如果缺少密钥,我可以在字典列表中使用列表推导吗？

例如,此代码适用于key1,因为所有字典都有密钥.

from collections import Counter
list_of_dicts = [
    {'key1': 'testing','key2': 'testing'},{'key1': 'prod',{'key1': 'testing',},]

print Counter(r['key1'] for r in list_of_dicts)

我得到了一个很好的结果

Counter({'testing': 4,'prod': 2})

但是,如果我将最后一次打印更改为：

print Counter(r['key2'] for r in list_of_dicts)

它失败了,因为在几个词典中缺少key2.

Traceback (most recent call last):
  File "test.py",line 11,in <module>
    print Counter(r['key2'] for r in list_of_dicts)
  File "h:Anacondalibcollections.py",line 453,in __init__
    self.update(iterable,**kwds)
  File "h:Anacondalibcollections.py",line 534,in update
    for elem in iterable:
  File "test.py",in <genexpr>
    print Counter(r['key2'] for r in list_of_dicts)
KeyError: 'key2'

我如何使用列表推导来计算key2的值,如果字典不包含密钥则不会失败？