<p>我们正在迭代<code>list</code>类型的数据结构。所以给变量起个有意义的名字。我给了<code>input_list =["d1", "d2", "d3", "d4"]</code></p>
<p>当我们迭代<code>list</code>即<code>input_list</code>变量时,列表中的一项将被迭代。所以我们不能做<code>if d ==d1 and d==d2:</code>,因为我们在比较一个项目和两个不同的条件,这两个条件总是错误的。你知道吗</p>
<p>我从你的代码中了解到,我们有一个列表,根据一些硬代码条件,我们必须执行一些场景。你知道吗</p>
<p>演示:</strong></p>
<pre><code>input_list =["d1", "d2", "d3", "d4"]
if "d1" in input_list and "d2" in input_list:
print 'Scenario:1 ==> d1 && d2'
elif "d3" in input_list and "d4" in input_list:
print 'Scenario:2 ==> d3 && d4'
elif "d1" in input_list:
print 'Scnario 3: ==> d1'
elif "d2" in input_list:
print 'Scnario 4: ==> d2'
elif "d3" in input_list:
print 'Scnario 5: ==> d3'
elif "d4" in input_list:
print 'Scnario 6: ==> d4'
</code></pre>
<hr/>
<p>假设列表包含字典项。你知道吗</p>
<p>演示</p>
<pre><code>d1 = {}
d2 = {}
d3 = {}
d4 = {}
input_list =[d1, d2, d3, d4]
for item in input_list:
if "d1" in item and "d2" in item:
print 'Scenario:1 ==> d1 && d2'
elif "d3" in item and "d4" in item:
print 'Scenario:2 ==> d3 && d4'
elif "d1" in item:
print 'Scnario 3: ==> d1'
elif "d2" in item:
print 'Scnario 4: ==> d2'
elif "d3" in item:
print 'Scnario 5: ==> d3'
elif "d4" in item:
print 'Scnario 6: ==> d4'
</code></pre>
<p><strong>备注:</strong></p>
<ol>
<li>不要使用Python的保留关键字。<code>dict</code>是Python中的关键字。你知道吗</li>
</ol>