Filling NA row values with nearest right side row value in R

Question

I want to convert the given dataframe from

             c1     c2   c3   c4    c5
    VEG PUFF <NA>    12  <NA>  <NA> 78.43
CHICKEN PUFF <NA>    16  <NA>  88.24 <NA>
BAKERY Total <NA>   <NA>  28   <NA> 84.04

to

             c1     c2  
    VEG PUFF 12     78.43   
CHICKEN PUFF 16     88.24    
BAKERY Total 28     84.04

I tried two methods but i didnt get accurate results it is sometimes taking left side row value

step1 <-  t(na.locf(t(df), fromLast=T))
step2 <-  t(na.locf(t(step1), fromLast=F))

library(dplyr)
MyReplace = function(data) {data %>% t %>% na.locf(.,,T) %>% na.locf %>% t

Answer 1

Update

As there was lot of confusion on the expected output, updating the answer as suggested by @DavidArenburg using a tidyverse solution

library(dplyr)
library(tidyr)
df %>%
  add_rownames() %>%
  gather(variable, value, -rowname) %>%
  filter(!is.na(value)) %>%
  group_by(rowname) %>%
  mutate(indx = row_number()) %>%
  select(-variable) %>%
  spread(indx, value)

#        rowname   `1`   `2`
#*        <chr> <dbl> <dbl>
#1 BAKERY_Total    28 84.04
#2 CHICKEN_PUFF    16 88.24
#3     VEG_PUFF    12 78.43

---

Another solution could be

library(data.table)
temp <- apply(df, 1, function(x) data.frame(matrix(x[!is.na(x)], nrow = 1)))
rbindlist(temp, fill = T)

---

Previous Answer

If I have understand you correctly, you are trying to replace NA values in a row with the latest non-NA value in the same row

We can use na.locf with fromLast set as TRUE

t(apply(df, 1, function(x) na.locf(x, fromLast = T, na.rm = F)))


#             c1 c2    c3    c4    c5
#VEG_PUFF     12 12 78.43 78.43 78.43
#CHICKEN_PUFF 16 16 88.24 88.24    NA
#BAKERY_Total 28 28 28.00 84.04 84.04

Answer 2

We can use na.omit

t(apply(df, 1, na.omit))
#             [,1]  [,2]
#VEG PUFF       12 78.43
#CHICKEN PUFF   16 88.24
#BAKERY Total   28 84.04

Update

Based on the excel data showed

lst <- apply(df, 1, na.omit)
df2 <- do.call(rbind, lapply(lst, `length<-`, max(lengths(lst))))
row.names(df2) <- row.names(df)

---

Or another option is melt/dcast from data.table

library(data.table)
dcast(melt(setDT(df1, keep.rownames=TRUE), id.var = 'rn', 
         na.rm = TRUE), rn~ paste0("c", rowid(rn)), value.var = "value")
#             rn c1    c2  c3
#1: BAKERY Total 28 84.04  NA
#2: CHICKEN PUFF 16 88.24 143
#3:     VEG PUFF 12 78.43  NA

To provide a reproducible example,

df1 <- structure(list(c1 = c(NA, NA, NA), c2 = c(12L, 16L, NA), c3 = c(NA, 
NA, 28L), c4 = c(NA, 88.24, NA), c5 = c(78.43, 143, 84.04)), .Names = c("c1", 
"c2", "c3", "c4", "c5"), class = "data.frame", row.names = c("VEG PUFF", 
"CHICKEN PUFF", "BAKERY Total"))

lst <- lapply(seq_len(nrow(df1)), function(i) {
               x1 <- unlist(df1[i,])
               x1[complete.cases(x1)]})
df2 <- do.call(rbind, lapply(lst, `length<-`, max(lengths(lst))))
row.names(df2) <- row.names(df1)

The above approach is similar to the apply method except that we can be always sure that this output a list (in the apply - it can vary. When the number of elements are the same after removing the NA, it will output a matrix, in other cases a list). So, we loop over the sequence of rows, remove the NA elements, pad NA at the end to make lengths of list elements same and then rbind

---

Or another option is which with arr.ind=TRUE

ind <- which(!is.na(df), arr.ind=TRUE)
matrix(df[ind[order(ind[,1]),]], ncol=2, byrow=TRUE, 
            dimnames = list(row.names(df), paste0("c", 1:2)))
#             c1    c2
#VEG PUFF     12 78.43
#CHICKEN PUFF 16 88.24
#BAKERY Total 28 84.04