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List, map function based on a condition
Conditional TableHow to replace an element in a list based on the value of the next element?Indexed Map ThreadChange some elements at the second level of a list based on some conditionsHow-to select an entry from a list of pairs that meets a condition depending the 2nd element of each pairInserting a condition inside a FOR loopFilter a nested list based on conditions on its elements
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ margin-bottom:0;
}
$begingroup$
I have the following list:
input = {{0, 0, 0, 0, 1, 1, 1, 0, 1, 0, 0, 0, 1, 1, 0, 1, 0, 0, 0, 1,
0, 1}, {0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 1, 1, 0, 1, 0, 0, 0, 1,
0, 1}, {1, 1, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 0, 1, 0, 0, 0,
1, 0, 1}}
there are two conditions:
- When a '0' is followed by '1', the value of '0' change to .5
- When a '1' is followed by '0', the value of '0' change to .5
the desired output is:
output = [{{0, 0, 0, .5, 1, 1, 1, .5, 1, .5, 0, .5, 1, 1, .5, 1, .5,
0, .5, 1, .5, 1}, {.5, 1, .5, 1, .5, .5, 1, .5, 0, 0, 0, .5, 1,
1, .5, 1, .5, 0, .5, 1, .5, 1}, {1, 1, .5, .5, 1, 1, 1, .5, 0, 0,
0, .5, 1, 1, .5, 1, 0, 0, .5, 1, .5, 1}}]
Who has a suggestion how to get the desired output
list-manipulation conditional map
asked 2 days ago
Michiel van MensMichiel van Mens
9576 silver badges19 bronze badges
$endgroup$
add a comment |
$begingroup$
I have the following list:
input = {{0, 0, 0, 0, 1, 1, 1, 0, 1, 0, 0, 0, 1, 1, 0, 1, 0, 0, 0, 1,
0, 1}, {0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 1, 1, 0, 1, 0, 0, 0, 1,
0, 1}, {1, 1, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 0, 1, 0, 0, 0,
1, 0, 1}}
there are two conditions:
- When a '0' is followed by '1', the value of '0' change to .5
- When a '1' is followed by '0', the value of '0' change to .5
the desired output is:
output = [{{0, 0, 0, .5, 1, 1, 1, .5, 1, .5, 0, .5, 1, 1, .5, 1, .5,
0, .5, 1, .5, 1}, {.5, 1, .5, 1, .5, .5, 1, .5, 0, 0, 0, .5, 1,
1, .5, 1, .5, 0, .5, 1, .5, 1}, {1, 1, .5, .5, 1, 1, 1, .5, 0, 0,
0, .5, 1, 1, .5, 1, 0, 0, .5, 1, .5, 1}}]
Who has a suggestion how to get the desired output
list-manipulation conditional map
asked 2 days ago
Michiel van MensMichiel van Mens
9576 silver badges19 bronze badges
$endgroup$
2
$begingroup$
should desired[[3]] be{1, 1, 0.5, 0.5, 1, 1, 1, 0.5, 0, 0, 0, 0.5, 1, 1, 0.5, 1, 0.5, 0, 0.5, 1, 0.5, 1}?
$endgroup$
– user1066
2 days ago
add a comment |
$begingroup$
I have the following list:
input = {{0, 0, 0, 0, 1, 1, 1, 0, 1, 0, 0, 0, 1, 1, 0, 1, 0, 0, 0, 1,
0, 1}, {0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 1, 1, 0, 1, 0, 0, 0, 1,
0, 1}, {1, 1, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 0, 1, 0, 0, 0,
1, 0, 1}}
there are two conditions:
- When a '0' is followed by '1', the value of '0' change to .5
- When a '1' is followed by '0', the value of '0' change to .5
the desired output is:
output = [{{0, 0, 0, .5, 1, 1, 1, .5, 1, .5, 0, .5, 1, 1, .5, 1, .5,
0, .5, 1, .5, 1}, {.5, 1, .5, 1, .5, .5, 1, .5, 0, 0, 0, .5, 1,
1, .5, 1, .5, 0, .5, 1, .5, 1}, {1, 1, .5, .5, 1, 1, 1, .5, 0, 0,
0, .5, 1, 1, .5, 1, 0, 0, .5, 1, .5, 1}}]
Who has a suggestion how to get the desired output
list-manipulation conditional map
asked 2 days ago
Michiel van MensMichiel van Mens
9576 silver badges19 bronze badges
$endgroup$
I have the following list:
input = {{0, 0, 0, 0, 1, 1, 1, 0, 1, 0, 0, 0, 1, 1, 0, 1, 0, 0, 0, 1,
0, 1}, {0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 1, 1, 0, 1, 0, 0, 0, 1,
0, 1}, {1, 1, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 0, 1, 0, 0, 0,
1, 0, 1}}
there are two conditions:
- When a '0' is followed by '1', the value of '0' change to .5
- When a '1' is followed by '0', the value of '0' change to .5
the desired output is:
output = [{{0, 0, 0, .5, 1, 1, 1, .5, 1, .5, 0, .5, 1, 1, .5, 1, .5,
0, .5, 1, .5, 1}, {.5, 1, .5, 1, .5, .5, 1, .5, 0, 0, 0, .5, 1,
1, .5, 1, .5, 0, .5, 1, .5, 1}, {1, 1, .5, .5, 1, 1, 1, .5, 0, 0,
0, .5, 1, 1, .5, 1, 0, 0, .5, 1, .5, 1}}]
Who has a suggestion how to get the desired output
list-manipulation conditional map
list-manipulation conditional map
asked 2 days ago
Michiel van MensMichiel van Mens
9576 silver badges19 bronze badges
asked 2 days ago
Michiel van MensMichiel van Mens
9576 silver badges19 bronze badges
asked 2 days ago
Michiel van MensMichiel van Mens
9576 silver badges19 bronze badges
asked 2 days ago
Michiel van MensMichiel van Mens
9576 silver badges19 bronze badges
asked 2 days ago
Michiel van MensMichiel van Mens
9576 silver badges19 bronze badges
9576 silver badges19 bronze badges
2
$begingroup$
should desired[[3]] be{1, 1, 0.5, 0.5, 1, 1, 1, 0.5, 0, 0, 0, 0.5, 1, 1, 0.5, 1, 0.5, 0, 0.5, 1, 0.5, 1}?
