cycle through list to match value with multiple matches possible












0















I am interested in cycling through a list of diagnosis codes and populating a new variable with a previously calculated risk score if the value matches, and if there are multiple matches populate the new variable with the highest risk score.



I am hoping to take the long form of the original data set and for each ID match the proc number with the highest risk score and store both the proc number and risk score in separate variables.



I have some experience using if loops to do similar things in wide data, but cannot figure out how to do it this way. I have no experience matching and then storing the highest value, so don't even know where to start with this.



Data to see what I am getting at:



Here is the data from for the diagnosis codes



dz <-c("disease_1", "disease_2", "disease_3", "disease_4")
code <-c(124, 546, 890, 898)
risk_score <-c(10, 122, 45, 98)
df <-data.frame(dz, code, risk_score)


And the simulated data set I am interested in



 id <- c(1,1,1,2,2,2,2,3,3,4,4,4,4,4,4,5,5,5)
proc <-c(244,546,234,345,890,123,434,634,233,345,124,234,634,546,789,890,567,124)
proc<-as.character(proc)
data<-data.frame(id, proc)


so what I want to achieve is something like this



id<-c(1,2,3,4,5)
code_match<-c(546,890,124,546,890)
highest_risk_score <-c(122,45,10,122,45)
output_df<-data.frame(id, code_match, highest_risk_score)


with this output



  id code_match highest_risk_score
1 1 546 122
2 2 890 45
3 3 124 10
4 4 546 122
5 5 890 45


with id being the identifier, the code_match being the code with the highest risk score, and the highest_risk_score being the value of the risk score (the highest value for that id).










share|improve this question

























  • you are correct

    – Tim Feeney
    Nov 25 '18 at 20:51











  • It's true, because I am not interested in those codes. the data I am working with will have multiple codes per ID that I have no interest in.

    – Tim Feeney
    Nov 25 '18 at 21:00











  • correct, because I don't know how to do it. I thought an example of the output I am interested in would illustrate what I am hoping to achieve.

    – Tim Feeney
    Nov 25 '18 at 21:02











  • Not sure why all the -1s today but I ticked it back to 0. This had code + data and was reproducible.

    – hrbrmstr
    Nov 26 '18 at 0:23
















0















I am interested in cycling through a list of diagnosis codes and populating a new variable with a previously calculated risk score if the value matches, and if there are multiple matches populate the new variable with the highest risk score.



I am hoping to take the long form of the original data set and for each ID match the proc number with the highest risk score and store both the proc number and risk score in separate variables.



I have some experience using if loops to do similar things in wide data, but cannot figure out how to do it this way. I have no experience matching and then storing the highest value, so don't even know where to start with this.



Data to see what I am getting at:



Here is the data from for the diagnosis codes



dz <-c("disease_1", "disease_2", "disease_3", "disease_4")
code <-c(124, 546, 890, 898)
risk_score <-c(10, 122, 45, 98)
df <-data.frame(dz, code, risk_score)


And the simulated data set I am interested in



 id <- c(1,1,1,2,2,2,2,3,3,4,4,4,4,4,4,5,5,5)
proc <-c(244,546,234,345,890,123,434,634,233,345,124,234,634,546,789,890,567,124)
proc<-as.character(proc)
data<-data.frame(id, proc)


so what I want to achieve is something like this



id<-c(1,2,3,4,5)
code_match<-c(546,890,124,546,890)
highest_risk_score <-c(122,45,10,122,45)
output_df<-data.frame(id, code_match, highest_risk_score)


with this output



  id code_match highest_risk_score
1 1 546 122
2 2 890 45
3 3 124 10
4 4 546 122
5 5 890 45


with id being the identifier, the code_match being the code with the highest risk score, and the highest_risk_score being the value of the risk score (the highest value for that id).










share|improve this question

























  • you are correct

    – Tim Feeney
    Nov 25 '18 at 20:51











  • It's true, because I am not interested in those codes. the data I am working with will have multiple codes per ID that I have no interest in.

