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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;
$begingroup$
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
asked 10 hours ago
CA TrevillianCA Trevillian
5741 gold badge2 silver badges12 bronze badges
$endgroup$
|
show 6 more comments
$begingroup$
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
asked 10 hours ago
CA TrevillianCA Trevillian
5741 gold badge2 silver badges12 bronze badges
$endgroup$
1
$begingroup$
Select[ a, ! MemberQ[ 4, 5, 6, #] & ]
$endgroup$
– LouisB
10 hours ago
1
$begingroup$
Complement[ a, 4, 5, 6 ]
$endgroup$
– LouisB
9 hours ago
1
$begingroup$
Cheating a little, we could write(a[[#]] = b) & /@ 1 ;; 3, 7 ;; 10.
$endgroup$
– WReach
8 hours ago
1
$begingroup$
If it's the values, then inTable[n^2,n,10]the initial segment would be the span1 ;; 1and the final segment would be3 ;; 10and the excluded segment (the positions not to be changed) would be2 ;; 2, if the excluded values were 4, 5, 6 -- no?
$endgroup$
– Michael E2
6 hours ago
1
$begingroup$
Thanks. I posted an answer. See if I understood correctly.
$endgroup$
– Michael E2
6 hours ago
|
show 6 more comments
$begingroup$
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
asked 10 hours ago
CA TrevillianCA Trevillian
5741 gold badge2 silver badges12 bronze badges
$endgroup$
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
asked 10 hours ago
CA TrevillianCA Trevillian
5741 gold badge2 silver badges12 bronze badges
asked 10 hours ago
CA TrevillianCA Trevillian
5741 gold badge2 silver badges12 bronze badges
asked 10 hours ago
CA TrevillianCA Trevillian
5741 gold badge2 silver badges12 bronze badges
asked 10 hours ago
CA TrevillianCA Trevillian
5741 gold badge2 silver badges12 bronze badges
asked 10 hours ago
CA TrevillianCA Trevillian
5741 gold badge2 silver badges12 bronze badges
5741 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
9 hours ago
1
$begingroup$
Cheating a little, we could write(a[[#]] = b) & /@ 1 ;; 3, 7 ;; 10.
$endgroup$
– WReach
8 hours ago
1
$begingroup$
If it's the values, then inTable[n^2,n,10]the initial segment would be the span1 ;; 1and the final segment would be3 ;; 10and the excluded segment (the positions not to be changed) would be2 ;; 2, if the excluded values were 4, 5, 6 -- no?
$endgroup$
– Michael E2
6 hours ago
1
$begingroup$
Thanks. I posted an answer. See if I understood correctly.
$endgroup$
– Michael E2
6 hours ago
|
show 6 more comments
1
$begingroup$
Select[ a, ! MemberQ[ 4, 5, 6, #] & ]
$endgroup$
– LouisB
10 hours ago
1
$begingroup$
Complement[ a, 4, 5, 6 ]
$endgroup$
– LouisB
9 hours ago
1
$begingroup$
Cheating a little, we could write(a[[#]] = b) & /@ 1 ;; 3, 7 ;; 10.
$endgroup$
– WReach
8 hours ago
1
$begingroup$
If it's the values, then inTable[n^2,n,10]the initial segment would be the span1 ;; 1and the final segment would be3 ;; 10and the excluded segment (the positions not to be changed) would be2 ;; 2, if the excluded values were 4, 5, 6 -- no?
$endgroup$
– Michael E2
6 hours ago
1
$begingroup$
Thanks. I posted an answer. See if I understood correctly.
$endgroup$
– Michael E2
6 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
9 hours ago
$begingroup$
Complement[ a, 4, 5, 6 ]$endgroup$
– LouisB
9 hours ago
1
1
$begingroup$
Cheating a little, we could write
(a[[#]] = b) & /@ 1 ;; 3, 7 ;; 10.$endgroup$
– WReach
8 hours ago
$begingroup$
Cheating a little, we could write
(a[[#]] = b) & /@ 1 ;; 3, 7 ;; 10.$endgroup$
– WReach
8 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
6 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
6 hours ago
1
1
$begingroup$
Thanks. I posted an answer. See if I understood correctly.
$endgroup$
– Michael E2
6 hours ago
$begingroup$
Thanks. I posted an answer. See if I understood correctly.
$endgroup$
– Michael E2
6 hours ago
|
show 6 more comments
5 Answers
5
active
oldest
votes
$begingroup$
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"
answered 10 hours ago
kglrkglr
210k10 gold badges241 silver badges481 bronze badges
$endgroup$
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 doa[[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
10 hours ago
add a comment |
$begingroup$
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*)
answered 10 hours ago
CA TrevillianCA Trevillian
5741 gold badge2 silver badges12 bronze badges
$endgroup$
add a comment |
$begingroup$
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]
answered 8 hours ago
WReachWReach
54.8k2 gold badges118 silver badges217 bronze badges
$endgroup$
$begingroup$
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,Scanneeds more use these days, imo.
