将字典键传递给新词典

Question

为什么此代码无法正常工作？它为key3返回一个关键错误。 adict拥有key3，除了我将它们在循环中彼此传递之外，它与我向空词典添加键的方式似乎没有什么不同。如何将密钥和值从一个字典传递到具有相同主键的新密钥？

adict = {'key1':{'a':.078, 'b':1000, 'c':100}, 
          'key2':{'a':.0645, 'b':10, 'c':5}, 
          'key3':{'a':.0871, 'b':250, 'c':45}, 
          'key4':{'a':.0842, 'b':200, 'c':37}, 
          'key5':{'a':.054, 'b':409, 'c':82}, 
          'key6':{'a':.055, 'b':350, 'c':60}} 

another_dict = {} 
for k in adict: 
    another_dict[k]['transferred'] = adict[k]['b'] 


>>> Traceback (most recent call last): 
     File "C:\Python27\test.py", line 26, in <module> 
     another_dict[k]['transferred'] = adict[k]['b'] 
     KeyError: 'key3' 

Answer 1

使用another_dict[k]['transferred']，您试图访问'transferred'键的key k字典，这是尚未建立。

你可以在这里使用defaultdict： -

from collections import defaultdict 
another_dict = defaultdict(dict) 

修改后的代码： -

>>> from collections import defaultdict 
>>> adict = {'key1':{'a':.078, 'b':1000, 'c':100}, 
...       'key2':{'a':.0645, 'b':10, 'c':5}, 
...       'key3':{'a':.0871, 'b':250, 'c':45}, 
...       'key4':{'a':.0842, 'b':200, 'c':37}, 
...       'key5':{'a':.054, 'b':409, 'c':82}, 
...       'key6':{'a':.055, 'b':350, 'c':60}} 
... 
... another_dict = defaultdict(dict) 
... for k in adict: 
...  another_dict[k]['transferred'] = adict[k]['b'] 

>>> another_dict 
5: defaultdict(<type 'dict'>, {'key3': {'transferred': 250}, 
          'key2': {'transferred': 10}, 
          'key1': {'transferred': 1000}, 
          'key6': {'transferred': 350}, 
          'key5': {'transferred': 409}, 
          'key4': {'transferred': 200}}) 

Answer 2

问题是another_dict[k]实际上还不存在和你想的东西，是做another_dict[k]['transferred']甚至没有初始化。所以，你需要先初始化：

In [35]: adict = {'key1':{'a':.078, 'b':1000, 'c':100}, 
          'key2':{'a':.0645, 'b':10, 'c':5}, 
          'key3':{'a':.0871, 'b':250, 'c':45}, 
          'key4':{'a':.0842, 'b':200, 'c':37}, 
          'key5':{'a':.054, 'b':409, 'c':82}, 
          'key6':{'a':.055, 'b':350, 'c':60}} 

In [36]: another_dict={} 

In [37]: for k in adict: 
    another_dict[k]={}       #initialize another_dict[k] 
    another_dict[k]['transferred']=adict[k]['b'] 

In [38]: another_dict 
Out[38]: 
{'key1': {'transferred': 1000}, 
'key2': {'transferred': 10}, 
'key3': {'transferred': 250}, 
'key4': {'transferred': 200}, 
'key5': {'transferred': 409}, 
'key6': {'transferred': 350}} 

Answer 3

你必须创建一个新的字典，another_dict[k]访问之前：

adict = {'key1':{'a':.078, 'b':1000, 'c':100}, 
         'key2':{'a':.0645, 'b':10, 'c':5}, 
         'key3':{'a':.0871, 'b':250, 'c':45}, 
         'key4':{'a':.0842, 'b':200, 'c':37}, 
         'key5':{'a':.054, 'b':409, 'c':82}, 
         'key6':{'a':.055, 'b':350, 'c':60}} 

another_dict = {} 
for k in adict: 
    if k not in another_dict: 
     another_dict[k] = {} 
    another_dict[k]['transferred'] = adict[k]['b'] 

Answer 4

如果我理解正确的话，这是相当简单的使用dict理解：

>>> pprint({k:{'transferred':v['b']} for k, v in adict.iteritems()}) 
{'key1': {'transferred': 1000}, 
'key2': {'transferred': 10}, 
'key3': {'transferred': 250}, 
'key4': {'transferred': 200}, 
'key5': {'transferred': 409}, 
'key6': {'transferred': 350}} 

你也可以做一个方便的功能，只保留某些子键（如果它们要命名相同）

>>> from operator import itemgetter 
>>> def dict_with_subkeys(odict, *keys): 
    return {k:dict(zip(keys, itemgetter(*keys)(odict[k]))) for k in odict} 

>>> pprint(dict_with_subkeys(adict, 'a', 'c')) 
{'key1': {'a': 0.078, 'c': 100}, 
'key2': {'a': 0.0645, 'c': 5}, 
'key3': {'a': 0.0871, 'c': 45}, 
'key4': {'a': 0.0842, 'c': 37}, 
'key5': {'a': 0.054, 'c': 82}, 
'key6': {'a': 0.055, 'c': 60}} 

Answer 5

谢谢@ Ashwini我明白你的观点。和@Jon。

正如我在Python和（？“字典解析”）获得更好的，另一种解决方案，我发现是这样的：

another_dict = {k:{'b':adict[k]['b']} for k in adict}