    – Tim Feeney
    Nov 25 '18 at 21:00











  • correct, because I don't know how to do it. I thought an example of the output I am interested in would illustrate what I am hoping to achieve.

    – Tim Feeney
    Nov 25 '18 at 21:02











  • Not sure why all the -1s today but I ticked it back to 0. This had code + data and was reproducible.

    – hrbrmstr
    Nov 26 '18 at 0:23














0












0








0








I am interested in cycling through a list of diagnosis codes and populating a new variable with a previously calculated risk score if the value matches, and if there are multiple matches populate the new variable with the highest risk score.



I am hoping to take the long form of the original data set and for each ID match the proc number with the highest risk score and store both the proc number and risk score in separate variables.



I have some experience using if loops to do similar things in wide data, but cannot figure out how to do it this way. I have no experience matching and then storing the highest value, so don't even know where to start with this.



Data to see what I am getting at:



Here is the data from for the diagnosis codes



dz <-c("disease_1", "disease_2", "disease_3", "disease_4")
code <-c(124, 546, 890, 898)
risk_score <-c(10, 122, 45, 98)
df <-data.frame(dz, code, risk_score)


And the simulated data set I am interested in



 id <- c(1,1,1,2,2,2,2,3,3,4,4,4,4,4,4,5,5,5)
proc <-c(244,546,234,345,890,123,434,634,233,345,124,234,634,546,789,890,567,124)
proc<-as.character(proc)
data<-data.frame(id, proc)


so what I want to achieve is something like this



id<-c(1,2,3,4,5)
code_match<-c(546,890,124,546,890)
highest_risk_score <-c(122,45,10,122,45)
output_df<-data.frame(id, code_match, highest_risk_score)


with this output



  id code_match highest_risk_score
1 1 546 122
2 2 890 45
3 3 124 10
4 4 546 122
5 5 890 45


with id being the identifier, the code_match being the code with the highest risk score, and the highest_risk_score being the value of the risk score (the highest value for that id).










share|improve this question
















I am interested in cycling through a list of diagnosis codes and populating a new variable with a previously calculated risk score if the value matches, and if there are multiple matches populate the new variable with the highest risk score.



I am hoping to take the long form of the original data set and for each ID match the proc number with the highest risk score and store both the proc number and risk score in separate variables.



I have some experience using if loops to do similar things in wide data, but cannot figure out how to do it this way. I have no experience matching and then storing the highest value, so don't even know where to start with this.



Data to see what I am getting at:



Here is the data from for the diagnosis codes



dz <-c("disease_1", "disease_2", "disease_3", "disease_4")
code <-c(124, 546, 890, 898)
risk_score <-c(10, 122, 45, 98)
df <-data.frame(dz, code, risk_score)


And the simulated data set I am interested in



 id <- c(1,1,1,2,2,2,2,3,3,4,4,4,4,4,4,5,5,5)
proc <-c(244,546,234,345,890,123,434,634,233,345,124,234,634,546,789,890,567,124)
proc<-as.character(proc)
data<-data.frame(id, proc)


so what I want to achieve is something like this



id<-c(1,2,3,4,5)
code_match<-c(546,890,124,546,890)
highest_risk_score <-c(122,45,10,122,45)
output_df<-data.frame(id, code_match, highest_risk_score)


with this output



  id code_match highest_risk_score
1 1 546 122
2 2 890 45
3 3 124 10
4 4 546 122
5 5 890 45


with id being the identifier, the code_match being the code with the highest risk score, and the highest_risk_score being the value of the risk score (the highest value for that id).







r






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 25 '18 at 20:51







Tim Feeney

















asked Nov 25 '18 at 20:32









Tim FeeneyTim Feeney

616




616













  • you are correct

    – Tim Feeney
    Nov 25 '18 at 20:51











  • It's true, because I am not interested in those codes. the data I am working with will have multiple codes per ID that I have no interest in.

