Add an example to address a common question of how to split iterators.

Doc/lib/libitertools.tex

@@ -321,13 +321,15 @@ Samuele
 
 \end{verbatim}
 
-This section has further examples of how itertools can be combined.
+This section shows how itertools can be combined to create other more
-Note that \function{enumerate()} and \method{iteritems()} already
+powerful itertools.  Note that \function{enumerate()} and \method{iteritems()}
-have highly efficient implementations in Python.  They are only
+already have efficient implementations in Python.  They are only included here
-included here to illustrate how higher level tools can be created
+to illustrate how higher level tools can be created from building blocks.
-from building blocks.
 
 \begin{verbatim}
+def take(n, seq):
+    return list(islice(seq, n))
+
 def enumerate(iterable):
     return izip(count(), iterable)
 
@@ -380,7 +382,18 @@ def window(seq, n=2):
         result = result[1:] + (elem,)
         yield result
 
-def take(n, seq):
+def tee(iterable):
-    return list(islice(seq, n))
+    "Return two independent iterators from a single iterable"
+    def gen(next, data={}, cnt=[0]):
+        dpop = data.pop
+        for i in count():
+            if i == cnt[0]:
+                item = data[i] = next()
+                cnt[0] += 1
+            else:
+                item = dpop(i)
+            yield item
+    next = iter(iterable).next
+    return (gen(next), gen(next))
 
 \end{verbatim}

Lib/test/test_itertools.py

@@ -487,6 +487,9 @@ Martin
 Walter
 Samuele
 
+>>> def take(n, seq):
+...     return list(islice(seq, n))
+
 >>> def enumerate(iterable):
 ...     return izip(count(), iterable)
 
@@ -539,12 +542,26 @@ Samuele
 ...         result = result[1:] + (elem,)
 ...         yield result
 
->>> def take(n, seq):
+>>> def tee(iterable):
-...     return list(islice(seq, n))
+...     "Return two independent iterators from a single iterable"
+...     def gen(next, data={}, cnt=[0]):
+...         dpop = data.pop
+...         for i in count():
+...             if i == cnt[0]:
+...                 item = data[i] = next()
+...                 cnt[0] += 1
+...             else:
+...                 item = dpop(i)
+...             yield item
+...     next = iter(iterable).next
+...     return (gen(next), gen(next))
 
 This is not part of the examples but it tests to make sure the definitions
 perform as purported.
 
+>>> take(10, count())
+[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
+
 >>> list(enumerate('abc'))
 [(0, 'a'), (1, 'b'), (2, 'c')]
 
@@ -590,8 +607,17 @@ False
 >>> dotproduct([1,2,3], [4,5,6])
 32
 
->>> take(10, count())
+>>> def irange(start, stop):
-[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
+...     for i in range(start, stop):
+...         yield i
+
+>>> x, y = tee(irange(2,10))
+>>> list(x), list(y)
+([2, 3, 4, 5, 6, 7, 8, 9], [2, 3, 4, 5, 6, 7, 8, 9])
+
+>>> x, y = tee(irange(2,10))
+>>> zip(x, y)
+[(2, 2), (3, 3), (4, 4), (5, 5), (6, 6), (7, 7), (8, 8), (9, 9)]
 
 """