<p>下面是一个具有双for循环的解决方案,可用于任意长度的列表:</p>
<pre><code>list_3_5 = [[1,2,3,4,5],[6,7,8,9,10],[11,12,13,14,15]]
newlist = []
for element in list_3_5:
newlist.append(element[:3])
for element in list_3_5:
newlist.append(element[3:])
newlist
Out: [[1, 2, 3], [6, 7, 8], [11, 12, 13], [4, 5], [9, 10], [14, 15]]
</code></pre>
<p>但是等等!还有更多:</p>
<pre><code>list_3_5 = [[1,2,3,4,5],[6,7,8,9,10],[11,12,13,14,15]]
newlist = list(range(2*len(list_3_5)))
for key, element in enumerate(list_3_5):
newlist[key] = element[:3]
newlist[key+len(list_3_5)] = element[3:]
newlist
[[1, 2, 3], [6, 7, 8], [11, 12, 13], [4, 5], [9, 10], [14, 15]]
</code></pre>
<p>以下部分回答了问题的更新部分:</p>
<pre><code>biglist = []
for i in list2:
biglist = biglist + i
#Here we put all small lists in the given list into a big list
print(biglist)
#construct an empty list to save the result as we go:
newlist = []
#add sections of 5 to newlist
while len(biglist) >= 5:
newlist.append(biglist[:5])
biglist = biglist[5:]
#add the remaining part of biglist to the newlist:
newlist.append(biglist)
print(newlist)
</code></pre>
<p>输出为:</p>
<pre><code>[[1, 2, 3, 4, 5], [6, 7, 8, 9, 10], [11, 12, 13, 14, 15], [16, 17, 18, 19, 20], [21, 22, 23, 24, 25], [26, 27, 28, 29, 30], [31, 32, 33, 34, 35], [36, 37, 38, 39, 40], [41, 42, 43, 44, 45], [46, 47, 48, 49, 50], [51, 52, 53, 54, 55], [56, 57, 58, 59, 60], [61, 62, 63, 64, 65], [66, 67, 68, 69, 70], [71, 72, 73, 74, 75], [76, 77, 78, 79, 80], [81, 82, 83, 84, 85], [86, 87, 88, 89, 90], [91, 92, 93, 94, 95], [96, 97, 98, 99]]
</code></pre>