在容差内找到Python中两个矩阵的交集？

Question

I'm looking for the most efficient way of finding the intersection of two different-sized matrices. Each matrix has three variables (columns) and a varying number of observations (rows). For example, matrix A:

a = np.matrix('1 5 1003; 2 4 1002; 4 3 1008; 8 1 2005')
b = np.matrix('7 9 1006; 4 4 1007; 7 7 1050; 8 2 2003'; 9 9 3000; 7 7 1000')

If I set the tolerance for each column as col1 = 1 , col2 = 2 , and col3 = 10 , I would want a function such that it would output the indices in a and b that are within their respective tolerance, for example:

[x1, x2] = func(a, b, col1, col2, col3)
print x1
>> [2 3]
print x2
>> [1 3]

You can see by the indices, that element 2 of a is within the tolerances of element 1 of b .

I'm thinking I could loop through each element of matrix a , check if it's within the tolerances of each element in b , and do it that way. But it seems inefficient for very large data sets.

Any suggestions for alternatives to a looping method for accomplishing this?

Answer 1

If you don't mind working with NumPy arrays, you could exploit broadcasting for a vectorized solution. Here's the implementation -

# Set tolerance values for each column
tol = [1, 2, 10]

# Get absolute differences between a and b keeping their columns aligned
diffs = np.abs(np.asarray(a[:,None]) - np.asarray(b))

# Compare each row with the triplet from `tol`.
# Get mask of all matching rows and finally get the matching indices
x1,x2 = np.nonzero((diffs < tol).all(2))

Sample run -

In [46]: # Inputs
    ...: a=np.matrix('1 5 1003; 2 4 1002; 4 3 1008; 8 1 2005')
    ...: b=np.matrix('7 9 1006; 4 4 1007; 7 7 1050; 8 2 2003; 9 9 3000; 7 7 1000')
    ...: 

In [47]: # Set tolerance values for each column
    ...: tol = [1, 2, 10]
    ...: 
    ...: # Get absolute differences between a and b keeping their columns aligned
    ...: diffs = np.abs(np.asarray(a[:,None]) - np.asarray(b))
    ...: 
    ...: # Compare each row with the triplet from `tol`.
    ...: # Get mask of all matching rows and finally get the matching indices
    ...: x1,x2 = np.nonzero((diffs < tol).all(2))
    ...: 

In [48]: x1,x2
Out[48]: (array([2, 3]), array([1, 3]))

---

Large datasizes case : If you are working with huge datasizes that cause memory issues and since you already know that the number of columns is a small number 3 , you might want to have a minimal loop of 3 iterations and save huge memory footprint, like so -

na = a.shape[0]
nb = b.shape[0]
accum = np.ones((na,nb),dtype=bool)
for i in range(a.shape[1]):
    accum &=  np.abs((a[:,i] - b[:,i].ravel())) < tol[i]
x1,x2 = np.nonzero(accum)