具有相同值的元组的子列表组

Question

我有以下列表：

o = [([1,1,1,1],[0.9,0.2,0.2,0.3]),([1,1,1],[0.7,0.6,0.7]),([1,1],[0.8,0.3]),([2,2,2,2,2],[0.8,0.9,0.2,0.3,0.1]),([2,2],[0.2,0.7]),([3,3,3],[0.9,0.4,0.3]),([3,3],[0.1,0.1]),([3,3,3,3],[0.9,0.4,0.3,0.1]),([4,4],[0.1,0.3]),([4,4],[0.6,0.7]),([4,4,4],[0.3,0.1,0.1])]

这是所需的输出：

output = [[([1,1,1,1],[0.9,0.2,0.2,0.3]),([1,1,1],[0.7,0.6,0.7]),([1,1],[0.8,0.3])],
[([2,2,2,2,2],[0.8,0.9,0.2,0.3,0.1]),([2,2],[0.2,0.7])],
[([3,3,3],[0.9,0.4,0.3]),([3,3],[0.1,0.1]),([3,3,3,3],[0.9,0.4,0.3,0.1])],
[([4,4],[0.1,0.3]),([4,4],[0.6,0.7]),([4,4,4],[0.3,0.1,0.1])]]

我想从初始列表创建子列表，按每个元组的第一个元素相等进行分组。到目前为止，我就是这么做的，但我不知道如何获得如上所示的子列表。

x=[]
for i in range(len(o)-1):
if o[i][0][0] == o[i+1][0][0]:
x.append(o[i])
else:
x[-1].append(o[i])

任何帮助都将不胜感激，谢谢！

Answer 1

x=[]
for i_o in range(len(o)):
for i_x in range(len(x)):
if o[i_o][0][0]==x[i_x][0][0][0]:
x[i_x].append(o[i_o])
break
else:
x.append([o[i_o]])