概率模拟
Probability simulation
我被要求模拟一天中的每个小时后(即 24 小时)打印机是否工作。如果打印机在一个小时后工作,那么它有 90% 的概率在工作,10% 的概率在下一个小时结束时坏掉。
如果坏了,那么它有 50% 的概率在下一个小时工作或坏掉。
假设随机均匀分布并且打印机工作的第一个小时。
我在 Python 中的代码如下:
Chance = []
Status = []
for i in range(24):
Chance.append(random.uniform(0,1))
Chance[0] = 1
Chance
for i in Chance:
if i > 0.1:
Status.append('Working')
else:
Status.append('Broken')
Chance, Status
我的问题是我无法根据之前的事件模拟当前事件,即如果之前的事件被打破,我如何将当前事件的概率调整为0.5。
你应该记住上一步(小时)打印机的状态,并调整计算随机二进制输出的阈值:
status = []
lastStatus = True # Assume it was working
for i in range(24):
threshold = 0.1 if lastStatus else 0.5
lastStatus = random.uniform(0, 1) > threshold
status.append('Working' if lastStatus else 'Broken')
你需要先定义你的概率table,像这样:
| NEXT STATE |
| WORKING | BROKEN |
CURRENT| WORKING | 0.9 | 0.1 |
STATE |---------|-------------------|--------------------|
| BROKEN | 0.5 | 0.5 |
为了能够很容易地更改值。您甚至可以轻松添加新状态。像这样使用它:
import random
# The table above
table = [[.9,.1],[.5,.5]]
def simulate():
# 0 means working, 1 means broken
current_state = 0
# Initialise variable
next_state = current_state
for i in range(24):
# Get a number between 0 and 1
chance = random.uniform(0, 1)
# Given my current state (line 0 or 1), what chance do I have
# to work the next hour ?
if chance <= table[current_state][0]:
# The chance variable has more chance to be under if the number
# in the table is high.
next_state = 0
else:
# BROKEN
next_state = 1
current_state = next_state
print("Current state " + ("Working" if current_state == 0 else "Broken"))
if __name__ == "__main__":
simulate()
使用该脚本:
import random
broken_probability = 0
for i in range(24):
if random.randrange(100) > broken_probability:
status = 'Working'
broken_probability += 10
else:
status = 'Broken '
broken_probability = 50
print('Hour: {} - Status: {} - Broken Probability: {}'.format(str(i).zfill(2), status, broken_probability))
我得到了以下结果:
Hour: 00 - Status: Working - Broken Probability: 10
Hour: 01 - Status: Working - Broken Probability: 20
Hour: 02 - Status: Broken - Broken Probability: 50
Hour: 03 - Status: Broken - Broken Probability: 50
Hour: 04 - Status: Broken - Broken Probability: 50
Hour: 05 - Status: Broken - Broken Probability: 50
Hour: 06 - Status: Working - Broken Probability: 60
Hour: 07 - Status: Broken - Broken Probability: 50
Hour: 08 - Status: Broken - Broken Probability: 50
Hour: 09 - Status: Working - Broken Probability: 60
Hour: 10 - Status: Broken - Broken Probability: 50
Hour: 11 - Status: Broken - Broken Probability: 50
Hour: 12 - Status: Working - Broken Probability: 60
Hour: 13 - Status: Broken - Broken Probability: 50
Hour: 14 - Status: Broken - Broken Probability: 50
Hour: 15 - Status: Working - Broken Probability: 60
Hour: 16 - Status: Working - Broken Probability: 70
Hour: 17 - Status: Broken - Broken Probability: 50
Hour: 18 - Status: Broken - Broken Probability: 50
Hour: 19 - Status: Broken - Broken Probability: 50
Hour: 20 - Status: Broken - Broken Probability: 50
Hour: 21 - Status: Working - Broken Probability: 60
Hour: 22 - Status: Broken - Broken Probability: 50
Hour: 23 - Status: Working - Broken Probability: 60
