问题:创建具有列表理解的字典

我喜欢Python列表理解语法。

它也可以用来创建字典吗?例如,通过遍历键和值对:

mydict = {(k,v) for (k,v) in blah blah blah}  # doesn't work

I like the Python list comprehension syntax.

Can it be used to create dictionaries too? For example, by iterating over pairs of keys and values:

mydict = {(k,v) for (k,v) in blah blah blah}  # doesn't work

回答 0

从Python 2.7和3开始,您应该只使用dict comprehension语法

{key: value for (key, value) in iterable}

在Python 2.6和更早版本中,dict内置函数可以接收键/值对的迭代,因此您可以将其传递给列表推导或生成器表达式。例如:

dict((key, func(key)) for key in keys)

但是,如果您已经具有可迭代的键和/或值,则根本不需要使用任何理解-最简单dict的方法是直接调用内置函数:

# consumed from any iterable yielding pairs of keys/vals
dict(pairs)

# "zipped" from two separate iterables of keys/vals
dict(zip(list_of_keys, list_of_values))

From Python 2.7 and 3 onwards, you should just use the dict comprehension syntax:

{key: value for (key, value) in iterable}

In Python 2.6 and earlier, the dict built-in can receive an iterable of key/value pairs, so you can pass it a list comprehension or generator expression. For example:

dict((key, func(key)) for key in keys)

However if you already have iterable(s) of keys and/or vals, you needn’t use a comprehension at all – it’s simplest just call the dict built-in directly:

# consumed from any iterable yielding pairs of keys/vals
dict(pairs)

# "zipped" from two separate iterables of keys/vals
dict(zip(list_of_keys, list_of_values))

回答 1

在Python 3和Python 2.7+中,字典理解如下所示:

d = {k:v for k, v in iterable}

对于Python 2.6或更早版本,请参见fortran的答案

In Python 3 and Python 2.7+, dictionary comprehensions look like the below:

d = {k:v for k, v in iterable}

For Python 2.6 or earlier, see fortran’s answer.


回答 2

实际上,如果它已经包含某种映射,则您甚至不需要遍历可迭代对象,dict构造函数会为您轻松地做到这一点:

>>> ts = [(1, 2), (3, 4), (5, 6)]
>>> dict(ts)
{1: 2, 3: 4, 5: 6}
>>> gen = ((i, i+1) for i in range(1, 6, 2))
>>> gen
<generator object <genexpr> at 0xb7201c5c>
>>> dict(gen)
{1: 2, 3: 4, 5: 6}

In fact, you don’t even need to iterate over the iterable if it already comprehends some kind of mapping, the dict constructor doing it graciously for you:

>>> ts = [(1, 2), (3, 4), (5, 6)]
>>> dict(ts)
{1: 2, 3: 4, 5: 6}
>>> gen = ((i, i+1) for i in range(1, 6, 2))
>>> gen
<generator object <genexpr> at 0xb7201c5c>
>>> dict(gen)
{1: 2, 3: 4, 5: 6}

回答 3

在Python 2.7中,它类似于:

>>> list1, list2 = ['a', 'b', 'c'], [1,2,3]
>>> dict( zip( list1, list2))
{'a': 1, 'c': 3, 'b': 2}

压缩他们

In Python 2.7, it goes like:

>>> list1, list2 = ['a', 'b', 'c'], [1,2,3]
>>> dict( zip( list1, list2))
{'a': 1, 'c': 3, 'b': 2}

Zip them!


回答 4

在Python中创建具有列表理解的字典

我喜欢Python列表理解语法。

它也可以用来创建字典吗?例如,通过遍历键和值对:

mydict = {(k,v) for (k,v) in blah blah blah}

您正在寻找“ dict comprehension”一词-实际上是:

mydict = {k: v for k, v in iterable}

假设blah blah blah是两个元组的迭代-您是如此亲密。让我们创建一些类似的“ blah”:

blahs = [('blah0', 'blah'), ('blah1', 'blah'), ('blah2', 'blah'), ('blah3', 'blah')]

Dict理解语法:

现在,这里的语法是映射部分。使它成为dict理解而不是set理解(这是您的伪代码近似值)的是冒号,:如下所示:

mydict = {k: v for k, v in blahs}

而且我们看到它起作用了,并且应该保留Python 3.7的插入顺序:

>>> mydict
{'blah0': 'blah', 'blah1': 'blah', 'blah2': 'blah', 'blah3': 'blah'}

在Python 2和3.6以下版本中,不能保证顺序:

>>> mydict
{'blah0': 'blah', 'blah1': 'blah', 'blah3': 'blah', 'blah2': 'blah'}

添加过滤器:

所有的理解都具有一个映射组件和一个过滤组件,您可以为它们提供任意表达式。

因此,您可以在末尾添加一个过滤器部分:

>>> mydict = {k: v for k, v in blahs if not int(k[-1]) % 2}
>>> mydict
{'blah0': 'blah', 'blah2': 'blah'}

在这里,我们只是测试最后符是否可被2整除以在映射键和值之前过滤掉数据。

Create a dictionary with list comprehension in Python

I like the Python list comprehension syntax.

