线性回归预测以列表形式提供多个特征答案

【问题标题】：LinearRegression predicting providing multiple features as a list线性回归预测以列表形式提供多个特征
【发布时间】：2019-07-22 14:10:08
【问题描述】：

我正在使用简单的 LinearRegression() 来预测第二天的收盘价。我知道它不可靠，但至少我正在尝试理解和训练 LR。

我只是提供最新的open, high, low, close 值作为功能。我真正想做的是提供最近 10 天的open, high, low, close 价格。您可以在下面找到我到目前为止所做的事情：

import pandas as pd
import matplotlib.pyplot as plt
import numpy as np
from sklearn.linear_model import LinearRegression
from sklearn.model_selection import train_test_split
from sklearn.metrics import mean_squared_error, mean_absolute_error, median_absolute_error
pd.options.mode.chained_assignment = None

sym = "EURUSD"
period = "1d"
fl = "./feed/{} {}.csv".format(period, sym)
dg = 0.0001 # pip digits

df = pd.read_csv(fl)
df['Date'] = pd.to_datetime(df['Date'])
df['NextClose'] = df['Close'].shift(-1)
cols = ['Open', 'High', 'Low', 'Close'] # features
prd_cols = ['NextClose'] # prediction
real_prd = df[cols].iloc[-1:] # predict this after training
hour = df['Date'].iloc[-1]
df.dropna(inplace=True)

test_p = 20 # percent of test size
total = len(df) # total in DataFrame
test_size = int(total * test_p / 100)
X = df[cols]
y = df[prd_cols]
X_train, X_test, y_train, y_test = train_test_split(X, y, shuffle=True)

lr = LinearRegression()
lr.fit(X_train, y_train)
y_pred = lr.predict(X_test)
mse = mean_squared_error(y_test, y_pred)
rmse = int(np.sqrt(mse) / dg)
mae = int(mean_absolute_error(y_test, y_pred) / dg)
print("")
print("Pair:", sym, "Time:", period, hour)
print("Mean Sq. Err: {:.10f}".format(mse))
print("Root Mean Sq. Err:", rmse, "pips")
print("Mean Abs. Err:", mae, "pips")
print("Score: {:.2f}%".format(np.round(lr.score(X_test, y_test) * 100, 2)))
# predict last value
real_pred = lr.predict(real_prd)
print("Prediction:", np.round(real_pred[0][0], 5))

plt.scatter(y_pred, y_test)
plt.show()

编辑 输出：

Pair: EURUSD Time: 1d 2019-02-28 00:00:00
Mean Sq. Err: 0.0000594541
Root Mean Sq. Err: 77 pips
Mean Abs. Err: 56 pips
Score: 99.78%
Prediction: 1.14154

样本来源：

         Open     High      Low    Close
4921  1.14087  1.14092  1.13610  1.13658
4922  1.13658  1.13678  1.13245  1.13383
4923  1.13385  1.13509  1.13213  1.13222
4924  1.13199  1.13251  1.13189  1.13241
4925  1.13243  1.13303  1.12675  1.12787
4926  1.12785  1.13397  1.12580  1.13340
4927  1.13336  1.13417  1.12495  1.12648
4928  1.12650  1.13099  1.12501  1.12950
4929  1.12950  1.13064  1.12343  1.12922
4930  1.12916  1.12960  1.12898  1.12952
4931  1.12950  1.13341  1.12940  1.13118
4932  1.13116  1.13576  1.12757  1.13401
4933  1.13399  1.13714  1.13251  1.13464
4934  1.13465  1.13665  1.13208  1.13400
4935  1.13399  1.13558  1.13163  1.13331
4936  1.13362  1.13424  1.13292  1.13424
4937  1.13420  1.13677  1.13369  1.13651
4938  1.13651  1.14028  1.13454  1.13931
4939  1.13936  1.14037  1.13624  1.13799
4940  1.13793  1.14198  1.13675  1.14116

剧情：

我怎样才能将最后 10 行作为一项功能部署并仍然得到一个结果？

【问题讨论】：

请发布您的数据样本，并进一步解释您的问题（“部署”在这里是什么意思？）；此外，在没有输出的代码中包含一大堆 print 和 plot 语句可能无助于解决问题的本质，只会造成混乱......
用样本数据、绘图和输出编辑了问题

标签： python machine-learning scikit-learn linear-regression

【解决方案1】：

我已经编写了以下解决方案

cols = []
for i in range(1, 11):
    cl = "Open_{}".format(i)
    cols.append(cl)
    df[cl] = df['Open'].shift(i)
    cl = "High_{}".format(i)
    cols.append(cl)
    df[cl] = df['High'].shift(i)
    cl = "Low_{}".format(i)
    cols.append(cl)
    df[cl] = df['Low'].shift(i)
    cl = "Close_{}".format(i)
    cols.append(cl)
    df[cl] = df['Close'].shift(i)

【讨论】：