递归第 n 个孩子 dict.get() - 效率？

Question

我需要从一些大型嵌套字典中获取一些值。出于懒惰，我决定编写一个递归调用自身的函数，直到找到最后一个孩子，或者叶子为空。

由于弹出了字典，并且每次新调用都会构建一个新字典，我想知道这有多有效。

有什么建议吗？

def recursive_dict_get(item, string, default=False):
    if not isinstance(item, dict):
        return default

    print "called with ", item, "and string", string
    if "." in string:
        attrs = string.split(".")
        parent = attrs.pop(0)
        rest = ".".join(attrs)
        result = item.get(parent, None)
        if result is None:
            return default
        else:
            return recursive_dict_get(item.get(parent, default), rest, default)
    else:
        return item.get(string, default)

---

foo = {
            "1": {
                "2": {
                    "3": {
                        "4":{
                            "5": {
                                "6": {
                                    "7": "juice"
                                }
                            }
                        }
                    }
                }
            }
        }

print recursive_dict_get(foo, "1.2.3.4.5.6.7", False)
print "*" * 3       
print recursive_dict_get(foo, "1.2.3.4.5.6", False)
print "*" * 3
print recursive_dict_get(foo, "1.3", False)

Answer 1

我的一个建议是给split() 第二个参数。你可以做一些更简单的事情，比如：

parent, rest = string.split(".", 1)

除此之外，我认为代码没有直接问题。

你也可以不用递归来做到这一点：

def recursive_dict_get(item, string, default=False):
    for s in string.split('.'):
        if (isinstance(item, dict) and s in item):
            item = item[s]
        else:
            return default
    return item

Answer 2

是的，您的实现效率相当低，即使它没有构建任何新的字典，但 bur 可能会返回许多现有的字典。无论如何，您可以将接受的答案调整为Access python nested dictionary items via a list of keys，以减少您的访问功能到一行代码。这类似于 J.F. Sebastian (@jfs) 在his comment 中提到的内容。我的看法是这样的：

def nonrecursive_dict_get(item, key_string, default=False):
    return reduce(lambda d, k: d.get(k, default), key_string.split('.'), item)

print "*" * 3, 'using nonrecursive_dict_get()'
print nonrecursive_dict_get(foo, "1.2.3.4.5.6.7")
print "*" * 3
print nonrecursive_dict_get(foo, "1.2.3.4.5.6")
print "*" * 3
print nonrecursive_dict_get(foo, "1.3")

更新：

每当关注效率时，最好的办法通常是对各种方法进行基准测试。这是我用过很多次的：

global_setup = """
    foo = {
            "1": {
                "2": {
                    "3": {
                        "4": {
                            "5": {
                                "6": {
                                    "7": "juice"
                                     }
                                 }
                             }
                         }
                     }
                 }
          }
"""

testcases = {
"jay":
    { 'setup' : """
        def recursive_dict_get(item, string, default=False):
            if not isinstance(item, dict):
                return default
            if "." in string:
                attrs = string.split(".")
                parent = attrs.pop(0)
                rest = ".".join(attrs)
                result = item.get(parent, None)
                if result is None:
                    return default
                else:
                    return recursive_dict_get(item.get(parent, default), rest, default)
            else:
                return item.get(string, default)
                """,
      'code' : """
        recursive_dict_get(foo, "1.2.3.4.5.6.7", False)
        recursive_dict_get(foo, "1.2.3.4.5.6", False)
        recursive_dict_get(foo, "1.3", False)
        """,
    },

"martineau":
    { 'setup' : """
        def nonrecursive_dict_get(nested_dict, key_string, default=False):
            return reduce(lambda d, k: d.get(k, default), key_string.split('.'), nested_dict)
            """,
      'code' : """
        nonrecursive_dict_get(foo, "1.2.3.4.5.6.7", False)
        nonrecursive_dict_get(foo, "1.2.3.4.5.6", False)
        nonrecursive_dict_get(foo, "1.3", False)
        """,
    },

"J.F. Sebastian":
    { 'setup' : """
        # modified to support 'default' keyword argument
        def quick_n_dirty(nested_dict, key_string, default=False):
            reduced = reduce(dict.get, key_string.split('.'), nested_dict)
            return default if reduced is None else reduced
            """,
      'code' : """
        quick_n_dirty(foo, "1.2.3.4.5.6.7", False)
        quick_n_dirty(foo, "1.2.3.4.5.6", False)
        quick_n_dirty(foo, "1.3", False)
        """,
    },

"arshajii":
    { 'setup' : """
        def recursive_dict_get(item, string, default=False):
            for s in string.split('.'):
                if (isinstance(item, dict) and s in item):
                    item = item[s]
                else:
                    return default
            return item
            """,
      'code' : """
        recursive_dict_get(foo, "1.2.3.4.5.6.7", False)
        recursive_dict_get(foo, "1.2.3.4.5.6", False)
        recursive_dict_get(foo, "1.3", False)
        """,
    },

"Brionius":
    { 'setup' : """
        def getVal(d, keys, default):
            keys = keys.split(".")
            for key in keys:
                try:
                    d = d[key]
                except KeyError:
                    return default
            return d
            """,
      'code' : """
        getVal(foo, "1.2.3.4.5.6.7", False)
        getVal(foo, "1.2.3.4.5.6", False)
        getVal(foo, "1.3", False)
        """,
    },
}

import sys
from textwrap import dedent
import timeit
N = 100000
R = 3

# remove leading whitespace from all code fragments
global_setup = dedent(global_setup)
for testcase in testcases.itervalues():
    for label, fragment in testcase.iteritems():
        testcase[label] = dedent(fragment)

timings = [(name,
            min(timeit.repeat(testcases[name]['code'],
                              setup=global_setup + testcases[name]['setup'],
                              repeat=R, number=N)),
           ) for name in testcases]

longest_name = max(len(t[0]) for t in timings)

print('fastest to slowest timings:\n'
      '  ({:,d} calls, best of {:d} repetitions)\n'.format(N, R))

ranked = sorted(timings, key=lambda t: t[1])  # sort by speed (fastest first)
for timing in ranked:
    print("{:>{width}} : {:.6f} secs ({rel:>8.6f}x)".format(
          timing[0], timing[1], rel=timing[1]/ranked[0][1], width=longest_name))

输出：

fastest to slowest timings:
  (100,000 calls, best of 3 repetitions)

J.F. Sebastian : 1.287209 secs (1.000000x)
      Brionius : 1.420099 secs (1.103239x)
      arshajii : 1.431521 secs (1.112112x)
     martineau : 2.031539 secs (1.578251x)
           jay : 7.817713 secs (6.073384x)

如您所见，J.F. Sebastian 的建议是最快的，即使我进行了修改以使其与其他建议相同。

Answer 3

这是另一种方式：

def getVal(d, keys, default):
    keys = keys.split(".")  # You can avoid this first step if you're willing to use a list like ["1", "2", "3"...] as an input instead of a string like "1.2.3..."
    for key in keys:
        try:
            d = d[key]
        except KeyError:
            return default
    return d

如果您愿意，我可以对其进行分析 - 请告诉我。请记住，除非您遇到或有理由相信您会遇到瓶颈，否则优化是没有意义的。