    – Tim Feeney
    Nov 25 '18 at 21:00











  • correct, because I don't know how to do it. I thought an example of the output I am interested in would illustrate what I am hoping to achieve.

    – Tim Feeney
    Nov 25 '18 at 21:02











  • Not sure why all the -1s today but I ticked it back to 0. This had code + data and was reproducible.

    – hrbrmstr
    Nov 26 '18 at 0:23



















  • you are correct

    – Tim Feeney
    Nov 25 '18 at 20:51











  • It's true, because I am not interested in those codes. the data I am working with will have multiple codes per ID that I have no interest in.

    – Tim Feeney
    Nov 25 '18 at 21:00











  • correct, because I don't know how to do it. I thought an example of the output I am interested in would illustrate what I am hoping to achieve.

    – Tim Feeney
    Nov 25 '18 at 21:02











  • Not sure why all the -1s today but I ticked it back to 0. This had code + data and was reproducible.

    – hrbrmstr
    Nov 26 '18 at 0:23

















you are correct

– Tim Feeney
Nov 25 '18 at 20:51





you are correct

– Tim Feeney
Nov 25 '18 at 20:51













It's true, because I am not interested in those codes. the data I am working with will have multiple codes per ID that I have no interest in.

– Tim Feeney
Nov 25 '18 at 21:00





It's true, because I am not interested in those codes. the data I am working with will have multiple codes per ID that I have no interest in.

– Tim Feeney
Nov 25 '18 at 21:00













correct, because I don't know how to do it. I thought an example of the output I am interested in would illustrate what I am hoping to achieve.

– Tim Feeney
Nov 25 '18 at 21:02





correct, because I don't know how to do it. I thought an example of the output I am interested in would illustrate what I am hoping to achieve.

– Tim Feeney
Nov 25 '18 at 21:02













Not sure why all the -1s today but I ticked it back to 0. This had code + data and was reproducible.

– hrbrmstr
Nov 26 '18 at 0:23





Not sure why all the -1s today but I ticked it back to 0. This had code + data and was reproducible.

– hrbrmstr
Nov 26 '18 at 0:23












1 Answer
1






active

oldest

votes


















1














We'll use an alternate way of creating those data frames:



data.frame(
dz = c("disease_1", "disease_2", "disease_3", "disease_4"),
code = as.character(c(124, 546, 890, 898)),
risk_score = c(10, 122, 45, 98),
stringsAsFactors = FALSE
) -> df

data.frame(
id = c(1, 1, 1, 2, 2, 2, 2, 3, 3, 4, 4, 4, 4, 4, 4, 5, 5, 5),
proc = as.character(c(244, 546, 234, 345, 890, 123, 434, 634, 233, 345, 124, 234, 634, 546, 789, 890, 567, 124)),
stringsAsFactors = FALSE
) -> data


Here's one way (in tidyverse and base R) to do this:



57 compiled 📦 dependency tidyverse solution:



library(tidyverse)

filter(data, proc %in% df$code) %>%
left_join(df, by=c("proc"="code")) %>%
group_by(id) %>%
top_n(1) %>%
slice(1) %>%
select(id, code_match = proc, highest_risk_score = risk_score)
## # A tibble: 4 x 3
## # Groups: id [4]
## id code_match highest_risk_score
## <dbl> <chr> <dbl>
## 1 1. 546 122.
## 2 2. 890 45.
## 3 4. 546 122.
## 4 5. 890 45.


0 📦 (ok, 1 — stats — which comes along for the ride with base R) Base R solution



tmp <- merge(data[with(data, proc %in% df$code),], df, by.x = "proc", by.y = "code")

do.call(
rbind.data.frame,
lapply(
split(tmp, tmp$id),
function(x) {
x[which.max(x$risk_score),]
}
)
)[,-3] -> tmp

setNames(tmp[,c(2,1,3)], c("id", "code_match", "highest_risk_score"))
## id code_match highest_risk_score
## 1 1 546 122
## 2 2 890 45
## 4 4 546 122
## 5 5 890 45


You've not mentioned how to handle not-matches so this just ignores them.






share|improve this answer


























  • This works great and introduced me to top_n(), thanks. However, I ran into an issue where some IDs have multiple instances because the id had a the same code with risk listed more than once. How can I select out just one instance after the last line in the tidyverse method code?

