Create a new numpy array based on conditions set out in numpy array

Question

How can I create an array based on certain conditions in another array. For example, if I have an array that gives me a Base number, a start and end number, and then multiple other Base numbers. I want to create a new matrix that lists the Base number, the loop number (based on start/end) and then the other Base number associated with this, while ignoring 0's. I am trying to find a way to do this without using a for loop.

For example, how can I get array B from array A.

           Base Start End Base1 Base2 Base3
A=np.array([[100,  1,   2,  101,  102,  103],
            [101,  3,   4,  100,  103,    0]])


B=np.array[[100,1,101,1],
           [100,1,102,1],
           [100,1,103,1],
           [100,2,101,2],
           [100,2,102,2],
           [100,2,103,2],
           [101,3,100,3],
           [101,3,103,3],
           [101,4,100,4],
           [101,4,103,4]]

Thanks for the help!

You might be able to do this with a messy combination of numpy.tile and numpy.repeat, but why go to such lengths to not use a loop? — wflynny
– wflynny, Commented Mar 3, 2014 at 21:17
I don't understand your target format. Why does row 4 of B end with a 1 instead of a 2? If you post a slow version with explicit loops, that would make it easier for us to understand exactly what you want to do. — amaurea
– amaurea, Commented Mar 3, 2014 at 23:40
Sorry that was an error on my part, rows 4-6 should have ended in a 2. Has been edited. Is it clear now or would you like me to break it out more? — user3357979
– user3357979, Commented Mar 4, 2014 at 0:36

ysakamoto · Accepted Answer · 2014-03-04 01:37:10Z

2

Here you go... nasty list comprehension

A = array([[100,   1,   2, 101, 102, 103], [101,   3,   4, 100, 103,   0]])
B = [[A[i,0], b, c, b] for i in range(len(A)) for b in A[i,1:3] for c in A[i,3:6] if c != 0]

>>> B
[[100, 1, 101, 1],
 [100, 1, 102, 1],
 [100, 1, 103, 1],
 [100, 2, 101, 2],
 [100, 2, 102, 2],
 [100, 2, 103, 2],
 [101, 3, 100, 3],
 [101, 3, 103, 3],
 [101, 4, 100, 4],
 [101, 4, 103, 4]]

answered Mar 4, 2014 at 1:37

ysakamoto

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Eelco Hoogendoorn · Accepted Answer · 2014-03-09 19:22:04Z

2

Here you go: pure numpy, no for loops at the python level

C = np.empty((len(A), 2,3,4), np.int)
C[...,0] = A[:,0  ][:,None,None]
C[...,1] = A[:,1:3][:,:   ,None]
C[...,2] = A[:,3: ][:,None,:   ]
C[...,3] = np.arange(len(A)*2).reshape(len(A),2,1)+1

C = C.reshape(-1,4)
C = C[C[:,2]!=0]

print C

edit: minor cleanup

edit2: or for maximum brevity and crypticity (see comment):

C = np.empty((len(A), 2,3,2,2), np.int)
C[...,0,0] = A[:,0  ][:,None,None]
C[...,1,0] = A[:,3: ][:,None,:   ]
C[...,:,1] = A[:,1:3][:,:   ,None,None]

edited Mar 9, 2014 at 19:22

answered Mar 9, 2014 at 11:17

Eelco Hoogendoorn

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1 Comment

Eelco Hoogendoorn Over a year ago

note; the last column appears to me to be redundant, so ive added another method of generating the same pattern; but I am not sure it generalizes to all your use cases; you might just want to write the same pattern to column 3 as to column 1, if that is the expected data layout of some downstream code.

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