def pmt(*args):
rate, nper, pv = args[0], args[1], args[2]
return scipy.pmt(rate, nper, pv)
### manipulate matrix col and row
np.std(x, axis=1) ##std by col
x.sum()
x.sum(axis=0) ##sum by row
### simulation
ret = np.array(range(1, 100), float)/100 ##create a grid of return pct
rand_50 = np.random.rand(50) ##50 rand num btw 0 and 1
rnorm_100 = np.random.normal(size=100)
## 2. Scipy ------------
### Compute the NPV of a Cashflow
cf = [-100, 50, 40, 20, 10, 50]
sp.npv(rate=0.1, values=cf)
### Monthly payment: APR = 0.045, compounded monthly, mortgage = 250k, T=30yrs
sp.pmt(rate=0.045/12, nper=12*30, pv=250000)
### PV of one future cf
sp.pv(rate=0.1,nper=5,pmt=0,fv=100)
### PV of Annuity (series of payments)
sp.pv(rate=0.1, nper=5, pmt=100)
### Geometric mean returns
def geoMeanreturn(ret):
return pow(sp.prod(ret+1), 1./len(ret)) - 1
ret = np.array([0.1, 0.05, -0.02])
geoMeanreturn(ret)
### other fundamental stuff
sp.unique([1,1,2,3,4,1,1,1])
sp.median([1,1,2,3,4,1,1,1])
## 3. matplotlib ------------
''' The sp.pmt() function is used to answer the following question: What is the monthly cash flow to pay off a mortgage of $250,000 over 30 years with an annual percentage rate (APR) of 4.5 percent, compounded monthly? ''' import scipy as sp ''' 贷款:200'000.00 年利率：5% 还款周期：月 贷款年限：3年 每月还款：5'994.18 ''' payment2 = sp.pmt(0.05 / 12, 3 * 12, 200000, fv=0, when='end') print(payment2)
''' The sp.pmt() function is used to answer the following question: What is the monthly cash flow to pay off a mortgage of $250,000 over 30 years with an annual percentage rate (APR) of 4.5 percent, compounded monthly? ''' import scipy as sp ''' 贷款:200'000.00 年利率：5% 还款周期：月 贷款年限：3年 每月还款：5'994.18 ''' payment2 = sp.pmt(0.05/12,3*12,200000,fv=0,when='end') print (payment2)
import scipy as sp
n = float(raw_input('N = '))
iy = float(raw_input('I/Y = '))
pv = float(raw_input('PV = '))
pmt = float(raw_input('PMT = '))
fv = float(raw_input('FV = '))
sol = raw_input('Solve for (n,iy,pv,pmt,fv): ')
if sol == "n":
print("N = %f" % sp.nper(iy, pmt, pv, fv))
if sol == "iy":
print("I/Y = %f" % sp.rate(n, pmt, pv, fv))
if sol == "pv":
print("PV = %f" % sp.pv(iy, n, pmt, fv))
if sol == "pmt":
print("PMT = %f" % sp.pmt(iy, n, pv, fv))
if sol == "fv":
print("FV = %f" % sp.fv(iy, n, pmt, pv))
'''
print 'N = %f, ' % n
print 'I/Y = %f, ' % iy
print 'PV = %f, ' % pv
print 'PMT = %f, ' % pmt
print 'FV = %f, ' % fv
'''
import numpy.lib.financial as fin # signature of fv fucntion is (Rate,nper(number of years),pmnt (payment),pv(present value), when='end' ) # the following is for rate = %10, one year, 0 payment and $100 principle futVal_1 = fin.fv(0.1,1,0,100) help(fin.fv) futVal_2 = fin.fv(0.1,5,0,100) ''' Here is an example. John is planning to buy a used car with a price tag of $5,000. Assume that he would pay $1,000 as the down payment and borrow the rest. The annual interest rate for a car load is 1.9% compounded monthly. What is his monthly payment if he plans to retire his load in three years? We could calculate the monthly payment manually; see the following code: ''' # manual calculation r=0.019/12 pv=4000 n=3*12 pymnt = pv*r/(1-1/(1+r)**n) pymntPermonth = 200 # Using scipy sp.pmt(r,n,pv) sp.rate(n,pymntPermonth,-pv,0) round(sp.nper(r,-pymntPermonth,-pv,0)) fin.pmt(r,n,pv) fin.rate(n,pymntPermonth,-pv,0)
Python 3.7.3 (v3.7.3:ef4ec6ed12, Mar 25 2019, 21:26:53) [MSC v.1916 32 bit (Intel)] on win32 Type "help", "copyright", "credits" or "license()" for more information. >>> import scipy as sp >>> sp.fv(.0025,25,0,5000) -5322.057184363162 >>> import scipy as sp >>> sp.pv(.0541/2,20,55,0,1) -863.7816463604273 >>> import scipy as sp >>> sp.pmt(.0312,5,2400) -525.8478442817194 >>>