Question

我正在使用Kobe Bryant Dataset。我希望通过KnnRegressor来预测shot_made_flag。

我已经使用game_date提取了year和month功能：

# covert season to years
kobe_data_encoded['season'] = kobe_data_encoded['season'].apply(lambda x: int(re.compile('(\d+)-').findall(x)[0]))

# add year and month using game_date
kobe_data_encoded['year'] = kobe_data_encoded['game_date'].apply(lambda x: int(re.compile('(\d{4})').findall(x)[0]))
kobe_data_encoded['month'] = kobe_data_encoded['game_date'].apply(lambda x: int(re.compile('-(\d+)-').findall(x)[0]))
kobe_data_encoded = kobe_data_encoded.drop(columns=['game_date'])

，我希望使用season，year，month功能在距离函数中赋予它们更大的权重，因此与当前事件日期更近的事件将成为更近的邻居，但仍然与潜在的其他数据点保持合理的距离，例如，我不希望同一天发生的事件会因为日期功能而成为最近的邻居，但它会考虑其他功能，例如shot_range等等..
为了给它更大的权重，我尝试将metric参数与自定义距离函数一起使用，但是该函数的参数只是numpy数组而没有熊猫的列信息，因此我不确定我能做什么以及如何实现我想做的事情。

编辑：

对日期特征使用更大的权重，以找到cv中k上运行的[1, 100]为10的最优k：

from IPython.display import display
from sklearn.neighbors import KNeighborsClassifier
from sklearn.model_selection import StratifiedKFold
from sklearn.model_selection import cross_val_score

# scaling
min_max_scaler = preprocessing.MinMaxScaler()
scaled_features_df = kobe_data_encoded.copy()
column_names = ['loc_x', 'loc_y', 'minutes_remaining', 'period',
                'seconds_remaining', 'shot_distance', 'shot_type', 'shot_zone_range']
scaled_features = min_max_scaler.fit_transform(scaled_features_df[column_names])
scaled_features_df[column_names] = scaled_features

not_classified_df = scaled_features_df[scaled_features_df['shot_made_flag'].isnull()]
classified_df = scaled_features_df[scaled_features_df['shot_made_flag'].notnull()]
X = classified_df.drop(columns=['shot_made_flag'])
y = classified_df['shot_made_flag']
cv = StratifiedKFold(n_splits=10, shuffle=True)

neighbors = [x for x in range(1, 100)]
cv_scores = []

weight = np.ones((X.shape[1],))
weight[[X.columns.get_loc("season"),
 X.columns.get_loc("year"),
 X.columns.get_loc("month")
]] = 5
weight = weight/weight.sum()  #Normalize weights

def my_distance(x, y):
    dist = ((x-y)**2)
    return np.dot(dist, weight)

for k in neighbors:
    print('k: ', k)
    knn = KNeighborsClassifier(n_neighbors=k, metric=my_distance)
    cv_scores.append(np.mean(cross_val_score(knn, X, y, cv=cv, scoring='roc_auc')))

#optimal K
optimal_k_index = cv_scores.index(min(cv_scores))
optimal_k = neighbors[optimal_k_index]
print('best k: ', optimal_k)
plt.plot(neighbors, cv_scores)
plt.xlabel('Number of Neighbors K')
plt.ylabel('ROC AUC')
plt.show()

运行速度真的很慢，是否知道如何使其更快？加权特征的想法是找到更接近数据点日期的邻居，以避免数据泄漏和cv来寻找最佳k。

Answer 1

首先，您必须准备一个numpy 1D weight数组，为每个功能指定权重。您可以执行以下操作：

weight = np.ones((M,))  # M is no of features
weight[[1,7,10]] = 2    # Increase weight of 1st,7th and 10th features
weight = weight/weight.sum()  #Normalize weights

您可以使用kobe_data_encoded.columns在数据框中查找season，year，month功能的索引，以替换上面的第二行。

现在定义一个距离函数，根据指南它必须采用两个一维numpy数组。

def my_dist(x,y):
    global weight     #1D array, same shape as x or y
    dist = ((x-y)**2) #1D array, same shape as x or y
    return np.dot(dist,weight)  # a scalar float

并将KNeighborsRegressor初始化为：

knn = KNeighborsRegressor(metric=my_dist)

编辑：为了提高效率，您可以预先计算距离矩阵，然后在KNN中重用它。由于此非矢量化的自定义python距离函数相当慢，因此应该通过减少对my_dist的调用来显着提高速度。所以现在-

dist = np.zeros((len(X),len(X)))  #Computing NXN distance matrix
for i in range(len(X)):           # You can halve this by using the fact that dist[i,j] = dist[j,i]
    for j in range(len(X)):
        dist[i,j] = my_dist(X[i],X[j])

for k in neighbors:
    print('k: ', k)
    knn = KNeighborsClassifier(n_neighbors=k, metric='precomputed') #Note: metric='precomputed' 
    cv_scores.append(np.mean(cross_val_score(knn, dist, y, cv=cv, scoring='roc_auc'))) #Note: passing dist instead of X

我无法测试它，所以让我知道是否有问题。

Answer 2

只需添加Shihab关于距离计算的答案即可。可以按照此post中的建议使用scipy pdist，它更快，更有效。

from scipy.spatial.distance import pdist, minkowski, squareform

# create the custom weight array
weight = ...
# calculate pairwise distances, using Minkowski norm with custom weights
distances = pdist(X, minkowski, 2, weight)
# reformat the result as a square matrix
distances_as_2d_matrix = squareform(distances)

Knn赋予距离特定特征更多的权重

2 个答案: