Question

我有以下代码，这是我的Python代码中的瓶颈：

def get_payoff(self,  actual, predicted):
    if abs(actual - 1.0) < 1e-5:  # if actual == 1
        if predicted < 0.5:
            return self.fn_payoff * (0.5 - predicted)
        elif predicted > 0.5:
            return self.tp_payoff * (predicted - 0.5)
        else:
            return 0
    else:
        if predicted < 0.5:
            return self.tn_payoff * (0.5 - predicted)
        elif predicted > 0.5:
            return self.fp_payoff * (predicted - 0.5)
        else:
            return 0

def get_total_payoff(self):
    total_payoff = 0
    for target_element, prediction_element in zip(np.nditer(self.target), np.nditer(predictions)):
        total_payoff += self.get_payoff(target_element, prediction_element)

fn_payoff，tp_payoff，tn_payoff和fp_payoff都是浮点数。 self.target和self.predictions都是numpy ndarrays。

我假设有一些方法可以用get_total_payoff替换for循环中的某种numpy向量化，但我不知道如何处理if / then语句来正确地进行向量化。 / p>

Answer 1

基于条件使用不同表达式的向量化函数的关键是使用np.choose。另外，在您的情况下，predict-0.5和0.5-predict可以替换为abs(predict-0.5)，加上predict==0.5的情况的特殊处理（我猜测特殊处理是针对的）正确处理NaN）。

import numpy as np

class A(object):
    def __init__(self):
        self.fn_payoff = 222.
        self.tn_payoff = 444.
        self.fp_payoff = 777.
        self.tp_payoff = 888.
        self.target = np.array([ 0.3, 1., 2. ])
        self.predictions = np.array([ 0.4, 0.5, 1.7 ])

    def get_payoff(self,  actual, predicted):
        if abs(actual - 1.0) < 1e-5:  # if actual == 1
            if predicted < 0.5:
                return self.fn_payoff * (0.5 - predicted)
            elif predicted > 0.5:
                return self.tp_payoff * (predicted - 0.5)
            else:
                return 0
        else:
            if predicted < 0.5:
                return self.tn_payoff * (0.5 - predicted)
            elif predicted > 0.5:
                return self.fp_payoff * (predicted - 0.5)
            else:
                return 0

    def get_total_payoff(self):
        total_payoff = 0
        for target_element, prediction_element in zip(np.nditer(self.target), np.nditer(self.predictions)):
            total_payoff += self.get_payoff(target_element, prediction_element)
        return total_payoff

    def get_total_payoff_VECTORIZED(self):
        actual_mask = np.abs(self.target - 1) < 1e-5
        predict_mask = self.predictions < 0.5
        payoff_n = np.choose(actual_mask, [ self.tn_payoff, self.fn_payoff ])
        payoff_p = np.choose(actual_mask, [ self.fp_payoff, self.tp_payoff ])
        payoff = np.choose(predict_mask, [ payoff_p, payoff_n ]) * abs(self.predictions-0.5)
        payoff[self.predictions==0.5] = 0
        return payoff.sum()

a = A()
print a.get_total_payoff()
=> 976.8
print a.get_total_payoff_VECTORIZED()
=> 976.8

Answer 2

def _get_payoff(self, actual, predicted):
    pred_factor = numpy.abs(0.5 - predicted)
    payoff_selector = 2*numpy.isclose(actual, 1) + (predicted < 0.5)
    payoff = numpy.choose(payoff_selector,
                          [
                              self.fp_payoff,
                              self.tn_payoff,
                              self.tp_payoff,
                              self.fn_payoff,
                          ])
    return numpy.sum(payoff * pred_factor)

def get_total_payoff(self):
    return self._get_payoff(self.target, predictions)

我们使用numpy.choose生成一个支付选择数组，并将其与0.5和预测值之间的绝对差值数组相乘，然后求和。 numpy.isclose用于测试actual值是否接近1.我们可以忽略predicted == 0.5情况，因为乘以numpy.abs(0.5 - predicted)无论如何都会得到正确的0结果。如果self.target和predictions保证为1D，则numpy.dot的效果可能优于单独的乘法和求和。

如何优化此NumPy代码？

2 个答案: