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How to generate the permutation of x values in y positions with fixed row sums in R?
To generate a permutation of x values in y positions, we can use expand.grid function. For example, if we want to generate three columns for the range of values 0 to 5 then it can be done in R by using the below command −
x<−expand.grid(rep(list(0:5),3));x[rowSums(x)==3,]
Output
Var1 Var2 Var3 4 3 0 0 9 2 1 0 14 1 2 0 19 0 3 0 39 2 0 1 44 1 1 1 49 0 2 1 74 1 0 2 79 0 1 2 109 0 0 3
Examples
x<−expand.grid(rep(list(0:2),3));x[rowSums(x)==2,]
Output
Var1 Var2 Var3 3 2 0 0 5 1 1 0 7 0 2 0 11 1 0 1 13 0 1 1 19 0 0 2
Example
x<−expand.grid(rep(list(0:2),5));x[rowSums(x)==2,]
Output
Var1 Var2 Var3 Var4 Var5 3 2 0 0 0 0 5 1 1 0 0 0 7 0 2 0 0 0 11 1 0 1 0 0 13 0 1 1 0 0 19 0 0 2 0 0 29 1 0 0 1 0 31 0 1 0 1 0 37 0 0 1 1 0 55 0 0 0 2 0 83 1 0 0 0 1 85 0 1 0 0 1 91 0 0 1 0 1 109 0 0 0 1 1 163 0 0 0 0 2
Example
x<−expand.grid(rep(list(0:5),2));x[rowSums(x)==5,]
Output
Var1 Var2 6 5 0 11 4 1 16 3 2 21 2 3 26 1 4 31 0 5
Example
x<−expand.grid(rep(list(0:5),3));x[rowSums(x)==5,]
Example
Var1 Var2 Var3 6 5 0 0 11 4 1 0 16 3 2 0 21 2 3 0 26 1 4 0 31 0 5 0 41 4 0 1 46 3 1 1 51 2 2 1 56 1 3 1 61 0 4 1 76 3 0 2 81 2 1 2 86 1 2 2 91 0 3 2 111 2 0 3 116 1 1 3 121 0 2 3 146 1 0 4 151 0 1 4 181 0 0 5
Example
x<−expand.grid(rep(list(0:3),3));x[rowSums(x)==3,]
Output
Var1 Var2 Var3 4 3 0 0 7 2 1 0 10 1 2 0 13 0 3 0 19 2 0 1 22 1 1 1 25 0 2 1 34 1 0 2 37 0 1 2 49 0 0 3
Example
x<−expand.grid(rep(list(0:3),4));x[rowSums(x)==3,]
Output
Var1 Var2 Var3 Var4 4 3 0 0 0 7 2 1 0 0 10 1 2 0 0 13 0 3 0 0 19 2 0 1 0 22 1 1 1 0 25 0 2 1 0 34 1 0 2 0 37 0 1 2 0 49 0 0 3 0 67 2 0 0 1 70 1 1 0 1 73 0 2 0 1 82 1 0 1 1 85 0 1 1 1 97 0 0 2 1 130 1 0 0 2 133 0 1 0 2 145 0 0 1 2 193 0 0 0 3
Example
x<−expand.grid(rep(list(0:3),2));x[rowSums(x)==3,]
Output
Var1 Var2 4 3 0 7 2 1 10 1 2 13 0 3
Example
x<−expand.grid(rep(list(0:1),5));x[rowSums(x)==1,]
Output
Var1 Var2 Var3 Var4 Var5 2 1 0 0 0 0 3 0 1 0 0 0 5 0 0 1 0 0 9 0 0 0 1 0 17 0 0 0 0 1
Updated on: 07-Nov-2020
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