$endgroup$
– user1066
2 days ago
add a comment |
2
$begingroup$
should desired[[3]] be{1, 1, 0.5, 0.5, 1, 1, 1, 0.5, 0, 0, 0, 0.5, 1, 1, 0.5, 1, 0.5, 0, 0.5, 1, 0.5, 1}?
$endgroup$
– user1066
2 days ago
2
2
$begingroup$
should desired[[3]] be
{1, 1, 0.5, 0.5, 1, 1, 1, 0.5, 0, 0, 0, 0.5, 1, 1, 0.5, 1, 0.5, 0, 0.5, 1, 0.5, 1}?$endgroup$
– user1066
2 days ago
$begingroup$
should desired[[3]] be
{1, 1, 0.5, 0.5, 1, 1, 1, 0.5, 0, 0, 0, 0.5, 1, 1, 0.5, 1, 0.5, 0, 0.5, 1, 0.5, 1}?$endgroup$
– user1066
2 days ago
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Using ReplaceRepeated (//.) and pattern matching:
input //. {{x___, 0, 1, y___} :> {x, 0.5, 1, y}, {x___, 1, 0, y___} :> {x, 1, 0.5, y}}
{{0, 0, 0, 0.5, 1, 1, 1, 0.5, 1, 0.5, 0, 0.5, 1, 1, 0.5, 1, 0.5, 0, 0.5, 1, 0.5, 1}, {0.5, 1, 0.5, 1, 0.5, 0.5, 1, 0.5, 0, 0, 0, 0.5, 1, 1, 0.5, 1, 0.5, 0, 0.5, 1, 0.5, 1}, {1, 1, 0.5, 0.5, 1, 1, 1, 0.5, 0, 0, 0, 0.5, 1, 1, 0.5, 1, 0.5, 0, 0.5, 1, 0.5, 1}}
Note that there is a slight difference between this and your suggested output, highlighted in bold. As @Nasser received this result by his method as well, I suspect that the original suggested output was in error.
answered 2 days ago
eyorbleeyorble
6,6681 gold badge11 silver badges30 bronze badges
$endgroup$
$begingroup$
Tricky. I tried it similar insideCasesbut only one substitution is made. What could be the reason? Thanks!
$endgroup$
– Ulrich Neumann
2 days ago
1
$begingroup$
/.only tries to replace once per complete match, as I understand, so since each list matches{x___, 1, 0, y___}or the other it only undergoes one substitution each.//.explicitly retries until no matches remain. I thinkCasesworks more like/.in that respect.
$endgroup$
– eyorble
2 days ago
$begingroup$
Thanks, I have to think about it.
$endgroup$
– Ulrich Neumann
2 days ago
add a comment |
$begingroup$
You can also use a combination of SequenceReplace and FixedPoint:
f = Map[SequenceReplace[{{0, 1} -> Sequence[.5, 1], {1, 0} -> Sequence[1, .5]}]],
FixedPoint[f, input]
{{0, 0, 0, 0.5, 1, 1, 1, 0.5, 1, 0.5, 0, 0.5, 1, 1, 0.5, 1, 0.5, 0,
0.5, 1, 0.5, 1},
{0.5, 1, 0.5, 1, 0.5, 0.5, 1, 0.5, 0, 0, 0, 0.5, 1,
1, 0.5, 1, 0.5, 0, 0.5, 1, 0.5, 1},
{1, 1, 0.5, 0.5, 1, 1, 1, 0.5,
0, 0, 0, 0.5, 1, 1, 0.5, 1, 0.5, 0, 0.5, 1, 0.5, 1}}
You can also use Nest in place of FixedPoint:
Nest[f , input, 2] == %
True
answered 2 days ago
kglrkglr
211k10 gold badges242 silver badges485 bronze badges
$endgroup$
add a comment |
$begingroup$
For a rewriting problem, use rewriting explicitly:
input //. {
{x___, 0, 1, y___} -> {x, 0.5, 1, y},
{x___, 1, 0, y___} -> {x, 1, 0.5, y}
}
(*
{{0, 0, 0, 0.5, 1, 1, 1, 0.5, 1, 0.5, 0, 0.5, 1, 1, 0.5, 1, 0.5, 0, 0.5, 1, 0.5, 1},
{0.5, 1, 0.5, 1, 0.5, 0.5, 1, 0.5, 0, 0, 0, 0.5, 1,1, 0.5, 1, 0.5, 0, 0.5, 1, 0.5, 1},
{1, 1, 0.5, 0.5, 1, 1, 1, 0.5, 0, 0, 0, 0.5, 1, 1, 0.5, 1, 0.5, 0, 0.5, 1, 0.5, 1}}
*)
answered 2 days ago
John DotyJohn Doty
8,7661 gold badge14 silver badges26 bronze badges
$endgroup$
add a comment |
$begingroup$
This is not a "functional" way to do it. But an old fashioned loop and few if's. But it gives the result you show
foo[input_List] := Module[{n, i, current, next, before},
n = Length[input];
(*handle edge cases*)
If[n == 1, Return[input, Module]];
If[n == 2,
Return[If[input[[1]] == 0 && input[[2]] == 1, {0.5, 1},
If[input[[2]] == 0 && input[[1]] == 1, {1, 0.5}, input]],
Module]];
(*general case for list of length [GreaterEqual] 3*)
Table[
current = input[[i]];
If[i == 1,
If[input[[i]] == 0 && input[[i + 1]] == 1, 0.5, input[[i]]]
,
If[i == n,
If[input[[n]] == 0 && input[[n - 1]] == 1, 0.5, input[[n]]]
,
before = input[[i - 1]];
next = input[[i + 1]];
If[current == 0 && next == 1, 0.5,
If[current == 0 && before == 1, 0.5, current]]
]
],
{i, 1, Length[input]}
]
]
input = {{0, 0, 0, 0, 1, 1, 1, 0, 1, 0, 0, 0, 1, 1, 0, 1, 0, 0, 0, 1,
0, 1}, {0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 1, 1, 0, 1, 0, 0, 0,
1, 0, 1}, {1, 1, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 0, 1, 0, 0,
0, 1, 0, 1}};
Now map foo on the input
result = foo[#] & /@ input
gives
{{0, 0, 0, 0.5, 1, 1, 1, 0.5, 1, 0.5, 0, 0.5, 1, 1, 0.5, 1, 0.5, 0,
0.5, 1, 0.5, 1}, {0.5, 1, 0.5, 1, 0.5, 0.5, 1, 0.5, 0, 0, 0, 0.5, 1,
1, 0.5, 1, 0.5, 0, 0.5, 1, 0.5, 1}, {1, 1, 0.5, 0.5, 1, 1, 1, 0.5,
0, 0, 0, 0.5, 1, 1, 0.5, 1, 0.5, 0, 0.5, 1, 0.5, 1}}
I suspect there might be a shorter way to do this if one works harder on it.
answered 2 days ago
NasserNasser
61.3k4 gold badges93 silver badges215 bronze badges
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Using ReplaceRepeated (//.) and pattern matching:
input //. {{x___, 0, 1, y___} :> {x, 0.5, 1, y}, {x___, 1, 0, y___} :> {x, 1, 0.5, y}}
{{0, 0, 0, 0.5, 1, 1, 1, 0.5, 1, 0.5, 0, 0.5, 1, 1, 0.5, 1, 0.5, 0, 0.5, 1, 0.5, 1}, {0.5, 1, 0.5, 1, 0.5, 0.5, 1, 0.5, 0, 0, 0, 0.5, 1, 1, 0.5, 1, 0.5, 0, 0.5, 1, 0.5, 1}, {1, 1, 0.5, 0.5, 1, 1, 1, 0.5, 0, 0, 0, 0.5, 1, 1, 0.5, 1, 0.5, 0, 0.5, 1, 0.5, 1}}
Note that there is a slight difference between this and your suggested output, highlighted in bold. As @Nasser received this result by his method as well, I suspect that the original suggested output was in error.
answered 2 days ago
eyorbleeyorble
6,6681 gold badge11 silver badges30 bronze badges
$endgroup$
$begingroup$
Tricky. I tried it similar insideCasesbut only one substitution is made. What could be the reason? Thanks!
$endgroup$
– Ulrich Neumann
2 days ago
1
$begingroup$
/.only tries to replace once per complete match, as I understand, so since each list matches{x___, 1, 0, y___}or the other it only undergoes one substitution each.//.explicitly retries until no matches remain. I thinkCasesworks more like/.in that respect.
$endgroup$
– eyorble
2 days ago
$begingroup$
Thanks, I have to think about it.
$endgroup$
– Ulrich Neumann
2 days ago
add a comment |
$begingroup$
Using ReplaceRepeated (//.) and pattern matching:
input //. {{x___, 0, 1, y___} :> {x, 0.5, 1, y}, {x___, 1, 0, y___} :> {x, 1, 0.5, y}}
{{0, 0, 0, 0.5, 1, 1, 1, 0.5, 1, 0.5, 0, 0.5, 1, 1, 0.5, 1, 0.5, 0, 0.5, 1, 0.5, 1}, {0.5, 1, 0.5, 1, 0.5, 0.5, 1, 0.5, 0, 0, 0, 0.5, 1, 1, 0.5, 1, 0.5, 0, 0.5, 1, 0.5, 1}, {1, 1, 0.5, 0.5, 1, 1, 1, 0.5, 0, 0, 0, 0.5, 1, 1, 0.5, 1, 0.5, 0, 0.5, 1, 0.5, 1}}
Note that there is a slight difference between this and your suggested output, highlighted in bold. As @Nasser received this result by his method as well, I suspect that the original suggested output was in error.
answered 2 days ago
eyorbleeyorble
6,6681 gold badge11 silver badges30 bronze badges
$endgroup$
$begingroup$
Tricky. I tried it similar insideCasesbut only one substitution is made. What could be the reason? Thanks!
$endgroup$
– Ulrich Neumann
2 days ago
1
$begingroup$
/.only tries to replace once per complete match, as I understand, so since each list matches{x___, 1, 0, y___}or the other it only undergoes one substitution each.//.explicitly retries until no matches remain. I thinkCasesworks more like/.in that respect.
$endgroup$
– eyorble
2 days ago
$begingroup$
Thanks, I have to think about it.
$endgroup$
– Ulrich Neumann
2 days ago
add a comment |
$begingroup$
Using ReplaceRepeated (//.) and pattern matching:
input //. {{x___, 0, 1, y___} :> {x, 0.5, 1, y}, {x___, 1, 0, y___} :> {x, 1, 0.5, y}}
{{0, 0, 0, 0.5, 1, 1, 1, 0.5, 1, 0.5, 0, 0.5, 1, 1, 0.5, 1, 0.5, 0, 0.5, 1, 0.5, 1}, {0.5, 1, 0.5, 1, 0.5, 0.5, 1, 0.5, 0, 0, 0, 0.5, 1, 1, 0.5, 1, 0.5, 0, 0.5, 1, 0.5, 1}, {1, 1, 0.5, 0.5, 1, 1, 1, 0.5, 0, 0, 0, 0.5, 1, 1, 0.5, 1, 0.5, 0, 0.5, 1, 0.5, 1}}
Note that there is a slight difference between this and your suggested output, highlighted in bold. As @Nasser received this result by his method as well, I suspect that the original suggested output was in error.
answered 2 days ago
eyorbleeyorble
6,6681 gold badge11 silver badges30 bronze badges
$endgroup$
Using ReplaceRepeated (//.) and pattern matching:
input //. {{x___, 0, 1, y___} :> {x, 0.5, 1, y}, {x___, 1, 0, y___} :> {x, 1, 0.5, y}}
{{0, 0, 0, 0.5, 1, 1, 1, 0.5, 1, 0.5, 0, 0.5, 1, 1, 0.5, 1, 0.5, 0, 0.5, 1, 0.5, 1}, {0.5, 1, 0.5, 1, 0.5, 0.5, 1, 0.5, 0, 0, 0, 0.5, 1, 1, 0.5, 1, 0.5, 0, 0.5, 1, 0.5, 1}, {1, 1, 0.5, 0.5, 1, 1, 1, 0.5, 0, 0, 0, 0.5, 1, 1, 0.5, 1, 0.5, 0, 0.5, 1, 0.5, 1}}
Note that there is a slight difference between this and your suggested output, highlighted in bold. As @Nasser received this result by his method as well, I suspect that the original suggested output was in error.
answered 2 days ago
eyorbleeyorble
6,6681 gold badge11 silver badges30 bronze badges
answered 2 days ago
eyorbleeyorble
6,6681 gold badge11 silver badges30 bronze badges
answered 2 days ago
eyorbleeyorble
6,6681 gold badge11 silver badges30 bronze badges
answered 2 days ago
eyorbleeyorble
6,6681 gold badge11 silver badges30 bronze badges
6,6681 gold badge11 silver badges30 bronze badges
$begingroup$
Tricky. I tried it similar insideCasesbut only one substitution is made. What could be the reason? Thanks!
$endgroup$
– Ulrich Neumann
2 days ago
1
$begingroup$
/.only tries to replace once per complete match, as I understand, so since each list matches{x___, 1, 0, y___}or the other it only undergoes one substitution each.//.explicitly retries until no matches remain. I thinkCasesworks more like/.in that respect.
$endgroup$
– eyorble
2 days ago
$begingroup$
Thanks, I have to think about it.
$endgroup$
– Ulrich Neumann
2 days ago
add a comment |
$begingroup$
Tricky. I tried it similar insideCasesbut only one substitution is made. What could be the reason? Thanks!
$endgroup$
– Ulrich Neumann
2 days ago
1
$begingroup$
/.only tries to replace once per complete match, as I understand, so since each list matches{x___, 1, 0, y___}or the other it only undergoes one substitution each.//.explicitly retries until no matches remain. I thinkCasesworks more like/.in that respect.
$endgroup$
– eyorble
2 days ago
$begingroup$
Thanks, I have to think about it.
$endgroup$
– Ulrich Neumann
2 days ago
$begingroup$
Tricky. I tried it similar inside
Cases but only one substitution is made. What could be the reason? Thanks!$endgroup$
– Ulrich Neumann
2 days ago
$begingroup$
Tricky. I tried it similar inside
Cases but only one substitution is made. What could be the reason? Thanks!$endgroup$
– Ulrich Neumann
2 days ago
1
1
$begingroup$
/. only tries to replace once per complete match, as I understand, so since each list matches {x___, 1, 0, y___} or the other it only undergoes one substitution each. //. explicitly retries until no matches remain. I think Cases works more like /. in that respect.$endgroup$
– eyorble
2 days ago
$begingroup$
/. only tries to replace once per complete match, as I understand, so since each list matches {x___, 1, 0, y___} or the other it only undergoes one substitution each. //. explicitly retries until no matches remain. I think Cases works more like /. in that respect.$endgroup$
– eyorble
2 days ago
$begingroup$
Thanks, I have to think about it.
$endgroup$
– Ulrich Neumann
2 days ago
$begingroup$
Thanks, I have to think about it.
$endgroup$
– Ulrich Neumann
2 days ago
add a comment |
$begingroup$
You can also use a combination of SequenceReplace and FixedPoint:
f = Map[SequenceReplace[{{0, 1} -> Sequence[.5, 1], {1, 0} -> Sequence[1, .5]}]],
FixedPoint[f, input]
{{0, 0, 0, 0.5, 1, 1, 1, 0.5, 1, 0.5, 0, 0.5, 1, 1, 0.5, 1, 0.5, 0,
0.5, 1, 0.5, 1},
{0.5, 1, 0.5, 1, 0.5, 0.5, 1, 0.5, 0, 0, 0, 0.5, 1,
1, 0.5, 1, 0.5, 0, 0.5, 1, 0.5, 1},
{1, 1, 0.5, 0.5, 1, 1, 1, 0.5,
0, 0, 0, 0.5, 1, 1, 0.5, 1, 0.5, 0, 0.5, 1, 0.5, 1}}
You can also use Nest in place of FixedPoint:
Nest[f , input, 2] == %
True
answered 2 days ago
kglrkglr
211k10 gold badges242 silver badges485 bronze badges
$endgroup$
add a comment |
$begingroup$
You can also use a combination of SequenceReplace and FixedPoint:
f = Map[SequenceReplace[{{0, 1} -> Sequence[.5, 1], {1, 0} -> Sequence[1, .5]}]],
FixedPoint[f, input]
{{0, 0, 0, 0.5, 1, 1, 1, 0.5, 1, 0.5, 0, 0.5, 1, 1, 0.5, 1, 0.5, 0,
0.5, 1, 0.5, 1},
{0.5, 1, 0.5, 1, 0.5, 0.5, 1, 0.5, 0, 0, 0, 0.5, 1,
1, 0.5, 1, 0.5, 0, 0.5, 1, 0.5, 1},
{1, 1, 0.5, 0.5, 1, 1, 1, 0.5,
0, 0, 0, 0.5, 1, 1, 0.5, 1, 0.5, 0, 0.5, 1, 0.5, 1}}
You can also use Nest in place of FixedPoint:
Nest[f , input, 2] == %
True
answered 2 days ago
kglrkglr
211k10 gold badges242 silver badges485 bronze badges
$endgroup$
add a comment |
$begingroup$
You can also use a combination of SequenceReplace and FixedPoint:
f = Map[SequenceReplace[{{0, 1} -> Sequence[.5, 1], {1, 0} -> Sequence[1, .5]}]],
FixedPoint[f, input]
{{0, 0, 0, 0.5, 1, 1, 1, 0.5, 1, 0.5, 0, 0.5, 1, 1, 0.5, 1, 0.5, 0,
0.5, 1, 0.5, 1},
{0.5, 1, 0.5, 1, 0.5, 0.5, 1, 0.5, 0, 0, 0, 0.5, 1,
1, 0.5, 1, 0.5, 0, 0.5, 1, 0.5, 1},
{1, 1, 0.5, 0.5, 1, 1, 1, 0.5,
0, 0, 0, 0.5, 1, 1, 0.5, 1, 0.5, 0, 0.5, 1, 0.5, 1}}
