Python 3.x python-按列表顺序从匹配列表项和dict值中获取密钥
我有以下代码:Python 3.x python-按列表顺序从匹配列表项和dict值中获取密钥,python-3.x,list,dictionary,key,Python 3.x,List,Dictionary,Key,我有以下代码: d = {'h' : 11111111, 't' : 1010101, 'e' : 10101111, 'n' : 1} my_list = [1010101, 11111111, 10101111, 1] get_keys = [k for k, v in d.items() if v in my_list] print(get_keys) 我得到的结果是: ['h', 't', 'e', 'n'] 但是,我希望它按照我的清单的顺序排列,以便: ['t', 'h',
d = {'h' : 11111111, 't' : 1010101, 'e' : 10101111, 'n' : 1}
my_list = [1010101, 11111111, 10101111, 1]
get_keys = [k for k, v in d.items() if v in my_list]
print(get_keys)
我得到的结果是:
['h', 't', 'e', 'n']
但是,我希望它按照我的清单的顺序排列,以便:
['t', 'h', 'e', 'n']
我该怎么做?谢谢大家! 给定(其中所有值也是唯一的):
d = {'h' : 11111111, 't' : 1010101, 'e' : 10101111, 'n' : 1}
my_list = [1010101, 11111111, 10101111, 1]
new_list = []
for i in my_list:
for key, value in d.items():
if value == i:
new_list.append(key)
print(new_list)
您可以反转该命令:
>>> d_inverted={v:k for k,v in d.items()}
然后按预期编制索引:
>>> [d_inverted[e] for e in my_list]
['t', 'h', 'e', 'n']
这适用于任何最新版本的Python
请注意,您发布的方法具有
O(n^2)
复杂性。这意味着执行代码的时间将随着元素数的平方而增加将元素加倍,执行时间将翻两番。结果不好
从视觉上看,如下所示:
相比之下,我发布的方法是O(n)
,或者与元素的数量成正比双倍数据等于双倍执行时间。更好的结果。(但不如O(1)
好,后者的执行时间与数据大小无关。)
如果要对它们进行比较,请执行以下操作:
def bad(d,l):
new_list = []
for i in l:
for key, value in d.items():
if value == i:
new_list.append(key)
return new_list
def better(d,l):
d_inverted={v:k for k,v in d.items()}
return [d_inverted[e] for e in my_list]
if __name__=='__main__':
import timeit
import random
for tgt in (5,10,20,40,80,160,320,640,1280):
d={chr(i):i for i in range(100,100+tgt)}
my_list=list(d.values())
random.shuffle(my_list)
print("Case of {} elements:".format(len(my_list)))
for f in (bad, better):
print("\t{:10s}{:.4f} secs".format(f.__name__, timeit.timeit("f(d,my_list)", setup="from __main__ import f, d, my_list", number=100)))
印刷品:
Case of 5 elements:
bad 0.0003 secs
better 0.0001 secs
Case of 10 elements:
bad 0.0006 secs
better 0.0002 secs
Case of 20 elements:
bad 0.0022 secs
better 0.0003 secs
Case of 40 elements:
bad 0.0071 secs
better 0.0004 secs
Case of 80 elements:
bad 0.0240 secs
better 0.0008 secs
Case of 160 elements:
bad 0.0912 secs
better 0.0018 secs
Case of 320 elements:
bad 0.3571 secs
better 0.0032 secs
Case of 640 elements:
bad 1.3704 secs
better 0.0053 secs
Case of 1280 elements:
bad 5.4443 secs
better 0.0107 secs
您可以看到,嵌套循环方法从
3x
开始变慢,并随着数据大小的增加而增加到500x
变慢。时间的增长与大O的预测密切相关。您可以想象数百万个元素会发生什么情况。python字典没有排序。您可能想使用这些值是否都是唯一的?否则,一些键可能会模棱两可。所有键都是唯一的。是的,我现在已经排序了,无论如何,干杯!对于my_列表中的每个项目
您正在dictd
中的每个项目上循环。换句话说,这具有O(n^2)复杂性。请不要用这个。。。。
Case of 5 elements:
bad 0.0003 secs
better 0.0001 secs
Case of 10 elements:
bad 0.0006 secs
better 0.0002 secs
Case of 20 elements:
bad 0.0022 secs
better 0.0003 secs
Case of 40 elements:
bad 0.0071 secs
better 0.0004 secs
Case of 80 elements:
bad 0.0240 secs
better 0.0008 secs
Case of 160 elements:
bad 0.0912 secs
better 0.0018 secs
Case of 320 elements:
bad 0.3571 secs
better 0.0032 secs
Case of 640 elements:
bad 1.3704 secs
better 0.0053 secs
Case of 1280 elements:
bad 5.4443 secs
better 0.0107 secs