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Right Regression Model to use
Regression Model for explained model(Details inside)Best regression model to use for sales predictionChose the right regression analysisMaking Use of the Target Values for RegressionIs removing poorly predicted data points a valid approach?Regression turned into classifficationLinear Model for Linear RegressionSelecting the right time series modelWhen to use an ordinal logistic regression modelCan I use Linear Regression to model a nonlinear function?
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I am trying to predict reservation count from a dataset with few features. Features are both categorical and continuous.
The dependent variable reservations looks like below: My dataset size is around 917 obs.
array([ 1, 7, 17, 2, 2, 13, 8, 11, 9, 4, 4, 3, 5, 2, 5, 7, 3,
12, 9, 13, 5, 2, 11, 13, 14, 19, 9, 11, 3, 6, 7, 10, 1, 6,
5, 10, 8, 5, 4, 3, 2, 10, 10, 10, 8, 13, 16, 6, 4, 6, 3,
11, 10, 1, 18, 7, 2, 12, 17, 4, 2, 19, 3, 4, 17, 13, 10, 2,
10, 1, 3, 4, 20, 3, 2, 1, 3, 5, 8, 8, 4, 3, 13, 3, 3,
5, 4, 17, 7, 6, 10, 5, 3, 9, 9, 8, 1, 5, 17, 5, 10, 9,
2, 7, 13, 2, 9, 1, 15, 13, 10, 4, 2, 4, 5, 4, 3, 3, 10,
4, 7, 5, 13, 12, 7, 5, 6, 9, 5, 11, 7, 1, 4, 12, 4, 3,
11, 1, 4, 4, 3, 7, 4, 11, 4, 1, 9, 2, 10, 10, 3, 4, 4,
3, 2, 7, 10, 7, 6, 1, 3, 19, 9, 3, 8, 20, 1, 12, 9, 13,
13, 2, 9, 4, 9, 2, 5, 6, 18, 3, 6, 8, 6, 4, 5, 13, 4,
8, 9, 5, 4, 8, 5, 2, 1, 6, 8, 3, 6, 4, 2, 6, 11, 5,
1, 5, 1, 5, 11, 11, 9, 3, 12, 2, 2, 9, 19, 7, 13, 13, 9,
2, 1, 1, 4, 3, 4, 9, 1, 25, 12, 8, 5, 18, 3, 1, 6, 17,
7, 4, 6, 9, 8, 10, 3, 8, 12, 5, 4, 4, 1, 9, 21, 4, 3,
3, 7, 13, 5, 12, 8, 8, 6, 3, 6, 7, 5, 3, 7, 3, 14, 3,
5, 2, 14, 16, 3, 8, 6, 13, 9, 3, 5, 4, 9, 4, 12, 12, 4,
9, 8, 11, 5, 13, 3, 2, 5, 4, 2, 1, 8, 8, 18, 11, 2, 5,
13, 4, 1, 2, 4, 1, 2, 2, 12, 2, 6, 19, 7, 20, 2, 10, 2,
9, 12, 9, 8, 1, 4, 8, 8, 12, 4, 8, 1, 3, 6, 9, 4, 3,
8, 2, 7, 15, 6, 5, 10, 6, 4, 3, 12, 5, 4, 13, 7, 2, 8,
5, 2, 4, 3, 14, 12, 3, 4, 3, 2, 15, 6, 14, 12, 11, 9, 5,
5, 7, 11, 10, 7, 9, 9, 7, 11, 5, 11, 3, 2, 5, 17, 5, 2,
6, 1, 10, 3, 13, 19, 5, 1, 3, 5, 3, 5, 6, 3, 9, 8, 2,
3, 2, 3, 7, 4, 9, 5, 1, 6, 14, 4, 8, 17, 13, 7, 1, 4,
5, 10, 5, 6, 2, 12, 5, 9, 3, 9, 9, 1, 5, 1, 2, 2, 5,
1, 4, 4, 13, 4, 25, 9, 10, 4, 3, 9, 13, 13, 2, 9, 2, 12,
4, 1, 20, 9, 10, 2, 5, 4, 10, 2, 6, 1, 7, 7, 7, 4, 8,
4, 3, 4, 13, 8, 3, 13, 12, 19, 9, 3, 2, 6, 7, 13, 8, 16,
7, 3, 11, 4, 10, 9, 12, 2, 8, 5, 2, 3, 4, 2, 1, 11, 5,
4, 2, 8, 12, 7, 5, 7, 7, 4, 6, 18, 2, 1, 6, 15, 11, 2,
5, 8, 3, 5, 9, 11, 5, 8, 6, 20, 1, 10, 3, 7, 1, 3, 5,
4, 4, 10, 11, 6, 1, 5, 4, 1, 2, 10, 4, 4, 11, 20, 5, 3,
2, 7, 8, 2, 10, 5, 1, 18, 5, 10, 5, 3, 8, 15, 2, 1, 14,
10, 7, 3, 5, 9, 3, 4, 21, 14, 1, 2, 1, 2, 4, 11, 9, 7,
6, 9, 18, 4, 6, 18, 12, 12, 4, 6, 3, 3, 9, 5, 12, 15, 3,
7, 3, 7, 4, 2, 15, 14, 7, 10, 5, 5, 5, 9, 3, 6, 3, 1,
11, 1, 5, 25, 8, 2, 24, 1, 12, 1, 6, 8, 5, 13, 4, 3, 3,
13, 4, 4, 18, 7, 13, 2, 8, 3, 4, 9, 2, 13, 12, 4, 5, 10,
9, 15, 1, 8, 8, 15, 10, 1, 9, 2, 2, 2, 2, 3, 6, 17, 7,
5, 5, 6, 12, 1, 8, 3, 1, 11, 4, 7, 8, 15, 6, 11, 9, 9,
13, 2, 3, 5, 3, 5, 12, 4, 4, 8, 7, 12, 2, 2, 4, 4, 12,
8, 11, 10, 6, 5, 1, 4, 2, 7, 3, 5, 15, 12, 12, 2, 9, 7,
4, 4, 5, 15, 5, 8, 13, 7, 2, 8, 12, 2, 13, 6, 24, 14, 3,
4, 1, 2, 8, 7, 5, 12, 8, 2, 6, 3, 7, 5, 2, 7, 3, 3,
1, 9, 9, 3, 12, 3, 2, 11, 11, 6, 3, 9, 12, 4, 8, 7, 5,
2, 10, 19, 1, 1, 10, 6, 2, 4, 2, 4, 4, 3, 7, 13, 9, 6,
2, 2, 2, 5, 13, 12, 2, 13, 12, 11, 10, 5, 8, 8, 15, 12, 3,
3, 9, 4, 6, 13, 15, 4, 7, 1, 12, 10, 9, 7, 3, 7, 4, 9,
2, 10, 2, 11, 10, 14, 3, 13, 8, 3, 12, 11, 10, 7, 5, 3, 3,
11, 3, 13, 9, 10, 20, 7, 12, 3, 6, 6, 18, 3, 10, 11, 10, 5,
6, 11, 4, 6, 7, 9, 13, 1, 14, 14, 13, 4, 3, 8, 5, 7, 14,
13, 13, 12, 8, 11, 12, 9, 8, 9, 4, 5, 4, 7, 5, 2, 3, 1,
7, 2, 1, 13, 5, 19, 9, 6, 9, 7])
When I plot the histogram of dependent variable I get this
So I used a log transform to remove some of the skewness.
as y=np.log(df["reservartions"].values)
Now the plot of distribution looks below:
Some of features.
type actual_price recommended_price num_videos image_ava text_length
1 67.85 59 5 0 7
0 100.70 53 5 0 224
0 74.00 74 4 1 21
0 135.00 75 1 0 184
0 59.36 53 2 1 31
Since actual_price and Recommended_price have huge correlation, I created a difference price of these two and dropped actual_price and recommended price.
But after running Linear regression or Random Forest Regression I get very poor results with R2 as 0.12 for both.
This shows the model is clearly not predicting and fitting well.
My dependent variable is clearly a positive variable. Is Linear Regression still right? Should I use Poisson regression? Log transformation makes sense?
EDIT:
Tried Poisson from Statsmodels. Gives worse results
import statsmodels.api as sm
poisson_mod = sm.Poisson(train_Y, train_X)
poisson_res = poisson_mod.fit(method="newton")
print(poisson_res.summary())
Optimization terminated successfully.
Current function value: 2.958960
Iterations 5
Poisson Regression Results
==============================================================================
Dep. Variable: y No. Observations: 637
Model: Poisson Df Residuals: 628
Method: MLE Df Model: 8
Date: Mon, 08 Oct 2018 Pseudo R-squ.: 0.09479
Time: 13:57:37 Log-Likelihood: -1884.9
converged: True LL-Null: -2082.2
LLR p-value: 2.506e-80
=================================================================================
coef std err z P>|z| [0.025 0.975]
---------------------------------------------------------------------------------
technology 0.0080 0.040 0.200 0.842 -0.070 0.086
street_parked 0.0014 0.030 0.046 0.963 -0.058 0.061
description 0.0002 0.000 0.884 0.377 -0.000 0.001
num_images_2 -0.0230 0.054 -0.430 0.667 -0.128 0.082
num_images_3 0.0619 0.053 1.160 0.246 -0.043 0.167
num_images_4 0.2234 0.050 4.501 0.000 0.126 0.321
num_images_5 0.2391 0.053 4.521 0.000 0.135 0.343
price_diff -0.0146 0.001 -16.300 0.000 -0.016 -0.013
Bias 2.1325 0.052 41.016 0.000 2.031 2.234
=================================================================================
machine-learning regression statistics
asked Oct 8 '18 at 3:44
BaktaawarBaktaawar
1164
$endgroup$
add a comment |
$begingroup$
I am trying to predict reservation count from a dataset with few features. Features are both categorical and continuous.
The dependent variable reservations looks like below: My dataset size is around 917 obs.
array([ 1, 7, 17, 2, 2, 13, 8, 11, 9, 4, 4, 3, 5, 2, 5, 7, 3,
12, 9, 13, 5, 2, 11, 13, 14, 19, 9, 11, 3, 6, 7, 10, 1, 6,
5, 10, 8, 5, 4, 3, 2, 10, 10, 10, 8, 13, 16, 6, 4, 6, 3,
11, 10, 1, 18, 7, 2, 12, 17, 4, 2, 19, 3, 4, 17, 13, 10, 2,
10, 1, 3, 4, 20, 3, 2, 1, 3, 5, 8, 8, 4, 3, 13, 3, 3,
5, 4, 17, 7, 6, 10, 5, 3, 9, 9, 8, 1, 5, 17, 5, 10, 9,
2, 7, 13, 2, 9, 1, 15, 13, 10, 4, 2, 4, 5, 4, 3, 3, 10,
4, 7, 5, 13, 12, 7, 5, 6, 9, 5, 11, 7, 1, 4, 12, 4, 3,
11, 1, 4, 4, 3, 7, 4, 11, 4, 1, 9, 2, 10, 10, 3, 4, 4,
3, 2, 7, 10, 7, 6, 1, 3, 19, 9, 3, 8, 20, 1, 12, 9, 13,
13, 2, 9, 4, 9, 2, 5, 6, 18, 3, 6, 8, 6, 4, 5, 13, 4,
8, 9, 5, 4, 8, 5, 2, 1, 6, 8, 3, 6, 4, 2, 6, 11, 5,
1, 5, 1, 5, 11, 11, 9, 3, 12, 2, 2, 9, 19, 7, 13, 13, 9,
2, 1, 1, 4, 3, 4, 9, 1, 25, 12, 8, 5, 18, 3, 1, 6, 17,
7, 4, 6, 9, 8, 10, 3, 8, 12, 5, 4, 4, 1, 9, 21, 4, 3,
3, 7, 13, 5, 12, 8, 8, 6, 3, 6, 7, 5, 3, 7, 3, 14, 3,
5, 2, 14, 16, 3, 8, 6, 13, 9, 3, 5, 4, 9, 4, 12, 12, 4,
9, 8, 11, 5, 13, 3, 2, 5, 4, 2, 1, 8, 8, 18, 11, 2, 5,
13, 4, 1, 2, 4, 1, 2, 2, 12, 2, 6, 19, 7, 20, 2, 10, 2,
9, 12, 9, 8, 1, 4, 8, 8, 12, 4, 8, 1, 3, 6, 9, 4, 3,
8, 2, 7, 15, 6, 5, 10, 6, 4, 3, 12, 5, 4, 13, 7, 2, 8,
5, 2, 4, 3, 14, 12, 3, 4, 3, 2, 15, 6, 14, 12, 11, 9, 5,
5, 7, 11, 10, 7, 9, 9, 7, 11, 5, 11, 3, 2, 5, 17, 5, 2,
6, 1, 10, 3, 13, 19, 5, 1, 3, 5, 3, 5, 6, 3, 9, 8, 2,
3, 2, 3, 7, 4, 9, 5, 1, 6, 14, 4, 8, 17, 13, 7, 1, 4,
5, 10, 5, 6, 2, 12, 5, 9, 3, 9, 9, 1, 5, 1, 2, 2, 5,
1, 4, 4, 13, 4, 25, 9, 10, 4, 3, 9, 13, 13, 2, 9, 2, 12,
4, 1, 20, 9, 10, 2, 5, 4, 10, 2, 6, 1, 7, 7, 7, 4, 8,
4, 3, 4, 13, 8, 3, 13, 12, 19, 9, 3, 2, 6, 7, 13, 8, 16,
7, 3, 11, 4, 10, 9, 12, 2, 8, 5, 2, 3, 4, 2, 1, 11, 5,
4, 2, 8, 12, 7, 5, 7, 7, 4, 6, 18, 2, 1, 6, 15, 11, 2,
5, 8, 3, 5, 9, 11, 5, 8, 6, 20, 1, 10, 3, 7, 1, 3, 5,
4, 4, 10, 11, 6, 1, 5, 4, 1, 2, 10, 4, 4, 11, 20, 5, 3,
2, 7, 8, 2, 10, 5, 1, 18, 5, 10, 5, 3, 8, 15, 2, 1, 14,
10, 7, 3, 5, 9, 3, 4, 21, 14, 1, 2, 1, 2, 4, 11, 9, 7,
6, 9, 18, 4, 6, 18, 12, 12, 4, 6, 3, 3, 9, 5, 12, 15, 3,
7, 3, 7, 4, 2, 15, 14, 7, 10, 5, 5, 5, 9, 3, 6, 3, 1,
11, 1, 5, 25, 8, 2, 24, 1, 12, 1, 6, 8, 5, 13, 4, 3, 3,
13, 4, 4, 18, 7, 13, 2, 8, 3, 4, 9, 2, 13, 12, 4, 5, 10,
9, 15, 1, 8, 8, 15, 10, 1, 9, 2, 2, 2, 2, 3, 6, 17, 7,
5, 5, 6, 12, 1, 8, 3, 1, 11, 4, 7, 8, 15, 6, 11, 9, 9,
13, 2, 3, 5, 3, 5, 12, 4, 4, 8, 7, 12, 2, 2, 4, 4, 12,
8, 11, 10, 6, 5, 1, 4, 2, 7, 3, 5, 15, 12, 12, 2, 9, 7,
4, 4, 5, 15, 5, 8, 13, 7, 2, 8, 12, 2, 13, 6, 24, 14, 3,
4, 1, 2, 8, 7, 5, 12, 8, 2, 6, 3, 7, 5, 2, 7, 3, 3,
1, 9, 9, 3, 12, 3, 2, 11, 11, 6, 3, 9, 12, 4, 8, 7, 5,
2, 10, 19, 1, 1, 10, 6, 2, 4, 2, 4, 4, 3, 7, 13, 9, 6,
2, 2, 2, 5, 13, 12, 2, 13, 12, 11, 10, 5, 8, 8, 15, 12, 3,
3, 9, 4, 6, 13, 15, 4, 7, 1, 12, 10, 9, 7, 3, 7, 4, 9,
2, 10, 2, 11, 10, 14, 3, 13, 8, 3, 12, 11, 10, 7, 5, 3, 3,
11, 3, 13, 9, 10, 20, 7, 12, 3, 6, 6, 18, 3, 10, 11, 10, 5,
6, 11, 4, 6, 7, 9, 13, 1, 14, 14, 13, 4, 3, 8, 5, 7, 14,
13, 13, 12, 8, 11, 12, 9, 8, 9, 4, 5, 4, 7, 5, 2, 3, 1,
7, 2, 1, 13, 5, 19, 9, 6, 9, 7])
When I plot the histogram of dependent variable I get this
So I used a log transform to remove some of the skewness.
as y=np.log(df["reservartions"].values)
Now the plot of distribution looks below:
Some of features.
type actual_price recommended_price num_videos image_ava text_length
1 67.85 59 5 0 7
0 100.70 53 5 0 224
0 74.00 74 4 1 21
0 135.00 75 1 0 184
0 59.36 53 2 1 31
Since actual_price and Recommended_price have huge correlation, I created a difference price of these two and dropped actual_price and recommended price.
But after running Linear regression or Random Forest Regression I get very poor results with R2 as 0.12 for both.
This shows the model is clearly not predicting and fitting well.
My dependent variable is clearly a positive variable. Is Linear Regression still right? Should I use Poisson regression? Log transformation makes sense?
EDIT:
Tried Poisson from Statsmodels. Gives worse results
import statsmodels.api as sm
poisson_mod = sm.Poisson(train_Y, train_X)
poisson_res = poisson_mod.fit(method="newton")
print(poisson_res.summary())
Optimization terminated successfully.
Current function value: 2.958960
Iterations 5
Poisson Regression Results
==============================================================================
Dep. Variable: y No. Observations: 637
Model: Poisson Df Residuals: 628
Method: MLE Df Model: 8
Date: Mon, 08 Oct 2018 Pseudo R-squ.: 0.09479
Time: 13:57:37 Log-Likelihood: -1884.9
converged: True LL-Null: -2082.2
LLR p-value: 2.506e-80
=================================================================================
coef std err z P>|z| [0.025 0.975]
---------------------------------------------------------------------------------
technology 0.0080 0.040 0.200 0.842 -0.070 0.086
street_parked 0.0014 0.030 0.046 0.963 -0.058 0.061
description 0.0002 0.000 0.884 0.377 -0.000 0.001
num_images_2 -0.0230 0.054 -0.430 0.667 -0.128 0.082
num_images_3 0.0619 0.053 1.160 0.246 -0.043 0.167
num_images_4 0.2234 0.050 4.501 0.000 0.126 0.321
num_images_5 0.2391 0.053 4.521 0.000 0.135 0.343
price_diff -0.0146 0.001 -16.300 0.000 -0.016 -0.013
Bias 2.1325 0.052 41.016 0.000 2.031 2.234
=================================================================================
machine-learning regression statistics
asked Oct 8 '18 at 3:44
BaktaawarBaktaawar
1164
$endgroup$
$begingroup$
I recommend you these tasks, after that, you can put the result again: 1. remove outlier if exist 2. normalize variables 3. put price and percent of off furthermore, big bias means your model could not predict target. you should think about feature engineering
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– parvij
Oct 9 '18 at 5:30
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These are feature engg. What do u mean by normalizations? I have already standardized the continuous features
$endgroup$
– Baktaawar
Oct 10 '18 at 17:22
$begingroup$
data for linear regression should be Gaussian. Box-Cox transformation changes your data. try it.
$endgroup$
– parvij
Oct 11 '18 at 5:25
$begingroup$
That is not true. You can have features of any distribution. It doesn't need to be gaussian. Its the error which is gaussian
$endgroup$
– Baktaawar
Oct 11 '18 at 23:00
add a comment |
$begingroup$
I am trying to predict reservation count from a dataset with few features. Features are both categorical and continuous.
The dependent variable reservations looks like below: My dataset size is around 917 obs.
array([ 1, 7, 17, 2, 2, 13, 8, 11, 9, 4, 4, 3, 5, 2, 5, 7, 3,
12, 9, 13, 5, 2, 11, 13, 14, 19, 9, 11, 3, 6, 7, 10, 1, 6,
5, 10, 8, 5, 4, 3, 2, 10, 10, 10, 8, 13, 16, 6, 4, 6, 3,
11, 10, 1, 18, 7, 2, 12, 17, 4, 2, 19, 3, 4, 17, 13, 10, 2,
10, 1, 3, 4, 20, 3, 2, 1, 3, 5, 8, 8, 4, 3, 13, 3, 3,
5, 4, 17, 7, 6, 10, 5, 3, 9, 9, 8, 1, 5, 17, 5, 10, 9,
2, 7, 13, 2, 9, 1, 15, 13, 10, 4, 2, 4, 5, 4, 3, 3, 10,
4, 7, 5, 13, 12, 7, 5, 6, 9, 5, 11, 7, 1, 4, 12, 4, 3,
11, 1, 4, 4, 3, 7, 4, 11, 4, 1, 9, 2, 10, 10, 3, 4, 4,
3, 2, 7, 10, 7, 6, 1, 3, 19, 9, 3, 8, 20, 1, 12, 9, 13,
13, 2, 9, 4, 9, 2, 5, 6, 18, 3, 6, 8, 6, 4, 5, 13, 4,
8, 9, 5, 4, 8, 5, 2, 1, 6, 8, 3, 6, 4, 2, 6, 11, 5,
1, 5, 1, 5, 11, 11, 9, 3, 12, 2, 2, 9, 19, 7, 13, 13, 9,
2, 1, 1, 4, 3, 4, 9, 1, 25, 12, 8, 5, 18, 3, 1, 6, 17,
7, 4, 6, 9, 8, 10, 3, 8, 12, 5, 4, 4, 1, 9, 21, 4, 3,
3, 7, 13, 5, 12, 8, 8, 6, 3, 6, 7, 5, 3, 7, 3, 14, 3,
5, 2, 14, 16, 3, 8, 6, 13, 9, 3, 5, 4, 9, 4, 12, 12, 4,
9, 8, 11, 5, 13, 3, 2, 5, 4, 2, 1, 8, 8, 18, 11, 2, 5,
13, 4, 1, 2, 4, 1, 2, 2, 12, 2, 6, 19, 7, 20, 2, 10, 2,
9, 12, 9, 8, 1, 4, 8, 8, 12, 4, 8, 1, 3, 6, 9, 4, 3,
8, 2, 7, 15, 6, 5, 10, 6, 4, 3, 12, 5, 4, 13, 7, 2, 8,
5, 2, 4, 3, 14, 12, 3, 4, 3, 2, 15, 6, 14, 12, 11, 9, 5,
5, 7, 11, 10, 7, 9, 9, 7, 11, 5, 11, 3, 2, 5, 17, 5, 2,
6, 1, 10, 3, 13, 19, 5, 1, 3, 5, 3, 5, 6, 3, 9, 8, 2,
3, 2, 3, 7, 4, 9, 5, 1, 6, 14, 4, 8, 17, 13, 7, 1, 4,
5, 10, 5, 6, 2, 12, 5, 9, 3, 9, 9, 1, 5, 1, 2, 2, 5,
1, 4, 4, 13, 4, 25, 9, 10, 4, 3, 9, 13, 13, 2, 9, 2, 12,
4, 1, 20, 9, 10, 2, 5, 4, 10, 2, 6, 1, 7, 7, 7, 4, 8,
4, 3, 4, 13, 8, 3, 13, 12, 19, 9, 3, 2, 6, 7, 13, 8, 16,
7, 3, 11, 4, 10, 9, 12, 2, 8, 5, 2, 3, 4, 2, 1, 11, 5,
4, 2, 8, 12, 7, 5, 7, 7, 4, 6, 18, 2, 1, 6, 15, 11, 2,
5, 8, 3, 5, 9, 11, 5, 8, 6, 20, 1, 10, 3, 7, 1, 3, 5,
4, 4, 10, 11, 6, 1, 5, 4, 1, 2, 10, 4, 4, 11, 20, 5, 3,
2, 7, 8, 2, 10, 5, 1, 18, 5, 10, 5, 3, 8, 15, 2, 1, 14,
10, 7, 3, 5, 9, 3, 4, 21, 14, 1, 2, 1, 2, 4, 11, 9, 7,
6, 9, 18, 4, 6, 18, 12, 12, 4, 6, 3, 3, 9, 5, 12, 15, 3,
7, 3, 7, 4, 2, 15, 14, 7, 10, 5, 5, 5, 9, 3, 6, 3, 1,
11, 1, 5, 25, 8, 2, 24, 1, 12, 1, 6, 8, 5, 13, 4, 3, 3,
13, 4, 4, 18, 7, 13, 2, 8, 3, 4, 9, 2, 13, 12, 4, 5, 10,
9, 15, 1, 8, 8, 15, 10, 1, 9, 2, 2, 2, 2, 3, 6, 17, 7,
5, 5, 6, 12, 1, 8, 3, 1, 11, 4, 7, 8, 15, 6, 11, 9, 9,
13, 2, 3, 5, 3, 5, 12, 4, 4, 8, 7, 12, 2, 2, 4, 4, 12,
8, 11, 10, 6, 5, 1, 4, 2, 7, 3, 5, 15, 12, 12, 2, 9, 7,
4, 4, 5, 15, 5, 8, 13, 7, 2, 8, 12, 2, 13, 6, 24, 14, 3,
4, 1, 2, 8, 7, 5, 12, 8, 2, 6, 3, 7, 5, 2, 7, 3, 3,
1, 9, 9, 3, 12, 3, 2, 11, 11, 6, 3, 9, 12, 4, 8, 7, 5,
2, 10, 19, 1, 1, 10, 6, 2, 4, 2, 4, 4, 3, 7, 13, 9, 6,
2, 2, 2, 5, 13, 12, 2, 13, 12, 11, 10, 5, 8, 8, 15, 12, 3,
3, 9, 4, 6, 13, 15, 4, 7, 1, 12, 10, 9, 7, 3, 7, 4, 9,
2, 10, 2, 11, 10, 14, 3, 13, 8, 3, 12, 11, 10, 7, 5, 3, 3,
11, 3, 13, 9, 10, 20, 7, 12, 3, 6, 6, 18, 3, 10, 11, 10, 5,
6, 11, 4, 6, 7, 9, 13, 1, 14, 14, 13, 4, 3, 8, 5, 7, 14,
13, 13, 12, 8, 11, 12, 9, 8, 9, 4, 5, 4, 7, 5, 2, 3, 1,
7, 2, 1, 13, 5, 19, 9, 6, 9, 7])
When I plot the histogram of dependent variable I get this
So I used a log transform to remove some of the skewness.
as y=np.log(df["reservartions"].values)
Now the plot of distribution looks below:
Some of features.
type actual_price recommended_price num_videos image_ava text_length
1 67.85 59 5 0 7
0 100.70 53 5 0 224
0 74.00 74 4 1 21
0 135.00 75 1 0 184
0 59.36 53 2 1 31
Since actual_price and Recommended_price have huge correlation, I created a difference price of these two and dropped actual_price and recommended price.
But after running Linear regression or Random Forest Regression I get very poor results with R2 as 0.12 for both.
This shows the model is clearly not predicting and fitting well.
My dependent variable is clearly a positive variable. Is Linear Regression still right? Should I use Poisson regression? Log transformation makes sense?
EDIT:
Tried Poisson from Statsmodels. Gives worse results
import statsmodels.api as sm
poisson_mod = sm.Poisson(train_Y, train_X)
poisson_res = poisson_mod.fit(method="newton")
print(poisson_res.summary())
Optimization terminated successfully.
Current function value: 2.958960
Iterations 5
Poisson Regression Results
==============================================================================
Dep. Variable: y No. Observations: 637
Model: Poisson Df Residuals: 628
Method: MLE Df Model: 8
Date: Mon, 08 Oct 2018 Pseudo R-squ.: 0.09479
Time: 13:57:37 Log-Likelihood: -1884.9
converged: True LL-Null: -2082.2
LLR p-value: 2.506e-80
=================================================================================
coef std err z P>|z| [0.025 0.975]
---------------------------------------------------------------------------------
technology 0.0080 0.040 0.200 0.842 -0.070 0.086
street_parked 0.0014 0.030 0.046 0.963 -0.058 0.061
description 0.0002 0.000 0.884 0.377 -0.000 0.001
num_images_2 -0.0230 0.054 -0.430 0.667 -0.128 0.082
num_images_3 0.0619 0.053 1.160 0.246 -0.043 0.167
num_images_4 0.2234 0.050 4.501 0.000 0.126 0.321
num_images_5 0.2391 0.053 4.521 0.000 0.135 0.343
price_diff -0.0146 0.001 -16.300 0.000 -0.016 -0.013
Bias 2.1325 0.052 41.016 0.000 2.031 2.234
=================================================================================
machine-learning regression statistics
asked Oct 8 '18 at 3:44
BaktaawarBaktaawar
1164
$endgroup$
I am trying to predict reservation count from a dataset with few features. Features are both categorical and continuous.
The dependent variable reservations looks like below: My dataset size is around 917 obs.
array([ 1, 7, 17, 2, 2, 13, 8, 11, 9, 4, 4, 3, 5, 2, 5, 7, 3,
12, 9, 13, 5, 2, 11, 13, 14, 19, 9, 11, 3, 6, 7, 10, 1, 6,
5, 10, 8, 5, 4, 3, 2, 10, 10, 10, 8, 13, 16, 6, 4, 6, 3,
11, 10, 1, 18, 7, 2, 12, 17, 4, 2, 19, 3, 4, 17, 13, 10, 2,
10, 1, 3, 4, 20, 3, 2, 1, 3, 5, 8, 8, 4, 3, 13, 3, 3,
5, 4, 17, 7, 6, 10, 5, 3, 9, 9, 8, 1, 5, 17, 5, 10, 9,
2, 7, 13, 2, 9, 1, 15, 13, 10, 4, 2, 4, 5, 4, 3, 3, 10,
4, 7, 5, 13, 12, 7, 5, 6, 9, 5, 11, 7, 1, 4, 12, 4, 3,
11, 1, 4, 4, 3, 7, 4, 11, 4, 1, 9, 2, 10, 10, 3, 4, 4,
3, 2, 7, 10, 7, 6, 1, 3, 19, 9, 3, 8, 20, 1, 12, 9, 13,
13, 2, 9, 4, 9, 2, 5, 6, 18, 3, 6, 8, 6, 4, 5, 13, 4,
8, 9, 5, 4, 8, 5, 2, 1, 6, 8, 3, 6, 4, 2, 6, 11, 5,
1, 5, 1, 5, 11, 11, 9, 3, 12, 2, 2, 9, 19, 7, 13, 13, 9,
2, 1, 1, 4, 3, 4, 9, 1, 25, 12, 8, 5, 18, 3, 1, 6, 17,
7, 4, 6, 9, 8, 10, 3, 8, 12, 5, 4, 4, 1, 9, 21, 4, 3,
3, 7, 13, 5, 12, 8, 8, 6, 3, 6, 7, 5, 3, 7, 3, 14, 3,
5, 2, 14, 16, 3, 8, 6, 13, 9, 3, 5, 4, 9, 4, 12, 12, 4,
9, 8, 11, 5, 13, 3, 2, 5, 4, 2, 1, 8, 8, 18, 11, 2, 5,
13, 4, 1, 2, 4, 1, 2, 2, 12, 2, 6, 19, 7, 20, 2, 10, 2,
9, 12, 9, 8, 1, 4, 8, 8, 12, 4, 8, 1, 3, 6, 9, 4, 3,
8, 2, 7, 15, 6, 5, 10, 6, 4, 3, 12, 5, 4, 13, 7, 2, 8,
5, 2, 4, 3, 14, 12, 3, 4, 3, 2, 15, 6, 14, 12, 11, 9, 5,
5, 7, 11, 10, 7, 9, 9, 7, 11, 5, 11, 3, 2, 5, 17, 5, 2,
6, 1, 10, 3, 13, 19, 5, 1, 3, 5, 3, 5, 6, 3, 9, 8, 2,
3, 2, 3, 7, 4, 9, 5, 1, 6, 14, 4, 8, 17, 13, 7, 1, 4,
5, 10, 5, 6, 2, 12, 5, 9, 3, 9, 9, 1, 5, 1, 2, 2, 5,
1, 4, 4, 13, 4, 25, 9, 10, 4, 3, 9, 13, 13, 2, 9, 2, 12,
4, 1, 20, 9, 10, 2, 5, 4, 10, 2, 6, 1, 7, 7, 7, 4, 8,
4, 3, 4, 13, 8, 3, 13, 12, 19, 9, 3, 2, 6, 7, 13, 8, 16,
7, 3, 11, 4, 10, 9, 12, 2, 8, 5, 2, 3, 4, 2, 1, 11, 5,
4, 2, 8, 12, 7, 5, 7, 7, 4, 6, 18, 2, 1, 6, 15, 11, 2,
5, 8, 3, 5, 9, 11, 5, 8, 6, 20, 1, 10, 3, 7, 1, 3, 5,
4, 4, 10, 11, 6, 1, 5, 4, 1, 2, 10, 4, 4, 11, 20, 5, 3,
2, 7, 8, 2, 10, 5, 1, 18, 5, 10, 5, 3, 8, 15, 2, 1, 14,
10, 7, 3, 5, 9, 3, 4, 21, 14, 1, 2, 1, 2, 4, 11, 9, 7,
6, 9, 18, 4, 6, 18, 12, 12, 4, 6, 3, 3, 9, 5, 12, 15, 3,
7, 3, 7, 4, 2, 15, 14, 7, 10, 5, 5, 5, 9, 3, 6, 3, 1,
11, 1, 5, 25, 8, 2, 24, 1, 12, 1, 6, 8, 5, 13, 4, 3, 3,
13, 4, 4, 18, 7, 13, 2, 8, 3, 4, 9, 2, 13, 12, 4, 5, 10,
9, 15, 1, 8, 8, 15, 10, 1, 9, 2, 2, 2, 2, 3, 6, 17, 7,
5, 5, 6, 12, 1, 8, 3, 1, 11, 4, 7, 8, 15, 6, 11, 9, 9,
13, 2, 3, 5, 3, 5, 12, 4, 4, 8, 7, 12, 2, 2, 4, 4, 12,
8, 11, 10, 6, 5, 1, 4, 2, 7, 3, 5, 15, 12, 12, 2, 9, 7,
4, 4, 5, 15, 5, 8, 13, 7, 2, 8, 12, 2, 13, 6, 24, 14, 3,
4, 1, 2, 8, 7, 5, 12, 8, 2, 6, 3, 7, 5, 2, 7, 3, 3,
1, 9, 9, 3, 12, 3, 2, 11, 11, 6, 3, 9, 12, 4, 8, 7, 5,
2, 10, 19, 1, 1, 10, 6, 2, 4, 2, 4, 4, 3, 7, 13, 9, 6,
2, 2, 2, 5, 13, 12, 2, 13, 12, 11, 10, 5, 8, 8, 15, 12, 3,
3, 9, 4, 6, 13, 15, 4, 7, 1, 12, 10, 9, 7, 3, 7, 4, 9,
2, 10, 2, 11, 10, 14, 3, 13, 8, 3, 12, 11, 10, 7, 5, 3, 3,
11, 3, 13, 9, 10, 20, 7, 12, 3, 6, 6, 18, 3, 10, 11, 10, 5,
6, 11, 4, 6, 7, 9, 13, 1, 14, 14, 13, 4, 3, 8, 5, 7, 14,
13, 13, 12, 8, 11, 12, 9, 8, 9, 4, 5, 4, 7, 5, 2, 3, 1,
7, 2, 1, 13, 5, 19, 9, 6, 9, 7])
When I plot the histogram of dependent variable I get this
So I used a log transform to remove some of the skewness.
as y=np.log(df["reservartions"].values)
Now the plot of distribution looks below:
Some of features.
type actual_price recommended_price num_videos image_ava text_length
1 67.85 59 5 0 7
0 100.70 53 5 0 224
0 74.00 74 4 1 21
0 135.00 75 1 0 184
0 59.36 53 2 1 31
Since actual_price and Recommended_price have huge correlation, I created a difference price of these two and dropped actual_price and recommended price.
But after running Linear regression or Random Forest Regression I get very poor results with R2 as 0.12 for both.
This shows the model is clearly not predicting and fitting well.
My dependent variable is clearly a positive variable. Is Linear Regression still right? Should I use Poisson regression? Log transformation makes sense?
EDIT:
Tried Poisson from Statsmodels. Gives worse results
import statsmodels.api as sm
poisson_mod = sm.Poisson(train_Y, train_X)
poisson_res = poisson_mod.fit(method="newton")
print(poisson_res.summary())
Optimization terminated successfully.
Current function value: 2.958960
Iterations 5
Poisson Regression Results
==============================================================================
Dep. Variable: y No. Observations: 637
Model: Poisson Df Residuals: 628
Method: MLE Df Model: 8
Date: Mon, 08 Oct 2018 Pseudo R-squ.: 0.09479
Time: 13:57:37 Log-Likelihood: -1884.9
converged: True LL-Null: -2082.2
LLR p-value: 2.506e-80
=================================================================================
coef std err z P>|z| [0.025 0.975]
---------------------------------------------------------------------------------
technology 0.0080 0.040 0.200 0.842 -0.070 0.086
street_parked 0.0014 0.030 0.046 0.963 -0.058 0.061
description 0.0002 0.000 0.884 0.377 -0.000 0.001
num_images_2 -0.0230 0.054 -0.430 0.667 -0.128 0.082
num_images_3 0.0619 0.053 1.160 0.246 -0.043 0.167
num_images_4 0.2234 0.050 4.501 0.000 0.126 0.321
num_images_5 0.2391 0.053 4.521 0.000 0.135 0.343
price_diff -0.0146 0.001 -16.300 0.000 -0.016 -0.013
Bias 2.1325 0.052 41.016 0.000 2.031 2.234
=================================================================================
machine-learning regression statistics
machine-learning regression statistics
asked Oct 8 '18 at 3:44
BaktaawarBaktaawar
1164
asked Oct 8 '18 at 3:44
BaktaawarBaktaawar
1164
edited Oct 8 '18 at 21:21
Baktaawar
asked Oct 8 '18 at 3:44
BaktaawarBaktaawar
1164
asked Oct 8 '18 at 3:44
BaktaawarBaktaawar
1164
asked Oct 8 '18 at 3:44
BaktaawarBaktaawar
1164
1164
$begingroup$
I recommend you these tasks, after that, you can put the result again: 1. remove outlier if exist 2. normalize variables 3. put price and percent of off furthermore, big bias means your model could not predict target. you should think about feature engineering
$endgroup$
– parvij
Oct 9 '18 at 5:30
$begingroup$
These are feature engg. What do u mean by normalizations? I have already standardized the continuous features
$endgroup$
– Baktaawar
Oct 10 '18 at 17:22
$begingroup$
data for linear regression should be Gaussian. Box-Cox transformation changes your data. try it.
$endgroup$
– parvij
Oct 11 '18 at 5:25
$begingroup$
That is not true. You can have features of any distribution. It doesn't need to be gaussian. Its the error which is gaussian
$endgroup$
– Baktaawar
Oct 11 '18 at 23:00
add a comment |
$begingroup$
I recommend you these tasks, after that, you can put the result again: 1. remove outlier if exist 2. normalize variables 3. put price and percent of off furthermore, big bias means your model could not predict target. you should think about feature engineering
$endgroup$
– parvij
Oct 9 '18 at 5:30
$begingroup$
These are feature engg. What do u mean by normalizations? I have already standardized the continuous features
$endgroup$
– Baktaawar
Oct 10 '18 at 17:22
$begingroup$
data for linear regression should be Gaussian. Box-Cox transformation changes your data. try it.
$endgroup$
– parvij
Oct 11 '18 at 5:25
$begingroup$
That is not true. You can have features of any distribution. It doesn't need to be gaussian. Its the error which is gaussian
$endgroup$
– Baktaawar
Oct 11 '18 at 23:00
$begingroup$
I recommend you these tasks, after that, you can put the result again: 1. remove outlier if exist 2. normalize variables 3. put price and percent of off furthermore, big bias means your model could not predict target. you should think about feature engineering
$endgroup$
– parvij
Oct 9 '18 at 5:30
$begingroup$
I recommend you these tasks, after that, you can put the result again: 1. remove outlier if exist 2. normalize variables 3. put price and percent of off furthermore, big bias means your model could not predict target. you should think about feature engineering
$endgroup$
– parvij
Oct 9 '18 at 5:30
$begingroup$
These are feature engg. What do u mean by normalizations? I have already standardized the continuous features
$endgroup$
– Baktaawar
Oct 10 '18 at 17:22
$begingroup$
These are feature engg. What do u mean by normalizations? I have already standardized the continuous features
$endgroup$
– Baktaawar
Oct 10 '18 at 17:22
$begingroup$
data for linear regression should be Gaussian. Box-Cox transformation changes your data. try it.
$endgroup$
– parvij
Oct 11 '18 at 5:25
$begingroup$
data for linear regression should be Gaussian. Box-Cox transformation changes your data. try it.
$endgroup$
– parvij
Oct 11 '18 at 5:25
$begingroup$
That is not true. You can have features of any distribution. It doesn't need to be gaussian. Its the error which is gaussian
$endgroup$
– Baktaawar
Oct 11 '18 at 23:00
$begingroup$
That is not true. You can have features of any distribution. It doesn't need to be gaussian. Its the error which is gaussian
$endgroup$
– Baktaawar
Oct 11 '18 at 23:00
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I think because your target is count so their distribution is Poisson and you should use Poisson regression.
I strongly recommend reading this, however, it's in R but will be helpful for you and also there is some similar python version in comments.
answered Oct 8 '18 at 7:29
parvijparvij
485214
$endgroup$
$begingroup$
Pls check the edit of Poisson I tried
$endgroup$
– Baktaawar
Oct 8 '18 at 21:20
add a comment |
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1 Answer
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$begingroup$
I think because your target is count so their distribution is Poisson and you should use Poisson regression.
I strongly recommend reading this, however, it's in R but will be helpful for you and also there is some similar python version in comments.
answered Oct 8 '18 at 7:29
parvijparvij
485214
$endgroup$
$begingroup$
Pls check the edit of Poisson I tried
$endgroup$
– Baktaawar
Oct 8 '18 at 21:20
add a comment |
$begingroup$
I think because your target is count so their distribution is Poisson and you should use Poisson regression.
I strongly recommend reading this, however, it's in R but will be helpful for you and also there is some similar python version in comments.
answered Oct 8 '18 at 7:29
parvijparvij
485214
$endgroup$
$begingroup$
Pls check the edit of Poisson I tried
$endgroup$
– Baktaawar
Oct 8 '18 at 21:20
add a comment |
$begingroup$
I think because your target is count so their distribution is Poisson and you should use Poisson regression.
I strongly recommend reading this, however, it's in R but will be helpful for you and also there is some similar python version in comments.
answered Oct 8 '18 at 7:29
parvijparvij
485214
$endgroup$
I think because your target is count so their distribution is Poisson and you should use Poisson regression.
I strongly recommend reading this, however, it's in R but will be helpful for you and also there is some similar python version in comments.
answered Oct 8 '18 at 7:29
parvijparvij
485214
edited Oct 9 '18 at 6:13
answered Oct 8 '18 at 7:29
parvijparvij
485214
answered Oct 8 '18 at 7:29
parvijparvij
485214
answered Oct 8 '18 at 7:29
parvijparvij
485214
485214
$begingroup$
Pls check the edit of Poisson I tried
$endgroup$
– Baktaawar
Oct 8 '18 at 21:20
add a comment |
$begingroup$
Pls check the edit of Poisson I tried
$endgroup$
– Baktaawar
Oct 8 '18 at 21:20
$begingroup$
Pls check the edit of Poisson I tried
$endgroup$
– Baktaawar
Oct 8 '18 at 21:20
$begingroup$
Pls check the edit of Poisson I tried
$endgroup$
– Baktaawar
Oct 8 '18 at 21:20
add a comment |
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$begingroup$
I recommend you these tasks, after that, you can put the result again: 1. remove outlier if exist 2. normalize variables 3. put price and percent of off furthermore, big bias means your model could not predict target. you should think about feature engineering
$endgroup$
– parvij
Oct 9 '18 at 5:30
$begingroup$
These are feature engg. What do u mean by normalizations? I have already standardized the continuous features
$endgroup$
– Baktaawar
Oct 10 '18 at 17:22
$begingroup$
data for linear regression should be Gaussian. Box-Cox transformation changes your data. try it.
$endgroup$
– parvij
Oct 11 '18 at 5:25
$begingroup$
That is not true. You can have features of any distribution. It doesn't need to be gaussian. Its the error which is gaussian
$endgroup$
– Baktaawar
Oct 11 '18 at 23:00