# Another solution approach to find if trailing zeros exist for each row id

The code below removes trailing zeros exist for each row id.

This question has been resolved here https://stackoverflow.com/questions/69452882/delete-column-depending-on-the-date-and-code-you-choose. However, I wouldn't want to use a long pipeline as done, using pivot_longer. Is there any reasonable way to do this? The code that does this is mat1

library(dplyr)
library(tidyverse)
library(lubridate)

df1 <- structure(
list(date1= c("2021-06-28","2021-06-28","2021-06-28","2021-06-28","2021-06-28",
"2021-06-28","2021-06-28","2021-06-28"),
date2 = c("2021-06-30","2021-06-30","2021-07-02","2021-07-07","2021-07-07","2021-07-09","2021-07-09","2021-07-09"),
Code = c("FDE","ABC","ABC","ABC","CDE","FGE","ABC","CDE"),
Week= c("Wednesday","Wednesday","Friday","Wednesday","Wednesday","Friday","Friday","Friday"),
DR1 = c(4,1,4,3,3,4,3,5),
DR01 = c(4,1,4,3,3,4,3,6), DR02= c(4,2,6,7,3,2,7,4),DR03= c(9,5,4,3,3,2,1,5),
DR04 = c(5,4,3,3,6,2,1,9),DR05 = c(5,4,5,3,6,2,1,9),
DR06 = c(2,4,3,3,5,6,7,8),DR07 = c(2,5,4,4,9,4,7,8),
DR08 = c(0,0,0,1,2,0,0,0),DR09 = c(0,0,0,0,0,0,0,0),DR010 = c(0,0,0,0,0,0,0,0),DR011 = c(4,0,0,0,0,0,0,0),
DR012 = c(0,0,0,3,0,0,0,5),DR013 = c(0,0,1,0,0,0,2,0),DR014 = c(0,0,0,0,0,2,0,0)),
class = "data.frame", row.names = c(NA, -8L))

> df1
date1      date2 Code      Week DR1 DR01 DR02 DR03 DR04 DR05 DR06 DR07 DR08 DR09 DR010 DR011 DR012 DR013 DR014
1 2021-06-28 2021-06-30  FDE Wednesday   4    4    4    9    5    5    2    2    0    0     0     4     0     0     0
2 2021-06-28 2021-06-30  ABC Wednesday   1    1    2    5    4    4    4    5    0    0     0     0     0     0     0
3 2021-06-28 2021-07-02  ABC    Friday   4    4    6    4    3    5    3    4    0    0     0     0     0     1     0
4 2021-06-28 2021-07-07  ABC Wednesday   3    3    7    3    3    3    3    4    1    0     0     0     3     0     0
5 2021-06-28 2021-07-07  CDE Wednesday   3    3    3    3    6    6    5    9    2    0     0     0     0     0     0
6 2021-06-28 2021-07-09  FGE    Friday   4    4    2    2    2    2    6    4    0    0     0     0     0     0     2
7 2021-06-28 2021-07-09  ABC    Friday   3    3    7    1    1    1    7    7    0    0     0     0     0     2     0
8 2021-06-28 2021-07-09  CDE    Friday   5    6    4    5    9    9    8    8    0    0     0     0     5     0     0

x<-df1 %>% select(starts_with("DR0"))

x<-cbind(df1, setNames(df1$DR1 - x, paste0(names(x), "_PV"))) PV<-select(x, date2,Week, Code, DR1, ends_with("PV")) med<-PV %>% group_by(Code,Week) %>% summarize(across(ends_with("PV"), median)) SPV<-df1%>% inner_join(med, by = c('Code', 'Week')) %>% mutate(across(matches("^DR0\\d+$"), ~.x +
get(paste0(cur_column(), '_PV')),
.names = '{col}_{col}_PV')) %>%
select(date1:Code, DR01_DR01_PV:last_col())

dmda<-"2021-07-07"
CodeChosse<-"CDE"

mat1 <- df1 %>%
filter(date2 == dmda, Code == CodeChosse) %>%
select(starts_with("DR0")) %>%
pivot_longer(cols = everything()) %>%
arrange(desc(row_number())) %>%
mutate(cs = cumsum(value)) %>%
filter(cs == 0) %>%
pull(name)

(dropnames <- paste0(mat1,"_",mat1, "_PV"))

SPV %>%
filter(date2 == dmda, Code == CodeChosse) %>%
select(-dropnames)

date1      date2 Code DR01_DR01_PV DR02_DR02_PV DR03_DR03_PV DR04_DR04_PV DR05_DR05_PV DR06_DR06_PV DR07_DR07_PV
1 2021-06-28 2021-07-07  CDE            3            3            3            3            3            3            3
DR08_DR08_PV
1            3