    – Tim Feeney
    Nov 26 '18 at 0:21











  • See the update. A slice(1) should work. (and I forgot an ungroup() at the end which you should def add.

    – hrbrmstr
    Nov 26 '18 at 0:22











  • worked great, thanks!

    – Tim Feeney
    Nov 26 '18 at 0:26











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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1














We'll use an alternate way of creating those data frames:



data.frame(
dz = c("disease_1", "disease_2", "disease_3", "disease_4"),
code = as.character(c(124, 546, 890, 898)),
risk_score = c(10, 122, 45, 98),
stringsAsFactors = FALSE
) -> df

data.frame(
id = c(1, 1, 1, 2, 2, 2, 2, 3, 3, 4, 4, 4, 4, 4, 4, 5, 5, 5),
proc = as.character(c(244, 546, 234, 345, 890, 123, 434, 634, 233, 345, 124, 234, 634, 546, 789, 890, 567, 124)),
stringsAsFactors = FALSE
) -> data


Here's one way (in tidyverse and base R) to do this:



57 compiled 📦 dependency tidyverse solution:



library(tidyverse)

filter(data, proc %in% df$code) %>%
left_join(df, by=c("proc"="code")) %>%
group_by(id) %>%
top_n(1) %>%
slice(1) %>%
select(id, code_match = proc, highest_risk_score = risk_score)
## # A tibble: 4 x 3
## # Groups: id [4]
## id code_match highest_risk_score
## <dbl> <chr> <dbl>
## 1 1. 546 122.
## 2 2. 890 45.
## 3 4. 546 122.
## 4 5. 890 45.


0 📦 (ok, 1 — stats — which comes along for the ride with base R) Base R solution



tmp <- merge(data[with(data, proc %in% df$code),], df, by.x = "proc", by.y = "code")

do.call(
rbind.data.frame,
lapply(
split(tmp, tmp$id),
function(x) {
x[which.max(x$risk_score),]
}
)
)[,-3] -> tmp

setNames(tmp[,c(2,1,3)], c("id", "code_match", "highest_risk_score"))
## id code_match highest_risk_score
## 1 1 546 122
## 2 2 890 45
## 4 4 546 122
## 5 5 890 45


You've not mentioned how to handle not-matches so this just ignores them.






share|improve this answer


























  • This works great and introduced me to top_n(), thanks. However, I ran into an issue where some IDs have multiple instances because the id had a the same code with risk listed more than once. How can I select out just one instance after the last line in the tidyverse method code?

    – Tim Feeney
    Nov 26 '18 at 0:21











  • See the update. A slice(1) should work. (and I forgot an ungroup() at the end which you should def add.

    – hrbrmstr
    Nov 26 '18 at 0:22











  • worked great, thanks!

    – Tim Feeney
    Nov 26 '18 at 0:26
















1














We'll use an alternate way of creating those data frames:



data.frame(
dz = c("disease_1", "disease_2", "disease_3", "disease_4"),
code = as.character(c(124, 546, 890, 898)),
risk_score = c(10, 122, 45, 98),
stringsAsFactors = FALSE
) -> df

data.frame(
id = c(1, 1, 1, 2, 2, 2, 2, 3, 3, 4, 4, 4, 4, 4, 4, 5, 5, 5),
proc = as.character(c(244, 546, 234, 345, 890, 123, 434, 634, 233, 345, 124, 234, 634, 546, 789, 890, 567, 124)),
stringsAsFactors = FALSE
) -> data


Here's one way (in tidyverse and base R) to do this:



57 compiled 📦 dependency tidyverse solution:



library(tidyverse)

filter(data, proc %in% df$code) %>%
left_join(df, by=c("proc"="code")) %>%
group_by(id) %>%
top_n(1) %>%
slice(1) %>%
select(id, code_match = proc, highest_risk_score = risk_score)
## # A tibble: 4 x 3
## # Groups: id [4]
## id code_match highest_risk_score
## <dbl> <chr> <dbl>
## 1 1. 546 122.
## 2 2. 890 45.
## 3 4. 546 122.
## 4 5. 890 45.


