Feature engineering suggestion required The 2019 Stack Overflow Developer Survey Results Are InFeature Extraction Technique - Summarizing a Sequence of DataPrepping Data For Usage ClusteringGround-truth and feature extraction for predictive modellingHow to use neural network's hidden layer output for feature engineering?Fix missing data by adding another feature instead of using the mean?What are best practices for collaborative feature engineering?How would knowing spammers email address improve spam detection algorithms?Is this a good practice of feature engineering?How do I develop a system to Recommend a marketing channel using data science?Cant get LSTM model to give required predictions
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Feature engineering suggestion required
The 2019 Stack Overflow Developer Survey Results Are InFeature Extraction Technique - Summarizing a Sequence of DataPrepping Data For Usage ClusteringGround-truth and feature extraction for predictive modellingHow to use neural network's hidden layer output for feature engineering?Fix missing data by adding another feature instead of using the mean?What are best practices for collaborative feature engineering?How would knowing spammers email address improve spam detection algorithms?Is this a good practice of feature engineering?How do I develop a system to Recommend a marketing channel using data science?Cant get LSTM model to give required predictions
$begingroup$
I am having a problem during feature engineering. Looking for some suggestions. Problem statement: I have usage data of multiple customers for 3 days. Some have just 1 day usage some 2 and some 3. Data is related to number of emails sent / contacts added on each day etc.
I am converting this time series data to column-wise ie., number of emails sent by a customer on day1 as one feature, number of emails sent by a customer on day2 as one feature and so on. But problem is that, the usage can be of either increasing order or decreasing order for different customers.
ie., example 1: customer 'A' --> 'number of emails sent on 1st . day' = 100 . ' number of emails sent on 2nd day'=0
example 2: customer 'B' --> 'number of emails sent on 1st . day' = 0 . ' number of emails sent on 2nd day'=100
example 3: customer 'C' --> 'number of emails sent on 1st . day' = 0 . ' number of emails sent on 2nd day'=0
example 4: customer 'D' --> 'number of emails sent on 1st . day' = 100 . ' number of emails sent on 2nd day'=100
In the first two cases => My new feature will have "-100" and "100" as values. Which I guess is good for differentiating. But the problem arises for 3rd and 4th columns when the new feature value will be "0" in both scenarios Can anyone suggest a way to handle this.
One way to handle this:
I can add "No change" in those scenarios, but I am confused about one thing. If I do that, I will have to make the new feature as categorical, which is not ideal as the other values will be continuous.
Instead, I can have absolute values in the new feature and indicate the trend as "+1" or increasing "-1" for decreasing "no change" for no change and "0" if both the values have been "0". Would that be a good approach though?
The end goal is to predict if a user would continue using the application or not. So it basically would be a two-class model. And I would want to capture even the scale of usage i.e., "A user sending 100 emails every day" should be different from "B user sending 10000 emails every day"
machine-learning feature-engineering data-science-model
asked 2 hours ago
SSuramSSuram
214
New contributor
SSuram is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I am having a problem during feature engineering. Looking for some suggestions. Problem statement: I have usage data of multiple customers for 3 days. Some have just 1 day usage some 2 and some 3. Data is related to number of emails sent / contacts added on each day etc.
I am converting this time series data to column-wise ie., number of emails sent by a customer on day1 as one feature, number of emails sent by a customer on day2 as one feature and so on. But problem is that, the usage can be of either increasing order or decreasing order for different customers.
ie., example 1: customer 'A' --> 'number of emails sent on 1st . day' = 100 . ' number of emails sent on 2nd day'=0
example 2: customer 'B' --> 'number of emails sent on 1st . day' = 0 . ' number of emails sent on 2nd day'=100
example 3: customer 'C' --> 'number of emails sent on 1st . day' = 0 . ' number of emails sent on 2nd day'=0
example 4: customer 'D' --> 'number of emails sent on 1st . day' = 100 . ' number of emails sent on 2nd day'=100
In the first two cases => My new feature will have "-100" and "100" as values. Which I guess is good for differentiating. But the problem arises for 3rd and 4th columns when the new feature value will be "0" in both scenarios Can anyone suggest a way to handle this.
One way to handle this:
I can add "No change" in those scenarios, but I am confused about one thing. If I do that, I will have to make the new feature as categorical, which is not ideal as the other values will be continuous.
Instead, I can have absolute values in the new feature and indicate the trend as "+1" or increasing "-1" for decreasing "no change" for no change and "0" if both the values have been "0". Would that be a good approach though?
The end goal is to predict if a user would continue using the application or not. So it basically would be a two-class model. And I would want to capture even the scale of usage i.e., "A user sending 100 emails every day" should be different from "B user sending 10000 emails every day"
machine-learning feature-engineering data-science-model
asked 2 hours ago
SSuramSSuram
214
New contributor
SSuram is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
could you explain a bit better what are you trying to predict? Your question is pretty well explained but the kind of model you plan do train might give some of us better ideas.
$endgroup$
– Pedro Henrique Monforte
2 hours ago
$begingroup$
I would want to predict if a user would continue using the application or not. So it basically would be a two-class model. Does that answer?
$endgroup$
– SSuram
1 hour ago
$begingroup$
Yes, just add it to your question and it will be perfect
$endgroup$
– Pedro Henrique Monforte
1 hour ago
add a comment |
$begingroup$
I am having a problem during feature engineering. Looking for some suggestions. Problem statement: I have usage data of multiple customers for 3 days. Some have just 1 day usage some 2 and some 3. Data is related to number of emails sent / contacts added on each day etc.
I am converting this time series data to column-wise ie., number of emails sent by a customer on day1 as one feature, number of emails sent by a customer on day2 as one feature and so on. But problem is that, the usage can be of either increasing order or decreasing order for different customers.
ie., example 1: customer 'A' --> 'number of emails sent on 1st . day' = 100 . ' number of emails sent on 2nd day'=0
example 2: customer 'B' --> 'number of emails sent on 1st . day' = 0 . ' number of emails sent on 2nd day'=100
example 3: customer 'C' --> 'number of emails sent on 1st . day' = 0 . ' number of emails sent on 2nd day'=0
example 4: customer 'D' --> 'number of emails sent on 1st . day' = 100 . ' number of emails sent on 2nd day'=100
In the first two cases => My new feature will have "-100" and "100" as values. Which I guess is good for differentiating. But the problem arises for 3rd and 4th columns when the new feature value will be "0" in both scenarios Can anyone suggest a way to handle this.
One way to handle this:
I can add "No change" in those scenarios, but I am confused about one thing. If I do that, I will have to make the new feature as categorical, which is not ideal as the other values will be continuous.
Instead, I can have absolute values in the new feature and indicate the trend as "+1" or increasing "-1" for decreasing "no change" for no change and "0" if both the values have been "0". Would that be a good approach though?
The end goal is to predict if a user would continue using the application or not. So it basically would be a two-class model. And I would want to capture even the scale of usage i.e., "A user sending 100 emails every day" should be different from "B user sending 10000 emails every day"
machine-learning feature-engineering data-science-model
asked 2 hours ago
SSuramSSuram
214
New contributor
SSuram is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I am having a problem during feature engineering. Looking for some suggestions. Problem statement: I have usage data of multiple customers for 3 days. Some have just 1 day usage some 2 and some 3. Data is related to number of emails sent / contacts added on each day etc.
I am converting this time series data to column-wise ie., number of emails sent by a customer on day1 as one feature, number of emails sent by a customer on day2 as one feature and so on. But problem is that, the usage can be of either increasing order or decreasing order for different customers.
ie., example 1: customer 'A' --> 'number of emails sent on 1st . day' = 100 . ' number of emails sent on 2nd day'=0
example 2: customer 'B' --> 'number of emails sent on 1st . day' = 0 . ' number of emails sent on 2nd day'=100
example 3: customer 'C' --> 'number of emails sent on 1st . day' = 0 . ' number of emails sent on 2nd day'=0
example 4: customer 'D' --> 'number of emails sent on 1st . day' = 100 . ' number of emails sent on 2nd day'=100
In the first two cases => My new feature will have "-100" and "100" as values. Which I guess is good for differentiating. But the problem arises for 3rd and 4th columns when the new feature value will be "0" in both scenarios Can anyone suggest a way to handle this.
