用聚类算法计算股票的支撑位和阻力位
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In [1]:
import pymongo
from pymongo import MongoClient
client_remote = MongoClient('mongodb://localhost:27017')
db_remote = client_remote['stocktdb']
collection_remote = db_remote.stock_data
In [2]:
mobj = collection_remote.find({'ticker':'GOOGL'}).sort([('_id',pymongo.DESCENDING)]).limit(60)
In [3]:
prices = []
for doc in mobj:
prices.append(doc['high'])
K-means聚类计算股票支撑位和阻力位
In [4]:
import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
from sklearn.cluster import AgglomerativeClustering
K-means
聚类，我们需要将数据转换为 Numpy 数组格式。
In [5]:
X = np.array(prices)
K-means
聚类来说，
K-means
聚类的数量非常重要。我们可以使用膝图找到最佳 K值，如下所示。
In [6]:
from sklearn.cluster import KMeans
import numpy as np
from kneed import KneeLocator
sum_of_sq_distances = []
K = range(1,10)
for k in K:
km = KMeans(n_clusters=k)
km = km.fit(X.reshape(-1,1))
sum_of_sq_distances.append(km.inertia_)
kn = KneeLocator(K, sum_of_sq_distances,S=1.0, curve="convex", direction="decreasing")
kn.plot_knee()
kn.knee
检查 K 的值
In [7]:
kn.knee
Out[7]:
3
In [8]:
kmeans = KMeans(n_clusters= kn.knee).fit(X.reshape(-1,1))
c = kmeans.predict(X.reshape(-1,1))
min_and_max = []
for i in range(kn.knee):
min_and_max.append([-np.inf,np.inf])
for i in range(len(X)):
cluster = c[i]
if X[i] > min_and_max[cluster][0]:
min_and_max[cluster][0] = X[i]
if X[i] < min_and_max[cluster][1]:
min_and_max[cluster][1] = X[i]
In [9]:
min_and_max
Out[9]:
[[2461.9099, 2365.55], [2687.98, 2508.0801], [2357.02, 2239.4399]]
2687.98
（当日最高点），正好也是52周来的最高点。因此，基于上述聚类，我们可以说
2687.98
是阻力位，下一个支撑水平是
2508.0801
。下一个支持级别是
2461.9099
、
2365.55
、
2357.02
、
2239.4399
。
In [10]:
mobj = collection_remote.find({'ticker':'GOOGL'}).sort([('_id',pymongo.DESCENDING)]).limit(60)
prices = []
for doc in mobj:
prices.append(doc['high'])
In [11]:
df = pd.DataFrame(prices)
max = df.rolling(20).max()
max.rename(columns={0: "price"},inplace=True)
min = df.rolling(20).min()
min.rename(columns={0: "price"},inplace=True)
In [12]:
maxdf = pd.concat([max,pd.Series(np.zeros(len(max))+1)],axis = 1)
mindf = pd.concat([min,pd.Series(np.zeros(len(min))+-1)],axis = 1)
maxdf.drop_duplicates('price',inplace = True)
mindf.drop_duplicates('price',inplace = True)
n_clusters=3
值来表示聚类数量。
In [13]:
F = maxdf.append(mindf).sort_index()
F = F[ F[0] != F[0].shift() ].dropna()
# Create [x,y] array where y is always 1
X = np.concatenate((F.price.values.reshape(-1,1),
(np.zeros(len(F))+1).reshape(-1,1)), axis = 1 )
cluster = AgglomerativeClustering(n_clusters=3,
affinity='euclidean', linkage='ward')
cluster.fit_predict(X)
F['clusters'] = cluster.labels_
F2 = F.loc[F.groupby('clusters')['price'].idxmax()]
# Plotit
fig, axis = plt.subplots()
for row in F2.itertuples():
axis.axhline( y = row.price,
color = 'green', ls = 'dashed' )
axis.plot( F.index.values, F.price.values )
plt.show()
2638.00
，我们可以说
2687.98
是阻力位，
2357.02
是支撑位。
In [14]:
F2
Out[14]:
In [15]:
F = maxdf.append(mindf).sort_index()
F = F[ F[0] != F[0].shift() ].dropna()
# Create [x,y] array where y is always 1
X = np.concatenate((F.price.values.reshape(-1,1),
(np.zeros(len(F))+1).reshape(-1,1)), axis = 1 )
cluster = AgglomerativeClustering(n_clusters=5,
affinity='euclidean', linkage='ward')
cluster.fit_predict(X)
F['clusters'] = cluster.labels_
F2 = F.loc[F.groupby('clusters')['price'].idxmax()]
# Plotit
fig, axis = plt.subplots()
for row in F2.itertuples():
axis.axhline( y = row.price,
color = 'green', ls = 'dashed' )
axis.plot( F.index.values, F.price.values )
plt.show()
In [16]:
F2
Out[16]:
2239.43
的价格获得了更多的聚类，这与今天
2638
的收盘价相去甚远。然而，基于3个聚类，阻力值看起来不错，为
2687.98
。
In [17]:
df = pd.DataFrame(prices)
max = df.rolling(10).max()
max.rename(columns={0: "price"},inplace=True)
min = df.rolling(10).min()
min.rename(columns={0: "price"},inplace=True)
maxdf = pd.concat([max,pd.Series(np.zeros(len(max))+1)],axis = 1)
mindf = pd.concat([min,pd.Series(np.zeros(len(min))+-1)],axis = 1)
maxdf.drop_duplicates('price',inplace = True)
mindf.drop_duplicates('price',inplace = True)
F = maxdf.append(mindf).sort_index()
F = F[ F[0] != F[0].shift() ].dropna()
# Create [x,y] array where y is always 1
X = np.concatenate((F.price.values.reshape(-1,1),
(np.zeros(len(F))+1).reshape(-1,1)), axis = 1 )
cluster = AgglomerativeClustering(n_clusters=5,
affinity='euclidean', linkage='ward')
cluster.fit_predict(X)
F['clusters'] = cluster.labels_
F2 = F.loc[F.groupby('clusters')['price'].idxmax()]
# Plotit
fig, axis = plt.subplots()
for row in F2.itertuples():
axis.axhline( y = row.price,
color = 'green', ls = 'dashed' )
axis.plot( F.index.values, F.price.values )
plt.show()
In [18]:
F2
Google
的阻力值约为
2687.98
，支撑位约为
2399.03
和
2412.8799
，这与支撑位约为
2400
相当接近。
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