Suppose I have
[[array([x1, y1]), z1]
[array([x2, y1]), z2]
......
[array([xn, yn]), zn]
]
And I want to find the index of array([x5, y5]). How can find effieciently using NumPy?
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Merlin
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poke19962008
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Why do you have a python list of 1D numpy arrays instead of just a 2D numpy array? – pretzlstyle Aug 11 '16 at 20:25
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I was making Tic Tac Toe using reinforment learning so every row contains the board configuration (every board can be distinguished by 2 attributes) and the V(s) score – poke19962008 Aug 11 '16 at 20:28
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1I don;t know if any numpy functions will save you since you have everything in an awkward mix of python lists and numpy arrays. Why not just standard python `.index()` – pretzlstyle Aug 11 '16 at 20:31
1 Answers
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To start off, owing to the mixed data format, I don't think you can extract the arrays in a vectorized manner. Thus, you can use loop comprehension to extract the first element corresponding to the arrays from each list element as a 2D array. So, let's say A is the input list, we would have -
arr = np.vstack([a[0] for a in A])
Then, simply do the comparison in a vectorized fashion using NumPy's broadcasting feature, as it will broadcast that comparison along all the rows and look all matching rows with np.all(axis=1). Finally, use np.flatnonzero to get the final indices. Thus, the final peace of the puzzle would be -
idx = np.flatnonzero((arr == search1D).all(1))
You can read up on the answers to this post to see other alternatives to get indices in such a 1D array searching in 2D array problem.
Sample run -
In [140]: A
Out[140]:
[[array([3, 4]), 11],
[array([2, 1]), 12],
[array([4, 2]), 16],
[array([2, 1]), 21]]
In [141]: search1D = [2,1]
In [142]: arr = np.vstack([a[0] for a in A]) # Extract 2D array
In [143]: arr
Out[143]:
array([[3, 4],
[2, 1],
[4, 2],
[2, 1]])
In [144]: np.flatnonzero((arr == search1D).all(1)) # Finally get indices
Out[144]: array([1, 3])
