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up vote 2 down vote favorite

This should be simple task but I am ashamed to admit I'm stuck.

I have a numpy array, called X:

X.shape is (10,3)and it looks like 

[[  0.   0.  13.  ]
 [  0.   0.   1.  ]
 [  0.   4.  16.  ]
 ..., 
 [  0.   0.   4.  ]
 [  0.   0.   2.  ]
 [  0.   0.   4.  ]]

I would like to select the 1, 2 and 3rd row of this array, using the indices in this other numpy array, called idx:

idx.shape is (3,) and it looks like [1 2 3]

When I try
new_array = X[idx] or variations on this, I get errors.

How does one index a numpy array using another numpy array that holds the indices?

Apologizes for such a basic question in advance.

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1 Answer

up vote 3 down vote accepted

I do it like this:

>>> import numpy as np
>>> x = np.arange(30).reshape((10, 3))
>>> x
array([[ 0,  1,  2],
       [ 3,  4,  5],
       [ 6,  7,  8],
       [ 9, 10, 11],
       [12, 13, 14],
       [15, 16, 17],
       [18, 19, 20],
       [21, 22, 23],
       [24, 25, 26],
       [27, 28, 29]])
>>> idx = np.array([1,2,3])
>>> x[idx, ...]
array([[ 3,  4,  5],
       [ 6,  7,  8],
       [ 9, 10, 11]])

Note that in this case, the ellipsis could be replaced by a simple slice if you'd rather:

x[idx, :]
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I haven't seen something like x[idx, ...] before. What are three dots doing in an array index?? I obviously need to spend a weekend studying numpy. –  Matt O'Brien Feb 9 at 4:10
    
@MattO'Brien -- ... is the python Ellipsis object. See this post for some links. stackoverflow.com/a/118395/748858 –  mgilson Feb 9 at 4:13
    
Err... apparently those links are broken. check this one instead. –  mgilson Feb 9 at 4:15
    
This works! (PS, what would I do if X were just a 1D array?) –  Matt O'Brien Feb 9 at 4:49
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