numpy check two array on multiple column

Question

Lets say edata and tdata two 2D numpy array with shape (x,6) and (y,6) where x and y are two arbitrary number and must not be equal. Now I have to check if there is any row in edata such that both 2nd and 3rd column value of that row equals to that of some row of tdata. If there is any such row then save them in another. For now I have written the following code. But I think numpy has better way to that instead of explicitly iterating and checking for values. Can you suggest me more efficient way?

res = np.array([], dtype=np.float64).reshape(0,6)
for line in edata:
    ind = line[1] == tdata[:,1]  
    ind = line[2] == tdata[ind,2]
    if np.any(ind):
        res = np.vstack((res,line))

Ben.T · Accepted Answer · 2018-11-22 15:37:21Z

1

You can use equal and outer to see if any value in a column of edata is in a column of tdata. You check for both the 2nd and 3rd columns, then use any on axis=1 to get edata rows as required such as:

res = edata[ (np.equal.outer(edata[:,1],tdata[:,1])
              &np.equal.outer(edata[:,2],tdata[:,2])).any(1) ,:]

for example, with simple input:

edata = np.arange(4*6).reshape(4,6)
tdata = np.arange(6*6).reshape(6,6) + 12

print (res)
array([[12, 13, 14, 15, 16, 17],
       [18, 19, 20, 21, 22, 23]])

which are the last two rows of edata that have the same value in column 2 and 3 than the first two rows of tdata

answered Nov 22, 2018 at 15:37

Ben.T

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4 Comments

Eular Over a year ago

check for this two [[1,2,3],[2,3,4],[4,5,6]] and [[1,2,3],[1,3,4],[7,5,6]], it should return the 1st row only

Ben.T Over a year ago

@Eular your method (like mine) return all the rows, and it makes sense because 2nd and 3rd column in both arrays are the same, no? if it is supposed to return only the 1st row then matching both 2nd and 3rd columns is not enough, but your question (and the code associated) seems to state only matching 2nd and 3rd columns

Eular Over a year ago

Sry, i was checking with 1st and 2nd column, my bad

Ben.T Over a year ago

@Eular no pb :)

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