我被要求模拟一天中的每个小时后(即 24 小时)打印机是否工作。如果打印机在一个小时后工作,那么它有 90% 的概率在工作,10% 的概率在下一个小时结束时坏掉。
如果坏了,那么它有 50% 的概率在下一个小时工作或坏掉。
假设随机均匀分布并且打印机工作的第一个小时。
我在 Python 中的代码如下:
Chance = []
Status = []
for i in range(24):
Chance.append(random.uniform(0,1))
Chance[0] = 1
Chance
for i in Chance:
if i > 0.1:
Status.append('Working')
else:
Status.append('Broken')
Chance, Status
我的问题是我无法根据之前的事件模拟当前事件,即如果之前的事件被打破,我如何将当前事件的概率调整为0.5。
你应该记住上一步(小时)打印机的状态,并调整计算随机二进制输出的阈值:
status = []
lastStatus = True # Assume it was working
for i in range(24):
threshold = 0.1 if lastStatus else 0.5
lastStatus = random.uniform(0, 1) > threshold
status.append('Working' if lastStatus else 'Broken')
你需要先定义你的概率table,像这样:
| NEXT STATE |
| WORKING | BROKEN |
CURRENT| WORKING | 0.9 | 0.1 |
STATE |---------|-------------------|--------------------|
| BROKEN | 0.5 | 0.5 |
为了能够很容易地更改值。您甚至可以轻松添加新状态。像这样使用它:
import random
# The table above
table = [[.9,.1],[.5,.5]]
def simulate():
# 0 means working, 1 means broken
current_state = 0
# Initialise variable
next_state = current_state
for i in range(24):
# Get a number between 0 and 1
chance = random.uniform(0, 1)
# Given my current state (line 0 or 1), what chance do I have
# to work the next hour ?
if chance <= table[current_state][0]:
# The chance variable has more chance to be under if the number
# in the table is high.
next_state = 0
else:
# BROKEN
next_state = 1
current_state = next_state
print("Current state " + ("Working" if current_state == 0 else "Broken"))
if __name__ == "__main__":
simulate()
使用该脚本:
import random
broken_probability = 0
for i in range(24):
if random.randrange(100) > broken_probability:
status = 'Working'
broken_probability += 10
else:
status = 'Broken '
broken_probability = 50
print('Hour: {} - Status: {} - Broken Probability: {}'.format(str(i).zfill(2), status, broken_probability))
我得到了以下结果:
Hour: 00 - Status: Working - Broken Probability: 10
Hour: 01 - Status: Working - Broken Probability: 20
Hour: 02 - Status: Broken - Broken Probability: 50
Hour: 03 - Status: Broken - Broken Probability: 50
Hour: 04 - Status: Broken - Broken Probability: 50
Hour: 05 - Status: Broken - Broken Probability: 50
Hour: 06 - Status: Working - Broken Probability: 60
Hour: 07 - Status: Broken - Broken Probability: 50
Hour: 08 - Status: Broken - Broken Probability: 50
Hour: 09 - Status: Working - Broken Probability: 60
Hour: 10 - Status: Broken - Broken Probability: 50
Hour: 11 - Status: Broken - Broken Probability: 50
Hour: 12 - Status: Working - Broken Probability: 60
Hour: 13 - Status: Broken - Broken Probability: 50
Hour: 14 - Status: Broken - Broken Probability: 50
Hour: 15 - Status: Working - Broken Probability: 60
Hour: 16 - Status: Working - Broken Probability: 70
Hour: 17 - Status: Broken - Broken Probability: 50
Hour: 18 - Status: Broken - Broken Probability: 50
Hour: 19 - Status: Broken - Broken Probability: 50
Hour: 20 - Status: Broken - Broken Probability: 50
Hour: 21 - Status: Working - Broken Probability: 60
Hour: 22 - Status: Broken - Broken Probability: 50
Hour: 23 - Status: Working - Broken Probability: 60