好吧,这件事让我困扰了几次,所以谢谢你Jayesh的提问。
上面的答案看起来像任何一个好的解决方案,但如果你在你的代码中使用这个,那么包装功能恕我直言是有意义的。 此外,这里有两种可能的用例:一种是您关心所有关键字是否都在原始字典中。 还有一个你没有的地方。 平等对待两者都会很好。
因此,对于我的两个百分之一的价值,我建议写一个字典的子类,例如
class my_dict(dict):
def subdict(self, keywords, fragile=False):
d = {}
for k in keywords:
try:
d[k] = self[k]
except KeyError:
if fragile:
raise
return d
现在你可以拿出一个子词典orig_dict.subdict(keywords)
用法示例:
#
## our keywords are letters of the alphabet
keywords = 'abcdefghijklmnopqrstuvwxyz'
#
## our dictionary maps letters to their index
d = my_dict([(k,i) for i,k in enumerate(keywords)])
print('Original dictionary:\n%r\n\n' % (d,))
#
## constructing a sub-dictionary with good keywords
oddkeywords = keywords[::2]
subd = d.subdict(oddkeywords)
print('Dictionary from odd numbered keys:\n%r\n\n' % (subd,))
#
## constructing a sub-dictionary with mixture of good and bad keywords
somebadkeywords = keywords[1::2] + 'A'
try:
subd2 = d.subdict(somebadkeywords)
print("We shouldn't see this message")
except KeyError:
print("subd2 construction fails:")
print("\toriginal dictionary doesn't contain some keys\n\n")
#
## Trying again with fragile set to false
try:
subd3 = d.subdict(somebadkeywords, fragile=False)
print('Dictionary constructed using some bad keys:\n%r\n\n' % (subd3,))
except KeyError:
print("We shouldn't see this message")
如果您运行以上所有代码,您应该看到(类似)以下输出(抱歉格式化):
原始字典:
{'a':0,'c':2,'b':1,'e':4,'d':3,'g':6,'f':5, 'i':8,'h':7,'k':10,'j':9,'m':12,'l':11,'o':14, 'n':13,'q':16,'p':15,'s':18,'r':17,'u':20, 't':19,'w':22,'v':21,'y':24,'x':23,'z':25}
奇数键的字典:
{'a':0,'c':2,'e':4,'g':6,'i':8,'k':10,'m':12,'o':14,' q':16,'s':18,'你':20,'w':22,'y':24}
subd2构造失败:
原始字典不包含一些键
使用一些坏键构造的字典:
{'b':1,'d':3,'f':5,'h':7,'j':9,'l':11,'n':13,'p':15,' r':17,'t':19,'v':21,'x':23,'z':25}