Fixed Couter so that Python3 can run the project

pytagcloud/lang/counter.py

@@ -2,23 +2,26 @@
 import re
 from pytagcloud.lang.stopwords import StopWords
 from operator import itemgetter
-from collections import defaultdict
 
 def get_tag_counts(text):
     """
     Search tags in a given text. The language detection is based on stop lists.
     This implementation is inspired by https://github.com/jdf/cue.language. Thanks Jonathan Feinberg.
     """
-    words = map(lambda x:x.lower(), re.findall(r"[\w']+", text, re.UNICODE))
-
-    s = StopWords()     
-    s.load_language(s.guess(words))
-
-    counted = defaultdict(int)
-
-    for word in words:
+
+    # words = map(lambda x:x.lower(), re.findall(r'\w+', text, re.UNICODE))
+    # the line above doesn't work on python 3.5
+
+    s = StopWords()
+    s.load_language(s.guess(map(lambda x:x.lower(), re.findall(r'\w+', text, re.UNICODE))))
+
+    counted = {}
+
+    for word in map(lambda x:x.lower(), re.findall(r'\w+', text, re.UNICODE)):
         if not s.is_stop_word(word) and len(word) > 1:
-            counted[word] += 1
-
-    return sorted(counted.iteritems(), key=itemgetter(1), reverse=True)
-
+            if word in counted: #no has_key() method on python 3, use in instead
+                counted[word] += 1
+            else:
+                counted[word] = 1
+
+    return sorted(counted.items(), key=itemgetter(1), reverse=True)