Menu
Say I have a 3 dimensional numpy array:
np.random.seed(1145)
A = np.random.random((5,5,5))
and I have two lists of indices corresponding to the 2nd and 3rd dimensions:
second = [1,2]
third = [3,4]
and I want to select the elements in the numpy array corresponding to
A[:][second][third]
so the shape of the sliced array would be (5,2,2) and
A[:][second][third].flatten()
would be equivalent to to:
In [226]:
for i in range(5):
for j in second:
for k in third:
print A[i][j][k]
0.556091074129
0.622016249651
0.622530505868
0.914954716368
0.729005532319
0.253214472335
0.892869371179
0.98279375528
0.814240066639
0.986060321906
0.829987410941
0.776715489939
0.404772469431
0.204696635072
0.190891168574
0.869554447412
0.364076117846
0.04760811817
0.440210532601
0.981601369658
Is there a way to slice a numpy array in this way? So far when I try A[:][second][third] I get IndexError: index 3 is out of bounds for axis 0 with size 2 because the [:] for the first dimension seems to be ignored.
914
For getting n-largest values from a NumPy array we have to first sort the NumPy array using numpy. argsort() function of NumPy then applying slicing concept with negative indexing. Return: [index_array, ndarray] Array of indices that sort arr along the specified axis.
The [:, :] stands for everything from the beginning to the end just like for lists. The difference is that the first : stands for first and the second : for the second dimension. a = numpy. zeros((3, 3)) In [132]: a Out[132]: array([[ 0., 0., 0.], [ 0., 0., 0.], [ 0., 0., 0.]])
Slicing arrays Slicing in python means taking elements from one given index to another given index. We pass slice instead of index like this: [start:end] . We can also define the step, like this: [start:end:step] .
Slice Two-dimensional Numpy ArraysTo slice elements from two-dimensional arrays, you need to specify both a row index and a column index as [row_index, column_index] . For example, you can use the index [1,2] to query the element at the second row, third column in precip_2002_2013 .
One way would be to use np.ix_:
>>> out = A[np.ix_(range(A.shape[0]),second, third)]
>>> out.shape
(5, 2, 2)
>>> manual = [A[i,j,k] for i in range(5) for j in second for k in third]
>>> (out.ravel() == manual).all()
True
Downside is that you have to specify the missing coordinate ranges explicitly, but you could wrap that into a function.
181
If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
Donate Us With
