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How to explain the strange behaviour about Function Nothing [duplicate]

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8

1

$begingroup$

There are some things wrong when i try to delete an element of a list by using function Nothing.
Here is a simple example:

a = Range[10], a[[1]], a[[1]] = Nothing, a(*1,2,3,4,5,6,7,8,9,10,1,2,3,4,5,6,7,8,9,10*)

It works,but when i try it again.

a[[1]], a[[1]] = Nothing, a(*2,2,3,4,5,6,7,8,9,10*)

It doesn't work.Finally i found that the first element of a could be deleted if i evaluate the code a[[2]]=Nothing,Any answer will be most appreciated

list-manipulation

share|improve this question

edited Sep 27 at 12:00

Lukas Lang

13.1k11 gold badge1414 silver badges4545 bronze badges

asked Sep 27 at 11:43

任天一任天一

37411 silver badge99 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Please do not use the Bugs tag for new questions - see the tags description for why
  $endgroup$
  – Lukas Lang
  Sep 27 at 12:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for the hint!@Lukas Lang
  $endgroup$
  – 任天一
  Sep 27 at 12:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Note: while this one is about Nothing rather than Sequence, the two are very similar, and the mechanism that is what puzzles the OP, is exactly the same. In this sense, this is still a duplicate, thus voting to close (apparently, I even mentioned Nothing in my answer there).
  $endgroup$
  – Leonid Shifrin
  Sep 27 at 23:20
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @LeonidShifrin Your answer is woefully ncomplete. A proper discussion of Nothing will never omit this
  $endgroup$
  – Daniel Lichtblau
  Nov 24 at 16:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @DanielLichtblau You are totally right. In fact, that link of yours is the answer, should be posted / accepted instead.
  $endgroup$
  – Leonid Shifrin
  Nov 24 at 18:33
  

  
  
  
  

add a comment
 | 

8

1

$begingroup$

There are some things wrong when i try to delete an element of a list by using function Nothing.
Here is a simple example:

a = Range[10], a[[1]], a[[1]] = Nothing, a(*1,2,3,4,5,6,7,8,9,10,1,2,3,4,5,6,7,8,9,10*)

It works,but when i try it again.

a[[1]], a[[1]] = Nothing, a(*2,2,3,4,5,6,7,8,9,10*)

It doesn't work.Finally i found that the first element of a could be deleted if i evaluate the code a[[2]]=Nothing,Any answer will be most appreciated

list-manipulation

share|improve this question

edited Sep 27 at 12:00

Lukas Lang

13.1k11 gold badge1414 silver badges4545 bronze badges

asked Sep 27 at 11:43

任天一任天一

37411 silver badge99 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Please do not use the Bugs tag for new questions - see the tags description for why
  $endgroup$
  – Lukas Lang
  Sep 27 at 12:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for the hint!@Lukas Lang
  $endgroup$
  – 任天一
  Sep 27 at 12:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Note: while this one is about Nothing rather than Sequence, the two are very similar, and the mechanism that is what puzzles the OP, is exactly the same. In this sense, this is still a duplicate, thus voting to close (apparently, I even mentioned Nothing in my answer there).
  $endgroup$
  – Leonid Shifrin
  Sep 27 at 23:20
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @LeonidShifrin Your answer is woefully ncomplete. A proper discussion of Nothing will never omit this
  $endgroup$
  – Daniel Lichtblau
  Nov 24 at 16:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @DanielLichtblau You are totally right. In fact, that link of yours is the answer, should be posted / accepted instead.
  $endgroup$
  – Leonid Shifrin
  Nov 24 at 18:33
  

  
  
  
  

add a comment
 | 

$begingroup$

There are some things wrong when i try to delete an element of a list by using function Nothing.
Here is a simple example:

a = Range[10], a[[1]], a[[1]] = Nothing, a(*1,2,3,4,5,6,7,8,9,10,1,2,3,4,5,6,7,8,9,10*)

It works,but when i try it again.

a[[1]], a[[1]] = Nothing, a(*2,2,3,4,5,6,7,8,9,10*)

