How to traverse the result of tf.unqiue? - tensorflow

Question

After invoking tf.unqiue, the shape of tensor will be unknown, but I want to traverse the result of tf.unqiue
Suppose tensor = tf.unqiue(...)
I have tried:
for i in tf.range(tf.shape(tensor)[0])
tf.unstack(tensor, num=tf.shape(tensor)[0])
tf.split(tensor, num_or_size_splits=tf.shape(tensor)[0])
All of them can't work, because these functions all need static shape or num/num_or_size_splits = integer? So how can I traverse tensor?
Update
Example
I have two 1-D tensor with the same shape
x=[1,3,2,1,3]
y=[3,6,5,8,9]
I want to do like this:
x_u = unique(x) # [1,3,2]
get bool_mask, and slice y
for i in x_u:
y[x == i]
when i=1, y[x==i]=y[[True,False,False,True,False]], and I can get y[0] and y[3]
when i=3, I can get y[1] and y[4]
when i=2, I can get y[2]
Solution
After some trials, that may be a solution.
Try to use tf.while_loop:
import tensorflow as tf
import numpy as np
x = tf.constant(np.array([1,3,2,1,3]), dtype='int32')
y = tf.constant(np.array([3,6,5,8,9]), dtype='int32')
x_u, _ = tf.unique(x)
n = tf.shape(x_u)[0]
for_i = tf.constant(0)
re = tf.constant([], dtype=tf.int32)
cond = lambda i, res: i<n
def body(i, res):
x_0 = tf.slice(x_u, [i], [1])
selected = tf.boolean_mask(y, tf.equal(x_0, x))
return i+1, tf.concat([res, selected], axis=0)
op = tf.while_loop(cond, body, [for_i, re], shape_invariants=[for_i.get_shape(), tf.TensorShape([None])])
print(op[1].shape)
with tf.Session() as sess:
print(sess.run(op[1]))

Answer 1

I just tried this:
import tensorflow as tf
import numpy as np
a = tf.constant(np.random.randn(200), dtype='float32')
b = tf.unique(a)
print b[0] #Tensor("Unique:0", shape=(?,), dtype=float32)
c = tf.map_fn(lambda x: x*x, b[0])
init = tf.global_variables_initializer()
sess = tf.Session()
sess.run(init)
d = sess.run(c)
print d
And it's working without knowing the shape of 'b' here. Be careful, tf.unique is returning a tuple (Tensor, Tensor) with the values and their indicies.
Update
This is the only way I found to do this, your result cannot have an insconsistent shape in Tensorflow.
import tensorflow as tf
import numpy as np
x = tf.constant(np.array([1,3,2,1,3]), dtype='int32')
y = tf.constant(np.array([3,6,5,8,9]), dtype='int32')
x_u = tf.unique(x)
eq = tf.equal(x, tf.expand_dims(x_u[0],1))
y_masked = y*tf.cast(eq, tf.int32)
init = tf.global_variables_initializer()
sess = tf.Session()
sess.run(init)
e = sess.run(y_masked)
print e
tf.boolean_mask can be used as well but you are going to get a flat output.
Last Update
This is what you want, way faster than what you proposed and it was already explained the line just above.
import tensorflow as tf
import numpy as np
x = tf.constant(np.array([1,3,2,1,3]), dtype='int32')
y = tf.constant(np.array([3,6,5,8,9]), dtype='int32')
x_u, _ = tf.unique(x)
eq = tf.equal(x, tf.expand_dims(x_u,1))
y_tiled = tf.tile(tf.expand_dims(y, 0), [tf.shape(x_u)[0], 1])
y_masked = tf.boolean_mask(y_tiled, eq)
init = tf.global_variables_initializer()
sess = tf.Session()
sess.run(init)
e = sess.run(y_masked)
print e