You can also use Nest in place of FixedPoint:
Nest[f , input, 2] == %
True
answered 2 days ago
kglrkglr
211k10 gold badges242 silver badges485 bronze badges
$endgroup$
You can also use a combination of SequenceReplace and FixedPoint:
f = Map[SequenceReplace[{{0, 1} -> Sequence[.5, 1], {1, 0} -> Sequence[1, .5]}]],
FixedPoint[f, input]
{{0, 0, 0, 0.5, 1, 1, 1, 0.5, 1, 0.5, 0, 0.5, 1, 1, 0.5, 1, 0.5, 0,
0.5, 1, 0.5, 1},
{0.5, 1, 0.5, 1, 0.5, 0.5, 1, 0.5, 0, 0, 0, 0.5, 1,
1, 0.5, 1, 0.5, 0, 0.5, 1, 0.5, 1},
{1, 1, 0.5, 0.5, 1, 1, 1, 0.5,
0, 0, 0, 0.5, 1, 1, 0.5, 1, 0.5, 0, 0.5, 1, 0.5, 1}}
You can also use Nest in place of FixedPoint:
Nest[f , input, 2] == %
True
answered 2 days ago
kglrkglr
211k10 gold badges242 silver badges485 bronze badges
edited 2 days ago
answered 2 days ago
kglrkglr
211k10 gold badges242 silver badges485 bronze badges
answered 2 days ago
kglrkglr
211k10 gold badges242 silver badges485 bronze badges
answered 2 days ago
kglrkglr
211k10 gold badges242 silver badges485 bronze badges
211k10 gold badges242 silver badges485 bronze badges
add a comment |
add a comment |
$begingroup$
For a rewriting problem, use rewriting explicitly:
input //. {
{x___, 0, 1, y___} -> {x, 0.5, 1, y},
{x___, 1, 0, y___} -> {x, 1, 0.5, y}
}
(*
{{0, 0, 0, 0.5, 1, 1, 1, 0.5, 1, 0.5, 0, 0.5, 1, 1, 0.5, 1, 0.5, 0, 0.5, 1, 0.5, 1},
{0.5, 1, 0.5, 1, 0.5, 0.5, 1, 0.5, 0, 0, 0, 0.5, 1,1, 0.5, 1, 0.5, 0, 0.5, 1, 0.5, 1},
{1, 1, 0.5, 0.5, 1, 1, 1, 0.5, 0, 0, 0, 0.5, 1, 1, 0.5, 1, 0.5, 0, 0.5, 1, 0.5, 1}}
*)
answered 2 days ago
John DotyJohn Doty
8,7661 gold badge14 silver badges26 bronze badges
$endgroup$
add a comment |
$begingroup$
For a rewriting problem, use rewriting explicitly:
input //. {
{x___, 0, 1, y___} -> {x, 0.5, 1, y},
{x___, 1, 0, y___} -> {x, 1, 0.5, y}
}
(*
{{0, 0, 0, 0.5, 1, 1, 1, 0.5, 1, 0.5, 0, 0.5, 1, 1, 0.5, 1, 0.5, 0, 0.5, 1, 0.5, 1},
{0.5, 1, 0.5, 1, 0.5, 0.5, 1, 0.5, 0, 0, 0, 0.5, 1,1, 0.5, 1, 0.5, 0, 0.5, 1, 0.5, 1},
{1, 1, 0.5, 0.5, 1, 1, 1, 0.5, 0, 0, 0, 0.5, 1, 1, 0.5, 1, 0.5, 0, 0.5, 1, 0.5, 1}}
*)
answered 2 days ago
John DotyJohn Doty
8,7661 gold badge14 silver badges26 bronze badges
$endgroup$
add a comment |
$begingroup$
For a rewriting problem, use rewriting explicitly:
input //. {
{x___, 0, 1, y___} -> {x, 0.5, 1, y},
{x___, 1, 0, y___} -> {x, 1, 0.5, y}
}
(*
{{0, 0, 0, 0.5, 1, 1, 1, 0.5, 1, 0.5, 0, 0.5, 1, 1, 0.5, 1, 0.5, 0, 0.5, 1, 0.5, 1},
{0.5, 1, 0.5, 1, 0.5, 0.5, 1, 0.5, 0, 0, 0, 0.5, 1,1, 0.5, 1, 0.5, 0, 0.5, 1, 0.5, 1},
{1, 1, 0.5, 0.5, 1, 1, 1, 0.5, 0, 0, 0, 0.5, 1, 1, 0.5, 1, 0.5, 0, 0.5, 1, 0.5, 1}}
*)
answered 2 days ago
John DotyJohn Doty
8,7661 gold badge14 silver badges26 bronze badges
$endgroup$
For a rewriting problem, use rewriting explicitly:
input //. {
{x___, 0, 1, y___} -> {x, 0.5, 1, y},
{x___, 1, 0, y___} -> {x, 1, 0.5, y}
}
(*
{{0, 0, 0, 0.5, 1, 1, 1, 0.5, 1, 0.5, 0, 0.5, 1, 1, 0.5, 1, 0.5, 0, 0.5, 1, 0.5, 1},
{0.5, 1, 0.5, 1, 0.5, 0.5, 1, 0.5, 0, 0, 0, 0.5, 1,1, 0.5, 1, 0.5, 0, 0.5, 1, 0.5, 1},
{1, 1, 0.5, 0.5, 1, 1, 1, 0.5, 0, 0, 0, 0.5, 1, 1, 0.5, 1, 0.5, 0, 0.5, 1, 0.5, 1}}
*)
answered 2 days ago
John DotyJohn Doty
8,7661 gold badge14 silver badges26 bronze badges
answered 2 days ago
John DotyJohn Doty
8,7661 gold badge14 silver badges26 bronze badges
