Tensorflow 和 Scikit 学习：相同的解决方案但不同的输出
Tensorflow and Scikit learn: Same solution but different outputs
我用 scikitlearn 和 tensorflow 实现了一个简单的线性回归。
我在 scikitlearn 中的解决方案看起来不错，但在 tensorflow 中，我的评估输出显示了一些疯狂的数字。
问题基本上是尝试根据多年的经验来预测薪水。
我不确定我在 Tensorflow 的代码中做错了什么。
谢谢！
ScikitLearn 解决方案
import pandas as pd
data = pd.read_csv('Salary_Data.csv')
X = data.iloc[:, :-1].values
y = data.iloc[:, 1].values
from sklearn.model_selection import train_test_split
X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.3, random_state=1)
from sklearn.linear_model import LinearRegression
regressor = LinearRegression()
regressor.fit(X_train, y_train)
y_pred = regressor.predict(X_test)
X_single_data = [[4.6]]
y_single_pred = regressor.predict(X_single_data)
print(f'Train score: {regressor.score(X_train, y_train)}')
print(f'Test score: {regressor.score(X_test, y_test)}')
Train score: 0.960775692121653
Test score: 0.9248580247217076
Tensorflow 解决方案
import tensorflow as tf
f_cols = [tf.feature_column.numeric_column(key='X', shape=[1])]
estimator = tf.estimator.LinearRegressor(feature_columns=f_cols)
train_input_fn = tf.estimator.inputs.numpy_input_fn(x={'X': X_train}, y=y_train,shuffle=False)
test_input_fn = tf.estimator.inputs.numpy_input_fn(x={'X': X_test}, y=y_test,shuffle=False)
train_spec = tf.estimator.TrainSpec(input_fn=train_input_fn)
eval_spec = tf.estimator.EvalSpec(input_fn=test_input_fn)
tf.estimator.train_and_evaluate(estimator, train_spec, eval_spec)
({'average_loss': 7675087400.0,
'label/mean': 84588.11,
'loss': 69075790000.0,
'prediction/mean': 5.0796494,
'global_step': 6},
[])
数据
YearsExperience,Salary
1.1,39343.00
1.3,46205.00
1.5,37731.00
2.0,43525.00
2.2,39891.00
2.9,56642.00
3.0,60150.00
3.2,54445.00
3.2,64445.00
3.7,57189.00
3.9,63218.00
4.0,55794.00
4.0,56957.00
4.1,57081.00
4.5,61111.00
4.9,67938.00
5.1,66029.00
5.3,83088.00
5.9,81363.00
6.0,93940.00
6.8,91738.00
7.1,98273.00
7.9,101302.00
8.2,113812.00
8.7,109431.00
9.0,105582.00
9.5,116969.00
9.6,112635.00
10.3,122391.00
10.5,121872.00
我不能在评论中放图片，所以放在这里。我怀疑这种关系可能是 S 形关系而不是线性关系，并发现以下 S 形方程和拟合统计量使用千单位作为薪水："y = a / (1.0 + exp(-(x-b)/c))"，拟合参数 a = 1.5535069418318591E+02，b = 5.4580059234664899E+00，和c = 3.7724942500630938E+00 给出 R 平方 = 0.96 和 RMSE = 5.30（千）
根据您在评论中的代码请求：虽然我已经使用我的在线曲线和曲面拟合网站 zunzun.com 在 http://zunzun.com/Equation/2/Sigmoidal/Sigmoid%20B/ 上对这个方程进行建模工作，但这里有一个图形源代码使用 scipy differential_evolution 遗传算法模块来估计初始参数估计值的示例。差分进化的 scipy 实现使用拉丁超立方体算法来确保彻底搜索参数 space，这需要在搜索范围内进行搜索 - 在本例中，这些范围取自数据的最大值和最小值, 并且拟合统计和参数值与网站上的几乎相同。
import numpy, scipy, matplotlib
import matplotlib.pyplot as plt
from scipy.optimize import curve_fit
from scipy.optimize import differential_evolution
import warnings
xData = numpy.array([ 1.1, 1.3, 1.5, 2.0, 2.2, 2.9, 3.0, 3.2, 3.2, 3.7, 3.9, 4.0, 4.0, 4.1, 4.5, 4.9, 5.1, 5.3, 5.9, 6.0, 6.8, 7.1, 7.9, 8.2, 8.7, 9.0, 9.5, 9.6, 10.3, 10.5])
yData = numpy.array([ 39.343, 46.205, 37.731, 43.525, 39.891, 56.642, 60.15, 54.445, 64.445, 57.189, 63.218, 55.794, 56.957, 57.081, 61.111, 67.938, 66.029, 83.088, 81.363, 93.94, 91.738, 98.273, 101.302, 113.812, 109.431, 105.582, 116.969, 112.635, 122.391, 121.872])
def func(x, a, b, c):
return a / (1.0 + numpy.exp(-(x-b)/c))
# function for genetic algorithm to minimize (sum of squared error)
def sumOfSquaredError(parameterTuple):
warnings.filterwarnings("ignore") # do not print warnings by genetic algorithm
val = func(xData, *parameterTuple)
return numpy.sum((yData - val) ** 2.0)
def generate_Initial_Parameters():
# min and max used for bounds
maxX = max(xData)
minX = min(xData)
maxY = max(yData)
minY = min(yData)
parameterBounds = []
