I have two tensors x for values and y for indexing. x.shape and y.shape are the same except the last dimension.
For example:
x=torch.tensor([[1, 6, 7, 5, 6],
[8, 6, 7, 8, 4],
[2, 8, 3, 5, 6]])
# x.shape:(3,5)
y=torch.tensor([[1, 2],
[2, 3],
[2, 2]])
# y.shape:(3,2)
Is there a simple way to slice it x[y]so that the result is:
torch.tensor([6,7],[7,8],[3,3])
What if x and y are higher dimensional tensors:
x.shape=(a,b,c,d)
y.shape=(a,b,c,e)
# a, b, c, d, e are positive integer
That's exactly what torch.gather does.
>> torch.gather(x, -1, y) # index along the last dimension
tensor([[6, 7],
[7, 8],
[3, 3]])
The exact same command works with your high-dimensional case as well.
Related
I have two numpy matrices (6 rows and 3 columns) :
a = np.array([[1,2,4],[3,6,2],[3,4,7],[9,7,7],[6,3,1],[3,5,9]])
b = np.array([[4,5,2],[9,2,5],[1,5,6],[4,5,6],[1,2,6],[6,4,3]])
a = array([[1, 2, 4],
[3, 6, 2],
[3, 4, 7],
[9, 7, 7],
[6, 3, 1],
[3, 5, 9]])
b = array([[4, 5, 2],
[9, 2, 5],
[1, 5, 6],
[4, 5, 6],
[1, 2, 6],
[6, 4, 3]])
I would like to calculate the pearson correlation coefficient between the first column of a and b, the second column of a and b and the third column of a and b.
The result would be a vector of 3 (3 correlation coeff).
One way using numpy.corrcoef and diagonal:
corr = np.corrcoef(a.T, b.T).diagonal(a.shape[1])
corr
Output:
array([-0.2324843 , -0.03631365, -0.18057878])
How can I retrieve a column vector from a 2d array given an indicator column vector?
Suppose I have
X = np.array([[1, 4, 6],
[8, 2, 9],
[0, 3, 7],
[6, 5, 1]])
and
S = np.array([0, 2, 1, 2])
Is there an elegant way to get from X and S the result array([1, 9, 3, 1]), which is equivalent to
np.array([x[s] for x, s in zip(X, S)])
You can achieve this using np.take_along_axis:
>>> np.take_along_axis(X, S[..., None], axis=1)
array([[1],
[9],
[3],
[1]])
You need to make sure both array arguments are of the same shape (or broadcasting can be applied), hence the S[..., None] broadcasting.
Of course your can reshape the returned value with a [:, 0] slice.
>>> np.take_along_axis(X, S[..., None], axis=1)[:, 0]
array([1, 9, 3, 1])
Alternatively you can just use indexing with an arangement:
>>> X[np.arange(len(S)), S[np.arange(len(S))]]
array([1, 9, 3, 1])
I believe this is also equivalent to np.diag(X[:, S]) but with unnecessary copying...
For 2d arrays
# Mention row numbers as one list and S which is column number as other
X[[0, 1, 2, 3], S]
# more general
X[np.indices(S.shape), S]
indexing_basics
I have a numpy array A as follows:
array([[1, 2, 3],
[4, 5, 6],
[7, 8, 9]])
and another numpy array column_indices_to_be_deleted as follows:
array([1, 0, 2])
I want to delete the element from every row of A specified by the column indices in column_indices_to_be_deleted. So, column index 1 from row 0, column index 0 from row 1 and column index 2 from row 2 in this case, to get a new array that looks like this:
array([[1, 3],
[5, 6],
[7, 8]])
What would be the simplest way of doing that?
One way with masking created with broadcatsed-comparison -
In [43]: a # input array
Out[43]:
array([[1, 2, 3],
[4, 5, 6],
[7, 8, 9]])
In [44]: remove_idx # indices to be removed from each row
Out[44]: array([1, 0, 2])
In [45]: n = a.shape[1]
In [46]: a[remove_idx[:,None]!=np.arange(n)].reshape(-1,n-1)
Out[46]:
array([[1, 3],
[5, 6],
[7, 8]])
Another mask based approach with the mask created with array-assignment -
In [47]: mask = np.ones(a.shape,dtype=bool)
In [48]: mask[np.arange(len(remove_idx)), remove_idx] = 0
In [49]: a[mask].reshape(-1,a.shape[1]-1)
Out[49]:
array([[1, 3],
[5, 6],
[7, 8]])
Another with np.delete -
In [64]: m,n = a.shape
In [66]: np.delete(a.flat,remove_idx+n*np.arange(m)).reshape(m,-1)
Out[66]:
array([[1, 3],
[5, 6],
[7, 8]])
I have tensor T of shape [batch_size, A] with values and tensor S of shape [batch_size] with shift parameters.
