How to take the beginning and end parts of a list with simpler syntax?Calling Table with custom...

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How to take the beginning and end parts of a list with simpler syntax?


Calling Table with custom iteratorExpectation over a list with nested-mapped expressionsDeleting powers of a common prime except the highest one from a list with further only primesChange orientation of GradientOrientationFilter from high value to lowHow to return the unevaluated variable names from a list of variables which have been declaredHow can I make ListPlot use my iteration as the x coord?Build a list from applying a recursive function on another listHow to split list into segments based on first element of sublist, and remove ineligible segments






.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ margin-bottom:0;
}







5












$begingroup$


So basically you have a table of some values, let's call it a:



a=Table[n,{n,10}];

(*{1, 2, 3, 4, 5, 6, 7, 8, 9, 10}*)


And you want to take the beginning and end parts of the table, within the same line, and do something with them. It is my intention to change them all to 0, but really you could change them to some other variable, say b. We want the parts of the list that do not include some inner values, we'll do {4,5,6} which happen to correspond to the indexes 4 through 6, convenient, no?



a[[Cases[a, (Except[Alternatives @@ Range[4, 6]])]]] = b; a

(*{b, b, b, 4, 5, 6, b, b, b, b}*)


What's a simpler syntax for this? How can we call the end and beginning values of a list in a simpler manner?



a[[Cases[a, (Except[Alternatives @@ Range[4, 6]])]]]

(*{1, 2, 3, 7, 8, 9, 10}*)


The ideal input would be, to me, something like



a[[1;;3;7;;10]]

(*{1, 2, 3, 7, 8, 9, 10}*)


But this gives



(*{7, 8, 9, 10}*)


As expected.



How can we simplify the syntax to call the beginning and end parts of a list, leaving out some chosen middle portion?






list-manipulation syntax







share|improve this question









$endgroup$










  • 1




    $begingroup$
    Select[ a, ! MemberQ[ {4, 5, 6}, #] & ]
    $endgroup$
    – LouisB
    10 hours ago






  • 1




    $begingroup$
    Complement[ a, {4, 5, 6} ]
    $endgroup$
    – LouisB
    10 hours ago






  • 1




    $begingroup$
    Cheating a little, we could write (a[[#]] = b) & /@ {1 ;; 3, 7 ;; 10}.
    $endgroup$
    – WReach
    9 hours ago






  • 1




    $begingroup$
    If it's the values, then in Table[n^2,{n,10}] the initial segment would be the span 1 ;; 1 and the final segment would be 3 ;; 10 and the excluded segment (the positions not to be changed) would be 2 ;; 2, if the excluded values were 4, 5, 6 -- no?
    $endgroup$
    – Michael E2
    7 hours ago






  • 1




    $begingroup$
    Thanks. I posted an answer. See if I understood correctly.
    $endgroup$
    – Michael E2
    7 hours ago


















5












$begingroup$


So basically you have a table of some values, let's call it a:



a=Table[n,{n,10}];

(*{1, 2, 3, 4, 5, 6, 7, 8, 9, 10}*)


And you want to take the beginning and end parts of the table, within the same line, and do something with them. It is my intention to change them all to 0, but really you could change them to some other variable, say b. We want the parts of the list that do not include some inner values, we'll do {4,5,6} which happen to correspond to the indexes 4 through 6, convenient, no?



a[[Cases[a, (Except[Alternatives @@ Range[4, 6]])]]] = b; a

(*{b, b, b, 4, 5, 6, b, b, b, b}*)


What's a simpler syntax for this? How can we call the end and beginning values of a list in a simpler manner?



a[[Cases[a, (Except[Alternatives @@ Range[4, 6]])]]]

(*{1, 2, 3, 7, 8, 9, 10}*)


The ideal input would be, to me, something like



a[[1;;3;7;;10]]

(*{1, 2, 3, 7, 8, 9, 10}*)


But this gives



(*{7, 8, 9, 10}*)


As expected.



How can we simplify the syntax to call the beginning and end parts of a list, leaving out some chosen middle portion?






list-manipulation syntax







share|improve this question









$endgroup$










  • 1




    $begingroup$
    Select[ a, ! MemberQ[ {4, 5, 6}, #] & ]
    $endgroup$
    – LouisB
    10 hours ago






  • 1




    $begingroup$
    Complement[ a, {4, 5, 6} ]
    $endgroup$
    – LouisB
    10 hours ago






  • 1




    $begingroup$
    Cheating a little, we could write (a[[#]] = b) & /@ {1 ;; 3, 7 ;; 10}.
    $endgroup$
    – WReach
    9 hours ago






  • 1




    $begingroup$
    If it's the values, then in Table[n^2,{n,10}] the initial segment would be the span 1 ;; 1 and the final segment would be 3 ;; 10 and the excluded segment (the positions not to be changed) would be 2 ;; 2, if the excluded values were 4, 5, 6 -- no?
    $endgroup$
    – Michael E2
    7 hours ago






  • 1




    $begingroup$
    Thanks. I posted an answer. See if I understood correctly.
    $endgroup$
    – Michael E2
    7 hours ago














5












5








5


1



$begingroup$


So basically you have a table of some values, let's call it a:



a=Table[n,{n,10}];

(*{1, 2, 3, 4, 5, 6, 7, 8, 9, 10}*)


And you want to take the beginning and end parts of the table, within the same line, and do something with them. It is my intention to change them all to 0, but really you could change them to some other variable, say b. We want the parts of the list that do not include some inner values, we'll do {4,5,6} which happen to correspond to the indexes 4 through 6, convenient, no?



a[[Cases[a, (Except[Alternatives @@ Range[4, 6]])]]] = b; a

(*{b, b, b, 4, 5, 6, b, b, b, b}*)


What's a simpler syntax for this? How can we call the end and beginning values of a list in a simpler manner?



a[[Cases[a, (Except[Alternatives @@ Range[4, 6]])]]]

(*{1, 2, 3, 7, 8, 9, 10}*)


The ideal input would be, to me, something like



a[[1;;3;7;;10]]

(*{1, 2, 3, 7, 8, 9, 10}*)


But this gives



(*{7, 8, 9, 10}*)


As expected.



How can we simplify the syntax to call the beginning and end parts of a list, leaving out some chosen middle portion?






list-manipulation syntax







share|improve this question









$endgroup$




So basically you have a table of some values, let's call it a:



a=Table[n,{n,10}];

(*{1, 2, 3, 4, 5, 6, 7, 8, 9, 10}*)


And you want to take the beginning and end parts of the table, within the same line, and do something with them. It is my intention to change them all to 0, but really you could change them to some other variable, say b. We want the parts of the list that do not include some inner values, we'll do {4,5,6} which happen to correspond to the indexes 4 through 6, convenient, no?



a[[Cases[a, (Except[Alternatives @@ Range[4, 6]])]]] = b; a

(*{b, b, b, 4, 5, 6, b, b, b, b}*)


What's a simpler syntax for this? How can we call the end and beginning values of a list in a simpler manner?



a[[Cases[a, (Except[Alternatives @@ Range[4, 6]])]]]

(*{1, 2, 3, 7, 8, 9, 10}*)


The ideal input would be, to me, something like



a[[1;;3;7;;10]]

(*{1, 2, 3, 7, 8, 9, 10}*)


But this gives



(*{7, 8, 9, 10}*)


As expected.



How can we simplify the syntax to call the beginning and end parts of a list, leaving out some chosen middle portion?






list-manipulation syntax




list-manipulation syntax






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked 11 hours ago









CA TrevillianCA Trevillian

5891 gold badge2 silver badges12 bronze badges




5891 gold badge2 silver badges12 bronze badges











  • 1




    $begingroup$
    Select[ a, ! MemberQ[ {4, 5, 6}, #] & ]
    $endgroup$
    – LouisB
    10 hours ago






  • 1




    $begingroup$
    Complement[ a, {4, 5, 6} ]
    $endgroup$
    – LouisB
    10 hours ago






  • 1




    $begingroup$
    Cheating a little, we could write (a[[#]] = b) & /@ {1 ;; 3, 7 ;; 10}.
    $endgroup$
    – WReach
    9 hours ago






  • 1




    $begingroup$
    If it's the values, then in Table[n^2,{n,10}] the initial segment would be the span 1 ;; 1 and the final segment would be 3 ;; 10 and the excluded segment (the positions not to be changed) would be 2 ;; 2, if the excluded values were 4, 5, 6 -- no?
    $endgroup$
    – Michael E2
    7 hours ago






  • 1




    $begingroup$
    Thanks. I posted an answer. See if I understood correctly.
    $endgroup$
    – Michael E2
    7 hours ago














  • 1




    $begingroup$
    Select[ a, ! MemberQ[ {4, 5, 6}, #] & ]
    $endgroup$
    – LouisB
    10 hours ago






  • 1




    $begingroup$
    Complement[ a, {4, 5, 6} ]
    $endgroup$
    – LouisB
    10 hours ago






  • 1




    $begingroup$
    Cheating a little, we could write (a[[#]] = b) & /@ {1 ;; 3, 7 ;; 10}.
    $endgroup$
    – WReach
    9 hours ago






  • 1




    $begingroup$
    If it's the values, then in Table[n^2,{n,10}] the initial segment would be the span 1 ;; 1 and the final segment would be 3 ;; 10 and the excluded segment (the positions not to be changed) would be 2 ;; 2, if the excluded values were 4, 5, 6 -- no?
    $endgroup$
    – Michael E2
    7 hours ago






  • 1




    $begingroup$
    Thanks. I posted an answer. See if I understood correctly.
    $endgroup$
    – Michael E2
    7 hours ago








1




1




$begingroup$
Select[ a, ! MemberQ[ {4, 5, 6}, #] & ]
$endgroup$
– LouisB
10 hours ago




$begingroup$
Select[ a, ! MemberQ[ {4, 5, 6}, #] & ]
$endgroup$
– LouisB
10 hours ago




1




1




$begingroup$
Complement[ a, {4, 5, 6} ]
$endgroup$
– LouisB
10 hours ago




$begingroup$
Complement[ a, {4, 5, 6} ]
$endgroup$
– LouisB
10 hours ago




1




1




$begingroup$
Cheating a little, we could write (a[[#]] = b) & /@ {1 ;; 3, 7 ;; 10}.
$endgroup$
– WReach
9 hours ago




$begingroup$
Cheating a little, we could write (a[[#]] = b) & /@ {1 ;; 3, 7 ;; 10}.
$endgroup$
– WReach
9 hours ago




1




1




$begingroup$
If it's the values, then in Table[n^2,{n,10}] the initial segment would be the span 1 ;; 1 and the final segment would be 3 ;; 10 and the excluded segment (the positions not to be changed) would be 2 ;; 2, if the excluded values were 4, 5, 6 -- no?
$endgroup$
– Michael E2
7 hours ago




$begingroup$
If it's the values, then in Table[n^2,{n,10}] the initial segment would be the span 1 ;; 1 and the final segment would be 3 ;; 10 and the excluded segment (the positions not to be changed) would be 2 ;; 2, if the excluded values were 4, 5, 6 -- no?
$endgroup$
– Michael E2
7 hours ago




1




1




$begingroup$
Thanks. I posted an answer. See if I understood correctly.
$endgroup$
– Michael E2
7 hours ago




$begingroup$
Thanks. I posted an answer. See if I understood correctly.
$endgroup$
– Michael E2
7 hours ago










5 Answers
5






active

oldest

votes


















7












$begingroup$

Update: You can also try MapAt:



a = Range[10];

a = MapAt[b &, a, {{;; 3}, {7 ;;}}]



{b, b, b, 4, 5, 6, b, b, b, b}




Or ReplaceAll



a = Range[10];

a /. Alternatives @@ Drop[a, 4;;6] -> b



{b, b, b, 4, 5, 6, b, b, b, b}




Original answer: Try Drop:



Drop[Range @ 10, {4, 6}]



{1, 2, 3, 7, 8, 9, 10}




Drop[Range @ 10, 4 ;; 6]




{1, 2, 3, 7, 8, 9, 10}





Drop[CharacterRange["a", "j"], {4, 6}]



{"a", "b", "c", "g", "h", "i", "j"}







share|improve this answer











$endgroup$











  • 1




    $begingroup$
    Hah, shot myself in the foot with my end question...That works to get the values, but I'm unsure of the readability when we do a[[Drop[Range@10, {4, 6}]]] = b; a. Though, it is a nice implementation and one I had not considered. I'm hoping to find something that is not too over-encumbered, in order to sit in a publication. This is much nicer than the original try of mine, and it only uses the definition of the neglected characters. Quite nice, actually, the more I debate it! Much more general of an approach. +1
    $endgroup$
    – CA Trevillian
    11 hours ago



















5












$begingroup$

If it is okay to perform two assignments instead of one, then we can write:



(a[[#]] = b) & /@ {1 ;; 3, 7 ;; 10}


Scan is probably better since it does not bother constructing the result list that we are just going to discard anyway:



