How to penalize for empty fields in a DataFrame?
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I have to calculate the consistency of racing car drivers during the whole season. My DataFrame consists of 10 columns (10 circuit names) and for each of those columns I have the standard deviation in lap time the driver posted in that circuit. In other words, how consistent the driver is from lap to lap. In races the driver did not finish the field is blank.
So far I have calculated their average season consistency by averaging all 10 columns. However, not finishing a race should affect a driver's consistency negatively and I do not know how to implement that.
pandas data
asked 3 hours ago
jatrp5jatrp5
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I have to calculate the consistency of racing car drivers during the whole season. My DataFrame consists of 10 columns (10 circuit names) and for each of those columns I have the standard deviation in lap time the driver posted in that circuit. In other words, how consistent the driver is from lap to lap. In races the driver did not finish the field is blank.
So far I have calculated their average season consistency by averaging all 10 columns. However, not finishing a race should affect a driver's consistency negatively and I do not know how to implement that.
pandas data
asked 3 hours ago
jatrp5jatrp5
111
New contributor
jatrp5 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 have to calculate the consistency of racing car drivers during the whole season. My DataFrame consists of 10 columns (10 circuit names) and for each of those columns I have the standard deviation in lap time the driver posted in that circuit. In other words, how consistent the driver is from lap to lap. In races the driver did not finish the field is blank.
So far I have calculated their average season consistency by averaging all 10 columns. However, not finishing a race should affect a driver's consistency negatively and I do not know how to implement that.
pandas data
asked 3 hours ago
jatrp5jatrp5
111
New contributor
jatrp5 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I have to calculate the consistency of racing car drivers during the whole season. My DataFrame consists of 10 columns (10 circuit names) and for each of those columns I have the standard deviation in lap time the driver posted in that circuit. In other words, how consistent the driver is from lap to lap. In races the driver did not finish the field is blank.
So far I have calculated their average season consistency by averaging all 10 columns. However, not finishing a race should affect a driver's consistency negatively and I do not know how to implement that.
pandas data
pandas data
asked 3 hours ago
jatrp5jatrp5
111
New contributor
jatrp5 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
jatrp5jatrp5
111
New contributor
jatrp5 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
jatrp5jatrp5
111
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jatrp5 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked 3 hours ago
jatrp5jatrp5
111
asked 3 hours ago
jatrp5jatrp5
111
111
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jatrp5 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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This heavily depends on the domain knowledge. A general approach would be to place
A multiplicative of the worst or average consistency at each circuit $c$, i.e. $(1 + m)text{max}(sigma_c)$ or $(1 + m)text{avg}(sigma_c)$ respectively, for the null values at that circuit, or
A multiplicative of the worst or average consistency of each driver $d$, i.e. $(1 + m)text{max}(sigma_d)$ or $(1 + m)text{avg}(sigma_d)$ respectively, for their unfinished races, or
A multiplicative of average of driver and circuit worst consistencies, i.e. $(1 + m)[text{max}(sigma_d) + text{max}(sigma_c)]/2$, for unfinished race of driver $d$ at circuit $c$, or some other combinations.
No matter which approach to choose, the choice of coefficient $m$ affects the final ranking and could be determined either
Subjectively by looking at the rankings from an expert point of view and selecting the one that makes more sense, or
By trying a range of values like $m in {-0.2, -0.1, 0, 0.1, 0.2, .., 0.5}$ and averaging the consistencies $sigma_d$ or rankings $R_d$ for each driver $d$. An advantage of this approach would be that when rank of a driver has a low variance over different values of $m$, it implies that driver's rank is insensitive to the choice of $m$, i.e. it is less controversial, and when rank changes a lot with different choices of $m$, the average rank is more controversial.
answered 2 hours ago
EsmailianEsmailian
2,222218
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1 Answer
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1 Answer
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$begingroup$
This heavily depends on the domain knowledge. A general approach would be to place
A multiplicative of the worst or average consistency at each circuit $c$, i.e. $(1 + m)text{max}(sigma_c)$ or $(1 + m)text{avg}(sigma_c)$ respectively, for the null values at that circuit, or
A multiplicative of the worst or average consistency of each driver $d$, i.e. $(1 + m)text{max}(sigma_d)$ or $(1 + m)text{avg}(sigma_d)$ respectively, for their unfinished races, or
A multiplicative of average of driver and circuit worst consistencies, i.e. $(1 + m)[text{max}(sigma_d) + text{max}(sigma_c)]/2$, for unfinished race of driver $d$ at circuit $c$, or some other combinations.
No matter which approach to choose, the choice of coefficient $m$ affects the final ranking and could be determined either
Subjectively by looking at the rankings from an expert point of view and selecting the one that makes more sense, or
By trying a range of values like $m in {-0.2, -0.1, 0, 0.1, 0.2, .., 0.5}$ and averaging the consistencies $sigma_d$ or rankings $R_d$ for each driver $d$. An advantage of this approach would be that when rank of a driver has a low variance over different values of $m$, it implies that driver's rank is insensitive to the choice of $m$, i.e. it is less controversial, and when rank changes a lot with different choices of $m$, the average rank is more controversial.
answered 2 hours ago
EsmailianEsmailian
2,222218
$endgroup$
add a comment |
$begingroup$
This heavily depends on the domain knowledge. A general approach would be to place
A multiplicative of the worst or average consistency at each circuit $c$, i.e. $(1 + m)text{max}(sigma_c)$ or $(1 + m)text{avg}(sigma_c)$ respectively, for the null values at that circuit, or
A multiplicative of the worst or average consistency of each driver $d$, i.e. $(1 + m)text{max}(sigma_d)$ or $(1 + m)text{avg}(sigma_d)$ respectively, for their unfinished races, or
A multiplicative of average of driver and circuit worst consistencies, i.e. $(1 + m)[text{max}(sigma_d) + text{max}(sigma_c)]/2$, for unfinished race of driver $d$ at circuit $c$, or some other combinations.
No matter which approach to choose, the choice of coefficient $m$ affects the final ranking and could be determined either
Subjectively by looking at the rankings from an expert point of view and selecting the one that makes more sense, or
By trying a range of values like $m in {-0.2, -0.1, 0, 0.1, 0.2, .., 0.5}$ and averaging the consistencies $sigma_d$ or rankings $R_d$ for each driver $d$. An advantage of this approach would be that when rank of a driver has a low variance over different values of $m$, it implies that driver's rank is insensitive to the choice of $m$, i.e. it is less controversial, and when rank changes a lot with different choices of $m$, the average rank is more controversial.
answered 2 hours ago
EsmailianEsmailian
2,222218
$endgroup$
add a comment |
$begingroup$
This heavily depends on the domain knowledge. A general approach would be to place
A multiplicative of the worst or average consistency at each circuit $c$, i.e. $(1 + m)text{max}(sigma_c)$ or $(1 + m)text{avg}(sigma_c)$ respectively, for the null values at that circuit, or
A multiplicative of the worst or average consistency of each driver $d$, i.e. $(1 + m)text{max}(sigma_d)$ or $(1 + m)text{avg}(sigma_d)$ respectively, for their unfinished races, or
A multiplicative of average of driver and circuit worst consistencies, i.e. $(1 + m)[text{max}(sigma_d) + text{max}(sigma_c)]/2$, for unfinished race of driver $d$ at circuit $c$, or some other combinations.
No matter which approach to choose, the choice of coefficient $m$ affects the final ranking and could be determined either
Subjectively by looking at the rankings from an expert point of view and selecting the one that makes more sense, or
By trying a range of values like $m in {-0.2, -0.1, 0, 0.1, 0.2, .., 0.5}$ and averaging the consistencies $sigma_d$ or rankings $R_d$ for each driver $d$. An advantage of this approach would be that when rank of a driver has a low variance over different values of $m$, it implies that driver's rank is insensitive to the choice of $m$, i.e. it is less controversial, and when rank changes a lot with different choices of $m$, the average rank is more controversial.
answered 2 hours ago
EsmailianEsmailian
2,222218
$endgroup$
This heavily depends on the domain knowledge. A general approach would be to place
A multiplicative of the worst or average consistency at each circuit $c$, i.e. $(1 + m)text{max}(sigma_c)$ or $(1 + m)text{avg}(sigma_c)$ respectively, for the null values at that circuit, or
A multiplicative of the worst or average consistency of each driver $d$, i.e. $(1 + m)text{max}(sigma_d)$ or $(1 + m)text{avg}(sigma_d)$ respectively, for their unfinished races, or
A multiplicative of average of driver and circuit worst consistencies, i.e. $(1 + m)[text{max}(sigma_d) + text{max}(sigma_c)]/2$, for unfinished race of driver $d$ at circuit $c$, or some other combinations.
No matter which approach to choose, the choice of coefficient $m$ affects the final ranking and could be determined either
Subjectively by looking at the rankings from an expert point of view and selecting the one that makes more sense, or
By trying a range of values like $m in {-0.2, -0.1, 0, 0.1, 0.2, .., 0.5}$ and averaging the consistencies $sigma_d$ or rankings $R_d$ for each driver $d$. An advantage of this approach would be that when rank of a driver has a low variance over different values of $m$, it implies that driver's rank is insensitive to the choice of $m$, i.e. it is less controversial, and when rank changes a lot with different choices of $m$, the average rank is more controversial.
answered 2 hours ago
EsmailianEsmailian
2,222218
edited 38 mins ago
answered 2 hours ago
EsmailianEsmailian
2,222218
answered 2 hours ago
EsmailianEsmailian
2,222218
answered 2 hours ago
EsmailianEsmailian
2,222218
2,222218
add a comment |
add a comment |
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