擅长:python、mysql、java
<pre><code>diffs=[]
for sub_b, sub_a in zip(b, a):
curr = []
for atom_b, atom_a in zip(sub_b, sub_a):
try:
curr.append(float(atom_b) - float(atom_a))
except ValueError:
curr.append('-'.join([atom_b, atom_a]))
diffs.append(curr)
ans1, ans2 = zip(*diffs)
</code></pre>
<p><code>zip</code>函数也可用于解压iterables。在</p>