Replace zeros in a list with last nonzero valueReplace “,” in a list with “.”Removing trailing zeros from a listReplace interior of matrix with zerosHow to copy with the last 1 with pattern matching method in a listremoving list containing zerosReplace elements of a matrix with zeros using another matrixIdentify the location of the last entry with a particular value in a list
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Replace zeros in a list with last nonzero value
Replace “,” in a list with “.”Removing trailing zeros from a listReplace interior of matrix with zerosHow to copy with the last 1 with pattern matching method in a listremoving list containing zerosReplace elements of a matrix with zeros using another matrixIdentify the location of the last entry with a particular value in a list
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty
margin-bottom:0;
.everyonelovesstackoverflowposition:absolute;height:1px;width:1px;opacity:0;top:0;left:0;pointer-events:none;
$begingroup$
This question is closely related to the problem of zero crossings.
Suppose I have a list l = -1, -1, 0, 1, 0, 0, -1, -1, 1, 1, 1, 0, 0, 0, -1
and I want to replace the zeros with the previous nonzero value, like this:
-1, -1, -1, 1, 1, 1, -1, -1, 1, 1, 1, 1, 1, 1, -1
A rather clunky way to do this is with:
f = TimeSeries[l /. 0 -> Missing[],
MissingDataMethod -> "Interpolation", InterpolationOrder -> 0]
Then indeed f[-1 + Range@Length@l]gives the required result.
However, there must be a more elegant way to achieve same, with patterns.
Can anyone supply a suggestion?
list-manipulation pattern-matching
asked Sep 25 at 12:11
Jonathan KinlayJonathan Kinlay
6273 silver badges11 bronze badges
$endgroup$
add a comment
|
$begingroup$
This question is closely related to the problem of zero crossings.
Suppose I have a list l = -1, -1, 0, 1, 0, 0, -1, -1, 1, 1, 1, 0, 0, 0, -1
and I want to replace the zeros with the previous nonzero value, like this:
-1, -1, -1, 1, 1, 1, -1, -1, 1, 1, 1, 1, 1, 1, -1
A rather clunky way to do this is with:
f = TimeSeries[l /. 0 -> Missing[],
MissingDataMethod -> "Interpolation", InterpolationOrder -> 0]
Then indeed f[-1 + Range@Length@l]gives the required result.
However, there must be a more elegant way to achieve same, with patterns.
Can anyone supply a suggestion?
list-manipulation pattern-matching
asked Sep 25 at 12:11
Jonathan KinlayJonathan Kinlay
6273 silver badges11 bronze badges
$endgroup$
4
$begingroup$
What should happen when the first element is a zero?
$endgroup$
– Sjoerd Smit
Sep 25 at 14:53
add a comment
|
$begingroup$
This question is closely related to the problem of zero crossings.
Suppose I have a list l = -1, -1, 0, 1, 0, 0, -1, -1, 1, 1, 1, 0, 0, 0, -1
and I want to replace the zeros with the previous nonzero value, like this:
-1, -1, -1, 1, 1, 1, -1, -1, 1, 1, 1, 1, 1, 1, -1
A rather clunky way to do this is with:
f = TimeSeries[l /. 0 -> Missing[],
MissingDataMethod -> "Interpolation", InterpolationOrder -> 0]
Then indeed f[-1 + Range@Length@l]gives the required result.
However, there must be a more elegant way to achieve same, with patterns.
Can anyone supply a suggestion?
list-manipulation pattern-matching
asked Sep 25 at 12:11
Jonathan KinlayJonathan Kinlay
6273 silver badges11 bronze badges
$endgroup$
This question is closely related to the problem of zero crossings.
Suppose I have a list l = -1, -1, 0, 1, 0, 0, -1, -1, 1, 1, 1, 0, 0, 0, -1
and I want to replace the zeros with the previous nonzero value, like this:
-1, -1, -1, 1, 1, 1, -1, -1, 1, 1, 1, 1, 1, 1, -1
A rather clunky way to do this is with:
f = TimeSeries[l /. 0 -> Missing[],
MissingDataMethod -> "Interpolation", InterpolationOrder -> 0]
Then indeed f[-1 + Range@Length@l]gives the required result.
However, there must be a more elegant way to achieve same, with patterns.
Can anyone supply a suggestion?
list-manipulation pattern-matching
list-manipulation pattern-matching
asked Sep 25 at 12:11
Jonathan KinlayJonathan Kinlay
6273 silver badges11 bronze badges
asked Sep 25 at 12:11
Jonathan KinlayJonathan Kinlay
6273 silver badges11 bronze badges
asked Sep 25 at 12:11
Jonathan KinlayJonathan Kinlay
6273 silver badges11 bronze badges
asked Sep 25 at 12:11
Jonathan KinlayJonathan Kinlay
6273 silver badges11 bronze badges
asked Sep 25 at 12:11
Jonathan KinlayJonathan Kinlay
6273 silver badges11 bronze badges
6273 silver badges11 bronze badges
4
$begingroup$
What should happen when the first element is a zero?
