擅长:python、mysql、java
<p>使用<code>itertools.groupby()</code>函数:</p>
<pre><code>import itertools
l = ['T46', 'T43', 'R45', 'R44', 'B46', 'B43', 'L45', 'L44', 'C46', 'C45']
coords_str = 'TRBLC' # Top, Right, Bottom, Left, Center
result = {}
for k,g in itertools.groupby(sorted(l, key=lambda x:x[1:]), key=lambda x:x[1:]):
items = list(_[0] for _ in g)
result[k] = [int(c in items) for c in coords_str]
print(result)
</code></pre>
<p>输出:</p>
<pre><code>{'44': [0, 1, 0, 1, 0], '45': [0, 1, 0, 1, 1], '43': [1, 0, 1, 0, 0], '46': [1, 0, 1, 0, 1]}
</code></pre>