擅长:python、mysql、java
<p>您不能访问具有元组的列表,元组就是<code>tilemap[x,y]</code>正在执行的操作,您需要:</p>
<pre><code>tilemap[x][y]
</code></pre>
<p>它从索引y的子列表中获取索引x处的元素:</p>
<pre><code>In [7]: tilemap = [
...: [9,3,2,2,2,4,2,2,2,3,9],
...: [8,11,10,10,10,10,10,10,10,11,5],
...: [8,1,6,9,9,9,9,9,8,1,5],
...: [8,1,5,9,9,9,9,9,7,1,5],
...: [8,1,6,9,9,9,9,9,7,1,5],
...: [8,1,5,9,9,9,9,9,8,1,5],
...: [8,1,12,4,2,3,2,4,13,1,5],
...: [7,11,10,10,10,11,10,10,10,11,6],
...: [9,9,9,9,7,1,6,9,9,9,9],
...: [9,9,9,9,7,1,6,9,9,9,9],
...: [9,9,9,9,7,1,6,9,9,9,9]
...: ]
In [8]: tilemap[0][0]
Out[8]: 9
In [9]: tilemap[1][0]
Out[9]: 8
</code></pre>
<p>也可以迭代tilemap列表:</p>
<pre><code>for ind, sub in enumerate(tilemap):
for i, y in enumerate(sub):
current_tile = textures[y]
screen.blit(current_tile, (ind * tilesize, i * tilesize))
</code></pre>