Python中2维数组/矩阵转化为字典(矩阵中第一列含有重复元素)??

1 >>> a=[[3,789],[1,100],[1,102],[2,102],[1,106],[2,456]];
2 >>> b=dict(a)
3 >>> b
4 {1: 106, 2: 456, 3: 789}
5 >>> 

 1 # 将列表转化为字典
 2 def listTodict():
 3     a=[[3,789],[1,100],[1,102],[2,102],[1,106],[2,456]];
 4     lenArr=len(a);
 5     # 列出第一列所有的元素值,包含重复元素
 6     firstCol=[];
 7     for i in range(lenArr):
 8         firstCol.append(a[i][0]); # [3, 1, 1, 2, 1, 2]
 9     # 列出第一列所有的不重复元素
10     firstColNotRep=set(firstCol);    # {1,2,3}集合中元素没有顺序
11     firstColNotRep=list(firstColNotRep); # firstColNotRep=[1,2,3]
12     # 统计第一列所有的不重复元素的个数
13     lenfirstColNotRep=len(firstColNotRep);
14     b=[];c=[];
15     for i in range(lenfirstColNotRep):
16         b.append([firstColNotRep[i]]);  # [[1], [2], [3]]
17     for i in range(lenfirstColNotRep): 
18         for j in range(lenArr):
19             if firstColNotRep[i]==a[j][0]: # firstColNotRep=[1,2,3]
20                 print(j);
21                 b[i].append(a[j][1]);
22     return b;
23 # 结果b=[[1,100,102,106],[2,102,456], [3,789]]
24 
25 # 怎样将[1,100,102,106]先转为[1,[100,102,106]]转化为[1:[100,102,106]]
26 def listToDict1():
27     d=[[1,100,102,106],[2,102,456], [3,789]];
28     e={};
29     for i in range(len(d)):
30         e[d[i][0]]=d[i][1:];
31     return e;
32