Using R `outer` with `%in%` operator

I am trying to perform the following outer operation:

x <- c(1, 11)

choices <- list(1:10, 10:20)



outer(x, choices, FUN=`%in%`)

I expect the following matrix:

      [,1]  [,2]

[1,]  TRUE FALSE

[2,] FALSE  TRUE

which would correspond to the following operations:

outer(x, choices, FUN=paste, sep=" %in% ")

     [,1]           [,2]           

[1,] "1 %in% 1:10"  "1 %in% 10:20" 

[2,] "11 %in% 1:10" "11 %in% 10:20"

But for some reason I am getting:

      [,1]  [,2]

[1,] FALSE FALSE

[2,] FALSE FALSE

What is happening?

asked Nov 26 '18 at 12:03

antoine-sac

2,78321342

3

outer only works with binary functions that are vectorized in both arguments. The table argument is not vectorized.

– Roland
Nov 26 '18 at 12:08

1

outer(x, choices, FUN = Vectorize(`%in%`)) works, but is of course only a hidden loop.

– Roland
Nov 26 '18 at 12:12

1

I actually thought outer was a (double) hidden loop, like all the apply functions. I understand better now: outer is basically expand.grid + FUN which only works if FUN is vectorized. Makes sense. Thanks!

– antoine-sac
Nov 26 '18 at 12:49

add a comment |

I am trying to perform the following outer operation:

x <- c(1, 11)

choices <- list(1:10, 10:20)



outer(x, choices, FUN=`%in%`)

I expect the following matrix:

      [,1]  [,2]

[1,]  TRUE FALSE

[2,] FALSE  TRUE

which would correspond to the following operations:

outer(x, choices, FUN=paste, sep=" %in% ")

     [,1]           [,2]           

[1,] "1 %in% 1:10"  "1 %in% 10:20" 

[2,] "11 %in% 1:10" "11 %in% 10:20"

But for some reason I am getting:

      [,1]  [,2]

[1,] FALSE FALSE

[2,] FALSE FALSE

What is happening?

asked Nov 26 '18 at 12:03

antoine-sac

2,78321342

3

outer only works with binary functions that are vectorized in both arguments. The table argument is not vectorized.

– Roland
Nov 26 '18 at 12:08

1

outer(x, choices, FUN = Vectorize(`%in%`)) works, but is of course only a hidden loop.

– Roland
Nov 26 '18 at 12:12

1

I actually thought outer was a (double) hidden loop, like all the apply functions. I understand better now: outer is basically expand.grid + FUN which only works if FUN is vectorized. Makes sense. Thanks!

– antoine-sac
Nov 26 '18 at 12:49

add a comment |

I am trying to perform the following outer operation:

x <- c(1, 11)

choices <- list(1:10, 10:20)



outer(x, choices, FUN=`%in%`)

I expect the following matrix:

      [,1]  [,2]

[1,]  TRUE FALSE

[2,] FALSE  TRUE

which would correspond to the following operations:

outer(x, choices, FUN=paste, sep=" %in% ")

     [,1]           [,2]           

[1,] "1 %in% 1:10"  "1 %in% 10:20" 

[2,] "11 %in% 1:10" "11 %in% 10:20"

But for some reason I am getting:

      [,1]  [,2]

[1,] FALSE FALSE

[2,] FALSE FALSE

What is happening?

asked Nov 26 '18 at 12:03

antoine-sac

2,78321342

I am trying to perform the following outer operation:

x <- c(1, 11)

choices <- list(1:10, 10:20)



outer(x, choices, FUN=`%in%`)

I expect the following matrix:

      [,1]  [,2]

[1,]  TRUE FALSE

[2,] FALSE  TRUE

which would correspond to the following operations:

outer(x, choices, FUN=paste, sep=" %in% ")

     [,1]           [,2]           

[1,] "1 %in% 1:10"  "1 %in% 10:20" 

[2,] "11 %in% 1:10" "11 %in% 10:20"

But for some reason I am getting:

      [,1]  [,2]

[1,] FALSE FALSE

[2,] FALSE FALSE

What is happening?

asked Nov 26 '18 at 12:03

antoine-sac

2,78321342

asked Nov 26 '18 at 12:03

antoine-sac

2,78321342

asked Nov 26 '18 at 12:03

antoine-sac

2,78321342

asked Nov 26 '18 at 12:03

antoine-sac

2,78321342

asked Nov 26 '18 at 12:03

antoine-sac

2,78321342

3

outer only works with binary functions that are vectorized in both arguments. The table argument is not vectorized.

– Roland
Nov 26 '18 at 12:08

1

outer(x, choices, FUN = Vectorize(`%in%`)) works, but is of course only a hidden loop.

– Roland
Nov 26 '18 at 12:12

1

I actually thought outer was a (double) hidden loop, like all the apply functions. I understand better now: outer is basically expand.grid + FUN which only works if FUN is vectorized. Makes sense. Thanks!

