PLH
tensorflow
type_error
ai_generated
true
RuntimeError: tf.placeholder() is not compatible with eager execution.
ID: tensorflow/placeholder-not-supported-eager
95%Fix Rate
90%Confidence
1Evidence
2023-10-05First Seen
Version Compatibility
| Version | Status | Introduced | Deprecated | Notes |
|---|---|---|---|---|
| tensorflow 2.15 | active | — | — | — |
| tensorflow 2.16 | active | — | — | — |
| tensorflow 2.17 | active | — | — | — |
Root Cause
Using tf.placeholder in TensorFlow 2.x where eager execution is enabled by default; placeholders are only for graph mode.
generic中文
在默认启用即时执行的 TensorFlow 2.x 中使用 tf.placeholder;占位符仅适用于图模式。
Official Documentation
https://www.tensorflow.org/guide/eagerWorkarounds
-
95% success Replace `tf.placeholder(tf.float32, shape=[None, 10])` with function arguments in eager mode. For example: `def model_fn(x): return tf.keras.layers.Dense(10)(x)`. Then call `model_fn(input_tensor)` directly.
Replace `tf.placeholder(tf.float32, shape=[None, 10])` with function arguments in eager mode. For example: `def model_fn(x): return tf.keras.layers.Dense(10)(x)`. Then call `model_fn(input_tensor)` directly.
-
90% success If graph mode is required, use `tf.function` with input signatures: `@tf.function(input_signature=[tf.TensorSpec(shape=[None, 10], dtype=tf.float32)])`.
If graph mode is required, use `tf.function` with input signatures: `@tf.function(input_signature=[tf.TensorSpec(shape=[None, 10], dtype=tf.float32)])`.
中文步骤
Replace `tf.placeholder(tf.float32, shape=[None, 10])` with function arguments in eager mode. For example: `def model_fn(x): return tf.keras.layers.Dense(10)(x)`. Then call `model_fn(input_tensor)` directly.
If graph mode is required, use `tf.function` with input signatures: `@tf.function(input_signature=[tf.TensorSpec(shape=[None, 10], dtype=tf.float32)])`.
Dead Ends
Common approaches that don't work:
-
50% fail
Disabling eager execution with `tf.compat.v1.disable_eager_execution()` works but loses TF2 benefits and is deprecated.
-
80% fail
Replacing placeholder with `tf.Variable` fails because variables hold specific values, not input slots.