$endgroup$
– Sjoerd Smit
Sep 25 at 14:53
add a comment
|
4
$begingroup$
What should happen when the first element is a zero?
$endgroup$
– Sjoerd Smit
Sep 25 at 14:53
4
4
$begingroup$
What should happen when the first element is a zero?
$endgroup$
– Sjoerd Smit
Sep 25 at 14:53
$begingroup$
What should happen when the first element is a zero?
$endgroup$
– Sjoerd Smit
Sep 25 at 14:53
add a comment
|
6 Answers
6
active
oldest
votes
$begingroup$
SequenceReplace
SequenceReplace[ p:a_, 0.. :> Sequence @@ (p /. 0 -> a)] @ l
-1, -1, -1, 1, 1, 1, -1, -1, 1, 1, 1, 1, 1, 1, -1
or
FixedPoint[SequenceReplace[a_, 0 :> Sequence[a, a]], l]
-1, -1, -1, 1, 1, 1, -1, -1, 1, 1, 1, 1, 1, 1, -1
ReplaceRepeated
An alternative way to use ReplaceRepeated with a single replacement rule:
l //. a___, b_, 0, c___ :> a, b, b, c
-1, -1, -1, 1, 1, 1, -1, -1, 1, 1, 1, 1, 1, 1, -1
Memoization
Memoize the last non-zero value (inspired by @WReach's answer):
f[x_] := (f[0] = x; x)
f /@ l
-1, -1, -1, 1, 1, 1, -1, -1, 1, 1, 1, 1, 1, 1, -1
Split + ReplaceAll
Split[l, #2 == 0 &] /. p : a_, 0 .. :> (p /. 0 -> a) // Flatten
-1, -1, -1, 1, 1, 1, -1, -1, 1, 1, 1, 1, 1, 1, -1
TimeSeries + MissingDataMethod
Using the "Values" property with OP's f:
f["Values"]
-1, -1, -1, 1, 1, 1, -1, -1, 1, 1, 1, 1, 1, 1, -1
answered Sep 25 at 12:18
kglrkglr
228k10 gold badges257 silver badges519 bronze badges
$endgroup$
1
$begingroup$
One of the reasons I asked this question even though I had a working (although inelegant) solution was to see all the creative ways in which MMA experts are able to use the Wolfram language to solve a problem. Outstanding!
$endgroup$
– Jonathan Kinlay
Sep 25 at 22:41
$begingroup$
What sort of (Mathematica) black magic is this!? My +1 sir.
$endgroup$
– Rebel-Scum
Sep 27 at 13:54
add a comment
|
$begingroup$
Define a simple function, using FoldList
op = FoldList[If[#2 == 0, #1, #2] &];
l = -1, -1, 0, 1, 0, 0, -1, -1, 1, 1, 1, 0, 0, 0, -1;
op@l
(* -1, -1, -1, 1, 1, 1, -1, -1, 1, 1, 1, 1, 1, 1, -1 *)
answered Sep 26 at 19:06
mikadomikado
9,2272 gold badges9 silver badges32 bronze badges
$endgroup$
$begingroup$
This must be the fastest (and simplest) of the lot. The one I knew existed, but never reached.
$endgroup$
– Suba Thomas
Sep 26 at 22:19
add a comment
|
$begingroup$
Perhaps something like this will appeal
l //. a___, 1, 0, b___ -> a, 1, 1, b, a___, -1, 0, b___ -> a, -1, -1, b
answered Sep 25 at 12:19
High Performance MarkHigh Performance Mark
1,1187 silver badges14 bronze badges
$endgroup$
$begingroup$
This is the kind of thing I was aiming for, but I couldn't find the right formulation. I find patterns such a challenge (although I do use them extensively). Anyway that you for a great solution.
$endgroup$
– Jonathan Kinlay
Sep 25 at 22:35
add a comment
|
$begingroup$
Given:
list = -1, -1, 0, 1, 0, 0, -1, -1, 1, 1, 1, 0, 0, 0, -1
Then:
Module[prev = 0, Replace[list, 0 :> prev, x_ :> (prev = x), 1]]
(* -1, -1, -1, 1, 1, 1, -1, -1, 1, 1, 1, 1, 1, 1, -1 *)
The initial value assigned by prev = 0 is only used for lists that start with a zero -- choose another value if desired.
answered Sep 25 at 15:47
WReachWReach
56.7k2 gold badges125 silver badges224 bronze badges
$endgroup$
add a comment
|
$begingroup$
A few more...
l = -1, -1, 0, 1, 0, 0, -1, -1, 1, 1, 1, 0, 0, 0, -1;
Do[If[l[[i]] == 0, l[[i]] = l[[i - 1]]], i, 2, Length[l]]
l
-1, -1, -1, 1, 1, 1, -1, -1, 1, 1, 1, 1, 1, 1, -1
fn[l_] := Block[l1 = l,
Set[Part[l1, #[[2]]], Part[l1, #[[1]]]] & /@ SequencePosition[l, _, 0];
l1
]
l = -1, -1, 0, 1, 0, 0, -1, -1, 1, 1, 1, 0, 0, 0, -1;
FixedPoint[fn, l]
-1, -1, -1, 1, 1, 1, -1, -1, 1, 1, 1, 1, 1, 1, -1
l = -1, -1, 0, 1, 0, 0, -1, -1, 1, 1, 1, 0, 0, 0, -1;
Fold[#1, If[#2 == 0, Last[#1], #2] &, l[[1 ;; 1]], l[[2 ;; -1]]] //Flatten
-1, -1, -1, 1, 1, 1, -1, -1, 1, 1, 1, 1, 1, 1, -1
answered Sep 26 at 17:30
Suba ThomasSuba Thomas
5,06911 silver badges20 bronze badges
$endgroup$
add a comment
|
$begingroup$
The creativity demonstrated by all these solutions is impressive.
Here’s a really tough follow-up challenge. Can anyone write a WL program that will generate the code for any one of these solutions?
answered Sep 26 at 22:03
Jonathan KinlayJonathan Kinlay
6273 silver badges11 bronze badges
$endgroup$
6
$begingroup$
This isn't an answer. It should either be an extension to the question, or a fresh question referring back to the original one.
$endgroup$
– High Performance Mark
Sep 27 at 9:38
add a comment
|
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6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
SequenceReplace
SequenceReplace[ p:a_, 0.. :> Sequence @@ (p /. 0 -> a)] @ l
-1, -1, -1, 1, 1, 1, -1, -1, 1, 1, 1, 1, 1, 1, -1
or
FixedPoint[SequenceReplace[a_, 0 :> Sequence[a, a]], l]
-1, -1, -1, 1, 1, 1, -1, -1, 1, 1, 1, 1, 1, 1, -1
ReplaceRepeated
An alternative way to use ReplaceRepeated with a single replacement rule:
l //. a___, b_, 0, c___ :> a, b, b, c
-1, -1, -1, 1, 1, 1, -1, -1, 1, 1, 1, 1, 1, 1, -1
Memoization
Memoize the last non-zero value (inspired by @WReach's answer):
f[x_] := (f[0] = x; x)
f /@ l
-1, -1, -1, 1, 1, 1, -1, -1, 1, 1, 1, 1, 1, 1, -1
Split + ReplaceAll
Split[l, #2 == 0 &] /. p : a_, 0 .. :> (p /. 0 -> a) // Flatten
-1, -1, -1, 1, 1, 1, -1, -1, 1, 1, 1, 1, 1, 1, -1
TimeSeries + MissingDataMethod
Using the "Values" property with OP's f:
f["Values"]
-1, -1, -1, 1, 1, 1, -1, -1, 1, 1, 1, 1, 1, 1, -1
answered Sep 25 at 12:18
kglrkglr
228k10 gold badges257 silver badges519 bronze badges
$endgroup$
1
$begingroup$
One of the reasons I asked this question even though I had a working (although inelegant) solution was to see all the creative ways in which MMA experts are able to use the Wolfram language to solve a problem. Outstanding!
$endgroup$
– Jonathan Kinlay
Sep 25 at 22:41
$begingroup$
What sort of (Mathematica) black magic is this!? My +1 sir.
$endgroup$
– Rebel-Scum
Sep 27 at 13:54
add a comment
|
$begingroup$
SequenceReplace
SequenceReplace[ p:a_, 0.. :> Sequence @@ (p /. 0 -> a)] @ l
-1, -1, -1, 1, 1, 1, -1, -1, 1, 1, 1, 1, 1, 1, -1
or
FixedPoint[SequenceReplace[a_, 0 :> Sequence[a, a]], l]
-1, -1, -1, 1, 1, 1, -1, -1, 1, 1, 1, 1, 1, 1, -1
ReplaceRepeated
An alternative way to use ReplaceRepeated with a single replacement rule:
l //. a___, b_, 0, c___ :> a, b, b, c
-1, -1, -1, 1, 1, 1, -1, -1, 1, 1, 1, 1, 1, 1, -1
Memoization
Memoize the last non-zero value (inspired by @WReach's answer):
f[x_] := (f[0] = x; x)
f /@ l
-1, -1, -1, 1, 1, 1, -1, -1, 1, 1, 1, 1, 1, 1, -1
Split + ReplaceAll
Split[l, #2 == 0 &] /. p : a_, 0 .. :> (p /. 0 -> a) // Flatten
-1, -1, -1, 1, 1, 1, -1, -1, 1, 1, 1, 1, 1, 1, -1
TimeSeries + MissingDataMethod
Using the "Values" property with OP's f:
f["Values"]
-1, -1, -1, 1, 1, 1, -1, -1, 1, 1, 1, 1, 1, 1, -1
answered Sep 25 at 12:18
kglrkglr
228k10 gold badges257 silver badges519 bronze badges
$endgroup$
1
$begingroup$
One of the reasons I asked this question even though I had a working (although inelegant) solution was to see all the creative ways in which MMA experts are able to use the Wolfram language to solve a problem. Outstanding!
