1 for i in range(10):
2     print(i)

1 fruits = ["apple", "banana", "pear"]
2 for ele in fruits:
3     print(ele)

1 fruits = ["apple", "banana", "pear"]
2 for i in range(len(fruits)):
3     print("%d: %s" % (i, fruits[i]))

1 fruits = ["apple", "banana", "pear"]
2 for i, ele in enumerate(fruits):
3     print("%d: %s" % (i, ele))

 1 fruits = ["apple", "banana", "pear"]
 2 letters = [len(fruit) for fruit in fruits]
 3 
 4 longest_name = None
 5 max_letters = 0
 6 
 7 for i in range(len(fruits)):
 8     if letters[i] > max_letters:
 9         max_letters = letters[i]
10         longest_name = fruits[i]
11 
12 print(max_letters)
13 print(longest_name)

 1 fruits = ["apple", "banana", "pear"]
 2 letters = [len(fruit) for fruit in fruits]
 3 
 4 longest_name = None
 5 max_letters = 0
 6 
 7 for i, fruit in enumerate(fruits):
 8     if letters[i] > max_letters:
 9         max_letters = letters[i]
10         longest_name = fruit
11 
12 print(max_letters)
13 print(longest_name)

 1 fruits = ["apple", "banana", "pear"]
 2 letters = [len(fruit) for fruit in fruits]
 3 
 4 longest_name = None
 5 max_letters = 0
 6 
 7 for fruit, length in zip(fruits, letters):
 8     if length > max_letters:
 9         max_letters = length
10         longest_name = fruit
11 
12 print(max_letters)
13 print(longest_name)

我正在使用AlamoFire库将JSON发送到REST服务器。

词典1

let dict1: [String: AnyObject] = [
            kFacebook: [
                kToken: token
            ]
        ]

词典2

let dict2: [String: AnyObject] = [
        kRequest: [
            kTargetUserId: userId
        ]
    ]

如何结合两个字典来实现如下？

let parameters: [String: AnyObject] = [
        kFacebook: [
            kToken: token
        ],kRequest: [
            kTargetUserId: userId
        ]
    ]

我已经尝试了dict1 = dict2但是有一个编译错误。

提前致谢！

var d1 = ["a": "b"]
var d2 = ["c": "e"]

extension Dictionary {
    mutating func merge(dict: [Key: Value]){
        for (k,v) in dict {
            updateValue(v,forKey: k)
        }
    }
}

d1.merge(d2)

参考美元& Cent项目
https://github.com/ankurp/Cent/blob/master/Sources/Dictionary.swift

from collections import defaultdict

d = defaultdict(int)
res = d[5]

print(d)
# defaultdict(<class ''int''>,{5: 0})
# we want this dictionary to remain empty

但是,我们通常只想在显式或隐式分配密钥时添加密钥：

d[8] = 1  # we want this key added
d[3] += 1 # we want this key added

一个用例是简单计数,以避免集合的较高开销.计数器,但这个功能一般也是可取的.

反例[原谅双关语]

这是我想要的功能：

from collections import Counter
c = Counter()
res = c[5]  # 0
print(c)  # Counter()

c[8] = 1  # key added successfully
c[3] += 1 # key added successfully

但Counter比defaultdict(int)明显慢.我发现性能下降通常比defaultdict(int)慢约2倍.

另外,显然Counter只能与defaultdict中的int参数相媲美,而defaultdict可以采用list,set等.

有没有办法有效地实现上述行为;例如,通过继承defaultdict？

基准测试示例

%timeit DwD(lst)           # 72 ms
%timeit dd(lst)            # 44 ms
%timeit counter_func(lst)  # 98 ms
%timeit af(lst)            # 72 ms

测试代码：

import numpy as np
from collections import defaultdict,Counter,UserDict

class DefaultDict(defaultdict):
    def get_and_forget(self,key):
        _sentinel = object()
        value = self.get(key,_sentinel)

        if value is _sentinel:
            return self.default_factory()
        return value

class DictWithDefaults(dict):
    __slots__ = [''_factory'']  # avoid using extra memory

    def __init__(self,factory,*args,**kwargs):
        self._factory = factory
        super().__init__(*args,**kwargs)

    def __missing__(self,key):
        return self._factory()

lst = np.random.randint(0,10,100000)

def DwD(lst):
    d = DictWithDefaults(int)
    for i in lst:
        d[i] += 1
    return d

def dd(lst):
    d = defaultdict(int)
    for i in lst:
        d[i] += 1
    return d

def counter_func(lst):
    d = Counter()
    for i in lst:
        d[i] += 1
    return d

def af(lst):
    d = DefaultDict(int)
    for i in lst:
        d[i] += 1
    return d

关于赏金评论的注意事项：

自Bounty提供以来,@Aran-Fey’s solution已经更新,所以请忽略Bounty的评论.

class DefaultDict(dict):
    def __init__(self,default_factory,**kwargs):
        super().__init__(**kwargs)

        self.default_factory = default_factory

    def __getitem__(self,key):
        try:
            return super().__getitem__(key)
        except KeyError:
            return self.default_factory()

这可以按照您想要的方式工作：

d = DefaultDict(int)

res = d[5]
d[8] = 1 
d[3] += 1

print(d)  # {8: 1,3: 1}

但是,对于可变类型,它可能会出乎意料地表现：

d = DefaultDict(list)
d[5].append(''foobar'')

print(d)  # output: {}

这可能是defaultdict在访问不存在的密钥时记住该值的原因.

另一种选择是扩展defaultdict并添加一个新方法来查找值而不记住它：

from collections import defaultdict

class DefaultDict(defaultdict):
    def get_and_forget(self,key):
        return self.get(key,self.default_factory())

请注意,无论密钥是否已存在于dict中,get_and_forget方法每次都会调用default_factory().如果这是不合需要的,您可以使用sentinel值来实现它：

class DefaultDict(defaultdict):
    def get_and_forget(self,_sentinel)

        if value is _sentinel:
            return self.default_factory()
        return value

这样可以更好地支持可变类型,因为它允许您选择是否应该将值添加到dict中.

当我尝试使用时：

plt.bar(range(len(cnt)),cnt.values(),align='center')
plt.xticks(range(len(cnt)),cnt.keys())
plt.show()

我收到以下错误：

ValueError: matplotlib display text must have all code points < 128 or use Unicode strings

这就是我试图将Counter字典键转换为Unicode的原因.

unidict = {k.decode('utf8'): v.decode('utf8') for k,v in strdict.items()}

对于旧版本：

unidict = dict((k.decode('utf8'),v.decode('utf8')) for k,v in strdict.items())

(这假设你的字符串是UTF-8,当然.)