Can it be used to create dictionaries too? For example, by iterating over pairs of keys and values:

mydict = {(k,v) for (k,v) in blah blah blah}

You’re looking for the phrase “dict comprehension” – it’s actually:

mydict = {k: v for k, v in iterable}

Assuming blah blah blah is an iterable of two-tuples – you’re so close. Let’s create some “blahs” like that:

blahs = [('blah0', 'blah'), ('blah1', 'blah'), ('blah2', 'blah'), ('blah3', 'blah')]

Dict comprehension syntax:

Now the syntax here is the mapping part. What makes this a dict comprehension instead of a set comprehension (which is what your pseudo-code approximates) is the colon, : like below:

mydict = {k: v for k, v in blahs}

And we see that it worked, and should retain insertion order as-of Python 3.7:

>>> mydict
{'blah0': 'blah', 'blah1': 'blah', 'blah2': 'blah', 'blah3': 'blah'}

In Python 2 and up to 3.6, order was not guaranteed:

>>> mydict
{'blah0': 'blah', 'blah1': 'blah', 'blah3': 'blah', 'blah2': 'blah'}

Adding a Filter:

All comprehensions feature a mapping component and a filtering component that you can provide with arbitrary expressions.

So you can add a filter part to the end:

>>> mydict = {k: v for k, v in blahs if not int(k[-1]) % 2}
>>> mydict
{'blah0': 'blah', 'blah2': 'blah'}

Here we are just testing for if the last character is divisible by 2 to filter out data before mapping the keys and values.


回答 5

Python版本<2.7 (RIP,2010年7月3日至2019年12月31日),请执行以下操作:

d = dict((i,True) for i in [1,2,3])

Python版本> = 2.7,请执行以下操作:

d = {i: True for i in [1,2,3]}

Python version < 2.7(RIP, 3 July 2010 – 31 December 2019), do the below:

d = dict((i,True) for i in [1,2,3])

Python version >= 2.7, do the below:

d = {i: True for i in [1,2,3]}

回答 6

如果要遍历键key_list列表和值列表,请添加到@fortran的答案中value_list

d = dict((key, value) for (key, value) in zip(key_list, value_list))

要么

d = {(key, value) for (key, value) in zip(key_list, value_list)}

To add onto @fortran’s answer, if you want to iterate over a list of keys key_list as well as a list of values value_list:

d = dict((key, value) for (key, value) in zip(key_list, value_list))

or

d = {(key, value) for (key, value) in zip(key_list, value_list)}

回答 7

这是使用dict理解创建字典的另一个示例:

我在这里要做的是在每对字典中创建一个字母字典。是英文字母及其在英文字母中的对应位置

>>> import string
>>> dict1 = {value: (int(key) + 1) for key, value in 
enumerate(list(string.ascii_lowercase))}
>>> dict1
{'a': 1, 'c': 3, 'b': 2, 'e': 5, 'd': 4, 'g': 7, 'f': 6, 'i': 9, 'h': 8, 
'k': 11, 'j': 10, 'm': 13, 'l': 12, 'o': 15, 'n': 14, 'q': 17, 'p': 16, 's': 
19, 'r': 18, 'u': 21, 't': 20, 'w': 23, 'v': 22, 'y': 25, 'x': 24, 'z': 26}
>>> 

请注意,此处使用枚举可获取列表中的字母及其索引,并交换字母和索引以生成字典的键值对

希望它对您的字典组合有个好主意,并鼓励您更多地使用它来使代码紧凑

Here is another example of dictionary creation using dict comprehension:

What i am tring to do here is to create a alphabet dictionary where each pair; is the english letter and its corresponding position in english alphabet

>>> import string
>>> dict1 = {value: (int(key) + 1) for key, value in 
enumerate(list(string.ascii_lowercase))}
>>> dict1
{'a': 1, 'c': 3, 'b': 2, 'e': 5, 'd': 4, 'g': 7, 'f': 6, 'i': 9, 'h': 8, 
'k': 11, 'j': 10, 'm': 13, 'l': 12, 'o': 15, 'n': 14, 'q': 17, 'p': 16, 's': 
19, 'r': 18, 'u': 21, 't': 20, 'w': 23, 'v': 22, 'y': 25, 'x': 24, 'z': 26}
>>> 

Notice the use of enumerate here to get a list of alphabets and their indexes in the list and swapping the alphabets and indices to generate the key value pair for dictionary

Hope it gives a good idea of dictionary comp to you and encourages you to use it more often to make your code compact


回答 8

尝试这个,

def get_dic_from_two_lists(keys, values):
    return { keys[i] : values[i] for i in range(len(keys)) }

假设我们有两个清单,国家首都

country = ['India', 'Pakistan', 'China']
capital = ['New Delhi', 'Islamabad', 'Beijing']

然后从两个列表中创建字典:

print get_dic_from_two_lists(country, capital)

输出是这样的,

{'Pakistan': 'Islamabad', 'China': 'Beijing', 'India': 'New Delhi'}

Try this,

def get_dic_from_two_lists(keys, values):
    return { keys[i] : values[i] for i in range(len(keys)) }