$endgroup$
– Jonathan Kinlay
Sep 25 at 22:41
$begingroup$
What sort of (Mathematica) black magic is this!? My +1 sir.
$endgroup$
– Rebel-Scum
Sep 27 at 13:54
add a comment
|
$begingroup$
SequenceReplace
SequenceReplace[ p:a_, 0.. :> Sequence @@ (p /. 0 -> a)] @ l
-1, -1, -1, 1, 1, 1, -1, -1, 1, 1, 1, 1, 1, 1, -1
or
FixedPoint[SequenceReplace[a_, 0 :> Sequence[a, a]], l]
-1, -1, -1, 1, 1, 1, -1, -1, 1, 1, 1, 1, 1, 1, -1
ReplaceRepeated
An alternative way to use ReplaceRepeated with a single replacement rule:
l //. a___, b_, 0, c___ :> a, b, b, c
-1, -1, -1, 1, 1, 1, -1, -1, 1, 1, 1, 1, 1, 1, -1
Memoization
Memoize the last non-zero value (inspired by @WReach's answer):
f[x_] := (f[0] = x; x)
f /@ l
-1, -1, -1, 1, 1, 1, -1, -1, 1, 1, 1, 1, 1, 1, -1
Split + ReplaceAll
Split[l, #2 == 0 &] /. p : a_, 0 .. :> (p /. 0 -> a) // Flatten
-1, -1, -1, 1, 1, 1, -1, -1, 1, 1, 1, 1, 1, 1, -1
TimeSeries + MissingDataMethod
Using the "Values" property with OP's f:
f["Values"]
-1, -1, -1, 1, 1, 1, -1, -1, 1, 1, 1, 1, 1, 1, -1
answered Sep 25 at 12:18
kglrkglr
228k10 gold badges257 silver badges519 bronze badges
$endgroup$
SequenceReplace
SequenceReplace[ p:a_, 0.. :> Sequence @@ (p /. 0 -> a)] @ l
-1, -1, -1, 1, 1, 1, -1, -1, 1, 1, 1, 1, 1, 1, -1
or
FixedPoint[SequenceReplace[a_, 0 :> Sequence[a, a]], l]
-1, -1, -1, 1, 1, 1, -1, -1, 1, 1, 1, 1, 1, 1, -1
ReplaceRepeated
An alternative way to use ReplaceRepeated with a single replacement rule:
l //. a___, b_, 0, c___ :> a, b, b, c
-1, -1, -1, 1, 1, 1, -1, -1, 1, 1, 1, 1, 1, 1, -1
Memoization
Memoize the last non-zero value (inspired by @WReach's answer):
f[x_] := (f[0] = x; x)
f /@ l
-1, -1, -1, 1, 1, 1, -1, -1, 1, 1, 1, 1, 1, 1, -1
Split + ReplaceAll
Split[l, #2 == 0 &] /. p : a_, 0 .. :> (p /. 0 -> a) // Flatten
-1, -1, -1, 1, 1, 1, -1, -1, 1, 1, 1, 1, 1, 1, -1
TimeSeries + MissingDataMethod
Using the "Values" property with OP's f:
f["Values"]
-1, -1, -1, 1, 1, 1, -1, -1, 1, 1, 1, 1, 1, 1, -1
answered Sep 25 at 12:18
kglrkglr
228k10 gold badges257 silver badges519 bronze badges
edited Sep 27 at 8:49
answered Sep 25 at 12:18
kglrkglr
228k10 gold badges257 silver badges519 bronze badges
answered Sep 25 at 12:18
kglrkglr
228k10 gold badges257 silver badges519 bronze badges
answered Sep 25 at 12:18
kglrkglr
228k10 gold badges257 silver badges519 bronze badges
228k10 gold badges257 silver badges519 bronze badges
1
$begingroup$
One of the reasons I asked this question even though I had a working (although inelegant) solution was to see all the creative ways in which MMA experts are able to use the Wolfram language to solve a problem. Outstanding!
$endgroup$
– Jonathan Kinlay
Sep 25 at 22:41
$begingroup$
What sort of (Mathematica) black magic is this!? My +1 sir.
$endgroup$
– Rebel-Scum
Sep 27 at 13:54
add a comment
|
1
$begingroup$
One of the reasons I asked this question even though I had a working (although inelegant) solution was to see all the creative ways in which MMA experts are able to use the Wolfram language to solve a problem. Outstanding!
$endgroup$
– Jonathan Kinlay
Sep 25 at 22:41
$begingroup$
What sort of (Mathematica) black magic is this!? My +1 sir.
$endgroup$
– Rebel-Scum
Sep 27 at 13:54
1
1
$begingroup$
One of the reasons I asked this question even though I had a working (although inelegant) solution was to see all the creative ways in which MMA experts are able to use the Wolfram language to solve a problem. Outstanding!
$endgroup$
– Jonathan Kinlay
Sep 25 at 22:41
$begingroup$
One of the reasons I asked this question even though I had a working (although inelegant) solution was to see all the creative ways in which MMA experts are able to use the Wolfram language to solve a problem. Outstanding!
$endgroup$
– Jonathan Kinlay
Sep 25 at 22:41
$begingroup$
What sort of (Mathematica) black magic is this!? My +1 sir.
$endgroup$
– Rebel-Scum
Sep 27 at 13:54
$begingroup$
What sort of (Mathematica) black magic is this!? My +1 sir.
$endgroup$
– Rebel-Scum
Sep 27 at 13:54
add a comment
|
$begingroup$
Define a simple function, using FoldList
op = FoldList[If[#2 == 0, #1, #2] &];
l = -1, -1, 0, 1, 0, 0, -1, -1, 1, 1, 1, 0, 0, 0, -1;
op@l
(* -1, -1, -1, 1, 1, 1, -1, -1, 1, 1, 1, 1, 1, 1, -1 *)
answered Sep 26 at 19:06
mikadomikado
9,2272 gold badges9 silver badges32 bronze badges
$endgroup$
$begingroup$
This must be the fastest (and simplest) of the lot. The one I knew existed, but never reached.
$endgroup$
– Suba Thomas
Sep 26 at 22:19
add a comment
|
$begingroup$
Define a simple function, using FoldList
op = FoldList[If[#2 == 0, #1, #2] &];
l = -1, -1, 0, 1, 0, 0, -1, -1, 1, 1, 1, 0, 0, 0, -1;
op@l
(* -1, -1, -1, 1, 1, 1, -1, -1, 1, 1, 1, 1, 1, 1, -1 *)
answered Sep 26 at 19:06
mikadomikado
9,2272 gold badges9 silver badges32 bronze badges
$endgroup$
$begingroup$
This must be the fastest (and simplest) of the lot. The one I knew existed, but never reached.
$endgroup$
– Suba Thomas
Sep 26 at 22:19
add a comment
|
$begingroup$
Define a simple function, using FoldList
op = FoldList[If[#2 == 0, #1, #2] &];
l = -1, -1, 0, 1, 0, 0, -1, -1, 1, 1, 1, 0, 0, 0, -1;
op@l
(* -1, -1, -1, 1, 1, 1, -1, -1, 1, 1, 1, 1, 1, 1, -1 *)
answered Sep 26 at 19:06
mikadomikado
9,2272 gold badges9 silver badges32 bronze badges
$endgroup$
Define a simple function, using FoldList
op = FoldList[If[#2 == 0, #1, #2] &];
l = -1, -1, 0, 1, 0, 0, -1, -1, 1, 1, 1, 0, 0, 0, -1;
op@l
(* -1, -1, -1, 1, 1, 1, -1, -1, 1, 1, 1, 1, 1, 1, -1 *)
answered Sep 26 at 19:06
mikadomikado
9,2272 gold badges9 silver badges32 bronze badges
answered Sep 26 at 19:06
mikadomikado
9,2272 gold badges9 silver badges32 bronze badges
answered Sep 26 at 19:06
mikadomikado
9,2272 gold badges9 silver badges32 bronze badges
answered Sep 26 at 19:06
mikadomikado
9,2272 gold badges9 silver badges32 bronze badges
9,2272 gold badges9 silver badges32 bronze badges
$begingroup$
This must be the fastest (and simplest) of the lot. The one I knew existed, but never reached.
$endgroup$
– Suba Thomas
Sep 26 at 22:19
add a comment
|
$begingroup$
This must be the fastest (and simplest) of the lot. The one I knew existed, but never reached.
$endgroup$
– Suba Thomas
Sep 26 at 22:19
$begingroup$
This must be the fastest (and simplest) of the lot. The one I knew existed, but never reached.
$endgroup$
– Suba Thomas
Sep 26 at 22:19
$begingroup$
This must be the fastest (and simplest) of the lot. The one I knew existed, but never reached.
$endgroup$
– Suba Thomas
Sep 26 at 22:19
add a comment
|
$begingroup$
Perhaps something like this will appeal
l //. a___, 1, 0, b___ -> a, 1, 1, b, a___, -1, 0, b___ -> a, -1, -1, b
answered Sep 25 at 12:19
High Performance MarkHigh Performance Mark
1,1187 silver badges14 bronze badges
$endgroup$
$begingroup$
This is the kind of thing I was aiming for, but I couldn't find the right formulation. I find patterns such a challenge (although I do use them extensively). Anyway that you for a great solution.
$endgroup$
– Jonathan Kinlay
Sep 25 at 22:35
add a comment
|
$begingroup$
Perhaps something like this will appeal
l //. a___, 1, 0, b___ -> a, 1, 1, b, a___, -1, 0, b___ -> a, -1, -1, b
answered Sep 25 at 12:19
High Performance MarkHigh Performance Mark
1,1187 silver badges14 bronze badges
$endgroup$
$begingroup$
This is the kind of thing I was aiming for, but I couldn't find the right formulation. I find patterns such a challenge (although I do use them extensively). Anyway that you for a great solution.
$endgroup$
– Jonathan Kinlay
Sep 25 at 22:35
add a comment
|
$begingroup$
Perhaps something like this will appeal
l //. a___, 1, 0, b___ -> a, 1, 1, b, a___, -1, 0, b___ -> a, -1, -1, b
answered Sep 25 at 12:19
High Performance MarkHigh Performance Mark
1,1187 silver badges14 bronze badges
$endgroup$
Perhaps something like this will appeal
l //. a___, 1, 0, b___ -> a, 1, 1, b, a___, -1, 0, b___ -> a, -1, -1, b
answered Sep 25 at 12:19
High Performance MarkHigh Performance Mark
1,1187 silver badges14 bronze badges
answered Sep 25 at 12:19
High Performance MarkHigh Performance Mark
1,1187 silver badges14 bronze badges
answered Sep 25 at 12:19
High Performance MarkHigh Performance Mark
1,1187 silver badges14 bronze badges
answered Sep 25 at 12:19
High Performance MarkHigh Performance Mark
1,1187 silver badges14 bronze badges
1,1187 silver badges14 bronze badges
$begingroup$
This is the kind of thing I was aiming for, but I couldn't find the right formulation. I find patterns such a challenge (although I do use them extensively). Anyway that you for a great solution.
$endgroup$
– Jonathan Kinlay
Sep 25 at 22:35
add a comment
|
$begingroup$
This is the kind of thing I was aiming for, but I couldn't find the right formulation. I find patterns such a challenge (although I do use them extensively). Anyway that you for a great solution.
$endgroup$
– Jonathan Kinlay
Sep 25 at 22:35
$begingroup$
This is the kind of thing I was aiming for, but I couldn't find the right formulation. I find patterns such a challenge (although I do use them extensively). Anyway that you for a great solution.
$endgroup$
– Jonathan Kinlay
Sep 25 at 22:35
$begingroup$
This is the kind of thing I was aiming for, but I couldn't find the right formulation. I find patterns such a challenge (although I do use them extensively). Anyway that you for a great solution.
$endgroup$
– Jonathan Kinlay
Sep 25 at 22:35
add a comment
|
$begingroup$
Given:
list = -1, -1, 0, 1, 0, 0, -1, -1, 1, 1, 1, 0, 0, 0, -1
Then:
Module[prev = 0, Replace[list, 0 :> prev, x_ :> (prev = x), 1]]
(* -1, -1, -1, 1, 1, 1, -1, -1, 1, 1, 1, 1, 1, 1, -1 *)
The initial value assigned by prev = 0 is only used for lists that start with a zero -- choose another value if desired.
answered Sep 25 at 15:47
WReachWReach
56.7k2 gold badges125 silver badges224 bronze badges
$endgroup$
add a comment
|
$begingroup$
Given:
list = -1, -1, 0, 1, 0, 0, -1, -1, 1, 1, 1, 0, 0, 0, -1
Then:
Module[prev = 0, Replace[list, 0 :> prev, x_ :> (prev = x), 1]]
(* -1, -1, -1, 1, 1, 1, -1, -1, 1, 1, 1, 1, 1, 1, -1 *)
The initial value assigned by prev = 0 is only used for lists that start with a zero -- choose another value if desired.
answered Sep 25 at 15:47
WReachWReach
56.7k2 gold badges125 silver badges224 bronze badges
$endgroup$
add a comment
|
$begingroup$
Given:
list = -1, -1, 0, 1, 0, 0, -1, -1, 1, 1, 1, 0, 0, 0, -1
Then:
Module[prev = 0, Replace[list, 0 :> prev, x_ :> (prev = x), 1]]
(* -1, -1, -1, 1, 1, 1, -1, -1, 1, 1, 1, 1, 1, 1, -1 *)
The initial value assigned by prev = 0 is only used for lists that start with a zero -- choose another value if desired.