One way to handle this:
I can add "No change" in those scenarios, but I am confused about one thing. If I do that, I will have to make the new feature as categorical, which is not ideal as the other values will be continuous.
Instead, I can have absolute values in the new feature and indicate the trend as "+1" or increasing "-1" for decreasing "no change" for no change and "0" if both the values have been "0". Would that be a good approach though?
The end goal is to predict if a user would continue using the application or not. So it basically would be a two-class model. And I would want to capture even the scale of usage i.e., "A user sending 100 emails every day" should be different from "B user sending 10000 emails every day"
machine-learning feature-engineering data-science-model
machine-learning feature-engineering data-science-model
asked 2 hours ago
SSuramSSuram
214
New contributor
SSuram is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
SSuramSSuram
214
New contributor
SSuram is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 1 hour ago
SSuram
asked 2 hours ago
SSuramSSuram
214
New contributor
SSuram is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
SSuramSSuram
214
asked 2 hours ago
SSuramSSuram
214
214
New contributor
SSuram is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
SSuram is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
SSuram is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
$begingroup$
could you explain a bit better what are you trying to predict? Your question is pretty well explained but the kind of model you plan do train might give some of us better ideas.
$endgroup$
– Pedro Henrique Monforte
2 hours ago
$begingroup$
I would want to predict if a user would continue using the application or not. So it basically would be a two-class model. Does that answer?
$endgroup$
– SSuram
1 hour ago
$begingroup$
Yes, just add it to your question and it will be perfect
$endgroup$
– Pedro Henrique Monforte
1 hour ago
add a comment |
1
$begingroup$
could you explain a bit better what are you trying to predict? Your question is pretty well explained but the kind of model you plan do train might give some of us better ideas.
$endgroup$
– Pedro Henrique Monforte
2 hours ago
$begingroup$
I would want to predict if a user would continue using the application or not. So it basically would be a two-class model. Does that answer?
$endgroup$
– SSuram
1 hour ago
$begingroup$
Yes, just add it to your question and it will be perfect
$endgroup$
– Pedro Henrique Monforte
1 hour ago
1
1
$begingroup$
could you explain a bit better what are you trying to predict? Your question is pretty well explained but the kind of model you plan do train might give some of us better ideas.
$endgroup$
– Pedro Henrique Monforte
2 hours ago
$begingroup$
could you explain a bit better what are you trying to predict? Your question is pretty well explained but the kind of model you plan do train might give some of us better ideas.
$endgroup$
– Pedro Henrique Monforte
2 hours ago
$begingroup$
I would want to predict if a user would continue using the application or not. So it basically would be a two-class model. Does that answer?
$endgroup$
– SSuram
1 hour ago
$begingroup$
I would want to predict if a user would continue using the application or not. So it basically would be a two-class model. Does that answer?
$endgroup$
– SSuram
1 hour ago
$begingroup$
Yes, just add it to your question and it will be perfect
$endgroup$
– Pedro Henrique Monforte
1 hour ago
$begingroup$
Yes, just add it to your question and it will be perfect
$endgroup$
– Pedro Henrique Monforte
1 hour ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Well, you want to identify change in usage you could try something like:
$$ f(day_1,day2) = fracday_2-day_1 + delta times Biggr|Biggr|fracday_2+day_1(day_2+day_1+1)(day_2-day_1+1)Biggl|Biggl| $$
where $delta$ is the eps of your machine (minimum value needed to be summed to differ it from other floats)
that will give you
$$f(100,0) approx -98.02$$
$$f(0,100) = 100$$
$$f(100,100) approx 0.995$$
$$f(0,0) = 0$$
You can look at my experiment here
This will map all non-changes from $[0,1]$ where $f(0,0)$ maps to $0$ and $f(infty,infty)$ maps to $1$
Where is it from? Just tuned the function manually. But I think this might suffice for your application
Explaining the ideia
You want to have a feature that packs a lot of information:
- Is the usage greater than zero?
