如果a具有类似以下内容的列表:
If a have a list like:
l = [1,2,3,4,5]我想在结尾处
min = 1 max = 5没有min(l)和max(l).
推荐答案我能想到的最快的方法是对原始列表进行排序，然后选择第一个和最后一个元素.这样可以避免多次循环，但是会破坏列表的原始结构.这可以通过简单地复制列表并仅对复制的列表进行排序来解决.我很好奇，这是否比使用此快速示例脚本仅使用max()和min()慢?
The fastest approach I can think of would be to sort the original list and then pick the first and last elements. This avoids looping multiple times, but it does destroy the original structure of your list. This can be solved by simply copying the list and sorting only the copied list. I was curious if this was slower than just using max() and min() with this quick example script:
import time l = [1,2,4,5,3] print "Run 1" t1 = time.time() print "Min =", min(l) print "Max =", max(l) print "time =", time.time() - t1 print "" print "l =", l print "" l = [1,2,4,5,3] l1 = list(l) print "Run 2" t1 = time.time() l1.sort() print "Min =", l1[0] print "Max =", l1[-1] print "time =", time.time() - t1 print "" print "l =", l print "l1 =", l1 print "" l = [1,2,4,5,3] print "Run 3" minimum = float('inf') maximum = float('-inf') for item in l: if item < minimum: minimum = item if item > maximum: maximum = item print "Min =", minimum print "Max =", maximum print "time =", time.time() - t1 print "" print "l =", l令人惊讶的是，第二种方法在我的计算机上速度提高了大约10毫秒.不确定使用非常大的列表会产生多大的效果，但是这种方法至少对于您提供的示例列表而言更快.
Surprisingly, the second approach is faster by about 10ms on my computer. Not sure how effective this would be with very large list, but this approach is faster for at least the example list you provided.
我在计时脚本中添加了@Martijn Pieters的简单循环算法. (因为时间安排是这个问题中值得探讨的唯一重要参数.)我的结果是:
I added @Martijn Pieters's simple loop algorithm to my timing script. (As timing would be the only important parameter worth exploring in this question.) My results are:
Run 1: 0.0199999809265s Run 2: 0.00999999046326s Run 3: 0.0299999713898s
包含用于计时的timeit模块.
Inclusion of timeit module for timing.
import timeit from random import shuffle l = range(10000) shuffle(l) def Run_1(): #print "Min =", min(l) #print "Max =", max(l) return min(l), max(l) def Run_2(): l1 = list(l) l1.sort() #print "Min =", l1[0] #print "Max =", l1[-1] return l1[0], l1[-1] def Run_3(): minimum = float('inf') maximum = float('-inf') for item in l: if item < minimum: minimum = item if item > maximum: maximum = item #print "Min =", minimum #print "Max =", maximum return minimum, maximum if __name__ == '__main__': num_runs = 10000 print "Run 1" run1 = timeit.Timer(Run_1) time_run1 = run1.repeat(3, num_runs) print "" print "Run 2" run2 = timeit.Timer(Run_2) time_run2 = run2.repeat(3,num_runs) print "" print "Run 3" run3 = timeit.Timer(Run_3) time_run3 = run3.repeat(3,num_runs) print "" print "Run 1" for each_time in time_run1: print "time =", each_time print "" print "Run 2" for each_time in time_run2: print "time =", each_time print "" print "Run 3" for each_time in time_run3: print "time =", each_time print ""我的结果是:
Run 1 time = 3.42100585452 time = 3.39309908229 time = 3.47903182233 Run 2 time = 26.5261287922 time = 26.2023346397 time = 26.7324208568 Run 3 time = 3.29800945144 time = 3.25067545773 time = 3.29783778232排序算法对于大型数组非常慢.
sort algorithm is very slow for large arrays.