Apply na.approx on row and output dataframe in R

Question

Given the following dataframe:

df1 <- data.frame(Company = c('A','B','C','D','E'), `X1980` = c(NA, 5, 3, 8, 13),
                                `X1981` = c(20, NA, 23, 11, 29),
                                `X1982` = c(NA, 32, NA, 41, 42),
                                `X1983` = c(45, 47, 53, 58, NA))

I would like to replace the NA's with values through interpolation over the rows resulting in the following data frame:

Company  1980  1981  1982  1983
   A      NA    20   32,5   45
   B      5    18,5   32    47
   C      3     23    38    53
   D      8     11    41    58
   E      13    29    42    NA

I tried using na.apporox in combination with apply:

df1[-1] <- t(apply(df1[-1], 1, FUN = na.approx))

But this results in the following error:

Fehler in h(simpleError(msg, call)) : 
  Fehler bei der Auswertung des Argumentes 'x' bei der Methodenauswahl für Funktion 't': dim(X) muss positive Länge haben

Thanks in advance for any help!!!

EDIT: I forgot to define how to treat the NA's in the na.approx function.

df1[-1] <- t(apply(df1[-1], 1, na.approx, na.rm=FALSE))

This results in the desired output!

Read about the `na.rm` argument in `?approx` – markus Nov 13 '20 at 12:31 — markus, Nov 13 '20 at 12:31
Thanks markus! I overlooked that part in the function. – Connor Uhl Nov 13 '20 at 12:53 — Connor Uhl, Nov 13 '20 at 12:53

Ricardo Semião e Castro · Accepted Answer · 2020-11-13T12:57:21.873

This interpolation methods, from what i know, don't deal with NA's at the end of the sequence, which is the case of entry Ex1983. So, na.approx (at least the one from zoo package) gives an optional argument na.rm for you to select if you want to remove those NA's or not.

By default it is TRUE, so it removes, giving your matrix a 4th column that only has 3 elements, not 4 (as the rest of the columns), so it produces that error. To deal with that you simply set that argument to FALSE:

df1[-1] <- t(apply(df1[-1], 1, FUN = zoo::na.approx, na.rm=FALSE))

Output:

> df1
  Company X1980 X1981 X1982 X1983
1       A    NA    20    NA    45
2       B     5    NA    32    47
3       C     3    23    NA    53
4       D     8    11    41    58
5       E    13    29    42    NA

To fill that NA, i can help you use the answers from here, if you need me to.

Apply na.approx on row and output dataframe in R

1 Answers1