It doesn't work.Finally i found that the first element of a could be deleted if i evaluate the code a[[2]]=Nothing,Any answer will be most appreciated

list-manipulation

share|improve this question

edited Sep 27 at 12:00

Lukas Lang

13.1k11 gold badge1414 silver badges4545 bronze badges

asked Sep 27 at 11:43

任天一任天一

37411 silver badge99 bronze badges

$endgroup$

a = Range[10], a[[1]], a[[1]] = Nothing, a(*1,2,3,4,5,6,7,8,9,10,1,2,3,4,5,6,7,8,9,10*)

a[[1]], a[[1]] = Nothing, a(*2,2,3,4,5,6,7,8,9,10*)

share|improve this question

edited Sep 27 at 12:00

Lukas Lang

13.1k11 gold badge1414 silver badges4545 bronze badges

asked Sep 27 at 11:43

任天一任天一

37411 silver badge99 bronze badges

share|improve this question

edited Sep 27 at 12:00

Lukas Lang

13.1k11 gold badge1414 silver badges4545 bronze badges

asked Sep 27 at 11:43

任天一任天一

37411 silver badge99 bronze badges

edited Sep 27 at 12:00

Lukas Lang

13.1k11 gold badge1414 silver badges4545 bronze badges

edited Sep 27 at 12:00

Lukas Lang

13.1k11 gold badge1414 silver badges4545 bronze badges

asked Sep 27 at 11:43

任天一任天一

37411 silver badge99 bronze badges

asked Sep 27 at 11:43

任天一任天一

37411 silver badge99 bronze badges

2 Answers
2

active

oldest

votes

11

$begingroup$

Part assignment performs in-place modification of an expression without evaluation of the result. At the same time, on the Documentation page for Nothing we read:

Nothing is removed as part of the standard evaluation process.

So after evaluation of a[[1]] = Nothing you still have a List of length 10 with first element being Nothing. You can replace Nothing with anything else in the same way again:

a[[1]] = Nothing;
Definition[a]
a[[1]] = Missing[];
Definition[a]

a=Nothing,2,3,4,5,6,7,8,9,10

a=Missing[],2,3,4,5,6,7,8,9,10

You can remove Nothing by evaluating the expression:

a[[1]] = Nothing;
Definition[a]
a = a;
Definition[a]

a=Nothing,2,3,4,5,6,7,8,9,10
a=2,3,4,5,6,7,8,9,10

---

Instead of Nothing you can use such functions as Delete, Drop, Take or ReplacePart for the same purpose.

share|improve this answer

edited Nov 23 at 7:24

answered Sep 27 at 13:53

Alexey PopkovAlexey Popkov

48.1k44 gold badges118118 silver badges282282 bronze badges

$endgroup$

add a comment
 | 

12

$begingroup$

Taking a look at ?a shows what's going on:

?a

(* Global`a *)

(* a=Nothing,2,3,4,5,6,7,8,9,10 *)

It seems that Part ([[...]]) does not apply the effect of Nothing after the replacement has been done, leaving you with a list that is still 10 elements long. So the second a[[1]]=... simply replaces the Nothing in the first element with Nothing again.

You case use Delete to do the deletion properly:

a = Range[10], a[[1]], a = Delete[a, 1], a
(* 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 1, 2, 3, 4, 5, 6, 7, 8, 9, 
 10, 2, 3, 4, 5, 6, 7, 8, 9, 10 *)

a[[1]], a = Delete[a, 1], a
(* 2, 3, 4, 5, 6, 7, 8, 9, 10, 3, 4, 5, 6, 7, 8, 9, 10 *)

share|improve this answer

answered Sep 27 at 12:00

Lukas LangLukas Lang

13.1k11 gold badge1414 silver badges4545 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for the answer!i get it.
  $endgroup$
  – 任天一
  Sep 27 at 12:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Interesting that a, b, Nothing, c, d, Nothing; (* new line *) % // Length gives 4.
  $endgroup$
  – Markhaim
  Sep 27 at 12:24
  
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  @Markhaim That happens because the list gets evaluated and after evaluation the Nothings will have disappeared. If you compare Length[Unevaluated[1, Nothing, 3]] with Length[1, Nothing, 3], you'll see the same effect.
  $endgroup$
  – Sjoerd Smit
  Sep 27 at 14:25
  

  
  
  
  

add a comment
 | 

11

$begingroup$

Part assignment performs in-place modification of an expression without evaluation of the result. At the same time, on the Documentation page for Nothing we read:

Nothing is removed as part of the standard evaluation process.

So after evaluation of a[[1]] = Nothing you still have a List of length 10 with first element being Nothing. You can replace Nothing with anything else in the same way again:

a[[1]] = Nothing;
Definition[a]
a[[1]] = Missing[];
Definition[a]

a=Nothing,2,3,4,5,6,7,8,9,10

a=Missing[],2,3,4,5,6,7,8,9,10

You can remove Nothing by evaluating the expression:

a[[1]] = Nothing;
Definition[a]
a = a;
Definition[a]

a=Nothing,2,3,4,5,6,7,8,9,10
a=2,3,4,5,6,7,8,9,10

---

Instead of Nothing you can use such functions as Delete, Drop, Take or ReplacePart for the same purpose.

share|improve this answer

edited Nov 23 at 7:24

answered Sep 27 at 13:53

Alexey PopkovAlexey Popkov

48.1k44 gold badges118118 silver badges282282 bronze badges

$endgroup$

add a comment
 | 

11

$begingroup$

Part assignment performs in-place modification of an expression without evaluation of the result. At the same time, on the Documentation page for Nothing we read:

Nothing is removed as part of the standard evaluation process.

So after evaluation of a[[1]] = Nothing you still have a List of length 10 with first element being Nothing. You can replace Nothing with anything else in the same way again:

a[[1]] = Nothing;
Definition[a]
a[[1]] = Missing[];
Definition[a]

a=Nothing,2,3,4,5,6,7,8,9,10

a=Missing[],2,3,4,5,6,7,8,9,10

You can remove Nothing by evaluating the expression:

a[[1]] = Nothing;
Definition[a]
a = a;
Definition[a]

a=Nothing,2,3,4,5,6,7,8,9,10
a=2,3,4,5,6,7,8,9,10

---

Instead of Nothing you can use such functions as Delete, Drop, Take or ReplacePart for the same purpose.

share|improve this answer

edited Nov 23 at 7:24

answered Sep 27 at 13:53

Alexey PopkovAlexey Popkov

48.1k44 gold badges118118 silver badges282282 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

Part assignment performs in-place modification of an expression without evaluation of the result. At the same time, on the Documentation page for Nothing we read:

Nothing is removed as part of the standard evaluation process.

So after evaluation of a[[1]] = Nothing you still have a List of length 10 with first element being Nothing. You can replace Nothing with anything else in the same way again:

a[[1]] = Nothing;
Definition[a]
a[[1]] = Missing[];
Definition[a]

a=Nothing,2,3,4,5,6,7,8,9,10

a=Missing[],2,3,4,5,6,7,8,9,10

You can remove Nothing by evaluating the expression:

a[[1]] = Nothing;
Definition[a]
a = a;
Definition[a]

a=Nothing,2,3,4,5,6,7,8,9,10
a=2,3,4,5,6,7,8,9,10

---

Instead of Nothing you can use such functions as Delete, Drop, Take or ReplacePart for the same purpose.

share|improve this answer

edited Nov 23 at 7:24

answered Sep 27 at 13:53

Alexey PopkovAlexey Popkov

48.1k44 gold badges118118 silver badges282282 bronze badges

$endgroup$

a[[1]] = Nothing;
Definition[a]
a[[1]] = Missing[];
Definition[a]

a[[1]] = Nothing;
Definition[a]
a = a;
Definition[a]

share|improve this answer

edited Nov 23 at 7:24

answered Sep 27 at 13:53

Alexey PopkovAlexey Popkov

48.1k44 gold badges118118 silver badges282282 bronze badges

answered Sep 27 at 13:53

Alexey PopkovAlexey Popkov

48.1k44 gold badges118118 silver badges282282 bronze badges

answered Sep 27 at 13:53

Alexey PopkovAlexey Popkov

48.1k44 gold badges118118 silver badges282282 bronze badges

12

$begingroup$

Taking a look at ?a shows what's going on:

?a

(* Global`a *)

(* a=Nothing,2,3,4,5,6,7,8,9,10 *)

It seems that Part ([[...]]) does not apply the effect of Nothing after the replacement has been done, leaving you with a list that is still 10 elements long. So the second a[[1]]=... simply replaces the Nothing in the first element with Nothing again.

You case use Delete to do the deletion properly:

a = Range[10], a[[1]], a = Delete[a, 1], a
(* 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 1, 2, 3, 4, 5, 6, 7, 8, 9, 
 10, 2, 3, 4, 5, 6, 7, 8, 9, 10 *)

a[[1]], a = Delete[a, 1], a
(* 2, 3, 4, 5, 6, 7, 8, 9, 10, 3, 4, 5, 6, 7, 8, 9, 10 *)

share|improve this answer

answered Sep 27 at 12:00

Lukas LangLukas Lang

13.1k11 gold badge1414 silver badges4545 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for the answer!i get it.
  $endgroup$
  – 任天一
  Sep 27 at 12:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Interesting that a, b, Nothing, c, d, Nothing; (* new line *) % // Length gives 4.
  $endgroup$
  – Markhaim
  Sep 27 at 12:24
  
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  @Markhaim That happens because the list gets evaluated and after evaluation the Nothings will have disappeared. If you compare Length[Unevaluated[1, Nothing, 3]] with Length[1, Nothing, 3], you'll see the same effect.
  $endgroup$
  – Sjoerd Smit
  Sep 27 at 14:25
  

  
  
  
  

add a comment
 | 

12

$begingroup$

Taking a look at ?a shows what's going on:

?a

(* Global`a *)

(* a=Nothing,2,3,4,5,6,7,8,9,10 *)

It seems that Part ([[...]]) does not apply the effect of Nothing after the replacement has been done, leaving you with a list that is still 10 elements long. So the second a[[1]]=... simply replaces the Nothing in the first element with Nothing again.

You case use Delete to do the deletion properly:

a = Range[10], a[[1]], a = Delete[a, 1], a
(* 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 1, 2, 3, 4, 5, 6, 7, 8, 9, 
 10, 2, 3, 4, 5, 6, 7, 8, 9, 10 *)

a[[1]], a = Delete[a, 1], a
(* 2, 3, 4, 5, 6, 7, 8, 9, 10, 3, 4, 5, 6, 7, 8, 9, 10 *)

share|improve this answer

answered Sep 27 at 12:00

Lukas LangLukas Lang

13.1k11 gold badge1414 silver badges4545 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for the answer!i get it.
  $endgroup$
  – 任天一
  Sep 27 at 12:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Interesting that a, b, Nothing, c, d, Nothing; (* new line *) % // Length gives 4.
  $endgroup$
  – Markhaim
  Sep 27 at 12:24
  
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  @Markhaim That happens because the list gets evaluated and after evaluation the Nothings will have disappeared. If you compare Length[Unevaluated[1, Nothing, 3]] with Length[1, Nothing, 3], you'll see the same effect.
  $endgroup$
  – Sjoerd Smit
  Sep 27 at 14:25
  

  
  
  
  

add a comment
 | 

$begingroup$

Taking a look at ?a shows what's going on:

?a

(* Global`a *)

(* a=Nothing,2,3,4,5,6,7,8,9,10 *)

It seems that Part ([[...]]) does not apply the effect of Nothing after the replacement has been done, leaving you with a list that is still 10 elements long. So the second a[[1]]=... simply replaces the Nothing in the first element with Nothing again.

You case use Delete to do the deletion properly:

a = Range[10], a[[1]], a = Delete[a, 1], a
(* 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 1, 2, 3, 4, 5, 6, 7, 8, 9, 
 10, 2, 3, 4, 5, 6, 7, 8, 9, 10 *)

a[[1]], a = Delete[a, 1], a
(* 2, 3, 4, 5, 6, 7, 8, 9, 10, 3, 4, 5, 6, 7, 8, 9, 10 *)

share|improve this answer

answered Sep 27 at 12:00

Lukas LangLukas Lang

13.1k11 gold badge1414 silver badges4545 bronze badges

$endgroup$

?a

(* Global`a *)

(* a=Nothing,2,3,4,5,6,7,8,9,10 *)

a = Range[10], a[[1]], a = Delete[a, 1], a
(* 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 1, 2, 3, 4, 5, 6, 7, 8, 9, 
 10, 2, 3, 4, 5, 6, 7, 8, 9, 10 *)

a[[1]], a = Delete[a, 1], a
(* 2, 3, 4, 5, 6, 7, 8, 9, 10, 3, 4, 5, 6, 7, 8, 9, 10 *)

share|improve this answer

answered Sep 27 at 12:00

Lukas LangLukas Lang

13.1k11 gold badge1414 silver badges4545 bronze badges

answered Sep 27 at 12:00

Lukas LangLukas Lang

13.1k11 gold badge1414 silver badges4545 bronze badges

answered Sep 27 at 12:00

Lukas LangLukas Lang

13.1k11 gold badge1414 silver badges4545 bronze badges