answered 2 days ago
John DotyJohn Doty
8,7661 gold badge14 silver badges26 bronze badges
answered 2 days ago
John DotyJohn Doty
8,7661 gold badge14 silver badges26 bronze badges
8,7661 gold badge14 silver badges26 bronze badges
add a comment |
add a comment |
$begingroup$
This is not a "functional" way to do it. But an old fashioned loop and few if's. But it gives the result you show
foo[input_List] := Module[{n, i, current, next, before},
n = Length[input];
(*handle edge cases*)
If[n == 1, Return[input, Module]];
If[n == 2,
Return[If[input[[1]] == 0 && input[[2]] == 1, {0.5, 1},
If[input[[2]] == 0 && input[[1]] == 1, {1, 0.5}, input]],
Module]];
(*general case for list of length [GreaterEqual] 3*)
Table[
current = input[[i]];
If[i == 1,
If[input[[i]] == 0 && input[[i + 1]] == 1, 0.5, input[[i]]]
,
If[i == n,
If[input[[n]] == 0 && input[[n - 1]] == 1, 0.5, input[[n]]]
,
before = input[[i - 1]];
next = input[[i + 1]];
If[current == 0 && next == 1, 0.5,
If[current == 0 && before == 1, 0.5, current]]
]
],
{i, 1, Length[input]}
]
]
input = {{0, 0, 0, 0, 1, 1, 1, 0, 1, 0, 0, 0, 1, 1, 0, 1, 0, 0, 0, 1,
0, 1}, {0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 1, 1, 0, 1, 0, 0, 0,
1, 0, 1}, {1, 1, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 0, 1, 0, 0,
0, 1, 0, 1}};
Now map foo on the input
result = foo[#] & /@ input
gives
{{0, 0, 0, 0.5, 1, 1, 1, 0.5, 1, 0.5, 0, 0.5, 1, 1, 0.5, 1, 0.5, 0,
0.5, 1, 0.5, 1}, {0.5, 1, 0.5, 1, 0.5, 0.5, 1, 0.5, 0, 0, 0, 0.5, 1,
1, 0.5, 1, 0.5, 0, 0.5, 1, 0.5, 1}, {1, 1, 0.5, 0.5, 1, 1, 1, 0.5,
0, 0, 0, 0.5, 1, 1, 0.5, 1, 0.5, 0, 0.5, 1, 0.5, 1}}
I suspect there might be a shorter way to do this if one works harder on it.
answered 2 days ago
NasserNasser
61.3k4 gold badges93 silver badges215 bronze badges
$endgroup$
add a comment |
$begingroup$
This is not a "functional" way to do it. But an old fashioned loop and few if's. But it gives the result you show
foo[input_List] := Module[{n, i, current, next, before},
n = Length[input];
(*handle edge cases*)
If[n == 1, Return[input, Module]];
If[n == 2,
Return[If[input[[1]] == 0 && input[[2]] == 1, {0.5, 1},
If[input[[2]] == 0 && input[[1]] == 1, {1, 0.5}, input]],
Module]];
(*general case for list of length [GreaterEqual] 3*)
Table[
current = input[[i]];
If[i == 1,
If[input[[i]] == 0 && input[[i + 1]] == 1, 0.5, input[[i]]]
,
If[i == n,
If[input[[n]] == 0 && input[[n - 1]] == 1, 0.5, input[[n]]]
,
before = input[[i - 1]];
next = input[[i + 1]];
If[current == 0 && next == 1, 0.5,
If[current == 0 && before == 1, 0.5, current]]
]
],
{i, 1, Length[input]}
]
]
input = {{0, 0, 0, 0, 1, 1, 1, 0, 1, 0, 0, 0, 1, 1, 0, 1, 0, 0, 0, 1,
0, 1}, {0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 1, 1, 0, 1, 0, 0, 0,
1, 0, 1}, {1, 1, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 0, 1, 0, 0,
0, 1, 0, 1}};
Now map foo on the input
result = foo[#] & /@ input
gives
{{0, 0, 0, 0.5, 1, 1, 1, 0.5, 1, 0.5, 0, 0.5, 1, 1, 0.5, 1, 0.5, 0,
0.5, 1, 0.5, 1}, {0.5, 1, 0.5, 1, 0.5, 0.5, 1, 0.5, 0, 0, 0, 0.5, 1,
1, 0.5, 1, 0.5, 0, 0.5, 1, 0.5, 1}, {1, 1, 0.5, 0.5, 1, 1, 1, 0.5,
0, 0, 0, 0.5, 1, 1, 0.5, 1, 0.5, 0, 0.5, 1, 0.5, 1}}
I suspect there might be a shorter way to do this if one works harder on it.