– antoine-sac
Nov 26 '18 at 12:49

add a comment |

3

outer only works with binary functions that are vectorized in both arguments. The table argument is not vectorized.

– Roland
Nov 26 '18 at 12:08

1

outer(x, choices, FUN = Vectorize(`%in%`)) works, but is of course only a hidden loop.

– Roland
Nov 26 '18 at 12:12

1

I actually thought outer was a (double) hidden loop, like all the apply functions. I understand better now: outer is basically expand.grid + FUN which only works if FUN is vectorized. Makes sense. Thanks!

– antoine-sac
Nov 26 '18 at 12:49

outer only works with binary functions that are vectorized in both arguments. The table argument is not vectorized.

– Roland
Nov 26 '18 at 12:08

outer(x, choices, FUN = Vectorize(`%in%`)) works, but is of course only a hidden loop.

– Roland
Nov 26 '18 at 12:12

I actually thought outer was a (double) hidden loop, like all the apply functions. I understand better now: outer is basically expand.grid + FUN which only works if FUN is vectorized. Makes sense. Thanks!

– antoine-sac
Nov 26 '18 at 12:49

add a comment |

2 Answers
2

active

oldest

votes

As expressed in the comments, the table argument of match (the function called by %in%) isn't intended to be a list (if it is, it gets coerced to a character). You should use vapply:

vapply(choices,function(y) x %in% y,logical(length(x)))

#      [,1]  [,2]

#[1,]  TRUE FALSE

#[2,] FALSE  TRUE

answered Nov 26 '18 at 12:13

nicola

18.9k21938

add a comment |

Another way that is close to your train of thought, would be to use expand.grid() to create the combinations, and then Map the two columns via %in% function, i.e.

d1 <- expand.grid(x, choices)

matrix(mapply(`%in%`, d1$Var1, d1$Var2), nrow = length(x))

#or you can use Map(`%in%`, ...) in order to keep results in a list

As @nicola suggests, in order to make things better,

d1 <- expand.grid(list(x), choices) 

mapply(%in%, d1$Var1, d1$Var2)

both giving,

      [,1]  [,2]

[1,]  TRUE FALSE

[2,] FALSE  TRUE

edited Nov 26 '18 at 13:13

answered Nov 26 '18 at 12:43

Sotos

31.1k51741

1

To make things a lot better, you should d1 <- expand.grid(list(x), choices) and then just mapply(%in%, d1$Var1, d1$Var2) leads to the correct result, with much less time and RAM usage.

– nicola
Nov 26 '18 at 13:07

@nicola I agree with everything you said. Only reason I posted this is because it is closer to OP's mindset for the specific problem. But yes, very 'RAM' consuming. Thank you for the alternative! Looks much better

– Sotos
Nov 26 '18 at 13:10

add a comment |

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

As expressed in the comments, the table argument of match (the function called by %in%) isn't intended to be a list (if it is, it gets coerced to a character). You should use vapply:

vapply(choices,function(y) x %in% y,logical(length(x)))

#      [,1]  [,2]

#[1,]  TRUE FALSE

#[2,] FALSE  TRUE

answered Nov 26 '18 at 12:13

nicola

18.9k21938

add a comment |

As expressed in the comments, the table argument of match (the function called by %in%) isn't intended to be a list (if it is, it gets coerced to a character). You should use vapply:

vapply(choices,function(y) x %in% y,logical(length(x)))

#      [,1]  [,2]

#[1,]  TRUE FALSE

#[2,] FALSE  TRUE

answered Nov 26 '18 at 12:13

nicola

18.9k21938

add a comment |

As expressed in the comments, the table argument of match (the function called by %in%) isn't intended to be a list (if it is, it gets coerced to a character). You should use vapply:

vapply(choices,function(y) x %in% y,logical(length(x)))

#      [,1]  [,2]

#[1,]  TRUE FALSE

#[2,] FALSE  TRUE

answered Nov 26 '18 at 12:13

nicola

18.9k21938

As expressed in the comments, the table argument of match (the function called by %in%) isn't intended to be a list (if it is, it gets coerced to a character). You should use vapply:

vapply(choices,function(y) x %in% y,logical(length(x)))

#      [,1]  [,2]

#[1,]  TRUE FALSE

#[2,] FALSE  TRUE

answered Nov 26 '18 at 12:13

nicola

18.9k21938

answered Nov 26 '18 at 12:13

nicola

18.9k21938

answered Nov 26 '18 at 12:13

nicola

18.9k21938

answered Nov 26 '18 at 12:13

nicola

18.9k21938

add a comment |

Another way that is close to your train of thought, would be to use expand.grid() to create the combinations, and then Map the two columns via %in% function, i.e.

d1 <- expand.grid(x, choices)

matrix(mapply(`%in%`, d1$Var1, d1$Var2), nrow = length(x))

#or you can use Map(`%in%`, ...) in order to keep results in a list

As @nicola suggests, in order to make things better,

d1 <- expand.grid(list(x), choices) 

mapply(%in%, d1$Var1, d1$Var2)

both giving,

      [,1]  [,2]

[1,]  TRUE FALSE

[2,] FALSE  TRUE

edited Nov 26 '18 at 13:13

answered Nov 26 '18 at 12:43

Sotos

31.1k51741

1

To make things a lot better, you should d1 <- expand.grid(list(x), choices) and then just mapply(%in%, d1$Var1, d1$Var2) leads to the correct result, with much less time and RAM usage.

– nicola
Nov 26 '18 at 13:07

@nicola I agree with everything you said. Only reason I posted this is because it is closer to OP's mindset for the specific problem. But yes, very 'RAM' consuming. Thank you for the alternative! Looks much better

– Sotos
Nov 26 '18 at 13:10

add a comment |

Another way that is close to your train of thought, would be to use expand.grid() to create the combinations, and then Map the two columns via %in% function, i.e.

d1 <- expand.grid(x, choices)

matrix(mapply(`%in%`, d1$Var1, d1$Var2), nrow = length(x))

#or you can use Map(`%in%`, ...) in order to keep results in a list

As @nicola suggests, in order to make things better,

d1 <- expand.grid(list(x), choices) 

mapply(%in%, d1$Var1, d1$Var2)

both giving,

      [,1]  [,2]

[1,]  TRUE FALSE

[2,] FALSE  TRUE

edited Nov 26 '18 at 13:13

answered Nov 26 '18 at 12:43

Sotos

31.1k51741

1

To make things a lot better, you should d1 <- expand.grid(list(x), choices) and then just mapply(%in%, d1$Var1, d1$Var2) leads to the correct result, with much less time and RAM usage.

– nicola
Nov 26 '18 at 13:07

@nicola I agree with everything you said. Only reason I posted this is because it is closer to OP's mindset for the specific problem. But yes, very 'RAM' consuming. Thank you for the alternative! Looks much better

– Sotos
Nov 26 '18 at 13:10

add a comment |

Another way that is close to your train of thought, would be to use expand.grid() to create the combinations, and then Map the two columns via %in% function, i.e.

d1 <- expand.grid(x, choices)

matrix(mapply(`%in%`, d1$Var1, d1$Var2), nrow = length(x))

#or you can use Map(`%in%`, ...) in order to keep results in a list

As @nicola suggests, in order to make things better,

d1 <- expand.grid(list(x), choices) 

mapply(%in%, d1$Var1, d1$Var2)

both giving,

      [,1]  [,2]

[1,]  TRUE FALSE

[2,] FALSE  TRUE

edited Nov 26 '18 at 13:13

answered Nov 26 '18 at 12:43

Sotos

31.1k51741

Another way that is close to your train of thought, would be to use expand.grid() to create the combinations, and then Map the two columns via %in% function, i.e.

d1 <- expand.grid(x, choices)

matrix(mapply(`%in%`, d1$Var1, d1$Var2), nrow = length(x))

#or you can use Map(`%in%`, ...) in order to keep results in a list

As @nicola suggests, in order to make things better,

d1 <- expand.grid(list(x), choices) 

mapply(%in%, d1$Var1, d1$Var2)

both giving,

      [,1]  [,2]

[1,]  TRUE FALSE

[2,] FALSE  TRUE

edited Nov 26 '18 at 13:13

answered Nov 26 '18 at 12:43

Sotos

31.1k51741

edited Nov 26 '18 at 13:13

answered Nov 26 '18 at 12:43

Sotos

31.1k51741

answered Nov 26 '18 at 12:43

Sotos

31.1k51741

answered Nov 26 '18 at 12:43

Sotos

31.1k51741

1

To make things a lot better, you should d1 <- expand.grid(list(x), choices) and then just mapply(%in%, d1$Var1, d1$Var2) leads to the correct result, with much less time and RAM usage.

– nicola
Nov 26 '18 at 13:07

@nicola I agree with everything you said. Only reason I posted this is because it is closer to OP's mindset for the specific problem. But yes, very 'RAM' consuming. Thank you for the alternative! Looks much better

– Sotos
Nov 26 '18 at 13:10

add a comment |

1

To make things a lot better, you should d1 <- expand.grid(list(x), choices) and then just mapply(%in%, d1$Var1, d1$Var2) leads to the correct result, with much less time and RAM usage.

– nicola
Nov 26 '18 at 13:07

@nicola I agree with everything you said. Only reason I posted this is because it is closer to OP's mindset for the specific problem. But yes, very 'RAM' consuming. Thank you for the alternative! Looks much better

– Sotos
Nov 26 '18 at 13:10

To make things a lot better, you should d1 <- expand.grid(list(x), choices) and then just mapply(%in%, d1$Var1, d1$Var2) leads to the correct result, with much less time and RAM usage.

– nicola
Nov 26 '18 at 13:07

@nicola I agree with everything you said. Only reason I posted this is because it is closer to OP's mindset for the specific problem. But yes, very 'RAM' consuming. Thank you for the alternative! Looks much better

– Sotos
Nov 26 '18 at 13:10

add a comment |

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