Ggtcf

How do you justify more code being written by following clean code practices?

Would a primitive species be able to learn English from reading books alone?

Friend wants my recommendation but I don't want to give it to him

Reasons for having MCU pin-states default to pull-up/down out of reset

Can you describe someone as luxurious? As in someone who likes luxurious things?

A seasonal riddle

Can a Knock spell open the door to Mordenkainen's Magnificent Mansion?

How to get directions in deep space?

Showing mass murder in a kid's book

Asserting that Atheism and Theism are both faith based positions

Connection Between Knot Theory and Number Theory

Put the phone down / Put down the phone

"Oh no!" in Latin

Output visual diagram of picture

Offset in split text content

Could a welfare state co-exist with mega corporations?

Should a narrator ever describe things based on a character's view instead of facts?

Checking @@ROWCOUNT failing

What (if any) is the reason to buy in small local stores?

Rendered textures different to 3D View

Highest stage count that are used one right after the other?

Make a Bowl of Alphabet Soup

Why does the frost depth increase when the surface temperature warms up?

Using an older 200A breaker panel on a 60A feeder circuit from house?

Find out where the given numbers are located in the series of vector in R

Test if a vector contains a given elementRscript: Determine path of the executing scriptCounting the number of elements with the values of x in a vectorDrop data frame columns by nameHow to disable scientific notation?How to unload a package without restarting R?Extracting specific columns from a data frameHow to find the amount of positive numbersHow can I view the source code for a function?data.table vs dplyr: can one do something well the other can't or does poorly?

0

I have three variables a1, a2 and a3.

a1 <- 1:10
a2 <- 11:20
a3 <- 21:30

then I have another variable called my.numbers <- c(1, 20, 22,11)

I want to find out where these numbers are located. So the result I want is:

1 in a1
20 in a2
22 in a3
11 in a2

Any suggestion on how it can be done easy way?

r

share|improve this question

asked Mar 7 at 20:34

MAPKMAPK

1,797835

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  
  L = mget(c("a1", "a2", "a3")); lapply(L, intersect, my.numbers), I guess.
  
  – Frank
  Mar 7 at 20:38
  

  
  
  
  

add a comment | 

0

I have three variables a1, a2 and a3.

a1 <- 1:10
a2 <- 11:20
a3 <- 21:30

then I have another variable called my.numbers <- c(1, 20, 22,11)

I want to find out where these numbers are located. So the result I want is:

1 in a1
20 in a2
22 in a3
11 in a2

Any suggestion on how it can be done easy way?

r

share|improve this question

asked Mar 7 at 20:34

MAPKMAPK

1,797835

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  
  L = mget(c("a1", "a2", "a3")); lapply(L, intersect, my.numbers), I guess.
  
  – Frank
  Mar 7 at 20:38
  

  
  
  
  

add a comment | 

I have three variables a1, a2 and a3.

a1 <- 1:10
a2 <- 11:20
a3 <- 21:30

then I have another variable called my.numbers <- c(1, 20, 22,11)

I want to find out where these numbers are located. So the result I want is:

1 in a1
20 in a2
22 in a3
11 in a2

Any suggestion on how it can be done easy way?

r

share|improve this question

asked Mar 7 at 20:34

MAPKMAPK

1,797835

a1 <- 1:10
a2 <- 11:20
a3 <- 21:30

1 in a1
20 in a2
22 in a3
11 in a2

share|improve this question

asked Mar 7 at 20:34

MAPKMAPK

1,797835

share|improve this question

asked Mar 7 at 20:34

MAPKMAPK

1,797835

asked Mar 7 at 20:34

MAPKMAPK

1,797835

asked Mar 7 at 20:34

MAPKMAPK

1,797835

2 Answers
2

active

oldest

votes

1

With a couple purrr::map functions, you can work across the numbers, then within that, across the a vectors.

I'm making a list of the a vectors with tibble::lst because it sets the names of the list as the names of the variables going into it—convenient for something like this where it's the name of the list item that's important.

library(tidyverse)

a_list <- lst(a1, a2, a3)

my.numbers %>%
 map_chr(function(num) 
 which_a <- map_lgl(a_list, ~(num %in% .))
 a_name <- names(a_list)[which_a]
 str_glue("num in a_name")
 )
#> [1] "1 in a1" "20 in a2" "22 in a3" "11 in a2"

You could use match or another function after the map_lgl instead—I left it verbose to make it a little more clear what's going on.

share|improve this answer

answered Mar 7 at 21:21

camillecamille

7,76631833

add a comment | 

1

For the record here is how you can get out the exact result in the question.

a1 <- 1:10
a2 <- 11:20
a3 <- 21:30

L<-list("a1"=a1,"a2"=a2,"a3"=a3)
my.numbers <- c(1, 20, 22, 11)

func<-function(item)
 my.numbers[which(my.numbers %in% item)]


Fin<-lapply(L, func)

for(i in 1:length(Fin))
Index<-unlist(Fin[i])
name<-paste("a",i, sep="")
 for(i in 1:length(Index))
 print(paste(Index[i], "in", name))
 


[1] "1 in a1"
[1] "20 in a2"
[1] "11 in a2"
[1] "22 in a3"

share|improve this answer

answered Mar 7 at 21:14

ChaboChabo

1,1981823

add a comment | 

Your Answer

StackExchange.ifUsing("editor", function ()
StackExchange.using("externalEditor", function ()
StackExchange.using("snippets", function ()
StackExchange.snippets.init();
);
);
, "code-snippets");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "1"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f55052382%2ffind-out-where-the-given-numbers-are-located-in-the-series-of-vector-in-r%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

1

With a couple purrr::map functions, you can work across the numbers, then within that, across the a vectors.

I'm making a list of the a vectors with tibble::lst because it sets the names of the list as the names of the variables going into it—convenient for something like this where it's the name of the list item that's important.

library(tidyverse)

a_list <- lst(a1, a2, a3)

my.numbers %>%
 map_chr(function(num) 
 which_a <- map_lgl(a_list, ~(num %in% .))
 a_name <- names(a_list)[which_a]
 str_glue("num in a_name")
 )
#> [1] "1 in a1" "20 in a2" "22 in a3" "11 in a2"

You could use match or another function after the map_lgl instead—I left it verbose to make it a little more clear what's going on.

share|improve this answer

answered Mar 7 at 21:21

camillecamille

7,76631833

add a comment | 

1

With a couple purrr::map functions, you can work across the numbers, then within that, across the a vectors.

I'm making a list of the a vectors with tibble::lst because it sets the names of the list as the names of the variables going into it—convenient for something like this where it's the name of the list item that's important.

library(tidyverse)

a_list <- lst(a1, a2, a3)

my.numbers %>%
 map_chr(function(num) 
 which_a <- map_lgl(a_list, ~(num %in% .))
 a_name <- names(a_list)[which_a]
 str_glue("num in a_name")
 )
#> [1] "1 in a1" "20 in a2" "22 in a3" "11 in a2"

You could use match or another function after the map_lgl instead—I left it verbose to make it a little more clear what's going on.

share|improve this answer

answered Mar 7 at 21:21

camillecamille

7,76631833

add a comment | 

With a couple purrr::map functions, you can work across the numbers, then within that, across the a vectors.