answered 2 days ago
NasserNasser
61.3k4 gold badges93 silver badges215 bronze badges
$endgroup$
add a comment |
$begingroup$
This is not a "functional" way to do it. But an old fashioned loop and few if's. But it gives the result you show
foo[input_List] := Module[{n, i, current, next, before},
n = Length[input];
(*handle edge cases*)
If[n == 1, Return[input, Module]];
If[n == 2,
Return[If[input[[1]] == 0 && input[[2]] == 1, {0.5, 1},
If[input[[2]] == 0 && input[[1]] == 1, {1, 0.5}, input]],
Module]];
(*general case for list of length [GreaterEqual] 3*)
Table[
current = input[[i]];
If[i == 1,
If[input[[i]] == 0 && input[[i + 1]] == 1, 0.5, input[[i]]]
,
If[i == n,
If[input[[n]] == 0 && input[[n - 1]] == 1, 0.5, input[[n]]]
,
before = input[[i - 1]];
next = input[[i + 1]];
If[current == 0 && next == 1, 0.5,
If[current == 0 && before == 1, 0.5, current]]
]
],
{i, 1, Length[input]}
]
]
input = {{0, 0, 0, 0, 1, 1, 1, 0, 1, 0, 0, 0, 1, 1, 0, 1, 0, 0, 0, 1,
0, 1}, {0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 1, 1, 0, 1, 0, 0, 0,
1, 0, 1}, {1, 1, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 0, 1, 0, 0,
0, 1, 0, 1}};
Now map foo on the input
result = foo[#] & /@ input
gives
{{0, 0, 0, 0.5, 1, 1, 1, 0.5, 1, 0.5, 0, 0.5, 1, 1, 0.5, 1, 0.5, 0,
0.5, 1, 0.5, 1}, {0.5, 1, 0.5, 1, 0.5, 0.5, 1, 0.5, 0, 0, 0, 0.5, 1,
1, 0.5, 1, 0.5, 0, 0.5, 1, 0.5, 1}, {1, 1, 0.5, 0.5, 1, 1, 1, 0.5,
0, 0, 0, 0.5, 1, 1, 0.5, 1, 0.5, 0, 0.5, 1, 0.5, 1}}
I suspect there might be a shorter way to do this if one works harder on it.
answered 2 days ago
NasserNasser
61.3k4 gold badges93 silver badges215 bronze badges
$endgroup$
This is not a "functional" way to do it. But an old fashioned loop and few if's. But it gives the result you show
foo[input_List] := Module[{n, i, current, next, before},
n = Length[input];
(*handle edge cases*)
If[n == 1, Return[input, Module]];
If[n == 2,
Return[If[input[[1]] == 0 && input[[2]] == 1, {0.5, 1},
If[input[[2]] == 0 && input[[1]] == 1, {1, 0.5}, input]],
Module]];
(*general case for list of length [GreaterEqual] 3*)
Table[
current = input[[i]];
If[i == 1,
If[input[[i]] == 0 && input[[i + 1]] == 1, 0.5, input[[i]]]
,
If[i == n,
If[input[[n]] == 0 && input[[n - 1]] == 1, 0.5, input[[n]]]
,
before = input[[i - 1]];
next = input[[i + 1]];
If[current == 0 && next == 1, 0.5,
If[current == 0 && before == 1, 0.5, current]]
]
],
{i, 1, Length[input]}
]
]
input = {{0, 0, 0, 0, 1, 1, 1, 0, 1, 0, 0, 0, 1, 1, 0, 1, 0, 0, 0, 1,
0, 1}, {0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 1, 1, 0, 1, 0, 0, 0,
1, 0, 1}, {1, 1, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 0, 1, 0, 0,
0, 1, 0, 1}};
Now map foo on the input
result = foo[#] & /@ input
gives
{{0, 0, 0, 0.5, 1, 1, 1, 0.5, 1, 0.5, 0, 0.5, 1, 1, 0.5, 1, 0.5, 0,
0.5, 1, 0.5, 1}, {0.5, 1, 0.5, 1, 0.5, 0.5, 1, 0.5, 0, 0, 0, 0.5, 1,
1, 0.5, 1, 0.5, 0, 0.5, 1, 0.5, 1}, {1, 1, 0.5, 0.5, 1, 1, 1, 0.5,
0, 0, 0, 0.5, 1, 1, 0.5, 1, 0.5, 0, 0.5, 1, 0.5, 1}}
I suspect there might be a shorter way to do this if one works harder on it.
answered 2 days ago
NasserNasser
61.3k4 gold badges93 silver badges215 bronze badges
edited 2 days ago
answered 2 days ago
NasserNasser
61.3k4 gold badges93 silver badges215 bronze badges
answered 2 days ago
NasserNasser
61.3k4 gold badges93 silver badges215 bronze badges
answered 2 days ago
NasserNasser
61.3k4 gold badges93 silver badges215 bronze badges
61.3k4 gold badges93 silver badges215 bronze badges
add a comment |
add a comment |
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$begingroup$
should desired[[3]] be
{1, 1, 0.5, 0.5, 1, 1, 1, 0.5, 0, 0, 0, 0.5, 1, 1, 0.5, 1, 0.5, 0, 0.5, 1, 0.5, 1}?$endgroup$
– user1066
2 days ago