擅长:python、mysql、java
<p>另一种方法是:</p>
<pre><code>from collections import defaultdict
repeat_count = len(alist)
ans_alist = [defaultdict(dict) for _ in range(repeat_count)]
for k, v in a_dict.items():
mul_times = repeat_count / len(v) # Use // if Python 3
extend_counter = repeat_count % (mul_times * len(v))
repeat_v = v * mul_times + v[:extend_counter]
unicode_k = u'{}'.format(k)
for index, val in enumerate(repeat_v):
ans_alist[index][unicode_k] = u'{}'.format(val)
</code></pre>