Assume we have two lists country and capital

country = ['India', 'Pakistan', 'China']
capital = ['New Delhi', 'Islamabad', 'Beijing']

Then create dictionary from the two lists:

print get_dic_from_two_lists(country, capital)

The output is like this,

{'Pakistan': 'Islamabad', 'China': 'Beijing', 'India': 'New Delhi'}

回答 9

>>> {k: v**3 for (k, v) in zip(string.ascii_lowercase, range(26))}

Python支持dict理解,它允许您使用类似的简洁语法在运行时表示字典的创建。

字典理解采用{key:(key,value)inerable中的值}的形式。该语法是在Python 3中引入的,并且一直移植到Python 2.7,因此,无论安装了哪个版本的Python,您都应该能够使用它。

一个典型的例子是获取两个列表并创建一个字典,其中第一个列表中每个位置的项成为键,而第二个列表中相应位置的项变为值。

此推导中使用的zip函数返回一个元组的迭代器,其中元组中的每个元素均取自每个输入可迭代对象中的相同位置。在上面的示例中,返回的迭代器包含元组(“ a”,1),(“ b”,2)等。

输出:

{‘i’:512,’e’:64,’o’:2744,’h’:343,’l’:1331,’s’:5832,’b’:1,’w’:10648,’ c’:8,’x’:12167,’y’:13824,’t’:6859,’p’:3375,’d’:27,’j’:729,’a’:0,’z’ :15625,’f’:125,’q’:4096,’u’:8000,’n’:2197,’m’:1728,’r’:4913,’k’:1000,’g’:216 ,’v’:9261}

>>> {k: v**3 for (k, v) in zip(string.ascii_lowercase, range(26))}

Python supports dict comprehensions, which allow you to express the creation of dictionaries at runtime using a similarly concise syntax.

A dictionary comprehension takes the form {key: value for (key, value) in iterable}. This syntax was introduced in Python 3 and backported as far as Python 2.7, so you should be able to use it regardless of which version of Python you have installed.

A canonical example is taking two lists and creating a dictionary where the item at each position in the first list becomes a key and the item at the corresponding position in the second list becomes the value.

The zip function used inside this comprehension returns an iterator of tuples, where each element in the tuple is taken from the same position in each of the input iterables. In the example above, the returned iterator contains the tuples (“a”, 1), (“b”, 2), etc.

Output:

{‘i’: 512, ‘e’: 64, ‘o’: 2744, ‘h’: 343, ‘l’: 1331, ‘s’: 5832, ‘b’: 1, ‘w’: 10648, ‘c’: 8, ‘x’: 12167, ‘y’: 13824, ‘t’: 6859, ‘p’: 3375, ‘d’: 27, ‘j’: 729, ‘a’: 0, ‘z’: 15625, ‘f’: 125, ‘q’: 4096, ‘u’: 8000, ‘n’: 2197, ‘m’: 1728, ‘r’: 4913, ‘k’: 1000, ‘g’: 216, ‘v’: 9261}


回答 10

此代码将使用列表推导为多个具有不同值的列表创建字典,这些字典可用于 pd.DataFrame()

#Multiple lists 
model=['A', 'B', 'C', 'D']
launched=[1983,1984,1984,1984]
discontinued=[1986, 1985, 1984, 1986]

#Dictionary with list comprehension
keys=['model','launched','discontinued']
vals=[model, launched,discontinued]
data = {key:vals[n] for n, key in enumerate(keys)}

enumerate将通过nvals使其与key列表匹配

This code will create dictionary using list comprehension for multiple lists with different values that can be used for pd.DataFrame()

#Multiple lists 
model=['A', 'B', 'C', 'D']
launched=[1983,1984,1984,1984]
discontinued=[1986, 1985, 1984, 1986]

#Dictionary with list comprehension
keys=['model','launched','discontinued']
vals=[model, launched,discontinued]
data = {key:vals[n] for n, key in enumerate(keys)}

enumerate will pass n to vals to match each key with its list


回答 11

仅举另一个例子。假设您有以下列表:

nums = [4,2,2,1,3]

并且您想将其变成字典,其中键是索引,值是列表中的元素。您可以使用以下代码行执行此操作:

{index:nums[index] for index in range(0,len(nums))}

Just to throw in another example. Imagine you have the following list:

nums = [4,2,2,1,3]

and you want to turn it into a dict where the key is the index and value is the element in the list. You can do so with the following line of code:

{index:nums[index] for index in range(0,len(nums))}

回答 12

您可以为每对创建一个新的字典并将其与上一个字典合并:

reduce(lambda p, q: {**p, **{q[0]: q[1]}}, bla bla bla, {})

显然,这种方法需要reduce来自functools

You can create a new dict for each pair and merge it with the previous dict:

reduce(lambda p, q: {**p, **{q[0]: q[1]}}, bla bla bla, {})

Obviously this approaches requires reduce from functools.


声明:本站所有文章,如无特殊说明或标注,均为本站原创发布。任何个人或组织,在未征得本站同意时,禁止复制、盗用、采集、发布本站内容到任何网站、书籍等各类媒体平台。如若本站内容侵犯了原著者的合法权益,可联系我们进行处理。