在Python上使用__add__是个坏主意吗？

Question

I'm looking to increment a value by one and Python does not have the ++ operator. Consider the following example:

# In a method called calculate(self, basecost, othertaxes=None)
# Returns the value of the tax (self) applied to basecost in relation to previous taxes
i = -1
basecost += sum((tax.calculate(basecost, othertaxes[:i.__add__(1)]) for tax in othertaxes))

Is the use of __add__ in this example a bad idea? Is there a better way to write this statement?

Cheers - D

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UPDATE

I have changed the answer because the for ... in ...: v += calc solution is much faster than the sum() method. 6 seconds faster over 10000 iterations given my setup but the performance difference is there. Bellow is my test setup:

class Tax(object):
    def __init__(self, rate):
        self.rate = rate

def calculate_inline(self, cost, other=[]):
    cost += sum((o.calculate(cost, other[:i]) for i, o in enumerate(other)))
    return cost * self.rate

def calculate_forloop(self, cost, other=[]):
    for i, o in enumerate(other):
        cost += o.calculate(cost, other[:i])
    return cost * self.rate

def test():
    tax1 = Tax(0.1)
    tax2 = Tax(0.2)
    tax3 = Tax(0.3)
    Tax.calculate =  calculate_inline # or calculate_forloop
    tax1.calculate(100.0, [tax2, tax3]) 

if __name__ == '__main__':
    from timeit import Timer
    t = Timer('test()', 'from __main__ import test; gc.enable()')
    print t.timeit()

With Tax.calculate = calculate_inline , the problem took 16.9 seconds, with calculate_forloop , it took 10.4 seconds.

Answer 1

Seems to be this:

basecost += sum((tax.calculate(basecost, othertaxes[:i]) 
                      for i,tax in enumerate(othertaxes))

Answer 2

If I'm reading that right:

for i,tax in enumerate(othertaxes):
    basecost += tax.calculate(basecost,othertaxes[:i])

Answer 3

In Python, integers are not mutable (neither are floats, booleans or strings).

You cannot change the value of i unless you write i += 1. i. add (1) does not change the value of i, it just returns a new integer which equals (i+1).

Answer 4

你通常会做一个lambda x: x+1而不是使用__add__