parameterBounds.append([minY, maxY]) # search bounds for a
parameterBounds.append([minX, maxX]) # search bounds for b
parameterBounds.append([minX, maxX]) # search bounds for c
# "seed" the numpy random number generator for repeatable results
result = differential_evolution(sumOfSquaredError, parameterBounds, seed=3)
return result.x
# by default, differential_evolution completes by calling curve_fit() using parameter bounds
geneticParameters = generate_Initial_Parameters()
# now call curve_fit without passing bounds from the genetic algorithm,
# just in case the best fit parameters are aoutside those bounds
fittedParameters, pcov = curve_fit(func, xData, yData, geneticParameters)
print('Fitted parameters:', fittedParameters)
print()
modelPredictions = func(xData, *fittedParameters)
absError = modelPredictions - yData
SE = numpy.square(absError) # squared errors
MSE = numpy.mean(SE) # mean squared errors
RMSE = numpy.sqrt(MSE) # Root Mean Squared Error, RMSE
Rsquared = 1.0 - (numpy.var(absError) / numpy.var(yData))
print()
print('RMSE:', RMSE)
print('R-squared:', Rsquared)
print()
##########################################################
# graphics output section
def ModelAndScatterPlot(graphWidth, graphHeight):
f = plt.figure(figsize=(graphWidth/100.0, graphHeight/100.0), dpi=100)
axes = f.add_subplot(111)
# first the raw data as a scatter plot
axes.plot(xData, yData, 'D')
# create data for the fitted equation plot
xModel = numpy.linspace(min(xData), max(xData))
yModel = func(xModel, *fittedParameters)
# now the model as a line plot
axes.plot(xModel, yModel)
axes.set_xlabel('Years of experience') # X axis data label
axes.set_ylabel('Salary in thousands') # Y axis data label
plt.show()
plt.close('all') # clean up after using pyplot
graphWidth = 800
graphHeight = 600
ModelAndScatterPlot(graphWidth, graphHeight)
我用 scikitlearn 和 tensorflow 实现了一个简单的线性回归。
我在 scikitlearn 中的解决方案看起来不错，但在 tensorflow 中，我的评估输出显示了一些疯狂的数字。
问题基本上是尝试根据多年的经验来预测薪水。
我不确定我在 Tensorflow 的代码中做错了什么。
谢谢！
ScikitLearn 解决方案
import pandas as pd
data = pd.read_csv('Salary_Data.csv')
X = data.iloc[:, :-1].values
y = data.iloc[:, 1].values
from sklearn.model_selection import train_test_split
X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.3, random_state=1)
from sklearn.linear_model import LinearRegression
regressor = LinearRegression()
regressor.fit(X_train, y_train)
y_pred = regressor.predict(X_test)
X_single_data = [[4.6]]
y_single_pred = regressor.predict(X_single_data)
print(f'Train score: {regressor.score(X_train, y_train)}')
print(f'Test score: {regressor.score(X_test, y_test)}')
Train score: 0.960775692121653
Test score: 0.9248580247217076
Tensorflow 解决方案
import tensorflow as tf
f_cols = [tf.feature_column.numeric_column(key='X', shape=[1])]
estimator = tf.estimator.LinearRegressor(feature_columns=f_cols)
train_input_fn = tf.estimator.inputs.numpy_input_fn(x={'X': X_train}, y=y_train,shuffle=False)
test_input_fn = tf.estimator.inputs.numpy_input_fn(x={'X': X_test}, y=y_test,shuffle=False)
train_spec = tf.estimator.TrainSpec(input_fn=train_input_fn)
eval_spec = tf.estimator.EvalSpec(input_fn=test_input_fn)
tf.estimator.train_and_evaluate(estimator, train_spec, eval_spec)
({'average_loss': 7675087400.0,
'label/mean': 84588.11,
'loss': 69075790000.0,
'prediction/mean': 5.0796494,
'global_step': 6},
[])
数据
YearsExperience,Salary
1.1,39343.00
1.3,46205.00
1.5,37731.00
2.0,43525.00
2.2,39891.00
2.9,56642.00
3.0,60150.00
3.2,54445.00
3.2,64445.00
3.7,57189.00
3.9,63218.00
4.0,55794.00
4.0,56957.00
4.1,57081.00
4.5,61111.00
4.9,67938.00
5.1,66029.00
5.3,83088.00
