Replace values into previous row with a condition - r

Question

I want to get data where ID column doesn´t start with 00 and append this value of ID column to the end of Description column in previous row.
Then replace the rest of values into after Name column in the previous row. How can I do that with R?
Here is source of dummy data: https://docs.google.com/spreadsheets/d/1SbmaM8hXck-z5nsNfDMbhwijvAGPkPPBgQ_eY4JAMC8/edit?usp=sharing
ID Year Description Name User Factor_1 Factor_2 Factor_3
0011 2016 blue colour AA James Xfac NA NA
is nice XXX XLM Yfac different Yfac NA NA
0024 2017 red colour DD Mark Zfac NA NA
is good YYY STM Lfac unique Zfac NA NA
What I want to have:
ID Year Description Name User Factor_1 Factor_2 Factor_3
0011 2016 blue colour is nice XXX XLM Yfac different Yfac
0024 2017 red colour is good YYY STM Lfac unique Zfac

Answer 1

There's the first part where you want to paste the descriptions together, and there's the part where you want to move your variables as well, as you want "XXX" and "YYY" in your "user" column.
Also, in Viveks answer all wrong lines are pasted with ALL "right" lines, which works in your example, but not if you have a few right lines, and then a wrong one.
Working with booleans (TRUE/FALSE) sometimes works fine, but in this case, I think you want to use an integer index, as that makes it easier to refer to "the previous line". Which gives me code:
rmlines <- which(!substr(df$ID,1,2)=="00")
df$Description[rmlines-1] <- paste(df$Description[rmlines-1], df[rmlines,1], sep=" ")
df[rmlines-1, 4:8] <- df[rmlines, 2:6]
df <- df[-rmlines,]
But there's one more problem to consider: what classes are your columns?
When I tried it out, I treated everything as a character, which means you can move columns around fine. In your data, some may be factors or something else, so you might want to change the classes. I think it's easiest to first change it all to character, and then change it (back) to the final class you want your columns to be.
# To change everything to character:
df <- as.data.frame(lapply(df, as.character), stringsAsFactors = FALSE)
# And to assign the right classes, you need to decide case-by-case:
df$Year <- as.integer(df$Year)
df$Factor_1 <- as.factor(df$Factor1) # Optionally provide levels

Answer 2

Here's a solution with dplyr:
library(dplyr)
df %>%
bind_cols(df %>% rename_all(function(x) paste0(x, "_dummy"))) %>%
mutate(
Description = ifelse(substr(lead(ID), 1, 2) != "00",
paste(Description, lead(ID)), Description),
Name = lead(Year_dummy),
User = lead(Description_dummy),
Factor_1 = lead(Name_dummy),
Factor_2 = lead(User_dummy),
Factor_3 = lead(Factor_1_dummy)
) %>% select(-ends_with("dummy")) %>%
filter(substr(ID, 1, 2) == "00")
Output:
ID Year Description Name User Factor_1 Factor_2 Factor_3
1 0011 2016 blue colour is nice XXX XLM Yfac different Yfac
2 0024 2017 red colour is good YYY STM Lfac unique Zfac
In case you're dealing with a large number of columns, a combination of dplyr and base R could do it:
library(dplyr)
df_combo <- cbind(df, df)
df$Description <- ifelse(substr(lead(df$ID), 1, 2) != "00",
paste(df$Description, lead(df$ID)), df$Description)
for (i in (ncol(df) + 4):ncol(df_combo)) {
df_combo[[i]] <- lead(df_combo[[i - ncol(df) - 2]])
}
df_combo <- subset(df_combo, substr(ID, 1, 2) == "00")
df_descr <- subset(df, substr(ID, 1, 2) == "00")
df_final <- df_combo[, (ncol(df) + 1):ncol(df_combo)]
df_final$Description <- df_descr$Description
rm(df_descr, df_combo)
Output:
ID Year Description Name User Factor_1 Factor_2 Factor_3
1: 0011 2016 blue colour is nice XXX XLM Yfac different Yfac
2: 0024 2017 red colour is good YYY STM Lfac unique Zfac

Answer 3

Use -
bools <- !substr(df$ID,1,2)=="00"
values <- df[bools,1]
df <- df[!bools,]
df$Description <- paste(df[substr(df$ID,1,2)=="00","Description"],values,sep=" ")
df
Output
ID Year Description Name User Factor_1 Factor_2
1 0011 2016 blue colour is nice AA James Xfac NA
3 0024 2017 red colour is good DD Mark Zfac NA
Factor_3
1 NA
3 NA