Converting a list of ints, tuples into an numpy array

Question

I have a list of [float, (float,float,float..) ] ... Which is basically an n-dimensional point along with a fitness value for each point. For eg.

4.3, (2,3,4)
3.2, (1,3,5)
.
.
48.2, (23,1,32)

I wish to randomly sample one point based upon the fitness values. I decided the best way to do this would be to use numpy.random.choice(range(n), 1, plist[:,:1,:1])

However, i need to convert this into an numpy array, for which i tried

>> pArr = np.array( plist ) 
ValueError: setting an array element with a sequence

I got the same error for np.asarray(plist) as well.. any suggestions??

Is it possible to convert it into a 3 dimensional array? or do i change the formation of this list itself to make it 3 dimensional? — Nicomoto
– Nicomoto, Commented Sep 3, 2012 at 20:11

Pierre GM · Accepted Answer · 2012-09-03 20:26:59Z

1

The following should work:

A = np.array([tuple(i) for i in initial_list],dtype=[('fitness',float),('point',(float,3))])

with initial_list = [[4.3, (2, 3, 4)], [3.2, (1, 3, 5)], ...]. Note that we need to transform each item of initial_list into a tuple for that trick to work, else NumPy cannot recognize the structure.

Your fitness entries are now accessible as A['fitness'], with the corresponding points as A['point']. If you select a list of actual fitness entries, indices, the corresponding points are given by A['point'][indices], which is a simple (n,3) array.

edited Sep 3, 2012 at 20:26

answered Sep 3, 2012 at 20:12

Pierre GM

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Nicomoto Over a year ago

That does work thanks!! but does not really help in this situation. Since i am trying to slice out the fitness values for random selection. Which brings me back to the initial problem of slicing one dimension in a multi dimensional array ...!

BrenBarn · Accepted Answer · 2012-09-03 20:11:43Z

1

Your question is difficult to understand. Is this what you're trying to do?

>>> x
[[4.3, (2, 3, 4)], [3.2, (1, 3, 5)], [48.2, (23, 1, 32)]]
>>> np.array([(a, b, c, d) for a, (b, c, d) in x])
array([[  4.3,   2. ,   3. ,   4. ],
       [  3.2,   1. ,   3. ,   5. ],
       [ 48.2,  23. ,   1. ,  32. ]])

answered Sep 3, 2012 at 20:11

BrenBarn

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2 Comments

Nicomoto Over a year ago

Well, the fitness is different from the coordinates, and i would like to keep it that way.. But I guess i could do this and add 1 where ever i wish to access the coordinates!

Nicomoto Over a year ago

The only problem being that, my dimensions are variable. So I may have (b,c,d) or (b,c,d,e...) in the tuple, can this be generalized?

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