I'm making a list of the a vectors with tibble::lst because it sets the names of the list as the names of the variables going into it—convenient for something like this where it's the name of the list item that's important.

library(tidyverse)

a_list <- lst(a1, a2, a3)

my.numbers %>%
 map_chr(function(num) 
 which_a <- map_lgl(a_list, ~(num %in% .))
 a_name <- names(a_list)[which_a]
 str_glue("num in a_name")
 )
#> [1] "1 in a1" "20 in a2" "22 in a3" "11 in a2"

You could use match or another function after the map_lgl instead—I left it verbose to make it a little more clear what's going on.

share|improve this answer

answered Mar 7 at 21:21

camillecamille

7,76631833

library(tidyverse)

a_list <- lst(a1, a2, a3)

my.numbers %>%
 map_chr(function(num) 
 which_a <- map_lgl(a_list, ~(num %in% .))
 a_name <- names(a_list)[which_a]
 str_glue("num in a_name")
 )
#> [1] "1 in a1" "20 in a2" "22 in a3" "11 in a2"

share|improve this answer

answered Mar 7 at 21:21

camillecamille

7,76631833

answered Mar 7 at 21:21

camillecamille

7,76631833

answered Mar 7 at 21:21

camillecamille

7,76631833

1

For the record here is how you can get out the exact result in the question.

a1 <- 1:10
a2 <- 11:20
a3 <- 21:30

L<-list("a1"=a1,"a2"=a2,"a3"=a3)
my.numbers <- c(1, 20, 22, 11)

func<-function(item)
 my.numbers[which(my.numbers %in% item)]


Fin<-lapply(L, func)

for(i in 1:length(Fin))
Index<-unlist(Fin[i])
name<-paste("a",i, sep="")
 for(i in 1:length(Index))
 print(paste(Index[i], "in", name))
 


[1] "1 in a1"
[1] "20 in a2"
[1] "11 in a2"
[1] "22 in a3"

share|improve this answer

answered Mar 7 at 21:14

ChaboChabo

1,1981823

add a comment | 

1

For the record here is how you can get out the exact result in the question.

a1 <- 1:10
a2 <- 11:20
a3 <- 21:30

L<-list("a1"=a1,"a2"=a2,"a3"=a3)
my.numbers <- c(1, 20, 22, 11)

func<-function(item)
 my.numbers[which(my.numbers %in% item)]


Fin<-lapply(L, func)

for(i in 1:length(Fin))
Index<-unlist(Fin[i])
name<-paste("a",i, sep="")
 for(i in 1:length(Index))
 print(paste(Index[i], "in", name))
 


[1] "1 in a1"
[1] "20 in a2"
[1] "11 in a2"
[1] "22 in a3"

share|improve this answer

answered Mar 7 at 21:14

ChaboChabo

1,1981823

add a comment | 

For the record here is how you can get out the exact result in the question.

a1 <- 1:10
a2 <- 11:20
a3 <- 21:30

L<-list("a1"=a1,"a2"=a2,"a3"=a3)
my.numbers <- c(1, 20, 22, 11)

func<-function(item)
 my.numbers[which(my.numbers %in% item)]


Fin<-lapply(L, func)

for(i in 1:length(Fin))
Index<-unlist(Fin[i])
name<-paste("a",i, sep="")
 for(i in 1:length(Index))
 print(paste(Index[i], "in", name))
 


[1] "1 in a1"
[1] "20 in a2"
[1] "11 in a2"
[1] "22 in a3"

share|improve this answer

answered Mar 7 at 21:14

ChaboChabo

1,1981823

a1 <- 1:10
a2 <- 11:20
a3 <- 21:30

L<-list("a1"=a1,"a2"=a2,"a3"=a3)
my.numbers <- c(1, 20, 22, 11)

func<-function(item)
 my.numbers[which(my.numbers %in% item)]


Fin<-lapply(L, func)

for(i in 1:length(Fin))
Index<-unlist(Fin[i])
name<-paste("a",i, sep="")
 for(i in 1:length(Index))
 print(paste(Index[i], "in", name))
 


[1] "1 in a1"
[1] "20 in a2"
[1] "11 in a2"
[1] "22 in a3"

share|improve this answer

answered Mar 7 at 21:14

ChaboChabo

1,1981823

answered Mar 7 at 21:14

ChaboChabo

1,1981823

answered Mar 7 at 21:14

ChaboChabo

1,1981823