在Numpy中将3列矩阵转换为N x N矩阵

Question

I have a 2D numpy array with 3 columns. Columns 1 and 2 are a list of connections between ID's. Column 3 is a the strength of that connection. I would like to transform this 3 column matrix into a weighted adjacency matrix (an N x N matrix where cells represent the strength of connection between each ID).

I have already done this in my code below. matrix is the 3 column 2D array and t1 is the weighted adjacency matrix. My problem is this code is very slow because I am using nested for loops. I am familiar with the pandas function melt which does this, but I am not able to use pandas . Is there a faster implementation not using pandas ?

import numpy as np
a = np.arange(2000)
np.random.shuffle(a)
b = np.arange(2000)
np.random.shuffle(b)
c = np.random.rand(2000,1)

matrix = np.column_stack((a,b,c))

#get unique value list of nm
flds = list(np.unique(matrix[:,0]))
flds.extend(list(np.unique(matrix[:,1])))
flds = np.asarray(flds)
flds = np.unique(flds)

#make lookup dict
lookup = dict(zip(np.arange(0,len(flds)), flds))
lookup_rev = dict(zip(flds, np.arange(0,len(flds))))

#make empty n by n matrix with unique lists
t1 = np.zeros([len(flds) , len(flds)])

#map values into the n by n matrix and make the rest 0
'''this takes a long time to run'''
#iterate through rows
for i in np.arange(0,len(lookup)):
    #iterate through columns
    for k in np.arange(0,len(lookup)):
        val = matrix[(matrix[:,0] == lookup[i]) & (matrix[:,1] == lookup[k])][:,2]  
        if val:
            t1[i,k] = sum(val)

Answer 1

Assuming that I understood the question correctly and that val is a scalar, you could use a vectorized approach that involves initializing with zeros and then indexing , like so -

out = np.zeros((len(flds),len(flds)))
out[matrix[:,0].astype(int),matrix[:,1].astype(int)] = matrix[:,2]

Please note that by my observation it looks like you can avoid using lookup .

Answer 2

The main acceleration you can get is by not iterating through each element of the NxN matrix but instead iterate trough your connection list, which is much smaller.

I tried to simplify your code a bit. It use the list.index method, which can be slow, but it should still be faster that what you had.

import numpy as np
a = np.arange(2000)
np.random.shuffle(a)
b = np.arange(2000)
np.random.shuffle(b)
c = np.random.rand(2000,1)

matrix = np.column_stack((a,b,c))

lookup = np.unique(matrix[:,:2]).tolist() # You can call unique only once

t1 = np.zeros((len(lookup),len(lookup)))

for i,j,val in matrix:
    t1[lookup.index(i),lookup.index(j)] = val # Fill the matrix

Answer 3

You need to iterate your matrix only once:

import numpy as np

size = 2000

a = np.arange(size)
np.random.shuffle(a)
b = np.arange(size)
np.random.shuffle(b)
c = np.random.rand(size,1)

matrix = np.column_stack((a,b,c))

#get unique value list of nm
fields = np.unique(matrix[:,:2])
n = len(fields)

#make reverse lookup dict
lookup = dict(zip(fields, range(n)))

#make empty n by n matrix
t1 = np.zeros([n, n])

for src, dest, val in matrix:
    i = lookup[src]
    j = lookup[dest]

    t1[i, j] += val