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is it bad to have many different measurements for the same target variable?
2019 Community Moderator ElectionBinary classification with unexplained dataDoes variation in data density over time affect regression models?Consistently inconsistent cross-validation results that are wildly different from original model accuracyHow to handle the target variable being in the featuresIs removing poorly predicted data points a valid approach?Is it valid to include your validation data in your vocabulary for NLP?How to apply machine learning model to new datasetClarification about Normalized Discounted Cumulative Gain (NDCG) together with Regression for Ranking?How important is it for each row of data to have the same number of features?How do I correctly build model on given data to predict target parameter?
$begingroup$
I'm working on a dataset that has repeated measurements for the same target variable.
When I don't change anything and create model, cross validation overfits with 0.99 score but in testset it gives around 0.39.
When I use mean, std, skew, quartiles for each measurement to have only one measurement for each feature, it gives a much better score.
Can anyone explain to me why? and when it is good to use the second method?
the original dataset looks like this (all numbers are fake):
id /measurement1/measurement2/.../target/
0-1/0.18283 /0.12855 /.../ 1 /
0-2/0.1141 /0.38484 /.../ 1 /
0-3/0.4475 /0.18374 /.../ 1 /
and transformed dataset looks like this:
id /meas1_avg/meas1_std/meas1_skew/meas2_avg/meas2_std/.../target/
0 /0.28747 /0.183848/ 0.198384 /0.18484 /0.28474 /.../ 1 /
machine-learning feature-engineering data-science-model
asked Mar 26 at 14:58
edunlimitedunlimit
203
$endgroup$
add a comment |
$begingroup$
I'm working on a dataset that has repeated measurements for the same target variable.
When I don't change anything and create model, cross validation overfits with 0.99 score but in testset it gives around 0.39.
When I use mean, std, skew, quartiles for each measurement to have only one measurement for each feature, it gives a much better score.
Can anyone explain to me why? and when it is good to use the second method?
the original dataset looks like this (all numbers are fake):
id /measurement1/measurement2/.../target/
0-1/0.18283 /0.12855 /.../ 1 /
0-2/0.1141 /0.38484 /.../ 1 /
0-3/0.4475 /0.18374 /.../ 1 /
and transformed dataset looks like this:
id /meas1_avg/meas1_std/meas1_skew/meas2_avg/meas2_std/.../target/
0 /0.28747 /0.183848/ 0.198384 /0.18484 /0.28474 /.../ 1 /
machine-learning feature-engineering data-science-model
asked Mar 26 at 14:58
edunlimitedunlimit
203
$endgroup$
add a comment |
$begingroup$
I'm working on a dataset that has repeated measurements for the same target variable.
When I don't change anything and create model, cross validation overfits with 0.99 score but in testset it gives around 0.39.
When I use mean, std, skew, quartiles for each measurement to have only one measurement for each feature, it gives a much better score.
Can anyone explain to me why? and when it is good to use the second method?
the original dataset looks like this (all numbers are fake):
id /measurement1/measurement2/.../target/
0-1/0.18283 /0.12855 /.../ 1 /
0-2/0.1141 /0.38484 /.../ 1 /
0-3/0.4475 /0.18374 /.../ 1 /
and transformed dataset looks like this:
id /meas1_avg/meas1_std/meas1_skew/meas2_avg/meas2_std/.../target/
0 /0.28747 /0.183848/ 0.198384 /0.18484 /0.28474 /.../ 1 /
machine-learning feature-engineering data-science-model
asked Mar 26 at 14:58
edunlimitedunlimit
203
$endgroup$
I'm working on a dataset that has repeated measurements for the same target variable.
When I don't change anything and create model, cross validation overfits with 0.99 score but in testset it gives around 0.39.
When I use mean, std, skew, quartiles for each measurement to have only one measurement for each feature, it gives a much better score.
Can anyone explain to me why? and when it is good to use the second method?
the original dataset looks like this (all numbers are fake):
id /measurement1/measurement2/.../target/
0-1/0.18283 /0.12855 /.../ 1 /
0-2/0.1141 /0.38484 /.../ 1 /
0-3/0.4475 /0.18374 /.../ 1 /
and transformed dataset looks like this:
id /meas1_avg/meas1_std/meas1_skew/meas2_avg/meas2_std/.../target/
0 /0.28747 /0.183848/ 0.198384 /0.18484 /0.28474 /.../ 1 /
machine-learning feature-engineering data-science-model
machine-learning feature-engineering data-science-model
asked Mar 26 at 14:58
edunlimitedunlimit
203
asked Mar 26 at 14:58
edunlimitedunlimit
203
asked Mar 26 at 14:58
edunlimitedunlimit
203
asked Mar 26 at 14:58
edunlimitedunlimit
203
asked Mar 26 at 14:58
edunlimitedunlimit
203
203
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Note that you are solving two different problems here.
In the first problem, you want to predict the target variable given one noisy measurement.
In the second problem, you want to predict the target variable given some statistics from a group of noisy measurements.
Your results show that the second problem is easier to solve which is intuitive, since the amount of noise (variance) for average of multiple measurements is less than only one measurement (closely related to Law of Large Numbers), thus the relation in the second problem is easier to find by the model.
Therefore, if both problems are equivalent to you, go with the second problem which is easier to solve.
answered Mar 26 at 15:06
EsmailianEsmailian
2,536318
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Note that you are solving two different problems here.
In the first problem, you want to predict the target variable given one noisy measurement.
In the second problem, you want to predict the target variable given some statistics from a group of noisy measurements.
Your results show that the second problem is easier to solve which is intuitive, since the amount of noise (variance) for average of multiple measurements is less than only one measurement (closely related to Law of Large Numbers), thus the relation in the second problem is easier to find by the model.
Therefore, if both problems are equivalent to you, go with the second problem which is easier to solve.
answered Mar 26 at 15:06
EsmailianEsmailian
2,536318
$endgroup$
add a comment |
$begingroup$
Note that you are solving two different problems here.
In the first problem, you want to predict the target variable given one noisy measurement.
In the second problem, you want to predict the target variable given some statistics from a group of noisy measurements.
Your results show that the second problem is easier to solve which is intuitive, since the amount of noise (variance) for average of multiple measurements is less than only one measurement (closely related to Law of Large Numbers), thus the relation in the second problem is easier to find by the model.
Therefore, if both problems are equivalent to you, go with the second problem which is easier to solve.
answered Mar 26 at 15:06
EsmailianEsmailian
2,536318
$endgroup$
add a comment |
$begingroup$
Note that you are solving two different problems here.
In the first problem, you want to predict the target variable given one noisy measurement.
In the second problem, you want to predict the target variable given some statistics from a group of noisy measurements.
Your results show that the second problem is easier to solve which is intuitive, since the amount of noise (variance) for average of multiple measurements is less than only one measurement (closely related to Law of Large Numbers), thus the relation in the second problem is easier to find by the model.
Therefore, if both problems are equivalent to you, go with the second problem which is easier to solve.
answered Mar 26 at 15:06
EsmailianEsmailian
2,536318
$endgroup$
Note that you are solving two different problems here.
In the first problem, you want to predict the target variable given one noisy measurement.
In the second problem, you want to predict the target variable given some statistics from a group of noisy measurements.
Your results show that the second problem is easier to solve which is intuitive, since the amount of noise (variance) for average of multiple measurements is less than only one measurement (closely related to Law of Large Numbers), thus the relation in the second problem is easier to find by the model.
Therefore, if both problems are equivalent to you, go with the second problem which is easier to solve.
answered Mar 26 at 15:06
EsmailianEsmailian
2,536318
edited Mar 26 at 18:09
answered Mar 26 at 15:06
EsmailianEsmailian
2,536318
answered Mar 26 at 15:06
EsmailianEsmailian
2,536318
answered Mar 26 at 15:06
EsmailianEsmailian
2,536318
2,536318
add a comment |
add a comment |
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