在python中映射的列表

Question

我有一个列表，我想把这个列表转换成地图

mylist = ["a",1,"b",2,"c",3]

mylist 等价于

mylist = [Key,Value,Key,Value,Key,Value]

所以输入：

mylist = ["a",1,"b",2,"c",3]

输出：

mymap = {"a":1,"b":2,"c":3}

P.S：我已经写了下面的函数做同样的工作，但我想使用 python 的迭代器工具：

def fun():
    mylist = ["a",1,"b",2,"c",3]
    mymap={}
    count = 0
    for value in mylist:
        if not count%2:
            mymap[value] = mylist[count+1]
        count = count+1
    return mymap        

Answer 1

使用iter 和dict-comprehension：

>>> mylist = ["a",1,"b",2,"c",3]
>>> it = iter(mylist)
>>> {k: next(it) for k in it}
{'a': 1, 'c': 3, 'b': 2}

使用zip 和iter：

>>> dict(zip(*[iter(mylist)]*2)) #use `itertools.izip` if the list is huge.
{'a': 1, 'c': 3, 'b': 2}

相关：How does zip(*[iter(s)]*n) work in Python

Answer 2

>>> mylist = ["a",1,"b",2,"c",3]
>>> zippedlist = zip(mylist [0::2],mylist [1::2]) #slicing magic
>>> zippedlist
[('a', 1), ('b', 2), ('c', 3)]
>>> dictedlist = dict(zippedlist)
>>> dictedlist 
{'a': 1, 'c': 3, 'b': 2}

这是因为切片[start:stop:skip]

列表从 0 开始，跳过 2

压缩包

列表从 1 开始，跳过 2

Answer 3

要从列表中获取所有键，您可以执行以下操作：

keys = mylist[::2]

mylist[::2] 通过迭代 mylist 中从索引 0 开始的每个第二个元素来创建一个新列表。

要从列表中获取所有值，您可以这样做：

vals = mylist[1::2]

mylist[1::2] 通过迭代 mylist 中从索引 1 开始的每个第二个元素来创建一个新列表。

然后你就可以使用 dict 和 zip：

dict(zip(keys,vals))  # or all in one line
dict(zip(mylist[::2], mylist[1::2]))

zip 需要两个列表，例如 ["a","b","c"] 和 [1,2,3]，然后将它们“压缩”在一起以给出类似

的列表

[("a",1), ("b", 2), ("c", 3)]

而dict 采用可迭代的对并将它们转换成字典。

Answer 4

使用压缩和循环：-

from itertools import compress
from itertools import cycle

def fun(mylist):
    keys = compress(mylist,cycle([1,0]))
    values = compress(mylist,cycle([0,1]))
    mymap = dict(zip(keys,values))
    return mymap

mylist = ["a",1,"b",2,"c",3]

fun(mylist)
{'a': 1, 'c': 3, 'b': 2}