Scan[(a[[#]] = b) &, {1 ;; 3, 7 ;; 10}]





share|improve this answer









$endgroup$















  • $begingroup$
    Well aaactually--that was the original reason for this question, and I probably could have written that....but I can't remember what cup of coffee I was on at that point! Regardless, this is a great implementation of a pure function applied to an index being used in a functional reassignment. This is technically in one line too, which satisfied my original criteria! Very cool. I wonder if this can be shorted at all for some sort-of code-golf-like...thing...answer? (Hint-hint, but it'd require some user thought) Oooooh scan, yessiree this is a great one, Scan needs more use these days, imo.
    $endgroup$
    – CA Trevillian
    9 hours ago



















5












$begingroup$

Perhaps this?:



ReplacePart[Range@10, i_ /; Not[4 <= i <= 6] :> b]

(*  {b, b, b, 4, 5, 6, b, b, b, b}  *)





share|improve this answer









$endgroup$















  • $begingroup$
    Wow nice!! I like the use of RuleDelayed, am I correct in understanding that you could use either Table[n,{n,10}] or Table[n^2,{n,10}] and we should see the 4th-6th indexed values preserved, as ReplacePart replaces the ith index? The use of Not provides a good deal of readability as well.
    $endgroup$
    – CA Trevillian
    7 hours ago






  • 1




    $begingroup$
    @CATrevillian Yes, it depends only on the index (part/position), not the value.
    $endgroup$
    – Michael E2
    6 hours ago



















4












$begingroup$

We can use Union[] to perform this operation in a simpler manner:



a[[Range[1, 3] [Union] Range[7, 10]]]

(*{1, 2, 3, 7, 8, 9, 10}*)


But this is still decently messy, and the readability is not too high. Perhaps someone else has a better method?



a[[Range[1, 3] [Union] Range[7, 10]]] = b; a

(*{b, b, b, 4, 5, 6, b, b, b, b}*)





share|improve this answer









$endgroup$























    4












    $begingroup$

    By far the simplest way is to do it in two lines:



    a = Table[RandomReal[], {n, 10}];
    
a[[1 ;; 3]] = b;
    
a[[7 ;; 10]] = b;
    
a


    This gives the output:




    {b, b, b, 0.378846, 0.475894, 0.533768, b, b, b, b}







    share|improve this answer











    $endgroup$


















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      5 Answers
      5






      active

      oldest

      votes








      5 Answers
      5






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      7












      $begingroup$

      Update: You can also try MapAt:



      a = Range[10];
      
a = MapAt[b &, a, {{;; 3}, {7 ;;}}]



      {b, b, b, 4, 5, 6, b, b, b, b}




      Or ReplaceAll



      a = Range[10];
      
a /. Alternatives @@ Drop[a, 4;;6] -> b



      {b, b, b, 4, 5, 6, b, b, b, b}




      Original answer: Try Drop:



      Drop[Range @ 10, {4, 6}]



      {1, 2, 3, 7, 8, 9, 10}




      Drop[Range @ 10, 4 ;; 6]




      {1, 2, 3, 7, 8, 9, 10}





      Drop[CharacterRange["a", "j"], {4, 6}]



      {"a", "b", "c", "g", "h", "i", "j"}







      share|improve this answer











      $endgroup$











      • 1




        $begingroup$
        Hah, shot myself in the foot with my end question...That works to get the values, but I'm unsure of the readability when we do a[[Drop[Range@10, {4, 6}]]] = b; a. Though, it is a nice implementation and one I had not considered. I'm hoping to find something that is not too over-encumbered, in order to sit in a publication. This is much nicer than the original try of mine, and it only uses the definition of the neglected characters. Quite nice, actually, the more I debate it! Much more general of an approach. +1
        $endgroup$
        – CA Trevillian
        11 hours ago
















      7












      $begingroup$

      Update: You can also try MapAt:



      a = Range[10];
      
a = MapAt[b &, a, {{;; 3}, {7 ;;}}]



      {b, b, b, 4, 5, 6, b, b, b, b}




      Or ReplaceAll



      a = Range[10];
      
a /. Alternatives @@ Drop[a, 4;;6] -> b



      {b, b, b, 4, 5, 6, b, b, b, b}




      Original answer: Try Drop:



      Drop[Range @ 10, {4, 6}]



      {1, 2, 3, 7, 8, 9, 10}




      Drop[Range @ 10, 4 ;; 6]




      {1, 2, 3, 7, 8, 9, 10}





      Drop[CharacterRange["a", "j"], {4, 6}]



