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i have a data frame with different variables. i have merged three columns of probabilities to my data frame. my question is how can i use these columns as probabilities in sample function so that for prob argument take each column as probability. for example for y= 1 take ncol (a) , for y=1 take ncol(b) and so on my codes are:

    a    b    c    y
1  0.090 0.12 0.10 1
2  0.015 0.13 0.09 1
3  0.034 0.20 0.34 1
4  0.440 0.44 0.70 1
5  0.090 0.12 0.10 2
6  0.015 0.13 0.09 2

mydata$mig<- sample( 1:3, size = 7, replace = TRUE, prob= ????)

any help would be appreciated

Chriss Paul
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mehmo
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  • Hi mehmo! Can you use `dput(mydata)` and copy paste the output into your question to make it easy to reproduce (for more detail: https://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example). If you can clarify by giving example output for your hypothetical function as well, that would be great. – verybadatthis Mar 30 '21 at 01:37

2 Answers2

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Using apply function per rows:

df <- read.table(header = TRUE, text="a    b    c    y
1  0.090 0.12 0.10 1
2  0.015 0.13 0.09 1
3  0.034 0.20 0.34 1
4  0.440 0.44 0.70 1
5  0.090 0.12 0.10 2
6  0.015 0.13 0.09 2")
set.seed(12344)
samples1<- apply(X = df[,-4], MARGIN = 1, # MARGIN = 1 indicates you are applying FUN per rows 
             FUN = function(x) sample( 1:3, 
                                 size = 7,
                                 replace= TRUE ,
                                 prob = x))
#You obtain six columns from samples with prob parameter in df's rows
samples1
     1 2 3 4 5 6
[1,] 2 3 3 1 3 2
[2,] 1 2 3 3 2 2
[3,] 2 3 3 3 1 3
[4,] 2 3 3 1 3 2
[5,] 2 2 3 2 3 2
[6,] 1 3 2 3 2 3
[7,] 3 3 3 2 1 2

Update: Given your comment on my answer, I update and propose a new solution using data.table. I leave the previous version for reference if there will be anyone interested.

library(data.table)
setDT(df)
set.seed(78787)
#Column V1 has your 7 samples per group y, with probs taken at random from a,b,c
df[, sample(1:.N,
            size = 7,
            replace = TRUE,
            prob = unlist(.SD)),
   by = y,
   .SDcols = sample(names(df)[-ncol(df)], 1)]

    y V1
 1: 1  4
 2: 1  3
 3: 1  4
 4: 1  3
 5: 1  4
 6: 1  4
 7: 1  4
 8: 2  2
 9: 2  1
10: 2  1
11: 2  1
12: 2  2
13: 2  1
14: 2  1
Chriss Paul
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  • i really appreciate for your help but in my sample function i want column "y" only take one of the a,b and c column. more clearly, if y=1 randomly take a,b or c column. – mehmo Mar 30 '21 at 14:01
  • I see I updated my answer considering your comment. Nevertheless, in your example when `y=1` you have in column `a` 4 numbers that will be your `prob` in `sample`, indeed `0.09, 0.015, 0.034 0.444`, but you have as `1:3` as the vector from which samples are drown. The `probs` vector should be same length as this vector. This causes confusion, and it' something @r2evans has pointed out too. Consider correct this issue. Also, you don't mention in your question about randomly select a, b or c ( included in my new updated answer) – Chriss Paul Mar 30 '21 at 17:37
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While the "normal" indexing of 2d matrices and frames is the [i,j] method, one can also provide a 2-column matrix to i alone to programmatically combine the rows and columns. We can use this to create a matrix whose first column is merely counting the rows (1:6 here), and the second column is taken directly from your y column:

cbind(seq_len(nrow(mydata)), mydata$y)
#      [,1] [,2]
# [1,]    1    1
# [2,]    2    1
# [3,]    3    1
# [4,]    4    1
# [5,]    5    2
# [6,]    6    2
mydata[cbind(seq_len(nrow(mydata)), mydata$y)]
# [1] 0.090 0.015 0.034 0.440 0.120 0.130

Note that in this case, your sample-ing code is not going to work:

  • true --> TRUE
  • the length of derived probabilities is not the same length as your 1:3
r2evans
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