NumPy的：打開散點圖成一個二維數組

Question

說我有以下幾點：

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() 

這給了我下面的結果：

而不是使用三個陣列來獲得2D圖像，是在Numpy（或Scipy等）中有沒有將它們表示爲（a，b）2D數組的方法？

Answer 1

編輯我在下面保留我的原始答案，但在相同主題上深入探討您的上一個問題，請遵循您所遵循的內容。請注意，它不處理重複的值，因此如果將多個值分配給同一位置，則只會保留其中一個值。此外，這會弄亂散點圖的規模，所以像我原來的答案可能更適合你的後續工作。但無論如何，下面的代碼：

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]]) 

原來的答案

如果dataX和dataY其中兩個整數數組，實現起來是非常簡單的。但是，因爲它們似乎不一定是，所以您需要做一些舍入操作，爲此您需要首先在每個方向上爲陣列選擇一個步長，然後可以執行下列操作：

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]]) 

當你這樣做時，你必須小心謹慎，確保你的舍入步驟足夠小，以避免數組中的相同位置沒有兩個值，因爲那樣會丟失信息。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]位置只有一個圖26示出，而不是在前面的例子中的相應的細胞中的18和26向上。