0 📦 (ok, 1 — stats — which comes along for the ride with base R) Base R solution



tmp <- merge(data[with(data, proc %in% df$code),], df, by.x = "proc", by.y = "code")

do.call(
rbind.data.frame,
lapply(
split(tmp, tmp$id),
function(x) {
x[which.max(x$risk_score),]
}
)
)[,-3] -> tmp

setNames(tmp[,c(2,1,3)], c("id", "code_match", "highest_risk_score"))
## id code_match highest_risk_score
## 1 1 546 122
## 2 2 890 45
## 4 4 546 122
## 5 5 890 45


You've not mentioned how to handle not-matches so this just ignores them.






share|improve this answer


























  • This works great and introduced me to top_n(), thanks. However, I ran into an issue where some IDs have multiple instances because the id had a the same code with risk listed more than once. How can I select out just one instance after the last line in the tidyverse method code?

    – Tim Feeney
    Nov 26 '18 at 0:21











  • See the update. A slice(1) should work. (and I forgot an ungroup() at the end which you should def add.

    – hrbrmstr
    Nov 26 '18 at 0:22











  • worked great, thanks!

    – Tim Feeney
    Nov 26 '18 at 0:26














1












1








1







We'll use an alternate way of creating those data frames:



data.frame(
dz = c("disease_1", "disease_2", "disease_3", "disease_4"),
code = as.character(c(124, 546, 890, 898)),
risk_score = c(10, 122, 45, 98),
stringsAsFactors = FALSE
) -> df

data.frame(
id = c(1, 1, 1, 2, 2, 2, 2, 3, 3, 4, 4, 4, 4, 4, 4, 5, 5, 5),
proc = as.character(c(244, 546, 234, 345, 890, 123, 434, 634, 233, 345, 124, 234, 634, 546, 789, 890, 567, 124)),
stringsAsFactors = FALSE
) -> data


Here's one way (in tidyverse and base R) to do this:



57 compiled 📦 dependency tidyverse solution:



library(tidyverse)

filter(data, proc %in% df$code) %>%
left_join(df, by=c("proc"="code")) %>%
group_by(id) %>%
top_n(1) %>%
slice(1) %>%
select(id, code_match = proc, highest_risk_score = risk_score)
## # A tibble: 4 x 3
## # Groups: id [4]
## id code_match highest_risk_score
## <dbl> <chr> <dbl>
## 1 1. 546 122.
## 2 2. 890 45.
## 3 4. 546 122.
## 4 5. 890 45.


0 📦 (ok, 1 — stats — which comes along for the ride with base R) Base R solution



tmp <- merge(data[with(data, proc %in% df$code),], df, by.x = "proc", by.y = "code")

do.call(
rbind.data.frame,
lapply(
split(tmp, tmp$id),
function(x) {
x[which.max(x$risk_score),]
}
)
)[,-3] -> tmp

setNames(tmp[,c(2,1,3)], c("id", "code_match", "highest_risk_score"))
## id code_match highest_risk_score
## 1 1 546 122
## 2 2 890 45
## 4 4 546 122
## 5 5 890 45


You've not mentioned how to handle not-matches so this just ignores them.






share|improve this answer















We'll use an alternate way of creating those data frames:



data.frame(
dz = c("disease_1", "disease_2", "disease_3", "disease_4"),
code = as.character(c(124, 546, 890, 898)),
risk_score = c(10, 122, 45, 98),
stringsAsFactors = FALSE
) -> df

data.frame(
id = c(1, 1, 1, 2, 2, 2, 2, 3, 3, 4, 4, 4, 4, 4, 4, 5, 5, 5),
proc = as.character(c(244, 546, 234, 345, 890, 123, 434, 634, 233, 345, 124, 234, 634, 546, 789, 890, 567, 124)),
stringsAsFactors = FALSE
) -> data


Here's one way (in tidyverse and base R) to do this:



57 compiled 📦 dependency tidyverse solution:



library(tidyverse)

filter(data, proc %in% df$code) %>%
left_join(df, by=c("proc"="code")) %>%
group_by(id) %>%
top_n(1) %>%
slice(1) %>%
select(id, code_match = proc, highest_risk_score = risk_score)
## # A tibble: 4 x 3
## # Groups: id [4]
## id code_match highest_risk_score
## <dbl> <chr> <dbl>
## 1 1. 546 122.
## 2 2. 890 45.
## 3 4. 546 122.
## 4 5. 890 45.


0 📦 (ok, 1 — stats — which comes along for the ride with base R) Base R solution



tmp <- merge(data[with(data, proc %in% df$code),], df, by.x = "proc", by.y = "code")

do.call(
rbind.data.frame,
lapply(
split(tmp, tmp$id),
function(x) {
x[which.max(x$risk_score),]
}
)
)[,-3] -> tmp

setNames(tmp[,c(2,1,3)], c("id", "code_match", "highest_risk_score"))
## id code_match highest_risk_score
## 1 1 546 122
## 2 2 890 45
## 4 4 546 122
## 5 5 890 45


You've not mentioned how to handle not-matches so this just ignores them.







share|improve this answer














share|improve this answer



share|improve this answer








edited Nov 26 '18 at 0:22

























answered Nov 25 '18 at 21:04









hrbrmstrhrbrmstr

61.6k693153




61.6k693153













  • This works great and introduced me to top_n(), thanks. However, I ran into an issue where some IDs have multiple instances because the id had a the same code with risk listed more than once. How can I select out just one instance after the last line in the tidyverse method code?

    – Tim Feeney
    Nov 26 '18 at 0:21











  • See the update. A slice(1) should work. (and I forgot an ungroup() at the end which you should def add.

    – hrbrmstr
    Nov 26 '18 at 0:22











  • worked great, thanks!

    – Tim Feeney
    Nov 26 '18 at 0:26



















  • This works great and introduced me to top_n(), thanks. However, I ran into an issue where some IDs have multiple instances because the id had a the same code with risk listed more than once. How can I select out just one instance after the last line in the tidyverse method code?

    – Tim Feeney
    Nov 26 '18 at 0:21











  • See the update. A slice(1) should work. (and I forgot an ungroup() at the end which you should def add.

    – hrbrmstr
    Nov 26 '18 at 0:22











  • worked great, thanks!

    – Tim Feeney
    Nov 26 '18 at 0:26

















This works great and introduced me to top_n(), thanks. However, I ran into an issue where some IDs have multiple instances because the id had a the same code with risk listed more than once. How can I select out just one instance after the last line in the tidyverse method code?

– Tim Feeney
Nov 26 '18 at 0:21





This works great and introduced me to top_n(), thanks. However, I ran into an issue where some IDs have multiple instances because the id had a the same code with risk listed more than once. How can I select out just one instance after the last line in the tidyverse method code?

– Tim Feeney
Nov 26 '18 at 0:21













See the update. A slice(1) should work. (and I forgot an ungroup() at the end which you should def add.

– hrbrmstr
Nov 26 '18 at 0:22





See the update. A slice(1) should work. (and I forgot an ungroup() at the end which you should def add.

– hrbrmstr
Nov 26 '18 at 0:22













worked great, thanks!

– Tim Feeney
Nov 26 '18 at 0:26





worked great, thanks!

– Tim Feeney
Nov 26 '18 at 0:26




















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