train.py 文件源码

def cluster(data):
    """
    Use k-means clustering on training data to find profitable patterns
    we can exploit
    """

    num_clusters = 100
    num_selected_clusters = 20

    # Split into 30, 60, and 120 min time intervals, cluster each
    split = lambda n: split_into_intervals(data, n)
    kmeans30 = sklearn.cluster.k_means(split(30), num_clusters)
    kmeans60 = sklearn.cluster.k_means(split(60), num_clusters)
    kmeans120 = sklearn.cluster.k_means(split(120), num_clusters)

    # Sort the clusters by performance
    hp30, hp60, hp120 = [], [], []
    for i in range(0, num_clusters):
        hp30.append((i,kmeans30[0][i,-1]))
        hp60.append((i,kmeans60[0][i,-1]))
        hp120.append((i,kmeans120[0][i,-1]))

    hp30 = sorted(hp30, reverse=True, key=lambda x: x[1])[0:num_selected_clusters]
    hp60 = sorted(hp60, reverse=True, key=lambda x: x[1])[0:num_selected_clusters]
    hp60 = sorted(hp120, reverse=True, key=lambda x: x[1])[0:num_selected_clusters]

    # Select the highest performing clusters
    top30 = np.zeros((num_selected_clusters,181))
    top60 = np.zeros((num_selected_clusters,361))
    top120 = np.zeros((num_selected_clusters,721))

    for i in range(0, num_selected_clusters):
        top30[i,0:181] = kmeans30[0][hp30[i][0],0:181]
        top60[i,0:361] = kmeans60[0][hp60[i][0],0:361]
        top120[i,0:721] = kmeans120[0][hp120[i][0],0:721]

    # Then normalize the clusters so we can use the faster similarity function
    #    from S&Z to compare instead of L2 norm
    scaler = sklearn.preprocessing.StandardScaler()
    for i in range(0, num_selected_clusters):
        top30[i,0:180] = scaler.fit_transform(top30[i,0:180])
        top60[i,0:360] = scaler.fit_transform(top60[i,0:360])
        top120[i,0:720] = scaler.fit_transform(top120[i,0:720])

    return [top30, top60, top120]