Question

我现在有代码尝试让我按任意深度键（如mongo）对字典进行排序，但它要求我对密钥的深度进行硬编码。

#This is the code for inside the function, you need to make a function to receive the  arguments
#and the dictionary. the arguments should be listed in order of layers. It then splits the argument string
#at the "." and assigns to each of the items.From there it should return the "l[]".
#you need to set it up to pass the arguments to these appropriate spots. so the list of dicts goes to
#list and the arguments go to argstring and it should be taken care of from there.


#splitting the argument
argstring="author.age"
arglist = argstring.split(".")

x=(5-len(arglist))#need to set this number to be the most you want to accept
while x>0:
    arglist.append('')
    x-=1

#test list
list = [
{'author' : {'name':'JKRowling','age':47,'bestseller':{'series':'harrypotter','copiessold':12345}}},


{'author' : {'name':'Tolkien','age':81,'bestseller':{'series':'LOTR','copiessold':5678}}},


{'author' : {'name':'GeorgeMartin','age':64,'bestseller':{'series':'Fire&Ice','copiessold':12}}},


{'author' : {'name':'UrsulaLeGuin','age':83,'bestseller':{'series':'EarthSea', 'copiessold':444444}}}
]
l=[]#the list for returning


#determining sort algorythm
l = sorted(list, key=lambda e: e[arglist[0]][arglist[1]])#need add as many of these as necesarry to match the number above
print()

这很有效，但是必须手动指定arglist中的参数似乎很愚蠢。如果我需要5深，我需要手动指定e 5次.. 有没有办法使用列表推导或for循环自动包含任意元素深度？

Answer 1

使用reduce()：

sorted(list, key=lambda e: reduce(lambda m, k: m[k], argslist, e))

reduce()接受一个函数，一个输入列表和一个可选的初始值，然后将该函数重新应用于next元素和最后一次调用的返回值（从初始值开始）。因此，它运行m[k0][k1][k2]..[kn]，其中k的连续argslist值取自>>> e = {'author' : {'name':'JKRowling','age':47,'bestseller':{'series':'harrypotter','copiessold':12345}}} >>> argslist = ['author', 'age'] >>> reduce(lambda m, k: m[k], argslist, e) 47。

简短演示：

{{1}}

Answer 2

使用for循环来遍历args

没有错

>>> e = {'author' : {'name':'JKRowling','age':47,'bestseller':{'series':'harrypotter','copiessold':12345}}}
>>> argslist = ['author', 'age']
>>> result = e
>>> for arg in argslist:
...     result = result[arg]
... 
>>> result
47

这种方式非常容易调试，你可以放try/except，print，断点等。

如何迭代列表推导中的子元素？

2 个答案: