Question

我想计算预测值和真值的平均绝对百分比误差（MAPE）。我找到了来自iTracing App的解决方案，但这会产生错误并在行mask = a <> 0中显示无效语法

    def mape_vectorized_v2(a, b): 
    mask = a <> 0
    return (np.fabs(a - b)/a)[mask].mean() 

   def mape_vectorized_v2(a, b): 
       File "<ipython-input-5-afa5c1162e83>", line 1
         def mape_vectorized_v2(a, b):
                                       ^
     SyntaxError: unexpected EOF while parsing

我正在使用spyder3。我的预测值是类型np.array，真值是dataframe

type(predicted)
Out[7]: numpy.ndarray
type(y_test)
Out[8]: pandas.core.frame.DataFrame

如何清除此错误并继续进行MAPE计算？

编辑：

predicted.head()
Out[22]: 
   Total_kWh
0   7.163627
1   6.584960
2   6.638057
3   7.785487
4   6.994427

y_test.head()
Out[23]: 
     Total_kWh
79         7.2
148        6.7
143        6.7
189        7.2
17         6.4

np.abs(y_test[['Total_kWh']] - predicted[['Total_kWh']]).head()
Out[24]: 
   Total_kWh
0        NaN
1        NaN
2        NaN
3        NaN
4   0.094427

Answer 1

在python中进行比较，不等于需要!=，而不是<>。

所以需要：

def mape_vectorized_v2(a, b): 
    mask = a != 0
    return (np.fabs(a - b)/a)[mask].mean()

stats.stackexchange的另一个解决方案：

def mean_absolute_percentage_error(y_true, y_pred): 
    y_true, y_pred = np.array(y_true), np.array(y_pred)
    return np.mean(np.abs((y_true - y_pred) / y_true)) * 100

Answer 2

新版 scikit-learn (v0.24) 有一个计算 MAPE 的函数。 sklearn.metrics.mean_absolute_percentage_error

您只需要两个类似数组的变量：y_true 存储实际/真实值，y_pred 存储预测值。

可以参考官方文档here。

Answer 3

两个解决方案都无法使用零值。这是我的工作方式：

def percentage_error(actual, predicted):
    res = np.empty(actual.shape)
    for j in range(actual.shape[0]):
        if actual[j] != 0:
            res[j] = (actual[j] - predicted[j]) / actual[j]
        else:
            res[j] = predicted[j] / np.mean(actual)
    return res

def mean_absolute_percentage_error(y_true, y_pred): 
    return np.mean(np.abs(percentage_error(np.asarray(y_true), np.asarray(y_pred)))) * 100

希望对您有帮助。

Answer 4

由于实际值也可以为零，因此我将分母中的实际值取平均值，而不是实际值：

Error = np.sum(np.abs(np.subtract(data_4['y'],data_4['pred'])))
Average = np.sum(data_4['y'])
MAPE = Error/Average

Answer 5

这是一个注意零的改进版本：

    #Mean Absolute Percentage error 
def mape(y_true, y_pred,sample_weight=None,multioutput='uniform_average'):
    y_type, y_true, y_pred, multioutput = _check_reg_targets(y_true, y_pred, multioutput)
    epsilon = np.finfo(np.float64).eps
    mape = np.abs(y_pred - y_true) / np.maximum(np.abs(y_true), epsilon)
    output_errors = np.average(mape,weights=sample_weight, axis=0)
    if isinstance(multioutput, str):
        if multioutput == 'raw_values':
            return output_errors
        elif multioutput == 'uniform_average':
            # pass None as weights to np.average: uniform mean
            multioutput = None
    return np.average(output_errors, weights=multioutput)

def _check_reg_targets(y_true, y_pred, multioutput, dtype="numeric"):
    if y_true.ndim == 1:
        y_true = y_true.reshape((-1, 1))

    if y_pred.ndim == 1:
        y_pred = y_pred.reshape((-1, 1))

    if y_true.shape[1] != y_pred.shape[1]:
        raise ValueError("y_true and y_pred have different number of output "
                         "({0}!={1})".format(y_true.shape[1], y_pred.shape[1]))

    n_outputs = y_true.shape[1]
    allowed_multioutput_str = ('raw_values', 'uniform_average',
                               'variance_weighted')
    if isinstance(multioutput, str):
        if multioutput not in allowed_multioutput_str:
            raise ValueError("Allowed 'multioutput' string values are {}. "
                             "You provided multioutput={!r}".format(
                                 allowed_multioutput_str,
                                 multioutput))
    elif multioutput is not None:
        multioutput = check_array(multioutput, ensure_2d=False)
        if n_outputs == 1:
            raise ValueError("Custom weights are useful only in "
                             "multi-output cases.")
        elif n_outputs != len(multioutput):
            raise ValueError(("There must be equally many custom weights "
                              "(%d) as outputs (%d).") %
                             (len(multioutput), n_outputs))
    y_type = 'continuous' if n_outputs == 1 else 'continuous-multioutput'

    return y_type, y_true, y_pred, multioutput

python中的MAPE计算

5 个答案: