Reputation: 307
Run length sequence by time and ID
This problem does not seem to have been put out here before.
I want to find the number of subjects that score 1 for 6 consecutive hours. The subjects have not been scored every hour so if an hour is missing the hours are not consecutive and the output for that 6-hour period should be NA. The reason for assigning NA would be that we do not know how the subject has scored on the missing hour. This problem can be used to count consecutive hits, but only count it if a subject has participated.
My dataframe looks like this:
ID<-c(1,1,1,1,1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,2)
hour<-c(1,2,3,7,8,9,10,11,12,17,18,19,1,2,3,4,5,6,8,9,15)
A<-c(0,1,0,1,1,1,1,1,1,0,0,0,1,1,1,1,1,1,1,1,1)
df<-data.frame(ID,hour,A)
I have tried to use the rle function (I am sure its possible) but I cannot get it to condition on both hour and ID. The output would be like this:
ID<-c(1,1,1,1,1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,2)
hour<-c(1,2,3,7,8,9,10,11,12,17,18,19,1,2,3,4,5,6,8,9,15)
A<-c(0,1,0,1,1,1,1,1,1,0,0,0,1,1,1,1,1,1,1,1,1)
six<-c(NA,NA,NA,0,0,0,0,0,1,NA,NA,NA,0,0,0,0,0,1,NA,NA,NA)
df<-data.frame(ID,hour,A,six)
Thank you in advance.
I believe the original data set I gave was too small to make the solutions more generalizable.
I just tried the codes with this dataset and found that this will result in a wrong result.
ID<-c(1,1,1,1,1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,2,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,4,4,4,4,4,4,4,4)
hour<-c(1,2,3,7,8,9,10,12,13,17,18,19,1,2,3,4,5,6,8,9,15,1:23,27,28,29,30,31)
A<-c(0,1,0,1,1,1,1,1,1,0,0,0,1,1,1,1,1,1,1,1,1,rep(1,28))
df<-data.frame(ID,hour,A)
For the new dataset the output should be:
ID<-c(1,1,1,1,1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,2,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,4,4,4,4,4,4,4,4)
hour<-c(1,2,3,7,8,9,10,12,13,17,18,19,1,2,3,4,5,6,8,9,15,1:23,27,28,29,30,31)
A<-c(0,1,0,1,1,1,1,1,1,0,0,0,1,1,1,1,1,1,1,1,1,rep(1,28))
six<-c(NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,0,0,0,0,0,1,NA,NA,NA,0,0,0,0,0,1,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA)
df<-data.frame(ID,hour,A,six)
Upvotes: 8
Views: 348
Answers (3)
Reputation: 19726
Here is an approach in tidyverse with the updated data set:
library(tidyverse)
df %>%
group_by(ID) %>%
expand(hour = seq(min(hour), max(hour))) %>%
left_join(df) %>%
mutate(rle = rep(rle(A)$lengths, times = rle(A)$lengths)) %>%
group_by(ID, rle) %>%
mutate(sum = cumsum(A),
six = ifelse(rle >= 6 & A == 1, 0, NA),
six = ifelse(sum == 6, 1, ifelse(sum > 6, NA, six))) %>%
filter(!is.na(A)) %>%
ungroup() %>%
select(ID, hour, A, six) %>%
as.data.frame() -> df_out2
check requested output:
ID<-c(1,1,1,1,1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,2,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,4,4,4,4,4,4,4,4)
hour<-c(1,2,3,7,8,9,10,12,13,17,18,19,1,2,3,4,5,6,8,9,15,1:23,27,28,29,30,31)
A<-c(0,1,0,1,1,1,1,1,1,0,0,0,1,1,1,1,1,1,1,1,1,rep(1,28))
six<-c(NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,0,0,0,0,0,1,NA,NA,NA,0,0,0,0,0,1,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA)
df<-data.frame(ID,hour,A,six)
all.equal(df, df_out2)
#output
TRUE
The old answer:
df %>%
mutate(rle = rep(rle(A)$lengths, times = rle(A)$lengths)) %>%
group_by(ID, rle) %>%
mutate(sum = cumsum(A),
six = ifelse(rle >= 6 & A == 1, 0, NA),
six = ifelse(sum == 6, 1, ifelse(sum > 6, NA, six))) %>%
ungroup() %>%
select(ID, hour, A, six) %>%
as.data.frame() -> df_out2
Lets check if the result is like requested:
ID <- c(1,1,1,1,1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,2)
hour <- c(1,2,3,7,8,9,10,11,12,17,18,19,1,2,3,4,5,6,8,9,15)
A <- c(0,1,0,1,1,1,1,1,1,0,0,0,1,1,1,1,1,1,1,1,1)
six <- c(NA,NA,NA,0,0,0,0,0,1,NA,NA,NA,0,0,0,0,0,1,NA,NA,NA)
df1 <- data.frame(ID, hour, A, six)
df1 is requested output
all.equal(df1, df_out2)
#output
TRUE
some benchmarking:
library(microbenchmark)
library(data.table)
akrun <- function(df){
setDT(df)[, grp := rleid(A)][, Anew := A *((hour - shift(hour, fill = hour[1])) ==1), grp
][, sixnew :=if(sum(A)>=5) rep(c(0, 1), c(.N-1, 1)) else NA_real_,.(rleid(Anew), grp)]
i1 <- df[, .I[which(is.na(sixnew) & shift(sixnew == 0, type = 'lead'))], grp]$V1
df[i1, sixnew := 0][, c("Anew", "grp") := NULL][]
}
missuse <- function(df){
df %>%
mutate(rle = rep(rle(A)$lengths, times = rle(A)$lengths)) %>%
group_by(ID, rle) %>%
mutate(sum = cumsum(A),
six = ifelse(rle >= 6 & A == 1, 0, NA),
six = ifelse(sum == 6, 1, ifelse(sum > 6, NA, six))) %>%
ungroup() %>%
select(ID, hour, A, six)
}
Mike <- function(df){
ave(df$A,
cumsum(!(df$hour == shift(df$hour, fill = 0) + 1)),
FUN = function(x) {
if(all(x==1) & length(x) >= 6) return(c(rep(0, length(x) - 1), 1))
else return(rep(NA, length(x)))})
}
microbenchmark(Mike(df),
akrun(df),
missuse(df))
#output
Unit: microseconds
expr min lq mean median uq max neval
Mike(df) 491.291 575.7115 704.2213 597.7155 629.0295 9578.684 100
akrun(df) 6568.313 6725.5175 7867.4059 6843.5790 7279.2240 69790.755 100
missuse(df) 11042.822 11321.0505 12434.8671 11512.3200 12616.3485 43170.935 100
way to go Mike H.!
Upvotes: 7
Reputation: 14370
To get the groupings you can compare the current hour to the lagged hour to see if they are "consecutive" or 1 integer apart and then take the cumsum of that. Once you have the groupings you can use a simple ave to get the output you want.
library(data.table)
df$six <- ave(df$A,
cumsum(!(df$hour == shift(df$hour, fill = 0) + 1)),
FUN = function(x) {
if(all(x==1) & length(x) >= 6) return(c(rep(0, length(x) - 1), 1))
else return(rep(NA, length(x)))}
)
df
# ID hour A six
#1 1 1 0 NA
#2 1 2 1 NA
#3 1 3 0 NA
#4 1 7 1 0
#5 1 8 1 0
#6 1 9 1 0
#7 1 10 1 0
#8 1 11 1 0
#9 1 12 1 1
#10 1 17 0 NA
#11 1 18 0 NA
#12 1 19 0 NA
#13 2 1 1 0
#14 2 2 1 0
#15 2 3 1 0
#16 2 4 1 0
#17 2 5 1 0
#18 2 6 1 1
#19 2 8 1 NA
#20 2 9 1 NA
#21 2 15 1 NA
If you only want to select a patient at most once, this will select the last time period:
df$six_adj <- ave(df$six, df$ID, df$six, FUN = function(x) {
if(all(x==1)) return(c(rep(0, length(x) - 1), 1))
else return(x)}
)
Upvotes: 7
Reputation: 887541
We can use rleid from data.table. Create a grouping column with run-length-id of 'A' ('grp'). Take the difference of 'hour' with the next value of 'hour', check if it is equal to 1 and multiply with 'A' to create 'Anew'. Grouped by run-length-id of 'Anew' and 'grp', if the sum of 'A' is greater than a particular value, replicate with 0s and 1s or else return NA. In the last phase, assign some spillover NAs to 0 by creating an index ('i1')