answered Sep 25 at 15:47
WReachWReach
56.7k2 gold badges125 silver badges224 bronze badges
$endgroup$
Given:
list = -1, -1, 0, 1, 0, 0, -1, -1, 1, 1, 1, 0, 0, 0, -1
Then:
Module[prev = 0, Replace[list, 0 :> prev, x_ :> (prev = x), 1]]
(* -1, -1, -1, 1, 1, 1, -1, -1, 1, 1, 1, 1, 1, 1, -1 *)
The initial value assigned by prev = 0 is only used for lists that start with a zero -- choose another value if desired.
answered Sep 25 at 15:47
WReachWReach
56.7k2 gold badges125 silver badges224 bronze badges
answered Sep 25 at 15:47
WReachWReach
56.7k2 gold badges125 silver badges224 bronze badges
answered Sep 25 at 15:47
WReachWReach
56.7k2 gold badges125 silver badges224 bronze badges
answered Sep 25 at 15:47
WReachWReach
56.7k2 gold badges125 silver badges224 bronze badges
56.7k2 gold badges125 silver badges224 bronze badges
add a comment
|
add a comment
|
$begingroup$
A few more...
l = -1, -1, 0, 1, 0, 0, -1, -1, 1, 1, 1, 0, 0, 0, -1;
Do[If[l[[i]] == 0, l[[i]] = l[[i - 1]]], i, 2, Length[l]]
l
-1, -1, -1, 1, 1, 1, -1, -1, 1, 1, 1, 1, 1, 1, -1
fn[l_] := Block[l1 = l,
Set[Part[l1, #[[2]]], Part[l1, #[[1]]]] & /@ SequencePosition[l, _, 0];
l1
]
l = -1, -1, 0, 1, 0, 0, -1, -1, 1, 1, 1, 0, 0, 0, -1;
FixedPoint[fn, l]
-1, -1, -1, 1, 1, 1, -1, -1, 1, 1, 1, 1, 1, 1, -1
l = -1, -1, 0, 1, 0, 0, -1, -1, 1, 1, 1, 0, 0, 0, -1;
Fold[#1, If[#2 == 0, Last[#1], #2] &, l[[1 ;; 1]], l[[2 ;; -1]]] //Flatten
-1, -1, -1, 1, 1, 1, -1, -1, 1, 1, 1, 1, 1, 1, -1
answered Sep 26 at 17:30
Suba ThomasSuba Thomas
5,06911 silver badges20 bronze badges
$endgroup$
add a comment
|
$begingroup$
A few more...
l = -1, -1, 0, 1, 0, 0, -1, -1, 1, 1, 1, 0, 0, 0, -1;
Do[If[l[[i]] == 0, l[[i]] = l[[i - 1]]], i, 2, Length[l]]
l
-1, -1, -1, 1, 1, 1, -1, -1, 1, 1, 1, 1, 1, 1, -1
fn[l_] := Block[l1 = l,
Set[Part[l1, #[[2]]], Part[l1, #[[1]]]] & /@ SequencePosition[l, _, 0];
l1
]
l = -1, -1, 0, 1, 0, 0, -1, -1, 1, 1, 1, 0, 0, 0, -1;
FixedPoint[fn, l]
-1, -1, -1, 1, 1, 1, -1, -1, 1, 1, 1, 1, 1, 1, -1
l = -1, -1, 0, 1, 0, 0, -1, -1, 1, 1, 1, 0, 0, 0, -1;
Fold[#1, If[#2 == 0, Last[#1], #2] &, l[[1 ;; 1]], l[[2 ;; -1]]] //Flatten
-1, -1, -1, 1, 1, 1, -1, -1, 1, 1, 1, 1, 1, 1, -1
answered Sep 26 at 17:30
Suba ThomasSuba Thomas
5,06911 silver badges20 bronze badges
$endgroup$
add a comment
|
$begingroup$
A few more...
l = -1, -1, 0, 1, 0, 0, -1, -1, 1, 1, 1, 0, 0, 0, -1;
Do[If[l[[i]] == 0, l[[i]] = l[[i - 1]]], i, 2, Length[l]]
l
-1, -1, -1, 1, 1, 1, -1, -1, 1, 1, 1, 1, 1, 1, -1
fn[l_] := Block[l1 = l,
Set[Part[l1, #[[2]]], Part[l1, #[[1]]]] & /@ SequencePosition[l, _, 0];
l1
]
l = -1, -1, 0, 1, 0, 0, -1, -1, 1, 1, 1, 0, 0, 0, -1;
FixedPoint[fn, l]
-1, -1, -1, 1, 1, 1, -1, -1, 1, 1, 1, 1, 1, 1, -1
l = -1, -1, 0, 1, 0, 0, -1, -1, 1, 1, 1, 0, 0, 0, -1;
Fold[#1, If[#2 == 0, Last[#1], #2] &, l[[1 ;; 1]], l[[2 ;; -1]]] //Flatten
-1, -1, -1, 1, 1, 1, -1, -1, 1, 1, 1, 1, 1, 1, -1
answered Sep 26 at 17:30
Suba ThomasSuba Thomas
5,06911 silver badges20 bronze badges
$endgroup$
A few more...
