tf.foldl

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foldl on the list of tensors unpacked from elems on dimension 0. (deprecated argument values)

tf.foldl(
    fn,
    elems,
    initializer=None,
    parallel_iterations=10,
    back_prop=True,
    swap_memory=False,
    name=None
)

This foldl operator repeatedly applies the callable fn to a sequence of elements from first to last. The elements are made of the tensors unpacked from elems on dimension 0. The callable fn takes two tensors as arguments. The first argument is the accumulated value computed from the preceding invocation of fn, and the second is the value at the current position of elems. If initializer is None, elems must contain at least one element, and its first element is used as the initializer.

Suppose that elems is unpacked into values, a list of tensors. The shape of the result tensor is fn(initializer, values[0]).shape`.

This method also allows multi-arity elems and output of fn. If elems is a (possibly nested) list or tuple of tensors, then each of these tensors must have a matching first (unpack) dimension. The signature of fn may match the structure of elems. That is, if elems is (t1, [t2, t3, [t4, t5]]), then an appropriate signature for fn is: fn = lambda (t1, [t2, t3, [t4, t5]]):.

elems = tf.constant([1, 2, 3, 4, 5, 6])
sum = tf.foldl(lambda a, x: a + x, elems)
# sum == 21