$endgroup$
– CA Trevillian
8 hours ago
add a comment |
$begingroup$
Perhaps this?:
ReplacePart[Range@10, i_ /; Not[4 <= i <= 6] :> b]
(* b, b, b, 4, 5, 6, b, b, b, b *)
answered 6 hours ago
Michael E2Michael E2
157k13 gold badges215 silver badges511 bronze badges
$endgroup$
$begingroup$
Wow nice!! I like the use ofRuleDelayed, am I correct in understanding that you could use eitherTable[n,n,10]orTable[n^2,n,10]and we should see the 4th-6th indexed values preserved, asReplacePartreplaces the ith index? The use ofNotprovides a good deal of readability as well.
$endgroup$
– CA Trevillian
6 hours ago
1
$begingroup$
@CATrevillian Yes, it depends only on the index (part/position), not the value.
$endgroup$
– Michael E2
6 hours ago
add a comment |
$begingroup$
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
answered 3 hours ago
axsvl77axsvl77
4773 silver badges16 bronze badges
$endgroup$
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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"
answered 10 hours ago
kglrkglr
210k10 gold badges241 silver badges481 bronze badges
$endgroup$
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 doa[[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
10 hours ago
add a comment |
$begingroup$
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"
answered 10 hours ago
kglrkglr
210k10 gold badges241 silver badges481 bronze badges
$endgroup$
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 doa[[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
10 hours ago
add a comment |
$begingroup$
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"
answered 10 hours ago
kglrkglr
210k10 gold badges241 silver badges481 bronze badges
$endgroup$
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"
answered 10 hours ago
kglrkglr
210k10 gold badges241 silver badges481 bronze badges
edited 32 mins ago
answered 10 hours ago
kglrkglr
210k10 gold badges241 silver badges481 bronze badges
answered 10 hours ago
kglrkglr
210k10 gold badges241 silver badges481 bronze badges
answered 10 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 doa[[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
10 hours ago
add a comment |
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 doa[[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
10 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
10 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
10 hours ago
add a comment |
$begingroup$
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*)
answered 10 hours ago
CA TrevillianCA Trevillian
5741 gold badge2 silver badges12 bronze badges
$endgroup$
add a comment |
$begingroup$
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*)
answered 10 hours ago
CA TrevillianCA Trevillian
5741 gold badge2 silver badges12 bronze badges
$endgroup$
add a comment |
$begingroup$
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*)
answered 10 hours ago
CA TrevillianCA Trevillian
5741 gold badge2 silver badges12 bronze badges
$endgroup$
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*)
answered 10 hours ago
CA TrevillianCA Trevillian
5741 gold badge2 silver badges12 bronze badges
answered 10 hours ago
CA TrevillianCA Trevillian
5741 gold badge2 silver badges12 bronze badges
answered 10 hours ago
CA TrevillianCA Trevillian
5741 gold badge2 silver badges12 bronze badges
answered 10 hours ago
CA TrevillianCA Trevillian
5741 gold badge2 silver badges12 bronze badges
5741 gold badge2 silver badges12 bronze badges
add a comment |
add a comment |
$begingroup$
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]
answered 8 hours ago
WReachWReach
54.8k2 gold badges118 silver badges217 bronze badges
$endgroup$
$begingroup$
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,Scanneeds more use these days, imo.
$endgroup$
– CA Trevillian
8 hours ago
add a comment |
$begingroup$
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]
answered 8 hours ago
WReachWReach
54.8k2 gold badges118 silver badges217 bronze badges
$endgroup$
$begingroup$
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,Scanneeds more use these days, imo.
$endgroup$
– CA Trevillian
8 hours ago
add a comment |
$begingroup$
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]
answered 8 hours ago
WReachWReach
54.8k2 gold badges118 silver badges217 bronze badges
$endgroup$
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]
answered 8 hours ago
WReachWReach
54.8k2 gold badges118 silver badges217 bronze badges
answered 8 hours ago
WReachWReach
54.8k2 gold badges118 silver badges217 bronze badges
answered 8 hours ago
WReachWReach
54.8k2 gold badges118 silver badges217 bronze badges
answered 8 hours ago
WReachWReach
54.8k2 gold badges118 silver badges217 bronze badges
54.8k2 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,Scanneeds more use these days, imo.
$endgroup$
– CA Trevillian
8 hours ago
add a comment |
$begingroup$
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,Scanneeds more use these days, imo.
$endgroup$
– CA Trevillian
8 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
8 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
8 hours ago
add a comment |
$begingroup$
Perhaps this?:
ReplacePart[Range@10, i_ /; Not[4 <= i <= 6] :> b]
(* b, b, b, 4, 5, 6, b, b, b, b *)
answered 6 hours ago
Michael E2Michael E2
157k13 gold badges215 silver badges511 bronze badges
$endgroup$
$begingroup$
Wow nice!! I like the use ofRuleDelayed, am I correct in understanding that you could use eitherTable[n,n,10]orTable[n^2,n,10]and we should see the 4th-6th indexed values preserved, asReplacePartreplaces the ith index? The use ofNotprovides a good deal of readability as well.
$endgroup$
– CA Trevillian
6 hours ago
1
$begingroup$
@CATrevillian Yes, it depends only on the index (part/position), not the value.
$endgroup$
– Michael E2
6 hours ago
add a comment |
$begingroup$
Perhaps this?:
ReplacePart[Range@10, i_ /; Not[4 <= i <= 6] :> b]
(* b, b, b, 4, 5, 6, b, b, b, b *)
answered 6 hours ago
Michael E2Michael E2
157k13 gold badges215 silver badges511 bronze badges
$endgroup$
$begingroup$
Wow nice!! I like the use ofRuleDelayed, am I correct in understanding that you could use eitherTable[n,n,10]orTable[n^2,n,10]and we should see the 4th-6th indexed values preserved, asReplacePartreplaces the ith index? The use ofNotprovides a good deal of readability as well.
$endgroup$
– CA Trevillian
6 hours ago
1
$begingroup$
@CATrevillian Yes, it depends only on the index (part/position), not the value.
$endgroup$
– Michael E2
6 hours ago
add a comment |
$begingroup$
Perhaps this?:
ReplacePart[Range@10, i_ /; Not[4 <= i <= 6] :> b]
(* b, b, b, 4, 5, 6, b, b, b, b *)
answered 6 hours ago
Michael E2Michael E2
157k13 gold badges215 silver badges511 bronze badges
$endgroup$
Perhaps this?:
ReplacePart[Range@10, i_ /; Not[4 <= i <= 6] :> b]
(* b, b, b, 4, 5, 6, b, b, b, b *)
answered 6 hours ago
Michael E2Michael E2
157k13 gold badges215 silver badges511 bronze badges
answered 6 hours ago
Michael E2Michael E2
157k13 gold badges215 silver badges511 bronze badges
answered 6 hours ago
Michael E2Michael E2
157k13 gold badges215 silver badges511 bronze badges
answered 6 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 ofRuleDelayed, am I correct in understanding that you could use eitherTable[n,n,10]orTable[n^2,n,10]and we should see the 4th-6th indexed values preserved, asReplacePartreplaces the ith index? The use ofNotprovides a good deal of readability as well.
$endgroup$
– CA Trevillian
6 hours ago
1
$begingroup$
@CATrevillian Yes, it depends only on the index (part/position), not the value.
$endgroup$
– Michael E2
6 hours ago
add a comment |
$begingroup$
Wow nice!! I like the use ofRuleDelayed, am I correct in understanding that you could use eitherTable[n,n,10]orTable[n^2,n,10]and we should see the 4th-6th indexed values preserved, asReplacePartreplaces the ith index? The use ofNotprovides a good deal of readability as well.
$endgroup$
– CA Trevillian
6 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
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
6 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
add a comment |
$begingroup$
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
answered 3 hours ago
axsvl77axsvl77
4773 silver badges16 bronze badges
$endgroup$
add a comment |
$begingroup$
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
answered 3 hours ago
axsvl77axsvl77
4773 silver badges16 bronze badges
$endgroup$
add a comment |
$begingroup$
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
answered 3 hours ago
axsvl77axsvl77
4773 silver badges16 bronze badges
$endgroup$
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
answered 3 hours ago
axsvl77axsvl77
4773 silver badges16 bronze badges
edited 20 mins ago
answered 3 hours ago
axsvl77axsvl77
4773 silver badges16 bronze badges
answered 3 hours ago
axsvl77axsvl77
4773 silver badges16 bronze badges
answered 3 hours ago
axsvl77axsvl77
4773 silver badges16 bronze badges
4773 silver badges16 bronze badges
add a comment |
add a comment |
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1
$begingroup$
Select[ a, ! MemberQ[ 4, 5, 6, #] & ]$endgroup$
– LouisB
10 hours ago
1
$begingroup$
Complement[ a, 4, 5, 6 ]$endgroup$
– LouisB
9 hours ago
1
$begingroup$
Cheating a little, we could write
(a[[#]] = b) & /@ 1 ;; 3, 7 ;; 10.$endgroup$
– WReach
8 hours ago
1
$begingroup$
If it's the values, then in
Table[n^2,n,10]the initial segment would be the span1 ;; 1and the final segment would be3 ;; 10and the excluded segment (the positions not to be changed) would be2 ;; 2, if the excluded values were 4, 5, 6 -- no?$endgroup$
– Michael E2
6 hours ago
1
$begingroup$
Thanks. I posted an answer. See if I understood correctly.
$endgroup$
– Michael E2
6 hours ago