l = -1, -1, 0, 1, 0, 0, -1, -1, 1, 1, 1, 0, 0, 0, -1;
Do[If[l[[i]] == 0, l[[i]] = l[[i - 1]]], i, 2, Length[l]]
l
-1, -1, -1, 1, 1, 1, -1, -1, 1, 1, 1, 1, 1, 1, -1
fn[l_] := Block[l1 = l,
Set[Part[l1, #[[2]]], Part[l1, #[[1]]]] & /@ SequencePosition[l, _, 0];
l1
]
l = -1, -1, 0, 1, 0, 0, -1, -1, 1, 1, 1, 0, 0, 0, -1;
FixedPoint[fn, l]
-1, -1, -1, 1, 1, 1, -1, -1, 1, 1, 1, 1, 1, 1, -1
l = -1, -1, 0, 1, 0, 0, -1, -1, 1, 1, 1, 0, 0, 0, -1;
Fold[#1, If[#2 == 0, Last[#1], #2] &, l[[1 ;; 1]], l[[2 ;; -1]]] //Flatten
-1, -1, -1, 1, 1, 1, -1, -1, 1, 1, 1, 1, 1, 1, -1
answered Sep 26 at 17:30
Suba ThomasSuba Thomas
5,06911 silver badges20 bronze badges
edited Sep 26 at 17:38
answered Sep 26 at 17:30
Suba ThomasSuba Thomas
5,06911 silver badges20 bronze badges
answered Sep 26 at 17:30
Suba ThomasSuba Thomas
5,06911 silver badges20 bronze badges
answered Sep 26 at 17:30
Suba ThomasSuba Thomas
5,06911 silver badges20 bronze badges
5,06911 silver badges20 bronze badges
add a comment
|
add a comment
|
$begingroup$
The creativity demonstrated by all these solutions is impressive.
Here’s a really tough follow-up challenge. Can anyone write a WL program that will generate the code for any one of these solutions?
answered Sep 26 at 22:03
Jonathan KinlayJonathan Kinlay
6273 silver badges11 bronze badges
$endgroup$
6
$begingroup$
This isn't an answer. It should either be an extension to the question, or a fresh question referring back to the original one.
$endgroup$
– High Performance Mark
Sep 27 at 9:38
add a comment
|
$begingroup$
The creativity demonstrated by all these solutions is impressive.
Here’s a really tough follow-up challenge. Can anyone write a WL program that will generate the code for any one of these solutions?
answered Sep 26 at 22:03
Jonathan KinlayJonathan Kinlay
6273 silver badges11 bronze badges
$endgroup$
6
$begingroup$
This isn't an answer. It should either be an extension to the question, or a fresh question referring back to the original one.
$endgroup$
– High Performance Mark
Sep 27 at 9:38
add a comment
|
$begingroup$
The creativity demonstrated by all these solutions is impressive.
Here’s a really tough follow-up challenge. Can anyone write a WL program that will generate the code for any one of these solutions?
answered Sep 26 at 22:03
Jonathan KinlayJonathan Kinlay
6273 silver badges11 bronze badges
$endgroup$
The creativity demonstrated by all these solutions is impressive.
Here’s a really tough follow-up challenge. Can anyone write a WL program that will generate the code for any one of these solutions?
answered Sep 26 at 22:03
Jonathan KinlayJonathan Kinlay
6273 silver badges11 bronze badges
answered Sep 26 at 22:03
Jonathan KinlayJonathan Kinlay
6273 silver badges11 bronze badges
answered Sep 26 at 22:03
Jonathan KinlayJonathan Kinlay
6273 silver badges11 bronze badges
answered Sep 26 at 22:03
Jonathan KinlayJonathan Kinlay
6273 silver badges11 bronze badges
6273 silver badges11 bronze badges
6
$begingroup$
This isn't an answer. It should either be an extension to the question, or a fresh question referring back to the original one.
$endgroup$
– High Performance Mark
Sep 27 at 9:38
add a comment
|
6
$begingroup$
This isn't an answer. It should either be an extension to the question, or a fresh question referring back to the original one.
$endgroup$
– High Performance Mark
Sep 27 at 9:38
6
6
$begingroup$
This isn't an answer. It should either be an extension to the question, or a fresh question referring back to the original one.
$endgroup$
– High Performance Mark
Sep 27 at 9:38
$begingroup$
This isn't an answer. It should either be an extension to the question, or a fresh question referring back to the original one.
$endgroup$
– High Performance Mark
Sep 27 at 9:38
add a comment
|
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$begingroup$
What should happen when the first element is a zero?
$endgroup$
– Sjoerd Smit
Sep 25 at 14:53