I would like to shift values in T[b] by S[b] positions to the right, the last S[b] elements of T[b] should be dropped and new elements should be set to 0.
So basically want to do something like:
for i in range(batch_size):
T[i] = zeros[:S[i]] + T[i, :A-S[i]]
Example:
For:
T = [[1, 2, 3], [4, 5, 6]]
S = [1, 2]
Return:
T' = [[0, 1, 2], [0, 0, 4]]
Is there some easy way to do it?
You can use tf.concat and tf.stack for that purpose:
T_shift = tf.zeros((batch_size, A), tf.float32)
tmp = []
for i in xrange(batch_size):
tmp.append(tf.concat([T_shift[i, :S[i, 0]],T[i, :17 - S[i,0]]], axis = 0))
T_shift = tf.stack(tmp)
If you are working in Tensorflow 2, you can use the tf.roll for that purpose:
"The elements are shifted positively (towards larger indices) by the
offset of shift along the dimension of axis. Negative shift values
will shift elements in the opposite direction. Elements that roll
passed the last position will wrap around to the first and vice versa.
Multiple shifts along multiple axes may be specified."
tf.roll(
input, shift, axis, name=None
)
# 't' is [0, 1, 2, 3, 4]
roll(t, shift=2, axis=0) ==> [3, 4, 0, 1, 2]
# shifting along multiple dimensions
# 't' is [[0, 1, 2, 3, 4], [5, 6, 7, 8, 9]]
roll(t, shift=[1, -2], axis=[0, 1]) ==> [[7, 8, 9, 5, 6], [2, 3, 4, 0, 1]]
# shifting along the same axis multiple times
# 't' is [[0, 1, 2, 3, 4], [5, 6, 7, 8, 9]]
roll(t, shift=[2, -3], axis=[1, 1]) ==> [[1, 2, 3, 4, 0], [6, 7, 8, 9, 5]]
I was using dynamic_rnn with an LSTMCell, which put out an LSTMStateTuple containing the inner state. Calling reshape on this object (by my mistake) results in a tensor without causing any error at graph creation. I didn't get any error at runtime when feeding input through the graph, either.
Code:
cell = tf.contrib.rnn.LSTMCell(size, state_is_tuple=True, ...)
outputs, states = tf.nn.dynamic_rnn(cell, inputs, ...)
print(states) # state is an LSTMStateTuple
states = tf.reshape(states, [-1, size])
print(states) # state is a tensor of shape [?, size]
Is this a bug (I ask because it's not documented anywhere)? What is the reshaped tensor holding?
I have conducted a similar experiment which may gives you some hints:
>>> s = tf.constant([[0, 0, 0, 1, 1, 1],
[2, 2, 2, 3, 3, 3]])
>>> t = tf.constant([[4, 4, 4, 5, 5, 5],
[6, 6, 6, 7, 7, 7]])
>>> g = tf.reshape((s, t), [-1, 3]) # <tf.Tensor 'Reshape_1:0' shape=(8, 3) dtype=int32>
>>> sess.run(g)
array([[0, 0, 0],
[1, 1, 1],
[2, 2, 2],
[3, 3, 3],
[4, 4, 4],
[5, 5, 5],
[6, 6, 6],
[7, 7, 7]], dtype=int32)
We can see that it just concatenates the two tensors in the first dimension and performs the reshaping. Since the LSTMStateTuple is like a namedtuple then it has the same effect as tuple and I think this is also what happens in your case.
Let's go further,
>>> st = tf.contrib.rnn.LSTMStateTuple(s, t)
>>> gg = tf.reshape(st, [-1, 3])
>>> sess.run(gg)
array([[0, 0, 0],
[1, 1, 1],
[2, 2, 2],
[3, 3, 3],
[4, 4, 4],
[5, 5, 5],
[6, 6, 6],
[7, 7, 7]], dtype=int32)
We can see that if we create a LSTMStateTuple, the result verifies our assumption.