擅长:python、mysql、java
<p>列表中没有dict。您在dict中有列表。一旦您修复了拼写错误(<code>append</code>是一个函数,应该使用括号,而不是括号,并且您没有<code>b</code>和<code>b1</code>,您有<code>b1</code>和<code>b2</code>),您的代码将生成您描述的结构。它不是这样打印的,因为Python不是这样打印列表的,但数据都在那里:</p>
<pre><code>dictlist = { "g1" : [], "g2" : [] , "g3" : [] }
a = ["a", 23]
a1 = ["asd", 3]
a2 = ["asdf", 10]
a3 = ["adg", 5]
b1 = ["df", 5]
b2 = ["dfg", 1]
c = ["dfg", 50]
dictlist["g1"].append(a)
dictlist["g1"].append(a1)
dictlist["g1"].append(a2)
dictlist["g1"].append(a3)
dictlist["g2"].append(b1)
dictlist["g2"].append(b2)
dictlist["g3"].append(c)
print(dictlist)
for k, v in dictlist.items():
print(k)
for i in v:
print(i)
</code></pre>
<p>输出:</p>
<pre><code>timr@tims-gram:~/src$ python x.py
{'g1': [['a', 23], ['asd', 3], ['asdf', 10], ['adg', 5]], 'g2': [['df', 5], ['dfg', 1]], 'g3': [['dfg', 50]]}
g1
['a', 23]
['asd', 3]
['asdf', 10]
['adg', 5]
g2
['df', 5]
['dfg', 1]
g3
['dfg', 50]
timr@tims-gram:~/src$
</code></pre>