DOC: Fix a typo in description and add an example of numpy.tensordot

numpy/core/numeric.py

@@ -971,22 +971,42 @@ def tensordot(a, b, axes=2):
         * ``axes = 1`` : tensor dot product :math:`a\\cdot b`
         * ``axes = 2`` : (default) tensor double contraction :math:`a:b`
 
-    When `axes` is integer_like, the sequence for evaluation will be: first
-    the -Nth axis in `a` and 0th axis in `b`, and the -1th axis in `a` and
-    Nth axis in `b` last.
+    When `axes` is integer_like, the sequence of axes for evaluation
+    will be: from the -Nth axis to the -1th axis in `a`, 
+    and from the 0th axis to (N-1)th axis in `b`.
+    For example, ``axes = 2`` is the equal to
+    ``axes = [[-2, -1], [0, 1]]``.
+    When N-1 is smaller than 0, or when -N is larger than -1,
+    the element of 'a' and 'b' are defined as the 'axes'.
 
     When there is more than one axis to sum over - and they are not the last
     (first) axes of `a` (`b`) - the argument `axes` should consist of
     two sequences of the same length, with the first axis to sum over given
     first in both sequences, the second axis second, and so forth.
+    The calculation can be referred to enisum().
 
     The shape of the result consists of the non-contracted axes of the
     first tensor, followed by the non-contracted axes of the second.
 
     Examples
     --------
-    A "traditional" example:
-
+    "traditional" examples:
+    >>> # An example on integer_like
+    >>> a_0 = np.array([[1, 2], [3, 4]])
+    >>> b_0 = np.array([[5, 6], [7, 8]])
+    >>> c_0 = np.tensordot(a_0, b_0, axes=0)
+    >>> c_0.shape
+    (2, 2, 2, 2)
+    >>> c_0
+    array([[[[ 5,  6],
+             [ 7,  8]],
+            [[10, 12],
+             [14, 16]]],
+           [[[15, 18],
+             [21, 24]],
+            [[20, 24],
+             [28, 32]]]])
+    >>> # An example on array_like
     >>> a = np.arange(60.).reshape(3,4,5)
     >>> b = np.arange(24.).reshape(4,3,2)
     >>> c = np.tensordot(a,b, axes=([1,0],[0,1]))
@@ -998,6 +1018,7 @@ def tensordot(a, b, axes=2):
            [4664., 5018.],
            [4796., 5162.],
            [4928., 5306.]])
+
     >>> # A slower but equivalent way of computing the same...
     >>> d = np.zeros((5,2))
     >>> for i in range(5):