How to replace NA in multiple columns with value from corresponding columns

Question

I'm trying to clean up a data frame and I would like to replace NAs in one column, with a corresponding value from another column. I would also like to do this for multiple columns at once.

Example Data Frame.

set.seed(123) 

dates <- seq(as.Date("2016-01-01"), by = "day", length = 10)  
names <- rep(c("John Doe", "Jane Smith"), each = 5)  
var1_group <- runif(10)  
var2_group <- runif(10)  
var1_person <- runif(10)  
var2_person <- runif(10)  

myDF <- data.frame(names, var1_group, var2_group, var1_person, var2_person)  
myDF <- cbind(dates, myDF)  

After some manipulations using dplyr...

myDF <- myDF %>% mutate_each(funs(lag), contains("group"))  
myDF <- myDF %>% group_by(names) %>% mutate_each(funs(lag), contains("person"))  

I get a bunch of NAs...

        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          NA          NA  
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  

What I would like to do now, is replace the NAs from *_person columns with the corresponding value from the *_group column. (See row 6)

        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   0.1029246     
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 

This works for one column...

myDF$var1_person <- ifelse(is.na(myDF$var1_person), myDF$var1_group, myDF$var1_person)  

But I'd like to do it for all the columns at once. In my actual data frame, each group is around 20 columns. I've tried a bunch of other stuff, but I don't want to clutter this post with my nonsense.

*Bonus points if you can get the code to match for n variables based on column prefix.

var1_group > var1_person  
var2_group > var2_person
...
varn_group > varn_person   

Answer 1

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

Answer 2

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.

Answer 3

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