Dataframe with Monte Carlo Simulation calculation next row 问题
Dataframe with Monte Carlo Simulation calculation next row Problem
我想从头开始构建一个 Dataframe,并根据 Value before named Barrier 选项进行计算。我知道我可以使用 Monte Carlo 模拟来解决它,但它不会按照我想要的方式工作。
公式为:
Value in row before * np.exp((r-sigma**2/2)*T/TradingDays+sigma*np.sqrt(T/TradingDays)*z)
我写的第一个代码只是计算第一列。我知道我需要第二个循环但无法真正管理它。
结果应该是,对于每次模拟,它将使用之前的值计算一个新值,对于 500 天,意味着 S_1 应该是 S_500,总共 1000 次模拟。 (在使用公式之前,我需要根据值生成新列。)
与此类似:
所以对于 1.模拟 500 天,2.模拟 500 天等等...
import numpy as np
import pandas as pd
from scipy.stats import norm
import random as rd
import math
simulation = 0
S_0 = 42
T = 2
r = 0.02
sigma = 0.20
TradingDays = 500
df = pd.DataFrame()
for i in range (0,TradingDays):
z = norm.ppf(rd.random())
simulation = simulation + 1
S_1 = S_0*np.exp((r-sigma**2/2)*T/TradingDays+sigma*np.sqrt(T/TradingDays)*z)
df = df.append ({
'S_1':S_1,
'S_0':S_0
}, ignore_index=True)
df = df.round ({'Z':6,
'S_T':2
})
df.index += 1
df.index.name = 'Simulation'
print(df)
我在此处找到了另一个可能的代码,它确实解决了问题,但仅针对一行,下一行只是相同的计算。
如果我只是用我的公式替换它,我会遇到同样的问题。
正在替换:
exp(r - q * sqrt(sigma))*T+ (np.random.randn(nrows) * sqrt(deltaT)))
与:
exp((r-sigma**2/2)*T/nrows+sigma*np.sqrt(T/nrows)*z))
import numpy as np
import pandas as pd
from scipy.stats import norm
import random as rd
import math
S_0 = 42
T = 2
r = 0.02
sigma = 0.20
TradingDays = 50
Simulation = 100
df = pd.DataFrame({'s0': [S_0] * Simulation})
for i in range(1, TradingDays):
z = norm.ppf(rd.random())
df[f's{i}'] = df.iloc[:, -1] * np.exp((r-sigma**2/2)*T/TradingDays+sigma*np.sqrt(T/TradingDays)*z)
print(df)
我会更有可能使用最后的代码并用它解决问题。
在循环并将所有模拟保存在列表中时,用 S_1 的新值覆盖 S_0 的值怎么样?
像这样:
import numpy as np
import pandas as pd
import random
from scipy.stats import norm
S_0 = 42
T = 2
r = 0.02
sigma = 0.20
trading_days = 50
output = []
for i in range(trading_days):
z = norm.ppf(random.random())
value = S_0*np.exp((r - sigma**2 / 2) * T / trading_days + sigma * np.sqrt(T/trading_days) * z)
output.append(value)
S_0 = value
df = pd.DataFrame({'simulation': output})
也许我遗漏了什么,但我认为不需要第二个循环。
此外,这消除了在循环中调用 df.append()
,这应该避免。 (参见 )
根据bartaelterman的回答解决,非常感谢!
import numpy as np
import pandas as pd
from scipy.stats import norm
import random as rd
import math
#Dividing the list in chunks to later append it to the dataframe in the right order
def chunk_list(lst, chunk_size):
for i in range(0, len(lst), chunk_size):
yield lst[i:i + chunk_size]
def blackscholes():
d1 = ((math.log(S_0/K)+(r+sigma**2/2)*T)/(sigma*np.sqrt(2)))
d2 = ((math.log(S_0/K)+(r-sigma**2/2)*T)/(sigma*np.sqrt(2)))
preis_call_option = S_0*norm.cdf(d1)-K*np.exp(-r*T)*norm.cdf(d2)
return preis_call_option
K = 40
S_0 = 42
T = 2
r = 0.02
sigma = 0.2
U = 38
simulation = 10000
trading_days = 500
trading_days = trading_days -1
#creating 2 lists for the first and second loop
loop_simulation = []
loop_trading_days = []
#first loop calculates the first column in a list
for j in range (0,simulation):
print("Progressbar_1_2 {:2.2%}".format(j / simulation), end="\n\r")
S_Tag_new = 0
NORM_S_INV = norm.ppf(rd.random())
S_Tag = S_0*np.exp((r-sigma**2/2)*T/trading_days+sigma*np.sqrt(T/trading_days)*NORM_S_INV)
S_Tag_new = S_Tag
loop_simulation.append(S_Tag)
#second loop calculates the the rows for the columns in a list
for i in range (0,trading_days):
NORM_S_INV = norm.ppf(rd.random())
S_Tag = S_Tag_new*np.exp((r-sigma**2/2)*T/trading_days+sigma*np.sqrt(T/trading_days)*NORM_S_INV)
loop_trading_days.append(S_Tag)
S_Tag_new = S_Tag
#values from the second loop will be divided in number of Trading days per Simulation
loop_trading_days_chunked = list(chunk_list(loop_trading_days,trading_days))
#First dataframe with just the first results from the firstloop for each simulation
df1 = pd.DataFrame({'S_Tag 1': loop_simulation})
#Appending the the chunked list from the second loop to a second dataframe
df2 = pd.DataFrame(loop_trading_days_chunked)
#Merging both dataframe into one
df3 = pd.concat([df1, df2], axis=1)
我想从头开始构建一个 Dataframe,并根据 Value before named Barrier 选项进行计算。我知道我可以使用 Monte Carlo 模拟来解决它,但它不会按照我想要的方式工作。
公式为:
Value in row before * np.exp((r-sigma**2/2)*T/TradingDays+sigma*np.sqrt(T/TradingDays)*z)
我写的第一个代码只是计算第一列。我知道我需要第二个循环但无法真正管理它。
结果应该是,对于每次模拟,它将使用之前的值计算一个新值,对于 500 天,意味着 S_1 应该是 S_500,总共 1000 次模拟。 (在使用公式之前,我需要根据值生成新列。)
与此类似:
import numpy as np
import pandas as pd
from scipy.stats import norm
import random as rd
import math
simulation = 0
S_0 = 42
T = 2
r = 0.02
sigma = 0.20
TradingDays = 500
df = pd.DataFrame()
for i in range (0,TradingDays):
z = norm.ppf(rd.random())
simulation = simulation + 1
S_1 = S_0*np.exp((r-sigma**2/2)*T/TradingDays+sigma*np.sqrt(T/TradingDays)*z)
df = df.append ({
'S_1':S_1,
'S_0':S_0
}, ignore_index=True)
df = df.round ({'Z':6,
'S_T':2
})
df.index += 1
df.index.name = 'Simulation'
print(df)
我在此处找到了另一个可能的代码,它确实解决了问题,但仅针对一行,下一行只是相同的计算。
如果我只是用我的公式替换它,我会遇到同样的问题。
正在替换:
exp(r - q * sqrt(sigma))*T+ (np.random.randn(nrows) * sqrt(deltaT)))
与:
exp((r-sigma**2/2)*T/nrows+sigma*np.sqrt(T/nrows)*z))
import numpy as np
import pandas as pd
from scipy.stats import norm
import random as rd
import math
S_0 = 42
T = 2
r = 0.02
sigma = 0.20
TradingDays = 50
Simulation = 100
df = pd.DataFrame({'s0': [S_0] * Simulation})
for i in range(1, TradingDays):
z = norm.ppf(rd.random())
df[f's{i}'] = df.iloc[:, -1] * np.exp((r-sigma**2/2)*T/TradingDays+sigma*np.sqrt(T/TradingDays)*z)
print(df)
我会更有可能使用最后的代码并用它解决问题。
在循环并将所有模拟保存在列表中时,用 S_1 的新值覆盖 S_0 的值怎么样? 像这样:
import numpy as np
import pandas as pd
import random
from scipy.stats import norm
S_0 = 42
T = 2
r = 0.02
sigma = 0.20
trading_days = 50
output = []
for i in range(trading_days):
z = norm.ppf(random.random())
value = S_0*np.exp((r - sigma**2 / 2) * T / trading_days + sigma * np.sqrt(T/trading_days) * z)
output.append(value)
S_0 = value
df = pd.DataFrame({'simulation': output})
也许我遗漏了什么,但我认为不需要第二个循环。
此外,这消除了在循环中调用 df.append()
,这应该避免。 (参见
根据bartaelterman的回答解决,非常感谢!
import numpy as np
import pandas as pd
from scipy.stats import norm
import random as rd
import math
#Dividing the list in chunks to later append it to the dataframe in the right order
def chunk_list(lst, chunk_size):
for i in range(0, len(lst), chunk_size):
yield lst[i:i + chunk_size]
def blackscholes():
d1 = ((math.log(S_0/K)+(r+sigma**2/2)*T)/(sigma*np.sqrt(2)))
d2 = ((math.log(S_0/K)+(r-sigma**2/2)*T)/(sigma*np.sqrt(2)))
preis_call_option = S_0*norm.cdf(d1)-K*np.exp(-r*T)*norm.cdf(d2)
return preis_call_option
K = 40
S_0 = 42
T = 2
r = 0.02
sigma = 0.2
U = 38
simulation = 10000
trading_days = 500
trading_days = trading_days -1
#creating 2 lists for the first and second loop
loop_simulation = []
loop_trading_days = []
#first loop calculates the first column in a list
for j in range (0,simulation):
print("Progressbar_1_2 {:2.2%}".format(j / simulation), end="\n\r")
S_Tag_new = 0
NORM_S_INV = norm.ppf(rd.random())
S_Tag = S_0*np.exp((r-sigma**2/2)*T/trading_days+sigma*np.sqrt(T/trading_days)*NORM_S_INV)
S_Tag_new = S_Tag
loop_simulation.append(S_Tag)
#second loop calculates the the rows for the columns in a list
for i in range (0,trading_days):
NORM_S_INV = norm.ppf(rd.random())
S_Tag = S_Tag_new*np.exp((r-sigma**2/2)*T/trading_days+sigma*np.sqrt(T/trading_days)*NORM_S_INV)
loop_trading_days.append(S_Tag)
S_Tag_new = S_Tag
#values from the second loop will be divided in number of Trading days per Simulation
loop_trading_days_chunked = list(chunk_list(loop_trading_days,trading_days))
#First dataframe with just the first results from the firstloop for each simulation
df1 = pd.DataFrame({'S_Tag 1': loop_simulation})
#Appending the the chunked list from the second loop to a second dataframe
df2 = pd.DataFrame(loop_trading_days_chunked)
#Merging both dataframe into one
df3 = pd.concat([df1, df2], axis=1)