我需要帮助以最有效的方式将以下列表转换为字典:
l = ['A:1','B:2','C:3','D:4']
目前,我在做以下事情:
mydict = {}
for e in l:
k,v = e.split(':')
mydict[k] = v
不过,我相信应该有一种更有效的方法来实现同样的目标。有什么想法吗?
最佳答案
将 dict()
与生成器表达式一起使用:
>>> lis=['A:1','B:2','C:3','D:4']
>>> dict(x.split(":") for x in lis)
{'A': '1', 'C': '3', 'B': '2', 'D': '4'}
使用 dict-comprehension(如@PaoloMoretti 所建议):
>>> {k:v for k,v in (e.split(':') for e in lis)}
{'A': '1', 'C': '3', 'B': '2', 'D': '4'}
10**6 项的计时结果:
>>> from so import *
>>> %timeit case1()
1 loops, best of 3: 2.09 s per loop
>>> %timeit case2()
1 loops, best of 3: 2.03 s per loop
>>> %timeit case3()
1 loops, best of 3: 2.17 s per loop
>>> %timeit case4()
1 loops, best of 3: 2.39 s per loop
>>> %timeit case5()
1 loops, best of 3: 2.82 s per loop
so.py:
a = ["{0}:{0}".format(i**2) for i in xrange(10**6)]
def case1():
dc = {}
for i in a:
q, w = i.split(':')
dc[q]=w
def case2():
dict(x.split(":") for x in a)
def case3():
{k:v for k,v in (e.split(':') for e in a)}
def case4():
dict([x.split(":") for x in a])
def case5():
{x.split(":")[0] : x.split(":")[1] for x in a}
关于python - 将列表转换为字典的有效方法,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/16374540/