You cannot use a function like that, because the parameter x is a TensorFlow tensor, not a Python value. So, in order for that to work, you would have to turn your dictionary into a tensor as well, but it's not so simple because keys in the dictionary may not be sequential.
You can instead solve this problem without mapping, but instead doing something similar to what is proposed here for NumPy. In TensorFlow, you could implement it like this:
import tensorflow as tf
def replace_by_dict(x, d):
# Get keys and values from dictionary
keys, values = zip(*d.items())
keys = tf.constant(keys, x.dtype)
values = tf.constant(values, x.dtype)
# Make a sequence for the range of values in the input
v_min = tf.reduce_min(x)
v_max = tf.reduce_max(x)
r = tf.range(v_min, v_max + 1)
r_shape = tf.shape(r)
# Mask replacements that are out of the input range
mask = (keys >= v_min) & (keys <= v_max)
keys = tf.boolean_mask(keys, mask)
values = tf.boolean_mask(values, mask)
# Replace values in the sequence with the corresponding replacements
scatter_idx = tf.expand_dims(keys, 1) - v_min
replace_mask = tf.scatter_nd(
scatter_idx, tf.ones_like(values, dtype=tf.bool), r_shape)
replace_values = tf.scatter_nd(scatter_idx, values, r_shape)
replacer = tf.where(replace_mask, replace_values, r)
# Gather the replacement value or the same value if it was not modified
return tf.gather(replacer, x - v_min)
# Test
kv_dict = {1: 11, 2: 12, 3: 13}
with tf.Graph().as_default(), tf.Session() as sess:
a = tf.constant([1, 2, 3])
print(sess.run(replace_by_dict(a, kv_dict)))
# [11, 12, 13]
This will allow you to have values in the input tensor without replacements (left as they are), and also does not require to have all the replacement values in the tensor. It should be efficient unless the minimum and maximum values in your input are very far away.