擅长:python、mysql、java
<p>使用<code>zip(narrative, subject, activity, filer)</code>基本上可以转换矩阵(您的等长度列表组成矩阵)。然后通过这些枚举来查找标志为true的位置<code>n</code>,并为适当的变量编制索引。在</p>
<pre><code>narrative = [0, 0, 0, 0]
subject = [1, 1, 0, 1]
activity = [0, 0, 0, 1]
filer = [0, 1, 1, 0]
variables = ("narrative", "subject", "activity", "filer")
# ========================================================
new_list = [[variables[n] for n, flag in enumerate(indicators) if flag]
for indicators in zip(narrative, subject, activity, filer)]
>>> new_list
[['subject'], ['subject', 'filer'], ['filer'], ['subject', 'activity']]
</code></pre>
<p>要查看转置:</p>
^{pr2}$