Replacing the Zeros of One List by the (i-1)th Element of Another List
$begingroup$
My goal is to replace the zeros of one list with the (i-1)th element of a second list. For example, if
list1 = {0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 0}
and
list2 = {6, 1, 4, 7, 2, 9, 10, 8, 11, 3, 5, 0, 12}
the desired output is {0, 1, 1, 4, 7, 1, 1, 10, 1, 11, 3, 1, 0}. Note that the first element of the output is defined as 0 still.
My attempt to create a code for this is to first find the zeros of list1then run a for-loop for $i in$zeroslist1.
zeroslist1 = Flatten[Position[list1, 0]]
DeleteCases[
Flatten[Reap@
Do[Sow[ReplacePart[vtest1,
i -> vtest2[[i - 1]] & /@ zeroslist1]], {i, zeroslist1}], 2], Null]
The results of the output are:
{{List,1,1,0,0,1,1,0,1,0,0,1,0}, {0,1,1,4,0,1,1,0,1,0,0,1,0}, {0,1,1,0,7,1,1,0,1,0,0,1,0},{0,1,1,0,0,1,1,10,1,0,0,1,0}, {0,1,1,0,0,1,1,0,1,11,0,1,0},{0,1,1,0,0,1,1,0,1,0,3,1,0}, {0,1,1,0,0,1,1,0,1,0,0,1,0}}.
Either a cleaner way to code the desire output or a method of merging the output of my current for-loop to get the desired output would be great.
list-manipulation functions function-construction
asked 11 hours ago
WillWill
3347
$endgroup$
add a comment |
$begingroup$
My goal is to replace the zeros of one list with the (i-1)th element of a second list. For example, if
list1 = {0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 0}
and
list2 = {6, 1, 4, 7, 2, 9, 10, 8, 11, 3, 5, 0, 12}
the desired output is {0, 1, 1, 4, 7, 1, 1, 10, 1, 11, 3, 1, 0}. Note that the first element of the output is defined as 0 still.
My attempt to create a code for this is to first find the zeros of list1then run a for-loop for $i in$zeroslist1.
zeroslist1 = Flatten[Position[list1, 0]]
DeleteCases[
Flatten[Reap@
Do[Sow[ReplacePart[vtest1,
i -> vtest2[[i - 1]] & /@ zeroslist1]], {i, zeroslist1}], 2], Null]
The results of the output are:
{{List,1,1,0,0,1,1,0,1,0,0,1,0}, {0,1,1,4,0,1,1,0,1,0,0,1,0}, {0,1,1,0,7,1,1,0,1,0,0,1,0},{0,1,1,0,0,1,1,10,1,0,0,1,0}, {0,1,1,0,0,1,1,0,1,11,0,1,0},{0,1,1,0,0,1,1,0,1,0,3,1,0}, {0,1,1,0,0,1,1,0,1,0,0,1,0}}.
Either a cleaner way to code the desire output or a method of merging the output of my current for-loop to get the desired output would be great.
list-manipulation functions function-construction
asked 11 hours ago
WillWill
3347
$endgroup$
add a comment |
$begingroup$
My goal is to replace the zeros of one list with the (i-1)th element of a second list. For example, if
list1 = {0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 0}
and
list2 = {6, 1, 4, 7, 2, 9, 10, 8, 11, 3, 5, 0, 12}
the desired output is {0, 1, 1, 4, 7, 1, 1, 10, 1, 11, 3, 1, 0}. Note that the first element of the output is defined as 0 still.
My attempt to create a code for this is to first find the zeros of list1then run a for-loop for $i in$zeroslist1.
zeroslist1 = Flatten[Position[list1, 0]]
DeleteCases[
Flatten[Reap@
Do[Sow[ReplacePart[vtest1,
i -> vtest2[[i - 1]] & /@ zeroslist1]], {i, zeroslist1}], 2], Null]
The results of the output are:
{{List,1,1,0,0,1,1,0,1,0,0,1,0}, {0,1,1,4,0,1,1,0,1,0,0,1,0}, {0,1,1,0,7,1,1,0,1,0,0,1,0},{0,1,1,0,0,1,1,10,1,0,0,1,0}, {0,1,1,0,0,1,1,0,1,11,0,1,0},{0,1,1,0,0,1,1,0,1,0,3,1,0}, {0,1,1,0,0,1,1,0,1,0,0,1,0}}.
