擅长:python、mysql、java
<p>如果我理解正确,您可以使用<a href="https://docs.python.org/3.5/library/functions.html#zip" rel="nofollow noreferrer">^{<cd1>}</a>执行类似操作:</p>
<pre><code>ListID = [1,2,3,4,5]
List1 = [34,56,34,345,55]
list2 = [644,64,232,44,1]
list3 = [622,12,44,55,1]
objectList = [list(elem) for elem in zip(ListID,List1,list2,list3)]
print(objectList)
</code></pre>
<p>输出:</p>
<pre><code>[[1, 34, 644, 622], [2, 56, 64, 12], [3, 34, 232, 44], [4, 345, 44, 55], [5, 55, 1, 1]]
</code></pre>
<p>另一种方法是使用<a href="http://book.pythontips.com/en/latest/map_filter.html#map" rel="nofollow noreferrer">^{<cd2>}</a>函数和<code>zip</code>:</p>
<pre><code>objectList = list(map(lambda elem: list(elem), zip(ListID,List1,list2,list3)))
</code></pre>