      {"a", "b", "c", "g", "h", "i", "j"}







      share|improve this answer











      $endgroup$











      • 1




        $begingroup$
        Hah, shot myself in the foot with my end question...That works to get the values, but I'm unsure of the readability when we do a[[Drop[Range@10, {4, 6}]]] = b; a. Though, it is a nice implementation and one I had not considered. I'm hoping to find something that is not too over-encumbered, in order to sit in a publication. This is much nicer than the original try of mine, and it only uses the definition of the neglected characters. Quite nice, actually, the more I debate it! Much more general of an approach. +1
        $endgroup$
        – CA Trevillian
        11 hours ago














      7












      7








      7





      $begingroup$

      Update: You can also try MapAt:



      a = Range[10];
      
a = MapAt[b &, a, {{;; 3}, {7 ;;}}]



      {b, b, b, 4, 5, 6, b, b, b, b}




      Or ReplaceAll



      a = Range[10];
      
a /. Alternatives @@ Drop[a, 4;;6] -> b



      {b, b, b, 4, 5, 6, b, b, b, b}




      Original answer: Try Drop:



      Drop[Range @ 10, {4, 6}]



      {1, 2, 3, 7, 8, 9, 10}




      Drop[Range @ 10, 4 ;; 6]




      {1, 2, 3, 7, 8, 9, 10}





      Drop[CharacterRange["a", "j"], {4, 6}]



      {"a", "b", "c", "g", "h", "i", "j"}







      share|improve this answer











      $endgroup$



      Update: You can also try MapAt:



      a = Range[10];
      
a = MapAt[b &, a, {{;; 3}, {7 ;;}}]



      {b, b, b, 4, 5, 6, b, b, b, b}




      Or ReplaceAll



      a = Range[10];
      
a /. Alternatives @@ Drop[a, 4;;6] -> b



      {b, b, b, 4, 5, 6, b, b, b, b}




      Original answer: Try Drop:



      Drop[Range @ 10, {4, 6}]



      {1, 2, 3, 7, 8, 9, 10}




      Drop[Range @ 10, 4 ;; 6]




      {1, 2, 3, 7, 8, 9, 10}





      Drop[CharacterRange["a", "j"], {4, 6}]



      {"a", "b", "c", "g", "h", "i", "j"}








      share|improve this answer














      share|improve this answer



      share|improve this answer








      edited 1 hour ago

























      answered 11 hours ago









      kglrkglr

      210k10 gold badges241 silver badges481 bronze badges




      210k10 gold badges241 silver badges481 bronze badges











      • 1




        $begingroup$
        Hah, shot myself in the foot with my end question...That works to get the values, but I'm unsure of the readability when we do a[[Drop[Range@10, {4, 6}]]] = b; a. Though, it is a nice implementation and one I had not considered. I'm hoping to find something that is not too over-encumbered, in order to sit in a publication. This is much nicer than the original try of mine, and it only uses the definition of the neglected characters. Quite nice, actually, the more I debate it! Much more general of an approach. +1
        $endgroup$
        – CA Trevillian
        11 hours ago














      • 1




        $begingroup$
        Hah, shot myself in the foot with my end question...That works to get the values, but I'm unsure of the readability when we do a[[Drop[Range@10, {4, 6}]]] = b; a. Though, it is a nice implementation and one I had not considered. I'm hoping to find something that is not too over-encumbered, in order to sit in a publication. This is much nicer than the original try of mine, and it only uses the definition of the neglected characters. Quite nice, actually, the more I debate it! Much more general of an approach. +1
        $endgroup$
        – CA Trevillian
        11 hours ago








      1




      1




      $begingroup$
      Hah, shot myself in the foot with my end question...That works to get the values, but I'm unsure of the readability when we do a[[Drop[Range@10, {4, 6}]]] = b; a. Though, it is a nice implementation and one I had not considered. I'm hoping to find something that is not too over-encumbered, in order to sit in a publication. This is much nicer than the original try of mine, and it only uses the definition of the neglected characters. Quite nice, actually, the more I debate it! Much more general of an approach. +1
      $endgroup$
      – CA Trevillian
      11 hours ago




      $begingroup$
      Hah, shot myself in the foot with my end question...That works to get the values, but I'm unsure of the readability when we do a[[Drop[Range@10, {4, 6}]]] = b; a. Though, it is a nice implementation and one I had not considered. I'm hoping to find something that is not too over-encumbered, in order to sit in a publication. This is much nicer than the original try of mine, and it only uses the definition of the neglected characters. Quite nice, actually, the more I debate it! Much more general of an approach. +1
      $endgroup$
      – CA Trevillian
      11 hours ago













      5












      $begingroup$

      If it is okay to perform two assignments instead of one, then we can write:



      (a[[#]] = b) & /@ {1 ;; 3, 7 ;; 10}


      Scan is probably better since it does not bother constructing the result list that we are just going to discard anyway:



      Scan[(a[[#]] = b) &, {1 ;; 3, 7 ;; 10}]





      share|improve this answer









      $endgroup$















      • $begingroup$
        Well aaactually--that was the original reason for this question, and I probably could have written that....but I can't remember what cup of coffee I was on at that point! Regardless, this is a great implementation of a pure function applied to an index being used in a functional reassignment. This is technically in one line too, which satisfied my original criteria! Very cool. I wonder if this can be shorted at all for some sort-of code-golf-like...thing...answer? (Hint-hint, but it'd require some user thought) Oooooh scan, yessiree this is a great one, Scan needs more use these days, imo.
        $endgroup$
        – CA Trevillian
        9 hours ago
















      5












      $begingroup$

      If it is okay to perform two assignments instead of one, then we can write:



      (a[[#]] = b) & /@ {1 ;; 3, 7 ;; 10}


      Scan is probably better since it does not bother constructing the result list that we are just going to discard anyway:



      Scan[(a[[#]] = b) &, {1 ;; 3, 7 ;; 10}]





      share|improve this answer









      $endgroup$















      • $begingroup$
        Well aaactually--that was the original reason for this question, and I probably could have written that....but I can't remember what cup of coffee I was on at that point! Regardless, this is a great implementation of a pure function applied to an index being used in a functional reassignment. This is technically in one line too, which satisfied my original criteria! Very cool. I wonder if this can be shorted at all for some sort-of code-golf-like...thing...answer? (Hint-hint, but it'd require some user thought) Oooooh scan, yessiree this is a great one, Scan needs more use these days, imo.
        $endgroup$
        – CA Trevillian
        9 hours ago














      5












      5








      5





      $begingroup$

      If it is okay to perform two assignments instead of one, then we can write:



      (a[[#]] = b) & /@ {1 ;; 3, 7 ;; 10}


      Scan is probably better since it does not bother constructing the result list that we are just going to discard anyway:



      Scan[(a[[#]] = b) &, {1 ;; 3, 7 ;; 10}]





      share|improve this answer









      $endgroup$



      If it is okay to perform two assignments instead of one, then we can write:



      (a[[#]] = b) & /@ {1 ;; 3, 7 ;; 10}


      Scan is probably better since it does not bother constructing the result list that we are just going to discard anyway:



      Scan[(a[[#]] = b) &, {1 ;; 3, 7 ;; 10}]






      share|improve this answer












      share|improve this answer



      share|improve this answer










      answered 9 hours ago









      WReachWReach

      54.9k2 gold badges118 silver badges217 bronze badges




      54.9k2 gold badges118 silver badges217 bronze badges















      • $begingroup$
        Well aaactually--that was the original reason for this question, and I probably could have written that....but I can't remember what cup of coffee I was on at that point! Regardless, this is a great implementation of a pure function applied to an index being used in a functional reassignment. This is technically in one line too, which satisfied my original criteria! Very cool. I wonder if this can be shorted at all for some sort-of code-golf-like...thing...answer? (Hint-hint, but it'd require some user thought) Oooooh scan, yessiree this is a great one, Scan needs more use these days, imo.
        $endgroup$
        – CA Trevillian
        9 hours ago


















      • $begingroup$
        Well aaactually--that was the original reason for this question, and I probably could have written that....but I can't remember what cup of coffee I was on at that point! Regardless, this is a great implementation of a pure function applied to an index being used in a functional reassignment. This is technically in one line too, which satisfied my original criteria! Very cool. I wonder if this can be shorted at all for some sort-of code-golf-like...thing...answer? (Hint-hint, but it'd require some user thought) Oooooh scan, yessiree this is a great one, Scan needs more use these days, imo.
        $endgroup$
        – CA Trevillian
        9 hours ago
















      $begingroup$
      Well aaactually--that was the original reason for this question, and I probably could have written that....but I can't remember what cup of coffee I was on at that point! Regardless, this is a great implementation of a pure function applied to an index being used in a functional reassignment. This is technically in one line too, which satisfied my original criteria! Very cool. I wonder if this can be shorted at all for some sort-of code-golf-like...thing...answer? (Hint-hint, but it'd require some user thought) Oooooh scan, yessiree this is a great one, Scan needs more use these days, imo.
      $endgroup$
      – CA Trevillian
      9 hours ago




