Question

我正试图用一个现有的dict，例如，五个可能的条目，并将其条目移动到一个新的字典，“翻译”沿途的键。第一个字典的键的结构使得移动不能在程序上完成。无法保证所有条目都出现在第一个词典中。

为了避免KeyError，可以在自己的try块

中分配每个条目

first_dict = {
   'key_1': foo1,
   'key_two': foo2,
   'key_three': foo3,
   'key_number_4': foo4,
   'key_5': foo5
}

second_dict = {}

try: 
    second_dict['translated_key_1'] = first_dict['key_1']
except KeyError:
    pass

try:
    second_dict['translated_key_2'] = first_dict['key_two']
except KeyError:
    pass

...

try:
    second_dict['translated_key_5'] = first_dict['key_5']
except KeyError:
    pass


   'translated_key_2': first_dict['key_two'],
   'translated_key_3': first_dict['key_three'],
   'translated_key_4': first_dict['key_number_4'],
   'translated_key_5': first_dict['key_5'],
}

或许更好的方法是检查第一个字典中是否存在该条目，然后分配它。

if 'key_1' in first_dict:
    second_dict['translated_key_1'] = first_dict['key_1']
if 'key_two' in first_dict:
    second_dict['translated_key_2'] = first_dict['key_two']
...

在任何一种情况下，有没有办法压缩这个？当元素的数量很大时，似乎这会变得不必要地笨重。有没有办法迭代关系而不为每个创建try块？

Answer 1

你可以用循环来做到这一点：

# This is a simple mapping which associates the old keys with the new ones
translations = {'translated_key_1': 'key_one',
                'translated_key_2': 'key_two'}

# iterate through the map
for k,v in translations.iteritems():
    # you could also replace with try except
    if v in first_dict:
        second_dict[k] = first_dict[v]

凝聚连续尝试＆＃39;块

1 个答案: