Randomly insert NAs into dataframe proportionaly

Same result, using binomial distribution:

dd=dim(df)
nna=20/100 #overall
df1<-df
df1[matrix(rbinom(prod(dd), size=1,prob=nna)==1,nrow=dd[1])]<-NA
df1

df <- data.frame(A = 1:10, B = 11:20, c = 21:30)
head(df)
##   A  B  c
## 1 1 11 21
## 2 2 12 22
## 3 3 13 23
## 4 4 14 24
## 5 5 15 25
## 6 6 16 26

as.data.frame(lapply(df, function(cc) cc[ sample(c(TRUE, NA), prob = c(0.85, 0.15), size = length(cc), replace = TRUE) ]))
##     A  B  c
## 1   1 11 21
## 2   2 12 22
## 3   3 13 23
## 4   4 14 24
## 5   5 NA 25
## 6   6 16 26
## 7  NA 17 27
## 8   8 18 28
## 9   9 19 29
## 10 10 20 30

It's a random process, so it might not give 15% every time.

If you are in the mood to use purrr instead of lapply, you can also do it like this:

> library(purrr)
> df <- data.frame(A = 1:10, B = 11:20, C = 21:30)
> df
    A  B  C
1   1 11 21
2   2 12 22
3   3 13 23
4   4 14 24
5   5 15 25
6   6 16 26
7   7 17 27
8   8 18 28
9   9 19 29
10 10 20 30
> map_df(df, function(x) {x[sample(c(TRUE, NA), prob = c(0.8, 0.2), size = length(x), replace = TRUE)]})
# A tibble: 10 x 3
       A     B     C
   <int> <int> <int>
1      1    11    21
2      2    12    22
3     NA    13    NA
4      4    14    NA
5      5    15    25
6      6    16    26
7      7    17    27
8      8    NA    28
9      9    19    29
10    10    20    30

You can unlist the data.frame and then take a random sample, then put back in a data.frame.

df <- unlist(df)
n <- length(df) * 0.15
df[sample(df, n)] <- NA
as.data.frame(matrix(df, ncol=3))

It can be done a bunch of different ways using sample().