blocks|key|1549090|text|如果在概念上将数组视为像素矩阵，则可以使用Bresenhams直线绘制算法在这两个点之间画一条线。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1549091|entityMap|0|LINK|mutability|MUTABLE|url|https://en.wikipedia.org/wiki/Bresenham's_line_algorithm^0|L|G|0|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@$A|M|B|N|1|O]]|C|$]]|$1|D|3|-4|5|6|7|P|8|@]|9|@]|C|$]]]|E|$F|$5|G|H|I|C|$J|K]]]]

If you treat the array conceptually as a matrix of pixels, then you can use <a href="https://en.wikipedia.org/wiki/Bresenham&#39;s_line_algorithm" rel="nofollow noreferrer">Bresenhams line drawing algorithm</a> to draw a line between these two points.

blocks|key|437386|text|这与选择屏幕上的两个像素并决定在两个像素之间亮起以直线连接的问题是一样的。查看直线绘制算法和其他类似的教程，了解这一问题是如何解决的。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|437387|entityMap|0|LINK|mutability|MUTABLE|url|https://en.wikipedia.org/wiki/Line_drawing_algorithm^0|13|6|0|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@$A|M|B|N|1|O]]|C|$]]|$1|D|3|-4|5|6|7|P|8|@]|9|@]|C|$]]]|E|$F|$5|G|H|I|C|$J|K]]]]

This is the same problem as choosing two pixels on a screen and deciding which pixels in between should be lit up to connect them in a straight line. Check out <a href="https://en.wikipedia.org/wiki/Line_drawing_algorithm" rel="nofollow noreferrer">line drawing algorithms</a> and other similar tutorials for how this has already been solved.

blocks|key|1674364|text|考虑DDA线生成算法。它用于绘制两点之间的直线上的点。在这里，您必须填充数组中的位置，而不是画布。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1674365|以下是我对此的看法：|1674366|def+printarr():
++++for+row+in+board:
++++++++print+'+'.join([str(x)+for+x+in+row])+

def+makeline():
++++dx+=+point2[0]+-+point1[0]
++++dy+=+point2[1]+-+point1[1]
++++steps+=+dx+if+(dx>dy)+else+dy
++++xinc=+dx/float(steps)
++++yinc+=+dy/float(steps)

++++x+=+point1[0]
++++y+=+point1[1]
++++for+i+in+range(steps%2B1):
++++++++if+(board[abs(x),abs(y)]+==+0):
++++++++++++board[abs(x),abs(y)]=1
++++++++x%2B=xinc
++++++++y%2B=yinc

point1+=+[2,2]
point2+=+[10,12]+

board[point1[0],point1[1]]+=+2
board[point2[0],point2[1]]+=+2

makeline()
printarr()|code-block|syntax|javascript|1674367|解释：|1674368|找到两个轴(dx和dy)要覆盖的距离。|ordered-list-item|1674369|计算要在任意轴中覆盖的单元格的步骤->最大值no的数目。|1674370|找出你需要增加多少|1674371|循环表示步骤数，并绘制点和增量。您需要将其转换为绝对值，因为列表指针需要是整数。|1674372|0+0+0+1+0+0+00+0+0|blockquote|1674373|entityMap^0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|Z|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|10|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|11|8|@]|9|@]|A|$]]|$1|K|3|L|5|M|7|12|8|@]|9|@]|A|$]]|$1|N|3|O|5|M|7|13|8|@]|9|@]|A|$]]|$1|P|3|Q|5|M|7|14|8|@]|9|@]|A|$]]|$1|R|3|S|5|M|7|15|8|@]|9|@]|A|$]]|$1|T|3|U|5|V|7|16|8|@]|9|@]|A|$]]|$1|W|3|-4|5|6|7|17|8|@]|9|@]|A|$]]]|X|$]]

Consider DDA Line Generation Algorithm. It is used to plot points in a line between two points. Here, instead of a drawing canvas, you have to fill places in an array.

Here's my take to it:

<pre><code>def printarr():
 for row in board:
 print ' '.join([str(x) for x in row]) 

def makeline():
 dx = point2[0] - point1[0]
 dy = point2[1] - point1[1]
 steps = dx if (dx&gt;dy) else dy
 xinc= dx/float(steps)
 yinc = dy/float(steps)

 x = point1[0]
 y = point1[1]
 for i in range(steps+1):
 if (board[abs(x),abs(y)] == 0):
 board[abs(x),abs(y)]=1
 x+=xinc
 y+=yinc

point1 = [2,2]
point2 = [10,12] 

board[point1[0],point1[1]] = 2
board[point2[0],point2[1]] = 2

makeline()
printarr()
</code></pre>

Explanation:

<ol>
<li>Find the distance to be covered in both the axes (dx and dy)</li>
<li>Compute the number of steps -> max no of cells to be covered in any axes</li>
<li>Find how much you need to increment</li>
<li>Loop for the number of steps and plot the point and increment. 
You'll need to convert it to absolute value as list pointers need to be integers.</li>
</ol>

<blockquote>
<pre><code>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 2 1 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 1 0 0 0 0 0 0 0
0 0 0 0 0 0 1 1 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 1 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 1 0
0 0 0 0 0 0 0 0 0 0 0 0 2
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
</code></pre>
</blockquote>

blocks|key|437476|text|这是一个使用NumPy的简单实现，只需计算行方程：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|437477|import+numpy+as+np

def+draw_line(mat,+x0,+y0,+x1,+y1,+inplace=False):
++++if+not+(0+<=+x0+<+mat.shape[0]+and+0+<=+x1+<+mat.shape[0]+and
++++++++++++0+<=+y0+<+mat.shape[1]+and+0+<=+y1+<+mat.shape[1]):
++++++++raise+ValueError('Invalid+coordinates.')
++++if+not+inplace:
++++++++mat+=+mat.copy()
++++if+(x0,+y0)+==+(x1,+y1):
++++++++mat[x0,+y0]+=+2
++++++++return+mat+if+not+inplace+else+None
++++#+Swap+axes+if+Y+slope+is+smaller+than+X+slope
++++transpose+=+abs(x1+-+x0)+<+abs(y1+-+y0)
++++if+transpose:
++++++++mat+=+mat.T
++++++++x0,+y0,+x1,+y1+=+y0,+x0,+y1,+x1
++++#+Swap+line+direction+to+go+left-to-right+if+necessary
++++if+x0+>+x1:
++++++++x0,+y0,+x1,+y1+=+x1,+y1,+x0,+y0
++++#+Write+line+ends
++++mat[x0,+y0]+=+2
++++mat[x1,+y1]+=+2
++++#+Compute+intermediate+coordinates+using+line+equation
++++x+=+np.arange(x0+%2B+1,+x1)
++++y+=+np.round(((y1+-+y0)+/+(x1+-+x0))+*+(x+-+x0)+%2B+y0).astype(x.dtype)
++++#+Write+intermediate+coordinates
++++mat[x,+y]+=+1
++++if+not+inplace:
++++++++return+mat+if+not+transpose+else+mat.T|code-block|syntax|javascript|437478|一些测试：|437479|print(draw_line(np.zeros((5,+5)),+0,+3,+4,+2))
#[[0.+0.+0.+2.+0.]
#+[0.+0.+0.+1.+0.]
#+[0.+0.+1.+0.+0.]
#+[0.+0.+1.+0.+0.]
#+[0.+0.+2.+0.+0.]]
print(draw_line(np.zeros((5,+5)),+4,+2,+0,+3))
#[[0.+0.+0.+2.+0.]
#+[0.+0.+0.+1.+0.]
#+[0.+0.+1.+0.+0.]
#+[0.+0.+1.+0.+0.]
#+[0.+0.+2.+0.+0.]]
print(draw_line(np.zeros((5,+5)),+1,+0,+3,+4))
#[[0.+0.+0.+0.+0.]
#+[2.+0.+0.+0.+0.]
#+[0.+1.+1.+1.+0.]
#+[0.+0.+0.+0.+2.]
#+[0.+0.+0.+0.+0.]]|437480|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

This is one simple implementation with NumPy, just computing the line equation:

<pre><code>import numpy as np

def draw_line(mat, x0, y0, x1, y1, inplace=False):
 if not (0 &lt;= x0 &lt; mat.shape[0] and 0 &lt;= x1 &lt; mat.shape[0] and
 0 &lt;= y0 &lt; mat.shape[1] and 0 &lt;= y1 &lt; mat.shape[1]):
 raise ValueError('Invalid coordinates.')
 if not inplace:
 mat = mat.copy()
 if (x0, y0) == (x1, y1):
 mat[x0, y0] = 2
 return mat if not inplace else None
 # Swap axes if Y slope is smaller than X slope
 transpose = abs(x1 - x0) &lt; abs(y1 - y0)
 if transpose:
 mat = mat.T
 x0, y0, x1, y1 = y0, x0, y1, x1
 # Swap line direction to go left-to-right if necessary
 if x0 &gt; x1:
 x0, y0, x1, y1 = x1, y1, x0, y0
 # Write line ends
 mat[x0, y0] = 2
 mat[x1, y1] = 2
 # Compute intermediate coordinates using line equation
 x = np.arange(x0 + 1, x1)
 y = np.round(((y1 - y0) / (x1 - x0)) * (x - x0) + y0).astype(x.dtype)
 # Write intermediate coordinates
 mat[x, y] = 1
 if not inplace:
 return mat if not transpose else mat.T
</code></pre>

Some tests:

<pre><code>print(draw_line(np.zeros((5, 5)), 0, 3, 4, 2))
#[[0. 0. 0. 2. 0.]
# [0. 0. 0. 1. 0.]
# [0. 0. 1. 0. 0.]
# [0. 0. 1. 0. 0.]
# [0. 0. 2. 0. 0.]]
print(draw_line(np.zeros((5, 5)), 4, 2, 0, 3))
#[[0. 0. 0. 2. 0.]
# [0. 0. 0. 1. 0.]
# [0. 0. 1. 0. 0.]
# [0. 0. 1. 0. 0.]
# [0. 0. 2. 0. 0.]]
print(draw_line(np.zeros((5, 5)), 1, 0, 3, 4))
#[[0. 0. 0. 0. 0.]
# [2. 0. 0. 0. 0.]
# [0. 1. 1. 1. 0.]
# [0. 0. 0. 0. 2.]
# [0. 0. 0. 0. 0.]]
</code></pre>

blocks|key|437498|text|如果您想要覆盖所有被这条线所触及的方块，您可以实现步骤大小行方法。我编写了一个函数，将矩阵作为mat输入，2点p1(x0，y0)和p2(x1，y1)必须在矩阵中并具有整数值(在我的示例中没有检查这些条件)。最后一个值是您希望矩阵取的值。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|437499|++++def+draw_line(mat,+x0,+y0,+x1,+y1,+value):
++++++++step_number++=+int(abs(x0-x1)+%2B+abs(y0-y1))+#Number+of+steps
++++++++step_size+=+1.0/step_number+#Increment+size
++++++++p+=+[]+#Point+array+(you+can+return+this+and+not+modify+the+matrix+in+the+last+2+lines)
++++++++t+=+0.0+#Step+current+increment
++++++++for+i+in+range(step_number):
++++++++++++p.append([x0+*+t+%2B+x1+*+(1+-+t),+y0+*+t+%2B+y1+*+(1+-+t)])
++++++++++++t%2B=step_size
++++++++for+i+in+range(len(p)):
++++++++++++mat[int(round(p[i][0])),+int(round(p[i][1]))]+=+block_value|code-block|syntax|javascript|437500|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

If you want to cover all the squares that are touched by the line, you can implement the step size line method. I wrote a function that takes as input the matrix as mat, 2 points p1(x0,y0) and p2(x1,y1) which MUST be inside the matrix and have integer value(I did not check for these conditions in my example). The last value is the value you want your matrix to take.
<pre><code> def draw_line(mat, x0, y0, x1, y1, value):
 step_number = int(abs(x0-x1) + abs(y0-y1)) #Number of steps
 step_size = 1.0/step_number #Increment size
 p = [] #Point array (you can return this and not modify the matrix in the last 2 lines)
 t = 0.0 #Step current increment
 for i in range(step_number):
 p.append([x0 * t + x1 * (1 - t), y0 * t + y1 * (1 - t)])
 t+=step_size
 for i in range(len(p)):
 mat[int(round(p[i][0])), int(round(p[i][1]))] = block_value
</code></pre>

I have a matrix of size <code>N x N</code> defined as <code>M = np.zeros((N,N))</code> and two coordinates <code>[x0,y0]</code> and <code>[x1,y1]</code>. Now I want to connect these two points with ones.

An example for <code>N=5</code> would be:

Lets set the coordinates to <code>2</code> in the matrix

<pre><code>[x0,y0] = [0,3]
[x1,y1] = [4,2]
</code></pre>

Then the matrix should look something like:

<pre><code>M = 
 0 0 0 2 0
 0 0 0 1 0
 0 0 1 0 0
 0 0 1 0 0
 0 0 2 0 0
</code></pre>

What would be a simple approach to do that?

Python: Draw line between two coordinates in a matrix

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我有一个大小为N x N的矩阵，定义为M = np.zeros((N,N))和两个坐标[x0,y0]和[x1,y1]。现在我想把这两点和一点联系起来。N=5的一个例子是：让我们在矩阵中将坐标设置为2。[x0,y0] = [0,3][x1,y1] = [4,2]那么矩阵应该类似于：M =     0 0 0 2 0   ...

问Python:在矩阵中的两个坐标之间画线
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问Python:在矩阵中的两个坐标之间画线EN