library(data.table)
setDT(df)[, grp := rleid(A)][, Anew := A *((hour - shift(hour, fill = hour[1])) ==1), grp
][, sixnew :=if(sum(A)>=5) rep(c(0, 1), c(.N-1, 1)) else NA_real_,.(rleid(Anew), grp)]
i1 <- df[, .I[which(is.na(sixnew) & shift(sixnew == 0, type = 'lead'))], grp]$V1
df[i1, sixnew := 0][, c("Anew", "grp") := NULL][]
# ID hour A six sixnew
# 1: 1 1 0 NA NA
# 2: 1 2 1 NA NA
# 3: 1 3 0 NA NA
# 4: 1 7 1 0 0
# 5: 1 8 1 0 0
# 6: 1 9 1 0 0
# 7: 1 10 1 0 0
# 8: 1 11 1 0 0
# 9: 1 12 1 1 1
#10: 1 17 0 NA NA
#11: 1 18 0 NA NA
#12: 1 19 0 NA NA
#13: 2 1 1 0 0
#14: 2 2 1 0 0
#15: 2 3 1 0 0
#16: 2 4 1 0 0
#17: 2 5 1 0 0
#18: 2 6 1 1 1
#19: 2 8 1 NA NA
#20: 2 9 1 NA NA
#21: 2 15 1 NA NA
Or slightly more compact option would be
i1 <- setDT(df)[, if(sum(A)>= 5) .I[.N] , rleid(c(TRUE, diff(hour)==1), A)]$V1
df[i1, sixnew := 1][unlist(Map(`:`, i1-5, i1-1)), sixnew := 0]
df
# ID hour A six sixnew
# 1: 1 1 0 NA NA
# 2: 1 2 1 NA NA
# 3: 1 3 0 NA NA
# 4: 1 7 1 0 0
# 5: 1 8 1 0 0
# 6: 1 9 1 0 0
# 7: 1 10 1 0 0
# 8: 1 11 1 0 0
# 9: 1 12 1 1 1
#10: 1 17 0 NA NA
#11: 1 18 0 NA NA
#12: 1 19 0 NA NA
#13: 2 1 1 0 0
#14: 2 2 1 0 0
#15: 2 3 1 0 0
#16: 2 4 1 0 0
#17: 2 5 1 0 0
#18: 2 6 1 1 1
#19: 2 8 1 NA NA
#20: 2 9 1 NA NA
#21: 2 15 1 NA NA
Benchmarks
Created a slightly bigger dataset and tested the solutions
-data
ID<-c(1,1,1,1,1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,2)
hour<-c(1,2,3,7,8,9,10,11,12,17,18,19,1,2,3,4,5,6,8,9,15)
A<-c(0,1,0,1,1,1,1,1,1,0,0,0,1,1,1,1,1,1,1,1,1)
ID <- rep(1:5000, rep(c(12, 9), 2500))
A <- rep(A, 2500)
hour <- rep(hour, 2500)
dftest <- data.frame(ID, hour, A)
-functions
akrun <- function(df){
df1 <- copy(df)
setDT(df1)[, grp := rleid(A)][, Anew := A *((hour - shift(hour, fill = hour[1])) ==1), grp
][, sixnew :=if(sum(A)>=5) rep(c(0, 1), c(.N-1, 1))
else NA_real_,.(rleid(Anew), grp)]
i1 <- df1[, .I[which(is.na(sixnew) & shift(sixnew == 0, type = 'lead'))], grp]$V1
df1[i1, sixnew := 0][, c("Anew", "grp") := NULL][]
}
akrun2 <- function(df) {
df1 <- copy(df)
i1 <- setDT(df1)[, if(sum(A)>= 5) .I[.N] , rleid(c(TRUE, diff(hour)==1), A)]$V1
df1[i1, sixnew := 1][unlist(Map(`:`, i1-5, i1-1)), sixnew := 0]
}
missuse <- function(df){
df %>%
mutate(rle = rep(rle(A)$lengths, times = rle(A)$lengths)) %>%
group_by(ID, rle) %>%
mutate(sum = cumsum(A),
six = ifelse(rle >= 6 & A == 1, 0, NA),
six = ifelse(sum == 6, 1, ifelse(sum > 6, NA, six))) %>%
ungroup() %>%
select(ID, hour, A, six)
}
Mike <- function(df){
ave(df$A,
cumsum(!(df$hour == shift(df$hour, fill = 0) + 1)),
FUN = function(x) {
if(all(x==1) & length(x) >= 6) return(c(rep(0, length(x) - 1), 1))
else return(rep(NA, length(x)))})
}
-benchmark
microbenchmark(Mike(dftest),
akrun(dftest),
akrun2(dftest),
missuse(dftest),
times = 10L, unit = 'relative')
-output
#Unit: relative
# expr min lq mean median uq max neval cld
# Mike(dftest) 1.682794 1.754494 1.673811 1.68806 1.632765 1.640221 10 a
# akrun(dftest) 13.159245 12.950117 12.176965 12.33716 11.856271 11.095228 10 b
# akrun2(dftest) 1.000000 1.000000 1.000000 1.00000 1.000000 1.000000 10 a
# missuse(dftest) 37.899905 36.773837 34.726845 34.87672 33.155939 30.665840 10 c
Upvotes: 5
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