我在程序中遇到了一个特殊的错误。该代码背后的一般思想是，首先创建5个列表的数组（称为list_of_nums）。每个列表由1到10000之间的100个数字组成。该程序然后运行5个线程。每个人都会找到一个列表的平均值，并打印完成该列表所花费的时间。
您会注意到，在threading_function（）中print（）语句的末尾只有一个换行符。在显示“秒。\ n”的位置。
问题是，如果我将't.join（）'放在单独的'for'循环中（在我的代码的最底部），为了同时运行线程，有时会删除该换行符（显然）。如果我逐个运行线程，则只能100％地起作用。
我希望有人可以帮助我了解如何同时运行线程，同时使换行符仍然可靠。
让我用代码示例及其不正确的输出（仅在某些情况下发生）演示我的意思：
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()
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.
