擅长:python、mysql、java
<p>如果我正确理解了这个问题,你想沿着路线找到站点,那里有同一辆巴士到访的站点,基本上是找到巴士路线的重复。你知道吗</p>
<p>请看下面的代码。你知道吗</p>
<pre><code>bus_routes = {"Lentil": ["Chinook", "Orchard", "Valley", "Emerald","Providence", "Stadium", "Main", "Arbor", "Sunnyside", "Fountain", "Crestview", "Wheatland", "Walmart", "Bishop", "Derby", "Dilke"], "Wheat": ["Chinook", "Orchard", "Valley", "Maple","Aspen", "TerreView", "Clay", "Dismores", "Martin", "Bishop", "Walmart", "PorchLight", "Campus"]}
route_dup = {}
for x,y in bus_routes.items():
for z in y:
try:
if route_dup[z]:
route_dup[z].append(x)
except KeyError:
route_dup[z] = [x]
print(route_dup)
</code></pre>
<p>我们用<code>bus_routes.items()</code>迭代得到一个变量(<code>y</code>),其中<code>x</code>是路由名,<code>y</code>是停止名列表。然后,我们用<code>y</code>创建另一个迭代,并尝试查看具有该停止名称的键是否已经存在于<code>route_dup</code>中,如果它不存在,它将捕获<code>KeyError</code>,并使用列表中路由名称的值创建它,但是,如果该键确实存在,我们可以放心地说它将访问我们已经创建的列表,因此<code>append()</code>,使用下一个路由名称。你知道吗</p>
<p>希望这有帮助。你知道吗</p>