Question

我有一个列表，我想将此列表转换为地图

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

mylist相当于

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：我已经编写了以下函数来完成相同的工作，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

列表从1开始，跳过2

Answer 3

要从列表中获取所有密钥，您可以执行以下操作：

keys = mylist[::2]

其中mylist [:: 2]通过迭代从索引0开始的mylist中的每个第二个元素来创建新列表。

要从列表中获取所有值，您可以执行以下操作：

vals = mylist[1::2]

其中mylist[1::2]通过迭代从索引1开始的mylist中的每个第二个元素来创建新列表。

然后你可以使用dict和zip：

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

zip会将["a","b","c"]和[1,2,3]两个列表和“拉链”放在一起，以提供类似

的列表

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

和dict采用可迭代的对并将它们变成字典。

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}

要在python中映射的列表

4 个答案: