x_step, y_step = 50, 0.10 
... 
>>> newArray 
array([[ 22, 0, 0, 0, 9, 0, 0, 0, 777, 0, 0], 
     [ 9, 0, 0, 0, 0, 0, 8, 0, 0, 0, 0], 
     [ 0, 19, 0, 0, 0, 29, 0, 0, 0, 0, 0], 
     [ 5, 0, 26, 0, 0, 0, 14, 0, 0, 0, 0], 
     [ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 9], 
     [ 0, 0, 0, 0, 0, 10, 0, 0, 13, 0, 0], 
     [ 0, 0, 0, 0, 0, 0, 16, 11, 0, 0, 0], 
     [ 0, 4, 0, 0, 0, 0, 0, 25, 0, 0, 0], 
     [ 0, 0, 0, 17, 0, 0, 0, 0, 0, 0, 0], 
     [ 16, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]]) 

和[3, 2]位置只有一个图26示出，而不是在前面的例子中的相应的细胞中的18和26向上。