AttributeError:使用 linregress 时,“float”对象没有属性“shape”

4 python linear-regression scikit-learn

我想使用LinearRegressionlinregress来计算Intercept, X_Variable_1, R_SquareSignificance_F就像 Excel 中的回归分析一样。

当我使用这段代码来做的时候,没有错误。

from sklearn.linear_model import LinearRegression
import pandas as pd
import numpy as np
from scipy.stats import linregress
from decimal import *

def calculate_parameters():
    list_a=[['2018', '3', 'aa', 'aa', 93,1884.7746222667, 165.36153386251098], ['2018', '3', 'bb', 'bb', 62, 665.6392779848, 125.30386609565328], ['2018', '3', 'cc', 'cc', 89, 580.2259903521, 160.19280253775514]]
    df = pd.DataFrame(list_a)
    X = df.iloc[:, 5]
    y = df.iloc[:, 6]
    X1 = X.values.reshape(-1, 1)
    y1 = y.values.reshape(-1, 1)
    clf = LinearRegression()
    clf.fit(X1, y1)
    yhat = clf.predict(X1)
    para_Intercept = clf.intercept_[0]
    para_X_Variable_1 = clf.coef_[0][0]
    SS_Residual = sum((y1 - yhat) ** 2)
    SS_Total = sum((y1 - np.mean(y1)) ** 2)
    para_R_Square = 1 - (float(SS_Residual)) / SS_Total
    adjusted_r_squared = 1 - (1 - para_R_Square) * (len(y1) - 1) / (len(y1) - X1.shape[1] - 1)
    para_a = linregress(X, y)
    para_Significance_F = para_a[3]
    print("Intercept:"+str(para_Intercept))
    print("X_Variable_1:"+str(para_X_Variable_1))
    print("R_Square:" + str(para_R_Square[0]))
    print("Significance_F:" + str(para_Significance_F))

if __name__ == "__main__":
    calculate_parameters()
Run Code Online (Sandbox Code Playgroud)

输出是:

拦截:133.10871357512195

X_Variable_1:0.016460552337949654

R_Square:0.3039426453800934

意义_F:0.62825637186​​49847

但事实上,list_a喜欢这样的:

list_a = [['2018', '3', 'aa', 'aa', 93, Decimal('1884.7746222667'), 165.36153386251098],
          ['2018', '3', 'bb', 'bb', 62, Decimal('665.6392779848'), 125.30386609565328],
          ['2018', '3', 'cc', 'cc', 89, Decimal('580.2259903521'), 160.19280253775514]]
Run Code Online (Sandbox Code Playgroud)

第6列是十进制类型。

当我改变时list_a,喜欢这样:

from sklearn.linear_model import LinearRegression
import pandas as pd
import numpy as np
from scipy.stats import linregress
from decimal import *

def calculate_parameters():
    # list_a=[['2018', '3', 'aa', 'aa', 93,1884.7746222667, 165.36153386251098], ['2018', '3', 'bb', 'bb', 62, 665.6392779848, 125.30386609565328], ['2018', '3', 'cc', 'cc', 89, 580.2259903521, 160.19280253775514]]
    list_a=[['2018', '3', 'aa', 'aa', 93,Decimal('1884.7746222667'), 165.36153386251098], ['2018', '3', 'bb', 'bb', 62, Decimal('665.6392779848'), 125.30386609565328], ['2018', '3', 'cc', 'cc', 89, Decimal('580.2259903521'), 160.19280253775514]]
    df = pd.DataFrame(list_a)
    X = df.iloc[:, 5]
    y = df.iloc[:, 6]
    X1 = X.values.reshape(-1, 1)
    y1 = y.values.reshape(-1, 1)
    clf = LinearRegression()
    clf.fit(X1, y1)
    yhat = clf.predict(X1)
    para_Intercept = clf.intercept_[0]
    para_X_Variable_1 = clf.coef_[0][0]
    SS_Residual = sum((y1 - yhat) ** 2)
    SS_Total = sum((y1 - np.mean(y1)) ** 2)
    para_R_Square = 1 - (float(SS_Residual)) / SS_Total
    adjusted_r_squared = 1 - (1 - para_R_Square) * (len(y1) - 1) / (len(y1) - X1.shape[1] - 1)
    para_a = linregress(X, y)
    para_Significance_F = para_a[3]
    print("Intercept:"+str(para_Intercept))
    print("X_Variable_1:"+str(para_X_Variable_1))
    print("R_Square:" + str(para_R_Square[0]))
    print("Significance_F:" + str(para_Significance_F))

if __name__ == "__main__":
    calculate_parameters()
Run Code Online (Sandbox Code Playgroud)

错误是:

回溯(最近一次调用最后一次):

文件“E:/test_opencv/MyTest.py”,第 32 行,在calculate_parameters()中

文件“E:/test_opencv/MyTest.py”,第24行,在calculate_parameters para_a = linregress(X, y)中

文件“E:\ Anaconda3 \ lib \ site-packages \ scipy \ stats_stats_mstats_common.py”,第79行,在linregress ssxm,ssxym,ssyxm,ssym = np.cov(x,y,bias = 1).flat

文件“E:\Anaconda3\lib\site-packages\numpy\lib\function_base.py”,第 3085 行,在 cov avg 中,w_sum = 平均值(X,轴=1,权重=w,返回=True)

文件“E:\Anaconda3\lib\site-packages\numpy\lib\function_base.py”,第 1163 行,平均如果 scl.shape != avg.shape:

属性错误:“浮动”对象没有属性“形状”

如何修复错误?

sac*_*cuL 9

您可以通过简单地转换X为浮动来实现此目的:

para_a = linregress(X.astype(float), y)
>>> para_a
LinregressResult(slope=0.016460552337949654, intercept=133.10871357512195, rvalue=0.5513099358619372, pvalue=0.6282563718649847, stderr=0.024909849163985552)
Run Code Online (Sandbox Code Playgroud)

  • 这非常有帮助! (2认同)