How slice Numpy array by column value

Question

I have an array like this numpy array

 dd =[[0.567 2 0.611]
      [0.469 1 0.479]
      [0.220 2 0.269]
      [0.480 1 0.508]
      [0.324 1 0.324]]

I need 2 seperate array dd[:,1] ==1 and dd[:,1] ==2

These array are what I am after:

 na =[[0.469 1 0.479]
      [0.480 1 0.508]
      [0.324 1 0.324]]

 na2 =[[0.567 2 0.611]
       [0.220 2 0.269]]

I have tried np.where did really work

wim · Accepted Answer · 2011-09-28 01:47:48Z

7

You could use numpy fancy indexing:

[~/repo/py]
|32>dd[dd[:,1] == 1]
[32] 
array([[ 0.469,  1.   ,  0.479],
       [ 0.48 ,  1.   ,  0.508],
       [ 0.324,  1.   ,  0.324]])

[~/repo/py]
|33>dd[dd[:,1] == 2]
[33] 
array([[ 0.567,  2.   ,  0.611],
       [ 0.22 ,  2.   ,  0.269]])

Alternatively you could use a list comprehension:

[~/repo/py]
|21>np.array([row for row in dd if row[1] == 1])
[21] 
array([[ 0.469,  1.   ,  0.479],
       [ 0.48 ,  1.   ,  0.508],
       [ 0.324,  1.   ,  0.324]])

[~/repo/py]
|22>np.array([row for row in dd if row[1] == 2])
[22] 
array([[ 0.567,  2.   ,  0.611],
       [ 0.22 ,  2.   ,  0.269]])

edit:

how to time these things in ipython:

[~/repo/py]
|36>timeit dd[dd[:,1] == 1]
100000 loops, best of 3: 6 us per loop

[~/repo/py]
|37>timeit np.array([row for row in dd if row[1] == 1])
100000 loops, best of 3: 11.5 us per loop

edited Sep 28, 2011 at 1:47

answered Sep 28, 2011 at 1:41

wim

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1 Comment

wim Over a year ago

the first example (fancy indexing) is about twice as fast.

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