如何找到重复元素在列表中的位置，然后对这些元素求和？

Question

I want to find the position of the repeated elements in the column "1" and the column "2" in a specified matrix "x" (the matrix "x" has a nx3 size ("n" is the maximum number between the column "1" and the column "2", in the next example is max(3,4)=4)), for then in the principal diagonal (number 1 to "n") of a new matrix "z"(the matrix "z" has a "n"x"n" size) sum the values for each column (except the columns with values equal to zero), and put the respective values off-diagonal in the respective column. This is an example of a matrix "x":

        1 2 3
x=  1 [[1,2,4],
    2  [1,3,2],
    3  [2,3,1],
    4  [3,4,5]]

The matrix "z" is a symmetric matrix, meaning for example the element z[1,2] is equal to z[2,1] and the element z[1,3] is equal to z[3,1].This is the structure of the matrix "z":

z=  [[z[1],z[1 to 2],z[1 to 3],z[1 to 4]], 
     [z[2 to 1],z[2],z[2 to 3],z[2 to 4]], 
     [[z[3 to 1],z[3 to 2],z[3],z[3 to 4]],
     [[z[4 to 1],z[4 to 2],z[4 to 3],z[4]]

The elements in the diagonal principal( for this example are):

 z[1]=z[1 to 2]+z[1 to 3]+z[1 to 4]
 z[2]=z[2 to 1]+z[2 to 3]+z[2 to 4]
 z[3]=z[3 to 1]+z[3 to 2]+z[3 to 4]
 z[4]=z[4 to 1]+z[4 to 2]+z[4 to 3]

The desired matrix (for this example) is:

z=  [[4+2=6,4,2,0], 
     [4,4+1=5,1,0], 
     [2,1,2+1+5=8,5],
     [0,0,5,5]]

Note: The zeros in the matrix "z" represent if there aren´t connection with the other numbers. In this example the number "4" has only connection with the number "3"(the value is 5). Then the connection for the number 4 with the number 1 and number 2 are zero.

Thanks.

Answer 1

Your problem is clarified if you introduce graph-theoretic terminology. Your x could be thought of as describing a weighted graph. A row like [1,3,2] can be interpreted as describing the existence of an edge of weight 2 connecting node 1 and node 3. The matrix z contains the following information:

1) For each node i , the corresponding diagonal entry is the sum of the weights of all edges which are incident with i .

2) For each pair of distinct nodes i and j , the corresponding entry of z is either 0 if the nodes are not connected, or the weight of the edge connecting i and j if they are (or sum of the weights if there are multiple edges).

Given this interpretation, things are easy enough. The following assumes that the nodes in x begin with 1 . Note that the resulting matrix is 0-based even though the graph is 1-based, so to find the weight from node 2 to node 3 you will need z[1][2] :

def makeMatrix(edges):
    n = max(max(edge[:2]) for edge in edges)
    z = [[0]*n for i in range(n)]
    for edge in edges:
        i,j,w = edge
        z[i-1][i-1] += w
        z[j-1][j-1] += w
        z[i-1][j-1] += w
        z[j-1][i-1] += w
    return z

The following helps for debugging:

def pprint(matrix):
    for row in matrix:
        print(' '.join(str(i) for i in row))

For example,

>>> x = [[1,2,4],[1,3,2],[2,3,1],[3,4,5]]
>>> z = makeMatrix(x)
>>> pprint(z)
6 4 2 0
4 5 1 0
2 1 8 5
0 0 5 5