OrderedDict([(u'attributes', OrderedDict([(u'type', u'Campaign__c'), (u'url', u'/services/data/v29.0/sobjects/Campaign__c/a0B9000000I6CDUEA3')])), (u'clicks__c', 0.0)])
我有一个类似上面的有序字典,如何获取clicks__c
的值?
OrderedDict([(u'attributes', OrderedDict([(u'type', u'Campaign__c'), (u'url', u'/services/data/v29.0/sobjects/Campaign__c/a0B9000000I6CDUEA3')])), (u'clicks__c', 0.0)])
我有一个类似上面的有序字典,如何获取clicks__c
的值?
它仍然是一个字典,只需要使用键:
your_ordered_dict['clicks__c']
范例:
>>> from collections import OrderedDict
>>> od = OrderedDict([(u'attributes', OrderedDict([(u'type', u'Campaign__c'), (u'url', u'/services/data/v29.0/sobjects/Campaign__c/a0B9000000I6CDUEA3')])), (u'clicks__c', 0.0)])
>>> od.keys()
[u'attributes', u'clicks__c']
>>> od['clicks__c']
0.0
object["clicks__c"]
- UltraInstinct