Python线程（）是否删除换行符？

Question

我在程序中遇到了一个特殊的错误。代码背后的一般思想是，首先创建一个由5个列表组成的数组（称为lists\u of_nums）。每个列表由1到10000范围内的100个数字组成。然后程序运行5个线程。每一个都会找到单个列表的平均值，并打印执行此操作所需的时间

您会注意到threading_函数（）中print（）语句的末尾有一个换行符。就在上面写着“秒。\n”的地方

问题是，如果我将't.join（）'放在一个单独的'for'循环中（代码的最底层），以便同时运行线程，那么换行符有时会被删除（显然）。如果我逐个单独运行线程，它只能在100%的时间内工作

我希望有人能帮助我理解如何同时运行线程，并且仍然让新行可靠

让我用一个代码示例及其错误输出（只在某些时候发生）来说明我的意思：

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()

输出不正确，换行符似乎在打印语句3和4之间消失：

The mean value of the list number 1 is 5270.34
The time taken to find it was 0.00012170000000000236 seconds.

The mean value of the list number 2 is 4768.17
The time taken to find it was 9.239999999999943e-05 seconds.

The mean value of the list number 3 is 4766.67
The time taken to find it was 8.369999999999905e-05 seconds.
The mean value of the list number 4 is 4969.7
The time taken to find it was 9.880000000000305e-05 seconds.


The mean value of the list number 5 is 4686.21
The time taken to find it was 9.25000000000023e-05 seconds.

Answer 1

字符串中的换行符是完整的print默认情况下，在后面添加一个换行符。它以单独的方式完成这项工作。来自不同线程的两次写入（原始字符串和换行符）可以交错，因此有时还会看到双空行。如果从字符串中删除换行符，您将看到类似的奇怪效果

一种解决方案是在两个换行符中结束字符串，并阻止print添加它自己的换行符，即print("....\n\n", end="")

Answer 2

新行正在打印中。注意，在语句4和5之间有一行额外的内容。您的线程之间可能存在竞争条件。尝试用锁保护print函数。即

printLock = threading.Lock() # global variable

然后

# inside threading_function
with printLock:
    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.
        )