How to replace NA in multiple columns with value from corresponding columns
Here's a "tidyverse" approach. Note that as @Gregor commented, it helps to tidy your data. The following handles this for you and also returns a somewhat tidy data frame. I'll leave it to you to get back into the original format if necessary.
Note that I've used the mutate_cond() function that can be found here.
library(tidyverse)
library(stringr)
myDF %>%
gather(key = col, value = val, -dates, -names) %>%
mutate(col = str_replace(col, "var", "")) %>%
separate(col, into = c("var", "group")) %>%
spread(key = group, value = val) %>%
mutate_cond(is.na(person), person = group)
#> Source: local data frame [20 x 5]
#> Groups: names [2]
#>
#> dates names var group person
#> * <date> <fctr> <chr> <dbl> <dbl>
#> 1 2016-01-01 John Doe 1 NA NA
#> 2 2016-01-01 John Doe 2 NA NA
#> 3 2016-01-02 John Doe 1 0.28757752 0.8895393
#> 4 2016-01-02 John Doe 2 0.95683335 0.9630242
#> 5 2016-01-03 John Doe 1 0.78830514 0.6928034
#> 6 2016-01-03 John Doe 2 0.45333416 0.9022990
#> 7 2016-01-04 John Doe 1 0.40897692 0.6405068
#> 8 2016-01-04 John Doe 2 0.67757064 0.6907053
#> 9 2016-01-05 John Doe 1 0.88301740 0.9942698
#> 10 2016-01-05 John Doe 2 0.57263340 0.7954674
#> 11 2016-01-06 Jane Smith 1 0.94046728 0.9404673
#> 12 2016-01-06 Jane Smith 2 0.10292468 0.1029247
#> 13 2016-01-07 Jane Smith 1 0.04555650 0.7085305
#> 14 2016-01-07 Jane Smith 2 0.89982497 0.4777960
#> 15 2016-01-08 Jane Smith 1 0.52810549 0.5440660
#> 16 2016-01-08 Jane Smith 2 0.24608773 0.7584595
#> 17 2016-01-09 Jane Smith 1 0.89241904 0.5941420
#> 18 2016-01-09 Jane Smith 2 0.04205953 0.2164079
#> 19 2016-01-10 Jane Smith 1 0.55143501 0.2891597
#> 20 2016-01-10 Jane Smith 2 0.32792072 0.3181810
Everything but the last line is about tidying the data. The last line (mutate_cond()) handles the substitution of NA values. If your columns are all named this way, then this should extend to any n.
Here's an idea:
tag <- c("person", "group")
# Create a list of 2 elements, persons and groups.
lst <- lapply(tag, function(x) { myDF[grepl(x, colnames(myDF))] })
# Extract everything before the underscore "_" in the column names
ext <- lapply(seq_along(lst), function(x) {
stringi::stri_extract(colnames(lst[[x]]), regex = "^[^_]+(?=_)") })
# Find the common elements between the two
int <- intersect(ext[[1]], ext[[2]])
# Create a new list with only the matching subset
match_list <- lapply(lst, function(x) { select(x, matches(paste(int, collapse = "|"))) })
# Replace all NA values in 'person' by the corresponding values in 'group'
res <- mapply(function(x, y) { replace(x, is.na(x), y[is.na(x)]) },
match_list[[1]], match_list[[2]])
# Assign the result back to the original data.frame
myDF[, colnames(res)] <- res
This should ignore non-matching person/group pairs and only replace on matching vars
Here is one option using set from data.table which does the replacement in place
library(data.table)
#convert the data.frame to data.table
setDT(myDF)
#get the column name of 'group' and 'person' columns
nm1 <- grep("group", names(myDF), value = TRUE)
nm2 <- grep("person", names(myDF), value = TRUE)
#loop through the sequence of 'nm1'
for(j in seq_along(nm1)){
#set the elements in the row that are NA for each 'period' column
#with the corresponding row from 'group' column specified in the "value"
set(myDF, i = which(is.na(myDF[[nm2[j]]])), j = nm2[j],
value = myDF[[nm1[j]]][is.na(myDF[[nm2[j]]])])
}
myDF
# dates names var1_group var2_group var1_person var2_person
#1: 2016-01-01 John Doe NA NA NA NA
#2: 2016-01-02 John Doe 0.2875775 0.95683335 0.8895393 0.9630242
#3: 2016-01-03 John Doe 0.7883051 0.45333416 0.6928034 0.9022990
#4: 2016-01-04 John Doe 0.4089769 0.67757064 0.6405068 0.6907053
#5: 2016-01-05 John Doe 0.8830174 0.57263340 0.9942698 0.7954674
#6: 2016-01-06 Jane Smith 0.9404673 0.10292468 0.9404673 0.1029247
#7: 2016-01-07 Jane Smith 0.0455565 0.89982497 0.7085305 0.4777960
#8: 2016-01-08 Jane Smith 0.5281055 0.24608773 0.5440660 0.7584595
#9: 2016-01-09 Jane Smith 0.8924190 0.04205953 0.5941420 0.2164079
#10:2016-01-10 Jane Smith 0.5514350 0.32792072 0.2891597 0.3181810