How to find an index of the first matching element in TensorFlow

Question

I am looking for a TensorFlow way of implementing something similar to Python\'s list.index() function.

Given a matrix and a value to find, I want to know the first

Answer 1

Here is a solution which also considers the case the element is not included by the matrix (solution from github repository of DeepMind)

def get_first_occurrence_indices(sequence, eos_idx):
    '''
    args:
        sequence: [batch, length]
        eos_idx: scalar
    '''
    batch_size, maxlen = sequence.get_shape().as_list()
    eos_idx = tf.convert_to_tensor(eos_idx)
    tensor = tf.concat(
            [sequence, tf.tile(eos_idx[None, None], [batch_size, 1])], axis = -1)
    index_all_occurrences = tf.where(tf.equal(tensor, eos_idx))
    index_all_occurrences = tf.cast(index_all_occurrences, tf.int32)
    index_first_occurrences = tf.segment_min(index_all_occurrences[:, 1], 
index_all_occurrences[:, 0])
    index_first_occurrences.set_shape([batch_size])
    index_first_occurrences = tf.minimum(index_first_occurrences + 1, maxlen)
    
    return index_first_occurrences

And:

import tensorflow as tf
mat = tf.Variable([[1,2,3,4,5], [2,3,4,5,6], [3,4,5,6,7], [0,0,0,0,0]], dtype = tf.int32)
idx = 3
first_occurrences = get_first_occurrence_indices(mat, idx)

sess = tf.InteractiveSession()
sess.run(tf.global_variables_initializer())
sess.run(first_occurrence) # [3, 2, 1, 5]

Answer 2

Looks a little ugly but works (assuming m and val are both tensors):

idx = list()
for t in tf.unpack(m, axis=0):
    idx.append(tf.reduce_min(tf.where(tf.equal(t, val))))
idx = tf.pack(idx, axis=0)

EDIT: As Yaroslav Bulatov mentioned, you could achieve the same result with tf.map_fn:

def index1d(t):
    return tf.reduce_min(tf.where(tf.equal(t, val)))

idx = tf.map_fn(index1d, m, dtype=tf.int64)

Answer 3

It seems that tf.argmax works like np.argmax (according to the test), which will return the first index when there are multiple occurrences of the max value. You can use tf.argmax(tf.cast(tf.equal(m, val), tf.int32), axis=1) to get what you want. However, currently the behavior of tf.argmax is undefined in case of multiple occurrences of the max value.

If you are worried about undefined behavior, you can apply tf.argmin on the return value of tf.where as @Igor Tsvetkov suggested. For example,

# test with tensorflow r1.0
import tensorflow as tf

val = 3
m = tf.placeholder(tf.int32)
m_feed = [[0  ,   0, val,   0, val],
          [val,   0, val, val,   0],
          [0  , val,   0,   0,   0]]

tmp_indices = tf.where(tf.equal(m, val))
result = tf.segment_min(tmp_indices[:, 1], tmp_indices[:, 0])

with tf.Session() as sess:
    print(sess.run(result, feed_dict={m: m_feed})) # [2, 0, 1]

Note that tf.segment_min will raise InvalidArgumentError when there is some row containing no val. In your code row_elems.index(val) will raise exception too when row_elems don't contain val.

Answer 4

Here is another solution to the problem, assuming there is a hit on every row.

import tensorflow as tf

val = 3
m = tf.constant([
    [0  ,   0,   val,   0, val],
    [val,   0,   val, val,   0],
    [0  , val,     0,   0,   0]])

# replace all entries in the matrix either with its column index, or out-of-index-number
match_indices = tf.where(                          # [[5, 5, 2, 5, 4],
    tf.equal(val, m),                              #  [0, 5, 2, 3, 5],
    x=tf.range(tf.shape(m)[1]) * tf.ones_like(m),  #  [5, 1, 5, 5, 5]]
    y=(tf.shape(m)[1])*tf.ones_like(m))

result = tf.reduce_min(match_indices, axis=1)

with tf.Session() as sess:
    print(sess.run(result)) # [2, 0, 1]