- Is it increasing or decreasing?
- If it is stalled, how much is the usage?
Well, your usage vary in integer values so you can map the entire non-changing but above 0 case to a previously non-used interval.
The function above will map in $[0,1]$ all non-changing possibilities, in a exponential kind of way ($a^(-frac1usage)$) also you can extract the actual value from positive changes and the approximate value for negative change (been a better approximation when the drop is high)
This is not the perfect scenario but it is the maximum information I could compress into 1 variable with little loss.
$endgroup$
$begingroup$
I am not sure if it would answer --- "'''And I would want to capture even the scale of usage i.e., "A user sending 100 emails every day" should be different from "B user sending 10000 emails every day ""''---- part of the question. Could you please explain?
$endgroup$
– SSuram
1 hour ago
$begingroup$
What would you say about adding the below info to it f = (((d2-d1+eps)/abs(d2-d1+eps))*abs((d2+d1)/(d1+d2+1)*(d2-d1+1)))*(d2/1000)*(d1/1000) where "1000"-- would be max(usage).
$endgroup$
– SSuram
54 mins ago
$begingroup$
that will actually return zero for near every case
$endgroup$
– Pedro Henrique Monforte
43 mins ago
add a comment |
Your Answer
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1 Answer
1
active
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Well, you want to identify change in usage you could try something like:
$$ f(day_1,day2) = fracday_2-day_1 + delta times Biggr|Biggr|fracday_2+day_1(day_2+day_1+1)(day_2-day_1+1)Biggl|Biggl| $$
where $delta$ is the eps of your machine (minimum value needed to be summed to differ it from other floats)
that will give you
$$f(100,0) approx -98.02$$
$$f(0,100) = 100$$
$$f(100,100) approx 0.995$$
$$f(0,0) = 0$$
You can look at my experiment here
This will map all non-changes from $[0,1]$ where $f(0,0)$ maps to $0$ and $f(infty,infty)$ maps to $1$
Where is it from? Just tuned the function manually. But I think this might suffice for your application
Explaining the ideia
You want to have a feature that packs a lot of information:
- Is the usage greater than zero?
- Is it increasing or decreasing?
- If it is stalled, how much is the usage?
Well, your usage vary in integer values so you can map the entire non-changing but above 0 case to a previously non-used interval.
The function above will map in $[0,1]$ all non-changing possibilities, in a exponential kind of way ($a^(-frac1usage)$) also you can extract the actual value from positive changes and the approximate value for negative change (been a better approximation when the drop is high)
This is not the perfect scenario but it is the maximum information I could compress into 1 variable with little loss.
$endgroup$
$begingroup$
I am not sure if it would answer --- "'''And I would want to capture even the scale of usage i.e., "A user sending 100 emails every day" should be different from "B user sending 10000 emails every day ""''---- part of the question. Could you please explain?
$endgroup$
– SSuram
1 hour ago
$begingroup$
What would you say about adding the below info to it f = (((d2-d1+eps)/abs(d2-d1+eps))*abs((d2+d1)/(d1+d2+1)*(d2-d1+1)))*(d2/1000)*(d1/1000) where "1000"-- would be max(usage).
$endgroup$
– SSuram
54 mins ago
$begingroup$
that will actually return zero for near every case
$endgroup$
– Pedro Henrique Monforte
43 mins ago
add a comment |
$begingroup$
Well, you want to identify change in usage you could try something like:
$$ f(day_1,day2) = fracday_2-day_1 + delta times Biggr|Biggr|fracday_2+day_1(day_2+day_1+1)(day_2-day_1+1)Biggl|Biggl| $$
where $delta$ is the eps of your machine (minimum value needed to be summed to differ it from other floats)
that will give you
$$f(100,0) approx -98.02$$
$$f(0,100) = 100$$
$$f(100,100) approx 0.995$$
$$f(0,0) = 0$$
You can look at my experiment here
This will map all non-changes from $[0,1]$ where $f(0,0)$ maps to $0$ and $f(infty,infty)$ maps to $1$
Where is it from? Just tuned the function manually. But I think this might suffice for your application
Explaining the ideia
You want to have a feature that packs a lot of information:
- Is the usage greater than zero?
- Is it increasing or decreasing?
- If it is stalled, how much is the usage?
Well, your usage vary in integer values so you can map the entire non-changing but above 0 case to a previously non-used interval.
The function above will map in $[0,1]$ all non-changing possibilities, in a exponential kind of way ($a^(-frac1usage)$) also you can extract the actual value from positive changes and the approximate value for negative change (been a better approximation when the drop is high)
This is not the perfect scenario but it is the maximum information I could compress into 1 variable with little loss.
$endgroup$
$begingroup$
I am not sure if it would answer --- "'''And I would want to capture even the scale of usage i.e., "A user sending 100 emails every day" should be different from "B user sending 10000 emails every day ""''---- part of the question. Could you please explain?
$endgroup$
– SSuram
1 hour ago
$begingroup$
What would you say about adding the below info to it f = (((d2-d1+eps)/abs(d2-d1+eps))*abs((d2+d1)/(d1+d2+1)*(d2-d1+1)))*(d2/1000)*(d1/1000) where "1000"-- would be max(usage).
$endgroup$
– SSuram
54 mins ago
$begingroup$
that will actually return zero for near every case
$endgroup$
– Pedro Henrique Monforte
43 mins ago
add a comment |
$begingroup$
Well, you want to identify change in usage you could try something like:
$$ f(day_1,day2) = fracday_2-day_1 + delta times Biggr|Biggr|fracday_2+day_1(day_2+day_1+1)(day_2-day_1+1)Biggl|Biggl| $$
where $delta$ is the eps of your machine (minimum value needed to be summed to differ it from other floats)
that will give you
$$f(100,0) approx -98.02$$
$$f(0,100) = 100$$
$$f(100,100) approx 0.995$$
$$f(0,0) = 0$$
You can look at my experiment here
This will map all non-changes from $[0,1]$ where $f(0,0)$ maps to $0$ and $f(infty,infty)$ maps to $1$
Where is it from? Just tuned the function manually. But I think this might suffice for your application
Explaining the ideia
You want to have a feature that packs a lot of information:
- Is the usage greater than zero?
- Is it increasing or decreasing?
- If it is stalled, how much is the usage?
Well, your usage vary in integer values so you can map the entire non-changing but above 0 case to a previously non-used interval.
The function above will map in $[0,1]$ all non-changing possibilities, in a exponential kind of way ($a^(-frac1usage)$) also you can extract the actual value from positive changes and the approximate value for negative change (been a better approximation when the drop is high)
This is not the perfect scenario but it is the maximum information I could compress into 1 variable with little loss.
$endgroup$
Well, you want to identify change in usage you could try something like:
$$ f(day_1,day2) = fracday_2-day_1 + delta times Biggr|Biggr|fracday_2+day_1(day_2+day_1+1)(day_2-day_1+1)Biggl|Biggl| $$
where $delta$ is the eps of your machine (minimum value needed to be summed to differ it from other floats)
that will give you
$$f(100,0) approx -98.02$$
$$f(0,100) = 100$$
$$f(100,100) approx 0.995$$
$$f(0,0) = 0$$
You can look at my experiment here
This will map all non-changes from $[0,1]$ where $f(0,0)$ maps to $0$ and $f(infty,infty)$ maps to $1$
Where is it from? Just tuned the function manually. But I think this might suffice for your application
Explaining the ideia
You want to have a feature that packs a lot of information:
- Is the usage greater than zero?
- Is it increasing or decreasing?
- If it is stalled, how much is the usage?
Well, your usage vary in integer values so you can map the entire non-changing but above 0 case to a previously non-used interval.
The function above will map in $[0,1]$ all non-changing possibilities, in a exponential kind of way ($a^(-frac1usage)$) also you can extract the actual value from positive changes and the approximate value for negative change (been a better approximation when the drop is high)
This is not the perfect scenario but it is the maximum information I could compress into 1 variable with little loss.
edited 46 mins ago
community wiki
2 revs
Pedro Henrique Monforte
$begingroup$
I am not sure if it would answer --- "'''And I would want to capture even the scale of usage i.e., "A user sending 100 emails every day" should be different from "B user sending 10000 emails every day ""''---- part of the question. Could you please explain?
$endgroup$
– SSuram
1 hour ago
$begingroup$
What would you say about adding the below info to it f = (((d2-d1+eps)/abs(d2-d1+eps))*abs((d2+d1)/(d1+d2+1)*(d2-d1+1)))*(d2/1000)*(d1/1000) where "1000"-- would be max(usage).
$endgroup$
– SSuram
54 mins ago
$begingroup$
that will actually return zero for near every case
$endgroup$
– Pedro Henrique Monforte
43 mins ago
add a comment |
$begingroup$
I am not sure if it would answer --- "'''And I would want to capture even the scale of usage i.e., "A user sending 100 emails every day" should be different from "B user sending 10000 emails every day ""''---- part of the question. Could you please explain?
$endgroup$
– SSuram
1 hour ago
$begingroup$
What would you say about adding the below info to it f = (((d2-d1+eps)/abs(d2-d1+eps))*abs((d2+d1)/(d1+d2+1)*(d2-d1+1)))*(d2/1000)*(d1/1000) where "1000"-- would be max(usage).
$endgroup$
– SSuram
54 mins ago
$begingroup$
that will actually return zero for near every case
$endgroup$
– Pedro Henrique Monforte
43 mins ago
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I am not sure if it would answer --- "'''And I would want to capture even the scale of usage i.e., "A user sending 100 emails every day" should be different from "B user sending 10000 emails every day ""''---- part of the question. Could you please explain?
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– SSuram
1 hour ago
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I am not sure if it would answer --- "'''And I would want to capture even the scale of usage i.e., "A user sending 100 emails every day" should be different from "B user sending 10000 emails every day ""''---- part of the question. Could you please explain?
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– SSuram
1 hour ago
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What would you say about adding the below info to it f = (((d2-d1+eps)/abs(d2-d1+eps))*abs((d2+d1)/(d1+d2+1)*(d2-d1+1)))*(d2/1000)*(d1/1000) where "1000"-- would be max(usage).
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– SSuram
54 mins ago
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What would you say about adding the below info to it f = (((d2-d1+eps)/abs(d2-d1+eps))*abs((d2+d1)/(d1+d2+1)*(d2-d1+1)))*(d2/1000)*(d1/1000) where "1000"-- would be max(usage).
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– SSuram
54 mins ago
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that will actually return zero for near every case
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– Pedro Henrique Monforte
43 mins ago
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that will actually return zero for near every case
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– Pedro Henrique Monforte
43 mins ago
add a comment |
SSuram is a new contributor. Be nice, and check out our Code of Conduct.
SSuram is a new contributor. Be nice, and check out our Code of Conduct.
SSuram is a new contributor. Be nice, and check out our Code of Conduct.
SSuram is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
could you explain a bit better what are you trying to predict? Your question is pretty well explained but the kind of model you plan do train might give some of us better ideas.
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– Pedro Henrique Monforte
2 hours ago
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I would want to predict if a user would continue using the application or not. So it basically would be a two-class model. Does that answer?
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– SSuram
1 hour ago
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Yes, just add it to your question and it will be perfect
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– Pedro Henrique Monforte
1 hour ago