tf.cond(pred, fn1, fn2, name=None)

tf.cond(pred, fn1, fn2, name=None)

See the guide: Control Flow > Control Flow Operations

Return either fn1() or fn2() based on the boolean predicate pred.

fn1 and fn2 both return lists of output tensors. fn1 and fn2 must have the same non-zero number and type of outputs.

Note that the conditional execution applies only to the operations defined in fn1 and fn2. Consider the following simple program:

z = tf.multiply(a, b)
result = tf.cond(x < y, lambda: tf.add(x, z), lambda: tf.square(y))

If x < y, the tf.add operation will be executed and tf.square operation will not be executed. Since z is needed for at least one branch of the cond, the tf.mul operation is always executed, unconditionally. Although this behavior is consistent with the dataflow model of TensorFlow, it has occasionally surprised some users who expected a lazier semantics.

Args:

• pred: A scalar determining whether to return the result of fn1 or fn2.
• fn1: The callable to be performed if pred is true.
• fn2: The callable to be performed if pref is false.
• name: Optional name prefix for the returned tensors.

Returns:

Tensors returned by the call to either fn1 or fn2. If the callables return a singleton list, the element is extracted from the list.

Raises:

• TypeError: if fn1 or fn2 is not callable.
• ValueError: if fn1 and fn2 do not return the same number of tensors, or return tensors of different types.

Example:

x = tf.constant(2)
y = tf.constant(5)
def f1(): return tf.multiply(x, 17)
def f2(): return tf.add(y, 23)
r = tf.cond(tf.less(x, y), f1, f2)
# r is set to f1().
# Operations in f2 (e.g., tf.add) are not executed.

Defined in tensorflow/python/ops/control_flow_ops.py.