在Python中访问数组元素

Question

这是我第一次处理多维数组，我在访问元素时遇到了问题。我想获取一张图片中的红色像素，但只想要数组中的前8个元素。以下是我的代码：

import Image
import numpy as np

im = Image.open("C:\Users\Jones\Pictures\1.jpg")
pix = im.load() 

r, g, b = np.array(im).T
print r[0:8]

Answer 1

我觉得这个可以更简单一点。这个例子使用了一个随机矩阵（这就是你的 r 矩阵）：

In [7]: from pylab import *                 # convention

In [8]: r = randint(0,10,(10,10))           # this is your image

In [9]: r
array([[7, 9, 5, 5, 6, 8, 1, 4, 3, 4],
       [5, 4, 4, 4, 2, 6, 2, 6, 4, 2],
       [1, 4, 9, 9, 2, 6, 1, 9, 0, 6],
       [5, 9, 0, 7, 9, 9, 5, 2, 0, 7],
       [8, 3, 3, 9, 0, 0, 5, 9, 2, 2],
       [5, 3, 7, 8, 8, 1, 6, 3, 2, 0],
       [0, 2, 5, 7, 0, 1, 0, 2, 1, 2],
       [4, 0, 4, 5, 9, 9, 3, 8, 3, 7],
       [4, 6, 9, 9, 5, 9, 3, 0, 5, 1],
       [6, 9, 9, 0, 3, 4, 9, 7, 9, 6]])

接下来，提取前8列并进行一些操作

In [17]: r_8 = r[:,:8]              # extract columns

In [18]: r_8
Out[18]: 
array([[7, 9, 5, 5, 6, 8, 1, 4],
       [5, 4, 4, 4, 2, 6, 2, 6],
       [1, 4, 9, 9, 2, 6, 1, 9],
       [5, 9, 0, 7, 9, 9, 5, 2],
       [8, 3, 3, 9, 0, 0, 5, 9],
       [5, 3, 7, 8, 8, 1, 6, 3],
       [0, 2, 5, 7, 0, 1, 0, 2],
       [4, 0, 4, 5, 9, 9, 3, 8],
       [4, 6, 9, 9, 5, 9, 3, 0],
       [6, 9, 9, 0, 3, 4, 9, 7]])

In [19]: r_8 = r_8 * 2              # do something

In [20]: r_8
Out[20]: 
array([[14, 18, 10, 10, 12, 16,  2,  8],
       [10,  8,  8,  8,  4, 12,  4, 12],
       [ 2,  8, 18, 18,  4, 12,  2, 18],
       [10, 18,  0, 14, 18, 18, 10,  4],
       [16,  6,  6, 18,  0,  0, 10, 18],
       [10,  6, 14, 16, 16,  2, 12,  6],
       [ 0,  4, 10, 14,  0,  2,  0,  4],
       [ 8,  0,  8, 10, 18, 18,  6, 16],
       [ 8, 12, 18, 18, 10, 18,  6,  0],
       [12, 18, 18,  0,  6,  8, 18, 14]])

现在，这里有个小窍门。用 hstack 替换 r 中的前8列：

In [21]: r = hstack((r_8, r[:,8:]))             # it replaces the FISRT 8 columns, note the indexing notation 

In [22]: r
Out[22]: 
array([[14, 18, 10, 10, 12, 16,  2,  8,  3,  4],    # it does not touch the last 2 columns
       [10,  8,  8,  8,  4, 12,  4, 12,  4,  2],
       [ 2,  8, 18, 18,  4, 12,  2, 18,  0,  6],
       [10, 18,  0, 14, 18, 18, 10,  4,  0,  7],
       [16,  6,  6, 18,  0,  0, 10, 18,  2,  2],
       [10,  6, 14, 16, 16,  2, 12,  6,  2,  0],
       [ 0,  4, 10, 14,  0,  2,  0,  4,  1,  2],
       [ 8,  0,  8, 10, 18, 18,  6, 16,  3,  7],
       [ 8, 12, 18, 18, 10, 18,  6,  0,  5,  1],
       [12, 18, 18,  0,  6,  8, 18, 14,  9,  6]])

Answer 2

你可以这样做：

r[0][:8]

不过要注意，如果第一行的高度少于8个像素，这样做是行不通的。要解决这个问题，可以这样：

from itertools import chain
r = list(chain.from_iterable(r))
r[:8]

或者（如果你不想引入整个模块）：

r = [val for element in r for val in element]
r[:8]

Answer 3

因为你在处理图像，所以r是一个二维数组。要获取图像中的前8个像素，可以试试下面的代码：

r.flatten()[:8]

如果第一行的像素少于8个，这段代码会自动处理，不会出错。