      $begingroup$
      Well aaactually--that was the original reason for this question, and I probably could have written that....but I can't remember what cup of coffee I was on at that point! Regardless, this is a great implementation of a pure function applied to an index being used in a functional reassignment. This is technically in one line too, which satisfied my original criteria! Very cool. I wonder if this can be shorted at all for some sort-of code-golf-like...thing...answer? (Hint-hint, but it'd require some user thought) Oooooh scan, yessiree this is a great one, Scan needs more use these days, imo.
      $endgroup$
      – CA Trevillian
      9 hours ago











      5












      $begingroup$

      Perhaps this?:



      ReplacePart[Range@10, i_ /; Not[4 <= i <= 6] :> b]
      
(*  {b, b, b, 4, 5, 6, b, b, b, b}  *)





      share|improve this answer









      $endgroup$















      • $begingroup$
        Wow nice!! I like the use of RuleDelayed, am I correct in understanding that you could use either Table[n,{n,10}] or Table[n^2,{n,10}] and we should see the 4th-6th indexed values preserved, as ReplacePart replaces the ith index? The use of Not provides a good deal of readability as well.
        $endgroup$
        – CA Trevillian
        7 hours ago






      • 1




        $begingroup$
        @CATrevillian Yes, it depends only on the index (part/position), not the value.
        $endgroup$
        – Michael E2
        6 hours ago
















      5












      $begingroup$

      Perhaps this?:



      ReplacePart[Range@10, i_ /; Not[4 <= i <= 6] :> b]
      
(*  {b, b, b, 4, 5, 6, b, b, b, b}  *)





      share|improve this answer









      $endgroup$















      • $begingroup$
        Wow nice!! I like the use of RuleDelayed, am I correct in understanding that you could use either Table[n,{n,10}] or Table[n^2,{n,10}] and we should see the 4th-6th indexed values preserved, as ReplacePart replaces the ith index? The use of Not provides a good deal of readability as well.
        $endgroup$
        – CA Trevillian
        7 hours ago






      • 1




        $begingroup$
        @CATrevillian Yes, it depends only on the index (part/position), not the value.
        $endgroup$
        – Michael E2
        6 hours ago














      5












      5








      5





      $begingroup$

      Perhaps this?:



      ReplacePart[Range@10, i_ /; Not[4 <= i <= 6] :> b]
      
(*  {b, b, b, 4, 5, 6, b, b, b, b}  *)





      share|improve this answer









      $endgroup$



      Perhaps this?:



      ReplacePart[Range@10, i_ /; Not[4 <= i <= 6] :> b]
      
(*  {b, b, b, 4, 5, 6, b, b, b, b}  *)






      share|improve this answer












      share|improve this answer



      share|improve this answer










      answered 7 hours ago









      Michael E2Michael E2

      157k13 gold badges215 silver badges511 bronze badges




      157k13 gold badges215 silver badges511 bronze badges















      • $begingroup$
        Wow nice!! I like the use of RuleDelayed, am I correct in understanding that you could use either Table[n,{n,10}] or Table[n^2,{n,10}] and we should see the 4th-6th indexed values preserved, as ReplacePart replaces the ith index? The use of Not provides a good deal of readability as well.
        $endgroup$
        – CA Trevillian
        7 hours ago






      • 1




        $begingroup$
        @CATrevillian Yes, it depends only on the index (part/position), not the value.
        $endgroup$
        – Michael E2
        6 hours ago


















      • $begingroup$
        Wow nice!! I like the use of RuleDelayed, am I correct in understanding that you could use either Table[n,{n,10}] or Table[n^2,{n,10}] and we should see the 4th-6th indexed values preserved, as ReplacePart replaces the ith index? The use of Not provides a good deal of readability as well.
        $endgroup$
        – CA Trevillian
        7 hours ago






      • 1




        $begingroup$
        @CATrevillian Yes, it depends only on the index (part/position), not the value.
        $endgroup$
        – Michael E2
        6 hours ago
















      $begingroup$
      Wow nice!! I like the use of RuleDelayed, am I correct in understanding that you could use either Table[n,{n,10}] or Table[n^2,{n,10}] and we should see the 4th-6th indexed values preserved, as ReplacePart replaces the ith index? The use of Not provides a good deal of readability as well.
      $endgroup$
      – CA Trevillian
      7 hours ago




      $begingroup$
      Wow nice!! I like the use of RuleDelayed, am I correct in understanding that you could use either Table[n,{n,10}] or Table[n^2,{n,10}] and we should see the 4th-6th indexed values preserved, as ReplacePart replaces the ith index? The use of Not provides a good deal of readability as well.
      $endgroup$
      – CA Trevillian
      7 hours ago




      1




      1




      $begingroup$
      @CATrevillian Yes, it depends only on the index (part/position), not the value.
      $endgroup$
      – Michael E2
      6 hours ago




      $begingroup$
      @CATrevillian Yes, it depends only on the index (part/position), not the value.
      $endgroup$
      – Michael E2
      6 hours ago











      4












      $begingroup$

      We can use Union[] to perform this operation in a simpler manner:



      a[[Range[1, 3] [Union] Range[7, 10]]]
      
(*{1, 2, 3, 7, 8, 9, 10}*)


      But this is still decently messy, and the readability is not too high. Perhaps someone else has a better method?



      a[[Range[1, 3] [Union] Range[7, 10]]] = b; a
      
(*{b, b, b, 4, 5, 6, b, b, b, b}*)





      share|improve this answer









      $endgroup$




















        4












        $begingroup$

        We can use Union[] to perform this operation in a simpler manner:



        a[[Range[1, 3] [Union] Range[7, 10]]]
        
(*{1, 2, 3, 7, 8, 9, 10}*)


        But this is still decently messy, and the readability is not too high. Perhaps someone else has a better method?



        a[[Range[1, 3] [Union] Range[7, 10]]] = b; a
        
(*{b, b, b, 4, 5, 6, b, b, b, b}*)





        share|improve this answer









        $endgroup$


















          4












          4








          4





          $begingroup$

          We can use Union[] to perform this operation in a simpler manner:



          a[[Range[1, 3] [Union] Range[7, 10]]]
          
(*{1, 2, 3, 7, 8, 9, 10}*)


          But this is still decently messy, and the readability is not too high. Perhaps someone else has a better method?



          a[[Range[1, 3] [Union] Range[7, 10]]] = b; a
          
(*{b, b, b, 4, 5, 6, b, b, b, b}*)





          share|improve this answer









          $endgroup$



          We can use Union[] to perform this operation in a simpler manner:



          a[[Range[1, 3] [Union] Range[7, 10]]]
          
(*{1, 2, 3, 7, 8, 9, 10}*)


          But this is still decently messy, and the readability is not too high. Perhaps someone else has a better method?



          a[[Range[1, 3] [Union] Range[7, 10]]] = b; a
          
(*{b, b, b, 4, 5, 6, b, b, b, b}*)






          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered 11 hours ago









          CA TrevillianCA Trevillian

          5891 gold badge2 silver badges12 bronze badges




          5891 gold badge2 silver badges12 bronze badges


























              4












              $begingroup$

              By far the simplest way is to do it in two lines:



              a = Table[RandomReal[], {n, 10}];
              
a[[1 ;; 3]] = b;
              
a[[7 ;; 10]] = b;
              
a


              This gives the output:




              {b, b, b, 0.378846, 0.475894, 0.533768, b, b, b, b}







              share|improve this answer











              $endgroup$




















                4












                $begingroup$

                By far the simplest way is to do it in two lines:



                a = Table[RandomReal[], {n, 10}];
                
a[[1 ;; 3]] = b;
                
a[[7 ;; 10]] = b;
                
a


                This gives the output:




                {b, b, b, 0.378846, 0.475894, 0.533768, b, b, b, b}







                share|improve this answer











                $endgroup$


















                  4












                  4








                  4





                  $begingroup$

                  By far the simplest way is to do it in two lines:



                  a = Table[RandomReal[], {n, 10}];
                  
a[[1 ;; 3]] = b;
                  
a[[7 ;; 10]] = b;
                  
a


                  This gives the output:




                  {b, b, b, 0.378846, 0.475894, 0.533768, b, b, b, b}







                  share|improve this answer











                  $endgroup$



                  By far the simplest way is to do it in two lines:



                  a = Table[RandomReal[], {n, 10}];
                  
a[[1 ;; 3]] = b;
                  
a[[7 ;; 10]] = b;
                  
a


                  This gives the output:




                  {b, b, b, 0.378846, 0.475894, 0.533768, b, b, b, b}








                  share|improve this answer














                  share|improve this answer



                  share|improve this answer








                  edited 1 hour ago

























                  answered 4 hours ago









                  axsvl77axsvl77

                  4973 silver badges16 bronze badges




                  4973 silver badges16 bronze badges

































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