I have a tensorflow situation. I want to find the intersection of two 2-D tensors which have different shapes.
Example:
object_ids_ [[0 0]
[0 1]
[1 1]]
object_ids_more_07_ [[0 0]
[0 1]
[0 2]
[1 0]
[1 2]]
The output I am looking for is:
[[0,0],
[0,1]]
I came across "tf.sets.set_intersection", tensorflow page: https://www.tensorflow.org/api_docs/python/tf/sets/set_intersection
But couldn't perform it for tensors with different shapes. Another implementation I found is at:
Find the intersection of two tensors. Return the sorted, unique values that are in both of the input tensors
but had a hard time replicating it for 2D tensors.
Any help would be appreciated , thanks
One way to do is to subtract->abs->sum of all the combinations and then get indices where it matches zero. Can be achieved using broadcasting.
a = tf.constant([[0,0],[0,1],[1,1]])
b = tf.constant([[0, 0],[0, 1],[0,2],[1, 0],[1, 2]])
find_match = tf.reduce_sum(tf.abs(tf.expand_dims(b,0) - tf.expand_dims(a,1)),2)
indices = tf.transpose(tf.where(tf.equal(find_match, tf.zeros_like(find_match))))[0]
out = tf.gather(a, indices)
with tf.Session() as sess:
print(sess.run(out))
#Output
#[[0 0]
#[0 1]]
Related
I am looking for an efficient way to compute the indices of the binnings of bincount as a ndarray.
To illustrate:
>>> x = np.array([0, 1, 1, 0, 2])
>>> b = np.bincount(x)
>>> b
[2 2 1]
I am now looking for an ndarray that represents the indices of the elements of each bin:
[0 3 1 2 4]
I am looking for a fast numpy solution that should not contain loops. Anyone knows how to implement this? Thanks very much in advance!
I have trained a model based on transfer learning via Tensorflow Hub. I have been looking at many places for hints on producing Confusion matrix but I haven't been able to get to the right solution.
Does anyone know if this is possible?
The last thing I tried was to write the results in an Excel sheet but I couldn't find a formula for the multi-class computation of confusion matrix in Excel.
Any help will be great!!
You can try tf.math.confusion_matrix function.
It computes the confusion matrix from predictions and labels.
See https://www.tensorflow.org/api_docs/python/tf/math/confusion_matrix.
Example:
my_confusion_matrix = tf.math.confusion_matrix(labels=[1, 2, 4], predictions=[2, 2, 3])
with tf.Session() as sess:
print(sess.run(my_confusion_matrix))
# Prints #
[[0 0 0 0 0]
[0 0 1 0 0]
[0 0 1 0 0]
[0 0 0 0 0]
[0 0 0 1 0]]
# Labels are assumed to be [0, 1, 2, 3, 4] thus resulting in 5X5 confusion matrix
I am new in TensorFlow. If there is a [10,1] tensor, I want to find out all rows with the same value and their subscript.
For example, there is a tensor like [[1],[2],[3],[4],[5],[1],[2],[3],[4],[6]].
By comparing each element in the matrix, it is easy to get a dictionary structure like
{‘1’: [0,5], ‘2’: [1,6], ‘3’: [2, 7], ‘4’: [3, 8], ‘5’: [4], ‘6’: [9]} in python, which can record how many times each element occurs in the matrix.
I expect to achieve this result in TensorFlow. Could someone please give me a hand? Thanks a lot.
I think this is a longer method. Still the elements and indices are not associated in a data structure.
Other shorter methods must be there.
t = tf.constant([[1],[2],[3],[4],[5],[1],[2],[3],[4],[6]])
sess = tf.InteractiveSession()
sess.run(tf.global_variables_initializer())
y, idx, count =tf.unique_with_counts(tf.squeeze(t))
y1, idx1, count1 = sess.run([y,idx,count])
for i in range(len(y1)) :
print( sess.run( tf.where(tf.equal(t,y1[i]))[:2,-2]))
Output is
[0 5]
[1 6]
[2 7]
[3 8]
[4]
[9]
I have a matrix:
Params =
[[ 0 1 2 3]
[ 4 5 6 7]
[ 8 9 10 11]]
For each row I want to select some elements using column indices:
col_indices =
[[0 1]
[1 2]
[2 3]]
In Numpy, I can create row indices:
row_indices =
[[0 0]
[1 1]
[2 2]]
and do params[row_indices, col_indices]
In TenforFlow, I did this:
tf_params = tf.constant(params)
tf_col_indices = tf.constant(col_indices, dtype=tf.int32)
tf_row_indices = tf.constant(row_indices, dtype=tf.int32)
tf_params[row_indices, col_indices]
But there raised an error:
ValueError: Shape must be rank 1 but is rank 3
What does it mean? How should I do this kind of indexing properly?
Thanks!
Tensor rank (sometimes referred to as order or degree or n-dimension) is the number of dimensions of the tensor. For example, the following tensor (defined as a Python list) has a rank of 2:
t = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
A rank two tensor is what we typically think of as a matrix, a rank one tensor is a vector. For a rank two tensor you can access any element with the syntax t[i, j]. For a rank three tensor you would need to address an element with t[i, j, k]. See this for more details.
ValueError: Shape must be rank 1 but is rank 3 means you are trying to create a 3-tensor (cube of numbers) instead of a vector.
To see how you can declare tensor constants of different shape, you can see this.
Suppose a Tensor containing :
[[0 0 1]
[0 1 0]
[1 0 0]]
How to get the dense representation in a native way (without using numpy or iterations) ?
[2,1,0]
There is tf.one_hot() to do the inverse, there is also tf.sparse_to_dense() that seems to do it but I was not able to figure out how to use it.
tf.argmax(x, axis=1) should do the job.
vec = tf.constant([[0, 0, 1], [0, 1, 0], [1, 0, 0]])
locations = tf.where(tf.equal(vec, 1))
# This gives array of locations of "1" indices below
# => [[0, 2], [1, 1], [2, 0]])
# strip first column
indices = locations[:,1]
sess = tf.Session()
print(sess.run(indices))
# => [2 1 0]
TensorFlow does not have a native dense to sparse conversion function/helper. Given that the input array is a dense tensor, such as the one you provided, you can define a function to convert a dense tensor to a sparse tensor.
def dense_to_sparse(dense_tensor):
where_dense_non_zero = tf.where(tf.not_equal(dense_tensor, 0))
indices = where_dense_non_zero
values = tf.gather_nd(dense_tensor, where_dense_non_zero)
shape = dense_tensor.get_shape()
return tf.SparseTensor(
indices=indices,
values=values,
shape=shape
)
This helper function finds the indices and values where the Tensor is non-zero and outputs a Sparse tensor with those indices and values. Additionally, the shape is effectively copied over.
You do not want to use tf.sparse_to_dense as that gives you the opposite representation. If you want your output to be [2, 1, 0] instead, you'll need to index the indices. First, you'll need the indices where the array isn't 0:
indices = tf.where(tf.not_equal(dense_tensor, 0))
Then, you'll need to access the tensor using slicing/indicing:
output = indices[:, 1]
You might notice that 1 in the slice above is equivalent to the dimension of the tensor - 1. Therefore, to make these value generic, you could do something like:
output = indices[:, len(dense_tensor.get_shape()) - 1]
Although I'm not exactly sure what you'd do with these values (the value of the column where the value is). Hope this helped!
EDIT: Yaroslav's answer is better if you're looking for the indices/locations of where the input tensor if 1; it won't be extensible for tensors with non-1/0 values if that is required.