Question

我正在编写一个按年回报利息总额的测试。如何将我感兴趣的值附加到我的投资者类中数组interest = []内的每一步上？出于某种原因，这很难绕过我的脑袋。

代码：

class Test(unittest.TestCase):

'''Represents my test case(s) for the dojo.'''

def setUp(self):
    '''Defines the initial state of an investor.'''
    self.investor = Investor(10000, 7.2, 5)

def test_solve_for_interest(self):
    self.assertEqual(self.investor.get_interest(), 720.0)

def test_get_interest_by_year(self):
    '''Returns a value of interest associated by the year.'''
    self.assertEqual(self.investor.interest_by_year(1), 10720.0)


class Investor(object):

    def __init__(self, amount, rate, time):
        self.amount = amount
        self.rate = rate
        self.time = time

    def get_interest(self):
        return (self.amount * self.rate) / 100.0

    def interest_by_year(self, year):
        interest = []
        for i in xrange(1, (self.time+1)):
            # collect compounding interest each year
        return interest[year]

预期结果：

year 1 = 10,720.00 (interest[1])

year 2 = 11491.84  (interest[2])

year 3 = 12319.25  (interest[3])

year 4 = 13206.24  (interest[4])

year 5 = 14157.09  (interest[5])

Answer 1

Pandas可以更优雅地解决这个问题：

years = range(1,6)
rate = 0.072
rates = pandas.Series(index = years, data = rate)

请注意，这些年份的费率可能会有所不同

initialValue*((rates + 1).cumprod())

Answer 2

def interest_by_year(self, year):
    interest = [self.amount]
    for i in xrange(self.time):
        interest.append(interest[i]*self.rate)
    return interest[year]

测试：

ivt = Investor(10000, 1.072, 10)
for i in range(1, 6):
    print "year", i, "=", ivt.interest_by_year(i)

输出：

year 1 = 10720.0
year 2 = 11491.84
year 3 = 12319.25248
year 4 = 13206.2386586
year 5 = 14157.087842

使用Python来增加兴趣

2 个答案: