Numpy Compare unequal rows and make both array of same dimension

Question

a=   
[[43655,    1428,     0, 2554]      
[44580,  1428,     0,  2555]  
[44930,  1428,     0,  2556]  
[47708,  1428,     0,  2557]]     
b=   
[[41641,  1428,     0, 2554]  
[44075,  1428,     0,  2555]  
[44901,  1428,     1,  2556]  
[45377,  1428,     0,  2557]  
[48056,  1428,     0,  2558]]


New  b= 
[[41641,  1428,     0, 2554]  
[44075,  1428,     0,  2555]  
[44901,  1428,     1,  2556]  
[45377,  1428,     0,  2557]

I have two numpy array with unequal rows. For eg. Array a has 4 rows while array b has 5 rows.
Edit: No. of rows in array 'b' is greater than array 'a'. Every element of a[:,3] lies in b[:,3]. Is there any function that extract only the rows of array b whose b[:,3]=a[:,3]

Seems that the output is not what you have want? and how you want to compare the columns and reduce the b based on what logic? — Kasravnd
– Kasravnd, Commented Jul 22, 2015 at 15:45
Every element in a[:,3] is a subset of b[:,3]. So I was thinking to compare a[:,3] with b[:,3] and extract index of b[:,3] where it matches with a[:,3]. In above eg. index would be i=[0,1,2,3]. — HaWa
– HaWa, Commented Jul 22, 2015 at 22:53

Kasravnd · Accepted Answer · 2015-07-23 10:56:26Z

1

You can compare your the elements of 3rd column using zip and np.equal within a list comprehension then convert the result to a numpy array and get the desire rows from array b.

>>> b[np.array([np.equal(*I) for I in zip(a[:,3],b[:,3])])]
array([[41641,  1428,     0,  2554],
       [44075,  1428,     0,  2555],
       [44901,  1428,     1,  2556],
       [45377,  1428,     0,  2557]])

If the order is not important for you you can use np.in1d :

>>> b[np.in1d(b[:,3],a[:,3])]
array([[41641,  1428,     0,  2554],
       [44075,  1428,     0,  2555],
       [44901,  1428,     1,  2556],
       [45377,  1428,     0,  2557]])

>>> a=np.array([[100, 1], [101, 4], [106, 6], [104, 10]])
>>> b= np.array([[ 1, 1], [ 2, 2], [ 3, 3], [ 4, 4], [ 5, 5], [ 6, 6], [ 7, 7], [ 8, 8], [ 9, 9], [10, 10]])
>>> 
>>> b[np.in1d(b[:,1],a[:,1])]
array([[ 1,  1],
       [ 4,  4],
       [ 6,  6],
       [10, 10]])

edited Jul 23, 2015 at 10:56

answered Jul 23, 2015 at 6:20

Kasravnd

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2 Comments

HaWa Over a year ago

>>> a array([[100, 1], [101, 4], [106, 6], [104, 10]]) >>> b array([[ 1, 1], [ 2, 2], [ 3, 3], [ 4, 4], [ 5, 5], [ 6, 6], [ 7, 7], [ 8, 8], [ 9, 9], [10, 10]]) is not showing the same behavior. New b should be=b array([[ 1, 1], [ 4, 4], [ 6, 6], [10, 10]])

Kasravnd Over a year ago

@HaWa So the order is not important for you! let me update

rohanp · Accepted Answer · 2015-07-22 16:05:30Z

1

You can omit the last element of an array by doing a[:-1]

Therefore you can omit the last row of an array by doing;

a[:,3] 
b[:-1,3]

answered Jul 22, 2015 at 16:05

rohanp

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Comments

derricw · Accepted Answer · 2015-07-22 18:13:08Z

0

if you don't know how much longer b is than a then you can just use

b[:a.shape[0]]

This will grab enough rows of b to compare to a.

So to grab a particular column like in your example, it would be

b[:a.shape[0], 3]

This technique will even work in the other direction (if b is smaller than a).

edited Jul 22, 2015 at 18:13

answered Jul 22, 2015 at 18:07

derricw

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