Either a cleaner way to code the desire output or a method of merging the output of my current for-loop to get the desired output would be great.
list-manipulation functions function-construction
asked 11 hours ago
WillWill
3347
$endgroup$
My goal is to replace the zeros of one list with the (i-1)th element of a second list. For example, if
list1 = {0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 0}
and
list2 = {6, 1, 4, 7, 2, 9, 10, 8, 11, 3, 5, 0, 12}
the desired output is {0, 1, 1, 4, 7, 1, 1, 10, 1, 11, 3, 1, 0}. Note that the first element of the output is defined as 0 still.
My attempt to create a code for this is to first find the zeros of list1then run a for-loop for $i in$zeroslist1.
zeroslist1 = Flatten[Position[list1, 0]]
DeleteCases[
Flatten[Reap@
Do[Sow[ReplacePart[vtest1,
i -> vtest2[[i - 1]] & /@ zeroslist1]], {i, zeroslist1}], 2], Null]
The results of the output are:
{{List,1,1,0,0,1,1,0,1,0,0,1,0}, {0,1,1,4,0,1,1,0,1,0,0,1,0}, {0,1,1,0,7,1,1,0,1,0,0,1,0},{0,1,1,0,0,1,1,10,1,0,0,1,0}, {0,1,1,0,0,1,1,0,1,11,0,1,0},{0,1,1,0,0,1,1,0,1,0,3,1,0}, {0,1,1,0,0,1,1,0,1,0,0,1,0}}.
Either a cleaner way to code the desire output or a method of merging the output of my current for-loop to get the desired output would be great.
list-manipulation functions function-construction
list-manipulation functions function-construction
asked 11 hours ago
WillWill
3347
asked 11 hours ago
WillWill
3347
asked 11 hours ago
WillWill
3347
asked 11 hours ago
WillWill
3347
asked 11 hours ago
WillWill
3347
3347
add a comment |
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
You can use a multiplication instead of looping or conditionals:
list1 + (1 - list1) Prepend[Most[list2], 0]
{0, 1, 1, 4, 7, 1, 1, 10, 1, 11, 3, 1, 0}
The central point is that 0 and 1 in list1 aren't just symbols but numeric quantities.
answered 10 hours ago
RomanRoman
1,147511
$endgroup$
$begingroup$
What a simple way to do this. Thank you so much.
$endgroup$
– Will
10 hours ago
add a comment |
$begingroup$
idx = Random`Private`PositionsOf[Rest[list1], 0];
result = list1;
result[[idx + 1]] = list2[[idx]];
result
{0, 1, 1, 4, 7, 1, 1, 10, 1, 11, 3, 1, 0}
answered 11 hours ago
Henrik SchumacherHenrik Schumacher
52.9k471148
$endgroup$
add a comment |
$begingroup$
Here's another possibility:
Module[{tmp=list1},
With[{i = Pick[Range[Length[list1]-1], Rest @ list1, 0]},
tmp[[i+1]]=list2[[i]]
];
tmp
]
{0, 1, 1, 4, 7, 1, 1, 10, 1, 11, 3, 1, 0}
answered 11 hours ago
Carl WollCarl Woll
67.9k389176
$endgroup$
add a comment |
$begingroup$
Also MapIndexed works fine
(# /. List -> 0) & /@ MapIndexed[Replace[#1, 0 -> list2[[#2[[1]] - 1]]] &, list1]
The first piece just replaces the special case when a zero element is in the first entry (in Mathematica the 0th element is the Head, which in this case is List). Then the function MapIndexed does its job.
answered 11 hours ago
Mane.andreaMane.andrea
1366
$endgroup$
add a comment |
$begingroup$
MapThread[If[#1 == 0, #2, #1] &, {Join[list1, {0}], Join[{0}, list2]}] // Most
{0, 1, 1, 4, 7, 1, 1, 10, 1, 11, 3, 1, 0}
Your answer from above
{0, 1, 1, 4, 7, 1, 1, 10, 1, 11, 3, 1, 0}
answered 11 hours ago
MikeYMikeY
2,697412
$endgroup$
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
$begingroup$
You can use a multiplication instead of looping or conditionals:
list1 + (1 - list1) Prepend[Most[list2], 0]
{0, 1, 1, 4, 7, 1, 1, 10, 1, 11, 3, 1, 0}
The central point is that 0 and 1 in list1 aren't just symbols but numeric quantities.
answered 10 hours ago
RomanRoman
1,147511
$endgroup$
$begingroup$
What a simple way to do this. Thank you so much.
$endgroup$
– Will
10 hours ago
add a comment |
$begingroup$
You can use a multiplication instead of looping or conditionals:
list1 + (1 - list1) Prepend[Most[list2], 0]
{0, 1, 1, 4, 7, 1, 1, 10, 1, 11, 3, 1, 0}
The central point is that 0 and 1 in list1 aren't just symbols but numeric quantities.
answered 10 hours ago
RomanRoman
1,147511
$endgroup$
$begingroup$
What a simple way to do this. Thank you so much.
