Sklearn LogisticRegression求解器需要2类数据

时间:2019-02-25 19:33:02

标签: python scikit-learn

我正在尝试通过sklearn运行Logistic回归:

from sklearn.model_selection import train_test_split
from sklearn.linear_model import LogisticRegression
from sklearn import metrics
import datetime as dt
import pandas as pd
import numpy as np
import talib
import matplotlib.pyplot as plt
import seaborn as sns

col_names = ['dates','prices']
# load dataset
df = pd.read_csv("DJI2.csv", header=None, names=col_names)

df.drop('dates', axis=1, inplace=True)
print(df.shape)
df['3day MA'] = df['prices'].shift(1).rolling(window = 3).mean()
df['10day MA'] = df['prices'].shift(1).rolling(window = 10).mean()
df['30day MA'] = df['prices'].shift(1).rolling(window = 30).mean()
df['Std_dev']= df['prices'].rolling(5).std()
df['RSI'] = talib.RSI(df['prices'].values, timeperiod = 9)
df['Price_Rise'] = np.where(df['prices'].shift(-1) > df['prices'], 1, 0)
df = df.dropna()

xCols = ['3day MA', '10day MA', '30day MA', 'Std_dev', 'RSI', 'prices']
X = df[xCols]
X = X.astype('int')
Y = df['Price_Rise']
Y = Y.astype('int')

logreg = LogisticRegression()

for i in range(len(X)):
   #Without this case below I get: ValueError: Found array with 0 sample(s) (shape=(0, 6)) while a minimum of 1 is required.
    if(i == 0): 
       continue
    logreg.fit(X[:i], Y[:i])

但是,当我尝试运行此代码时,出现以下错误:

ValueError: 
This solver needs samples of at least 2 classes in the data, but the data contains only one class: 58

我的X数据的形状为:(27779, 6) 我的Y数据的形状为:(27779,)

下面是一个df.head(3)示例,用于查看我的数据:

     prices    3day MA  10day MA   30day MA   Std_dev        RSI  Price_Rise
30   58.11  57.973333    57.277  55.602333  0.247123  81.932338           1
31   58.42  58.043333    57.480  55.718667  0.213542  84.279674           1
32   58.51  58.216667    57.667  55.774000  0.249139  84.919586           0

我曾尝试搜索自己从何处获得此问题,但是我只设法找到了these two的答案,两个答案都将这个问题作为sklearn中的错误进行了讨论,但是它们都差不多。两岁,所以我不认为我遇到了同样的问题。

2 个答案:

答案 0 :(得分:0)

您应该确保Y [:i]中有两个唯一值。因此,在循环之前,请添加以下内容:

starting_i = 0
for i in range(len(X)):
   if np.unique(Y[:i]) == 2:
      starting_i = i

然后只需在运行主循环之前检查starting_i不为0。 甚至更简单,您可以找到第一个出现的Y [i]!= Y [0]。

答案 1 :(得分:0)

if i in range (0,3): 
    continue

解决了这个问题。 Y [:i]在i = 3之前不是唯一的。