5.9,81363.00
6.0,93940.00
6.8,91738.00
7.1,98273.00
7.9,101302.00
8.2,113812.00
8.7,109431.00
9.0,105582.00
9.5,116969.00
9.6,112635.00
10.3,122391.00
10.5,121872.00
我不能在评论中放图片，所以放在这里。我怀疑这种关系可能是 S 形关系而不是线性关系，并发现以下 S 形方程和拟合统计量使用千单位作为薪水："y = a / (1.0 + exp(-(x-b)/c))"，拟合参数 a = 1.5535069418318591E+02，b = 5.4580059234664899E+00，和c = 3.7724942500630938E+00 给出 R 平方 = 0.96 和 RMSE = 5.30（千）
根据您在评论中的代码请求：虽然我已经使用我的在线曲线和曲面拟合网站 zunzun.com 在 http://zunzun.com/Equation/2/Sigmoidal/Sigmoid%20B/ 上对这个方程进行建模工作，但这里有一个图形源代码使用 scipy differential_evolution 遗传算法模块来估计初始参数估计值的示例。差分进化的 scipy 实现使用拉丁超立方体算法来确保彻底搜索参数 space，这需要在搜索范围内进行搜索 - 在本例中，这些范围取自数据的最大值和最小值, 并且拟合统计和参数值与网站上的几乎相同。
import numpy, scipy, matplotlib
import matplotlib.pyplot as plt
from scipy.optimize import curve_fit
from scipy.optimize import differential_evolution
import warnings
xData = numpy.array([ 1.1, 1.3, 1.5, 2.0, 2.2, 2.9, 3.0, 3.2, 3.2, 3.7, 3.9, 4.0, 4.0, 4.1, 4.5, 4.9, 5.1, 5.3, 5.9, 6.0, 6.8, 7.1, 7.9, 8.2, 8.7, 9.0, 9.5, 9.6, 10.3, 10.5])
yData = numpy.array([ 39.343, 46.205, 37.731, 43.525, 39.891, 56.642, 60.15, 54.445, 64.445, 57.189, 63.218, 55.794, 56.957, 57.081, 61.111, 67.938, 66.029, 83.088, 81.363, 93.94, 91.738, 98.273, 101.302, 113.812, 109.431, 105.582, 116.969, 112.635, 122.391, 121.872])
def func(x, a, b, c):
return a / (1.0 + numpy.exp(-(x-b)/c))
# function for genetic algorithm to minimize (sum of squared error)
def sumOfSquaredError(parameterTuple):
warnings.filterwarnings("ignore") # do not print warnings by genetic algorithm
val = func(xData, *parameterTuple)
return numpy.sum((yData - val) ** 2.0)
def generate_Initial_Parameters():
# min and max used for bounds
maxX = max(xData)
minX = min(xData)
maxY = max(yData)
minY = min(yData)
parameterBounds = []
parameterBounds.append([minY, maxY]) # search bounds for a
parameterBounds.append([minX, maxX]) # search bounds for b
parameterBounds.append([minX, maxX]) # search bounds for c
# "seed" the numpy random number generator for repeatable results
result = differential_evolution(sumOfSquaredError, parameterBounds, seed=3)
return result.x
# by default, differential_evolution completes by calling curve_fit() using parameter bounds
geneticParameters = generate_Initial_Parameters()
# now call curve_fit without passing bounds from the genetic algorithm,
# just in case the best fit parameters are aoutside those bounds
fittedParameters, pcov = curve_fit(func, xData, yData, geneticParameters)
print('Fitted parameters:', fittedParameters)
print()
modelPredictions = func(xData, *fittedParameters)
absError = modelPredictions - yData
SE = numpy.square(absError) # squared errors
MSE = numpy.mean(SE) # mean squared errors
RMSE = numpy.sqrt(MSE) # Root Mean Squared Error, RMSE
Rsquared = 1.0 - (numpy.var(absError) / numpy.var(yData))
print()
print('RMSE:', RMSE)
print('R-squared:', Rsquared)
print()
##########################################################
# graphics output section
def ModelAndScatterPlot(graphWidth, graphHeight):
f = plt.figure(figsize=(graphWidth/100.0, graphHeight/100.0), dpi=100)
axes = f.add_subplot(111)
# first the raw data as a scatter plot
axes.plot(xData, yData, 'D')
# create data for the fitted equation plot
xModel = numpy.linspace(min(xData), max(xData))
yModel = func(xModel, *fittedParameters)
# now the model as a line plot
axes.plot(xModel, yModel)
axes.set_xlabel('Years of experience') # X axis data label
axes.set_ylabel('Salary in thousands') # Y axis data label
plt.show()
plt.close('all') # clean up after using pyplot
graphWidth = 800
graphHeight = 600
ModelAndScatterPlot(graphWidth, graphHeight)