Tensorflow: InvalidArgumentError: Input ... incompatible with expected float_ref

Question

The following code results in a very unhelpful error:

import tensorflow as tf

x = tf.Variable(tf.constant(0.), name="x")
with tf.Session() as s:
    val = s.run(x.assign(1))
    print(val)  # 1
    val = s.run(x, {x: 2})
    print(val)  # 2
    val = s.run(x.assign(1), {x: 0.})   # InvalidArgumentError

tensorflow.python.framework.errors_impl.InvalidArgumentError: Input 0 of node Assign_1 was passed float from _arg_x_0_0:0 incompatible with expected float_ref.

How did I get this error?
Why do I get this error?

Hi, Did you fix the issue?

Nael Marwan
– Nael Marwan

2018-07-05 08:23:54 +00:00
Commented Jul 5, 2018 at 8:23 — Nael Marwan
– Nael Marwan, Commented Jul 5, 2018 at 8:23

reubenjohn · Accepted Answer · 2018-07-28 05:01:57Z

2

Here's what I could infer.

How did I get this error?
This error is seen when attempting to perform the following two operations in a single session run:

A Tensorflow variable is assigned a value
That same variable is also passed a value as part of the feed_dict

This is why the first 2 runs succeed (they both don't simultaneously attempt to perform both these operations).

Why do I get this error?
I am not sure, but I don't think this was an intentional design choice by Google. Here's my explanation:

Firstly, the TF(TensorFlow) source code (basically) resolves x.assign(1) to tf.assign(x, 1) which gives us a hint for better understand the error message when it says Input 0.
The error message refers to x when it says Input 0 of the assign op. It goes on to say that the first argument of the assign op was passed float from _arg_x_0_0:0.

TLDR
Thus for a run where a TF variable is provided as a feed, that variable will no longer be treated as a variable (but instead as the value it was assigned), and thus any attempts at further assigning a value to it would be erroneous since only TF variables can be assigned a value in the graph.

Fix

If your graph has variable assignment operation, don't pass a value to that same variable in your feed_dict. ¯_(ツ)_/¯. Assuming you're using the feed_dict to provide an initial value, you could instead assign it a value in a prior session run. Or, leverage tf.control_dependencies when building your graph to assign it an initial value from a placeholder as shown below:

import tensorflow as tf

x = tf.Variable(tf.constant(0.), name="x")

initial_x = tf.placeholder(tf.float32)
assign_from_placeholder = x.assign(initial_x)
with tf.control_dependencies([assign_from_placeholder]):
    x_assign = x.assign(1)

with tf.Session() as s:
    val = s.run(x_assign, {initial_x: 0.})  # Success!

edited Jul 28, 2018 at 5:01

answered Mar 25, 2018 at 18:12

reubenjohn

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4 Comments

Nael Marwan Over a year ago

Hi, how to fix the issue?

reubenjohn Over a year ago

The solution is edited to explicitly specify the fix.

Nael Marwan Over a year ago

Do you know what does the error mean if we don't have assignment operation?

reubenjohn Over a year ago

Sorry, not sure. Even though you didn't create an assignment operator explicitly, It's likely created by something like the update step of the optimizer. You're probably additionally passing a value to that variable through feed_dict.

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