Remove use of defaultdict
Remove the use of defaultdict to preserve Python 2.4 compatibility.
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				|  | @ -23,7 +23,6 @@ import sys | |||
| import threading | ||||
| from Queue import Queue | ||||
| from xml.dom import minidom as DOM | ||||
| from collections import defaultdict | ||||
| try: | ||||
|     from urlparse import parse_qs | ||||
| except ImportError: | ||||
|  | @ -204,11 +203,14 @@ def closestServers(client): | |||
|         return None | ||||
|     uh.close() | ||||
|     root = DOM.parseString(serversxml) | ||||
|     servers = defaultdict(list) | ||||
|     servers = {} | ||||
|     for server in root.getElementsByTagName('server'): | ||||
|         attrib = dict(server.attributes.items()) | ||||
|         d = distance([float(client['lat']), float(client['lon'])], | ||||
|                      [float(attrib.get('lat')), float(attrib.get('lon'))]) | ||||
|         if d not in servers: | ||||
|             servers[d] = [attrib] | ||||
|         else: | ||||
|             servers[d].append(attrib) | ||||
| 
 | ||||
|     closest = [] | ||||
|  |  | |||
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