updated dict_set_solution with more concise way to update dictionary

codedragon · codedragon · commit 2964b02cd932 · 2014-11-01T13:53:56.000-07:00
diff --git a/Solutions/Session04/dict_set_solution.py b/Solutions/Session04/dict_set_solution.py
@@ -77,11 +77,14 @@
 
 
 # replacing the values in the original dict:
-
-for key, val in d.items():
-    d[key] = val.count('t')
+d.update(a_dict)
 print d
 
+# could also do it this way. 
+#for key, val in d.items():
+#    d[key] = val.count('t')
+#print d
+
 # Create sets s2, s3 and s4 that contain numbers from zero through
 # twenty, divisible 2, 3 and 4.
 
