TensorFlow: numpy.repeat() alternative

Just for 1-d tensors, I've made this function

def tf_repeat(y,repeat_num):   
        return tf.reshape(tf.tile(tf.expand_dims(y,axis=-1),[1,repeat_num]),[-1]) 

You can achieve the effect of np.repeat() using a combination of tf.tile() and tf.reshape():

idx = tf.range(len(yp))
idx = tf.reshape(idx, [-1, 1])    # Convert to a len(yp) x 1 matrix.
idx = tf.tile(idx, [1, len(yp)])  # Create multiple columns.
idx = tf.reshape(idx, [-1])       # Convert back to a vector.

You can simply compute jdx using tf.tile():

jdx = tf.range(len(yp))
jdx = tf.tile(jdx, [len(yp)])

For the indexing, you could try using tf.gather() to extract non-contiguous slices from the yp tensor:

s = tf.gather(yp, idx) - tf.gather(yp, jdx)

According to tf api document, tf.keras.backend.repeat_elements() does the same work with np.repeat() . For example,

x = tf.constant([1, 3, 3, 1], dtype=tf.float32)
rep_x = tf.keras.backend.repeat_elements(x, 5, axis=0)
# result: [1. 1. 1. 1. 1. 3. 3. 3. 3. 3. 3. 3. 3. 3. 3. 1. 1. 1. 1. 1.]

It looks like your question is so popular that people refer it on TF tracker. Sadly the same function is not still implemented in TF.

You can implement it by combining tf.tile, tf.reshape, tf.squeeze. Here is a way to convert examples from np.repeat:

import numpy as np
import tensorflow as tf

x = [[1,2],[3,4]]
print np.repeat(3, 4)
print np.repeat(x, 2)
print np.repeat(x, 3, axis=1)

x = tf.constant([[1,2],[3,4]])
with tf.Session() as sess:
    print sess.run(tf.tile([3], [4]))
    print sess.run(tf.squeeze(tf.reshape(tf.tile(tf.reshape(x, (-1, 1)), (1, 2)), (1, -1))))
    print sess.run(tf.reshape(tf.tile(tf.reshape(x, (-1, 1)), (1, 3)), (2, -1)))

In the last case where repeats are different for each element you most probably will need loops.