【求助，疑问】比较单线程与多线程操作的用时

早起的达先生

#1 Created at ... [Delete] [Delete and Lock User]

根据上文所讲，上了锁之后，多线程中一个线程执行完后才进行下一个线程，因此我猜想其效率应该和上了锁的单线程是一样的吧，因此我编写了如下程序，单线程实现存取2000000次，多线程（2线程）各实现存取1000000次，这样从工作量来讲，应该是一样的，在命令行模式下运行发现每次操作多线程都比单线程用时多1倍左右，请问这是为什么？

import time, threading
print('本程序将进行比较，单线程和多线程的用时')
balance=0
lock = threading.Lock()

def change_it(n):
    global balance
    lock.acquire()
    try:
        balance = balance + n
        balance = balance - n
    finally:
            lock.release()

def run_thread1(n):
    for i in range(2000000):
        change_it(n)

def run_thread2(n):
    for i in range(1000000):
        change_it(n)

balance=0
start1 = time.time()
td = threading.Thread(target=run_thread1, args=(5,))
td.start()
td.join()
end1 = time.time()
timeuse1=end1-start1
print('\nbalance最终结果:%d        单线程用时:%f 秒'%(balance,timeuse1))

balance=0
start2 = time.time()
t1 = threading.Thread(target=run_thread2, args=(5,))
t2 = threading.Thread(target=run_thread2, args=(5,))
t1.start()
t2.start()
t1.join()
t2.join()
end2 = time.time()
timeuse2=end2-start2
print('\nbalance最终结果:%d        多线程用时:%f 秒'%(balance,timeuse2))

问题2：如果我编写了如下程序，算不算单线程？

下面的用时比上面的单线程和多线程更少，比上面的单线程又少了一倍，那多线程的效率到底体现在哪里？

import time, threading
print('下面的操作属不属于单线程？为什么？')
balance1=0
def change_it1(n):
    global balance1
    balance1 = balance1 + n
    balance1 = balance1 - n

start1 = time.time()
for i in range(2000000):
    change_it1(5)
end1 = time.time()
timeuse1=end1-start1
print('balance1最终结果:%d        单线程用时:%f 秒'%(balance1,timeuse1))

早起的达先生

#2 Created at ... [Delete] [Delete and Lock User]

刚才我发的用的是一个网友的方法，即在change_it函数里面加锁，单线程用时1s多，双线程用时2s多，现在我用廖老师的方法来写比较用时函数，即在run_thread函数里面加锁，单线程用时0.1s多，双线程用时0.2s多......为什么同样是加锁，加锁的地方不一样，程序跑的时间会相差10倍之多？？

import time, threading
print('本程序将进行比较，单线程和多线程的用时')
balance=0
lock = threading.Lock()

def change_it(n):
    global balance
    balance = balance + n
    balance = balance - n


def run_thread1(n):
    for i in range(200000):
        lock.acquire()
        try:
            change_it(n)
        finally:
            lock.release()

def run_thread2(n):
    for i in range(100000):
        lock.acquire()
        try:
            change_it(n)
        finally:
            lock.release()

balance=0
start1 = time.time()
td = threading.Thread(target=run_thread1, args=(5,))
td.start()
td.join()
end1 = time.time()
timeuse1=end1-start1
print('\nbalance最终结果:%d        单线程用时:%f 秒'%(balance,timeuse1))

balance=0
start2 = time.time()
t1 = threading.Thread(target=run_thread2, args=(5,))
t2 = threading.Thread(target=run_thread2, args=(5,))
t1.start()
t2.start()
t1.join()
t2.join()
end2 = time.time()
timeuse2=end2-start2
print('\nbalance最终结果:%d        多线程用时:%f 秒'%(balance,timeuse2))