擅长:python、mysql、java
<pre><code>a = ["1a", "1b", "2b", "3c", "2d", "1g", "3d", "3g"]
coeffs = [a[0][0]] # collect coefficients
for i in a[1:]:
if i[0] not in coeffs: coeffs.append(i[0])
s = [[x for x in a if x[0] == i] for i in coeffs] # groups according to coefficients
# s = [["1a", "1b", "1g"], ["2b", "2d"], ["3c", "3d", "3g"]]
solution = [i[-1] for i in s if len(i)%2]
# ['1g', '3g']
</code></pre>