我在我的程序中遇到了一個(gè)特殊的錯(cuò)誤。代碼背后的總體思路是首先創(chuàng)建一個(gè)包含 5 個(gè)列表的數(shù)組（稱為 lists_of_nums）。每個(gè)列表由 1 到 10000 范圍內(nèi)的 100 個(gè)數(shù)字組成。程序然后運(yùn)行 5 個(gè)線程。每個(gè)人都找到一個(gè)列表的平均值，并打印出完成它所花費(fèi)的時(shí)間。您會(huì)注意到在 threading_function() 中的 print() 語句的末尾有一個(gè)換行符。就在它說“seconds.\n”的地方。問題是，如果我將 't.join()' 放在一個(gè)單獨(dú)的 'for' 循環(huán)中（在我的代碼的最底部），為了同時(shí)運(yùn)行線程，換行符有時(shí)會(huì)被刪除（顯然）。如果我一個(gè)一個(gè)地單獨(dú)運(yùn)行線程，它只能在 100% 的時(shí)間內(nèi)工作。我希望有人可以幫助我了解如何同時(shí)運(yùn)行線程并且仍然讓換行符可靠。讓我用代碼示例及其不正確的輸出（僅在某些時(shí)候發(fā)生）來說明我的意思：lists_of_nums = []def create_set():    # Function for creating a list of random integers and appending them to the lists_of_nums array.    random_nums = []    for _ in range(100):        random_nums.append(randint(1, 10000))    lists_of_nums.append(random_nums)for _ in range(5):    # Five of these lists get appended to the array, by using the create_set() function.    create_set()def threading_function(list_index):    # Function responsible for finding the mean value of a given list as well as the time taken to do it.    start = timeit.default_timer()    mean_value = mean(lists_of_nums[list_index])    end = timeit.default_timer() - start    print(        "The mean value of the list number " + str(list_index + 1) + " is " + str(mean_value) +        "\nThe time taken to find it was " + str(end) + " seconds.\n" # The abovementioned newline.        )threads = []for i in range(len(lists_of_nums)):    t = Thread(target = threading_function, args = [i])    t.start()    threads.append(t)for t in threads:    # If t.join() remains in a separate 'for' loop than the Thread() class, the newline occasionally disappears.    t.join()