tf.math.cumsum

TensorFlow 1 version

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Compute the cumulative sum of the tensor x along axis.

tf.math.cumsum(
    x, axis=0, exclusive=False, reverse=False, name=None
)

By default, this op performs an inclusive cumsum, which means that the first element of the input is identical to the first element of the output: For example:

# tf.cumsum([a, b, c])   # [a, a + b, a + b + c]
x = tf.constant([2, 4, 6, 8])
tf.cumsum(x)
<tf.Tensor: shape=(4,), dtype=int32,
numpy=array([ 2,  6, 12, 20], dtype=int32)>

# using varying `axis` values
y = tf.constant([[2, 4, 6, 8], [1,3,5,7]])
tf.cumsum(y, axis=0)
<tf.Tensor: shape=(2, 4), dtype=int32, numpy=
array([[ 2,  4,  6,  8],
       [ 3,  7, 11, 15]], dtype=int32)>
tf.cumsum(y, axis=1)
<tf.Tensor: shape=(2, 4), dtype=int32, numpy=
array([[ 2,  6, 12, 20],
       [ 1,  4,  9, 16]], dtype=int32)>

By setting the exclusive kwarg to True, an exclusive cumsum is performed instead:

# tf.cumsum([a, b, c], exclusive=True)  => [0, a, a + b]
x = tf.constant([2, 4, 6, 8])
tf.cumsum(x, exclusive=True)
<tf.Tensor: shape=(4,), dtype=int32,
numpy=array([ 0,  2,  6, 12], dtype=int32)>

By setting the reverse kwarg to True, the cumsum is performed in the opposite direction:

# tf.cumsum([a, b, c], reverse=True)  # [a + b + c, b + c, c]
x = tf.constant([2, 4, 6, 8])
tf.cumsum(x, reverse=True)
<tf.Tensor: shape=(4,), dtype=int32,
numpy=array([20, 18, 14,  8], dtype=int32)>

This is more efficient than using separate tf.reverse ops. The reverse and exclusive kwargs can also be combined:

# tf.cumsum([a, b, c], exclusive=True, reverse=True)  # [b + c, c, 0]
x = tf.constant([2, 4, 6, 8])
tf.cumsum(x, exclusive=True, reverse=True)
<tf.Tensor: shape=(4,), dtype=int32,
numpy=array([18, 14,  8,  0], dtype=int32)>