There is list of list of tuples:
[[(0, 0.5), (1, 0.6)], [(4, 0.01), (5, 0.005), (6, 0.002)], [(1,0.7)]]
I need to get matrix X x Y:
x = num of sublists
y = max among second eleme throught all pairs
elem[x,y] = second elem for x sublist if first elem==Y
| 0 | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| 0.5 | 0.6 | 0 | 0 | 0 | 0 | 0 |
| 0 | 0 | 0 | 0 | 0.01 | 0.005 | 0.002 |
| 0 | 0.7 | 0 | 0 | 0 | 0 | 0 |
You can figure out the array's dimensions the following way. The Y dimension is the number of sublists
>>> data = [[(0, 0.5), (1, 0.6)], [(4, 0.01), (5, 0.005), (6, 0.002)], [(1,0.7)]]
>>> dim_y = len(data)
>>> dim_y
3
The X dimension is the largest [0] index of all of the tuples, plus 1.
>>> dim_x = max(max(i for i,j in sub) for sub in data) + 1
>>> dim_x
7
So then initialize an array of all zeros with this size
>>> import numpy as np
>>> arr = np.zeros((dim_x, dim_y))
>>> arr
array([[0., 0., 0.],
[0., 0., 0.],
[0., 0., 0.],
[0., 0., 0.],
[0., 0., 0.],
[0., 0., 0.]])
Now to fill it, enumerate over your sublists to keep track of the y index. Then for each sublist use the [0] for the x index and the [1] for the value itself
for y, sub in enumerate(data):
for x, value in sub:
arr[x,y] = value
Then the resulting array should be populated (might want to transpose to look like your desired dimensions).
>>> arr.T
array([[0.5 , 0.6 , 0. , 0. , 0. , 0. , 0. ],
[0. , 0. , 0. , 0. , 0.01 , 0.005, 0.002],
[0. , 0.7 , 0. , 0. , 0. , 0. , 0. ]])
As I commented in the accepted answer, data is 'ragged' and can't be made into a array.
Now if the data had a more regular form, a no-loop solution is possible. But conversion to such a form requires the same double looping!
In [814]: [(i,j,v) for i,row in enumerate(data) for j,v in row]
Out[814]:
[(0, 0, 0.5),
(0, 1, 0.6),
(1, 4, 0.01),
(1, 5, 0.005),
(1, 6, 0.002),
(2, 1, 0.7)]
'transpose' and separate into 3 variables:
In [815]: I,J,V=zip(*_)
In [816]: I,J,V
Out[816]: ((0, 0, 1, 1, 1, 2), (0, 1, 4, 5, 6, 1), (0.5, 0.6, 0.01, 0.005, 0.002, 0.7))
I stuck with the list transpose here so as to not convert the integer indices to floats. It may also be faster, since making an array from a list isn't a time-trivial task.
Now we can assign values via numpy magic:
In [819]: arr = np.zeros((3,7))
In [820]: arr[I,J]=V
In [821]: arr
Out[821]:
array([[0.5 , 0.6 , 0. , 0. , 0. , 0. , 0. ],
[0. , 0. , 0. , 0. , 0.01 , 0.005, 0.002],
[0. , 0.7 , 0. , 0. , 0. , 0. , 0. ]])
I,J,V could also be used as input to a scipy.sparse.coo_matrix call, making a sparse matrix.
Speaking of a sparse matrix, here's what a sparse version of arr looks like:
In list-of-lists format:
In [822]: from scipy import sparse
In [823]: M = sparse.lil_matrix(arr)
In [824]: M
Out[824]:
<3x7 sparse matrix of type '<class 'numpy.float64'>'
with 6 stored elements in List of Lists format>
In [825]: M.A
Out[825]:
array([[0.5 , 0.6 , 0. , 0. , 0. , 0. , 0. ],
[0. , 0. , 0. , 0. , 0.01 , 0.005, 0.002],
[0. , 0.7 , 0. , 0. , 0. , 0. , 0. ]])
In [826]: M.rows
Out[826]: array([list([0, 1]), list([4, 5, 6]), list([1])], dtype=object)
In [827]: M.data
Out[827]:
array([list([0.5, 0.6]), list([0.01, 0.005, 0.002]), list([0.7])],
dtype=object)
and the more common coo format:
In [828]: Mc=M.tocoo()
In [829]: Mc.row
Out[829]: array([0, 0, 1, 1, 1, 2], dtype=int32)
In [830]: Mc.col
Out[830]: array([0, 1, 4, 5, 6, 1], dtype=int32)
In [831]: Mc.data
Out[831]: array([0.5 , 0.6 , 0.01 , 0.005, 0.002, 0.7 ])
and the csr used for most calculations:
In [832]: Mr=M.tocsr()
In [833]: Mr.data
Out[833]: array([0.5 , 0.6 , 0.01 , 0.005, 0.002, 0.7 ])
In [834]: Mr.indices
Out[834]: array([0, 1, 4, 5, 6, 1], dtype=int32)
In [835]: Mr.indptr
Out[835]: array([0, 2, 5, 6], dtype=int32)
