default count in increment_count to 1

fixes #26

Author: jtauber

README.md

@@ -86,7 +86,7 @@ And here is an example with a two-level hierarchy:
 >>> c.increment_count("2.1", "foo", 2)
 >>> c.increment_count("2.2", "foo", 2)
 >>> c.increment_count("2.1", "bar", 1)
->>> c.increment_count("2.2", "bar", 1)
+>>> c.increment_count("2.2", "bar")
 >>> c.get_counts()["foo"]
 10
 >>> c.get_counts()["bar"]
@@ -96,6 +96,8 @@ And here is an example with a two-level hierarchy:
 
 ```
 
+Note that if the `count` is `1` you can omit it.
+
 You can **prune** a HTDM to just `n` levels with the method `prune(n)`.
 
 You can iterate over the document-term counts at the leaves of the HTDM with the method `leaf_entries()` (this returns a generator yielding `(document_address, term, count)` tuples). This is effectively a traditional TDM (the document IDs will still reflect the hierarchy but the aggregate counts aren't present).

termdoc/htdm.py

@@ -26,7 +26,7 @@ def get_or_create_counter(self, depth):
             self.counters.append(collections.defaultdict(collections.Counter))
         return self.counters[depth]
 
-    def increment_count(self, address, term, count):
+    def increment_count(self, address, term, count=1):
         first = True
         while True:
             depth = self.depth(address)

tests.py

@@ -324,5 +324,18 @@ def test_multi_level_error(self):
         self.assertEqual(c.get_counts("2")["foo"], 4)
 
 
+class Test3(unittest.TestCase):
+    def test_two_arg_increment_count(self):
+        import termdoc
+
+        c = termdoc.HTDM()
+        c.increment_count("1", "foo")
+        c.increment_count("1", "bar", 2)
+        c.increment_count("2", "foo", 2)
+        c.increment_count("2", "bar")
+        self.assertEqual(c.get_counts()["foo"], 3)
+        self.assertEqual(c.get_counts()["bar"], 3)
+
+
 if __name__ == "__main__":
     unittest.main()