为什么列表理解不能产生与for/in循环相同的结果？

Question

生成一个随机字符串列表，然后使用for/in循环和列表理解表达式为最长字符串和该字符串的长度提供资金

这两种技术都能正确计算最大长度，但有时for/in循环会找到与列表理解相同的最长单词，有时则不会。为什么？逻辑错误是什么

import random
import string
def cobble_large_dataset(dataset_number_of_elements):
    '''
    Build a list of Lists, each List is a String of a random sequence of 1-10 characters
    '''
    myList = []         # Empty List   
    for i in range(0,dataset_number_of_elements):
        string_length = random.randint(1, 10)
        tmp = ''.join(random.choices(string.ascii_uppercase + string.digits, k=string_length))  # https://stackoverflow.com/questions/2257441/random-string-generation-with-upper-case-letters-and-digits
        tmp = [tmp]
        #print(tmp)
        myList.extend([tmp])
    return myList    

def list_comprehension_test(wordsList):
    '''
    Process a List of Lists using List Comprehension. 
    Each List in the List of Lists is a single String
    '''
    start_time = time.time()
   
    maximumWordLength, longest_word = max([(len(x[0]), x[0]) for x in wordsList]) # This works because x is a List of strings
    return ((time.time() - start_time), longest_word, maximumWordLength)

def brute_force_test(wordsList):
    '''
    Process a List of Lists using a brute-force for/in loop. 
    Each List in the List of Lists is a single String    
    '''
    start_time = time.time()
    maximumWordLength = 0
    for word in wordsList:
        tmp = word[0]
        #print(tmp)
        if (len(tmp) >= maximumWordLength):
            maximumWordLength = len(tmp)
            longest_word = tmp
            #print(tmp)
            #print(longest_word + " : " + str(maximumWordLength))
    return ((time.time() - start_time), longest_word, maximumWordLength)

import time
start_time = time.time()
dataset = cobble_large_dataset(100)
print (str(len(dataset)) + ' Strings generated in ' + str((time.time() - start_time)) + ' seconds.')

# Let's see if both techniques produce the same results:
result_brute_force = brute_force_test(dataset)
print('Results from Brute Force = ' + result_brute_force[1] + ', ' + str(result_brute_force[2]) + ' characters' )
result_list_comprehension = list_comprehension_test(dataset)
print('Results from List Comprehension = ' + result_list_comprehension[1] + ', ' + str(result_list_comprehension[2]) + ' characters' )
if (result_list_comprehension[1] == result_brute_force[1]):
    print("Techniques produced the same results.")
else:
    print("Techniques DID NOT PRODUCE the same results

Answer 1

您正在向max传递一个元组列表，并且没有key函数，因此max正在比较元组，而不仅仅是长度。当长度相等时，元组比较将继续比较第二个元素，即字符串本身，因此在长度相等的情况下，最大值是比较最大的字符串（通过字典代码点比较）

相反，在长度平局的情况下，循环选择最后出现的候选项。（如果您在if (len(tmp) >= maximumWordLength):中使用了>而不是>=，它将选择第一个候选者。）

（还有，你正在用tmp做一些奇怪的事情。你正在构建的1元素列表是没有意义的-cobble_large_dataset应该只返回一个简单的字符串列表。）

Answer 2

在列表理解案例中，您希望告诉max只对列表中每对值中的第一项进行操作。这相当于for循环的情况，因为它只考虑每个字符串的长度。所以你想要：

maximumWordLength, longest_word = max(
    [(len(x[0]), x[0]) for x in wordsList],
    key = lambda x: x[0])  # This works because x is a List of strings

正如其他人已经指出的，您还希望将暴力案例中的>=比较更改为>。如果您进行这两个更改，您将从这两个方法中获得相同的结果