$endgroup$
– Will
10 hours ago
add a comment |
$begingroup$
You can use a multiplication instead of looping or conditionals:
list1 + (1 - list1) Prepend[Most[list2], 0]
{0, 1, 1, 4, 7, 1, 1, 10, 1, 11, 3, 1, 0}
The central point is that 0 and 1 in list1 aren't just symbols but numeric quantities.
answered 10 hours ago
RomanRoman
1,147511
$endgroup$
You can use a multiplication instead of looping or conditionals:
list1 + (1 - list1) Prepend[Most[list2], 0]
{0, 1, 1, 4, 7, 1, 1, 10, 1, 11, 3, 1, 0}
The central point is that 0 and 1 in list1 aren't just symbols but numeric quantities.
answered 10 hours ago
RomanRoman
1,147511
edited 10 hours ago
answered 10 hours ago
RomanRoman
1,147511
answered 10 hours ago
RomanRoman
1,147511
answered 10 hours ago
RomanRoman
1,147511
1,147511
$begingroup$
What a simple way to do this. Thank you so much.
$endgroup$
– Will
10 hours ago
add a comment |
$begingroup$
What a simple way to do this. Thank you so much.
$endgroup$
– Will
10 hours ago
$begingroup$
What a simple way to do this. Thank you so much.
$endgroup$
– Will
10 hours ago
$begingroup$
What a simple way to do this. Thank you so much.
$endgroup$
– Will
10 hours ago
add a comment |
$begingroup$
idx = Random`Private`PositionsOf[Rest[list1], 0];
result = list1;
result[[idx + 1]] = list2[[idx]];
result
{0, 1, 1, 4, 7, 1, 1, 10, 1, 11, 3, 1, 0}
answered 11 hours ago
Henrik SchumacherHenrik Schumacher
52.9k471148
$endgroup$
add a comment |
$begingroup$
idx = Random`Private`PositionsOf[Rest[list1], 0];
result = list1;
result[[idx + 1]] = list2[[idx]];
result
{0, 1, 1, 4, 7, 1, 1, 10, 1, 11, 3, 1, 0}
answered 11 hours ago
Henrik SchumacherHenrik Schumacher
52.9k471148
$endgroup$
add a comment |
$begingroup$
idx = Random`Private`PositionsOf[Rest[list1], 0];
result = list1;
result[[idx + 1]] = list2[[idx]];
result
{0, 1, 1, 4, 7, 1, 1, 10, 1, 11, 3, 1, 0}
answered 11 hours ago
Henrik SchumacherHenrik Schumacher
52.9k471148
$endgroup$
idx = Random`Private`PositionsOf[Rest[list1], 0];
result = list1;
result[[idx + 1]] = list2[[idx]];
result
{0, 1, 1, 4, 7, 1, 1, 10, 1, 11, 3, 1, 0}
answered 11 hours ago
Henrik SchumacherHenrik Schumacher
52.9k471148
answered 11 hours ago
Henrik SchumacherHenrik Schumacher
52.9k471148
answered 11 hours ago
Henrik SchumacherHenrik Schumacher
52.9k471148
answered 11 hours ago
Henrik SchumacherHenrik Schumacher
52.9k471148
52.9k471148
add a comment |
add a comment |
$begingroup$
Here's another possibility:
Module[{tmp=list1},
With[{i = Pick[Range[Length[list1]-1], Rest @ list1, 0]},
tmp[[i+1]]=list2[[i]]
];
tmp
]
{0, 1, 1, 4, 7, 1, 1, 10, 1, 11, 3, 1, 0}
answered 11 hours ago
Carl WollCarl Woll
67.9k389176
$endgroup$
add a comment |
$begingroup$
Here's another possibility:
Module[{tmp=list1},
With[{i = Pick[Range[Length[list1]-1], Rest @ list1, 0]},
tmp[[i+1]]=list2[[i]]
];
tmp
]
{0, 1, 1, 4, 7, 1, 1, 10, 1, 11, 3, 1, 0}
answered 11 hours ago
Carl WollCarl Woll
67.9k389176
$endgroup$
add a comment |
$begingroup$
Here's another possibility:
Module[{tmp=list1},
With[{i = Pick[Range[Length[list1]-1], Rest @ list1, 0]},
tmp[[i+1]]=list2[[i]]
];
tmp
]
{0, 1, 1, 4, 7, 1, 1, 10, 1, 11, 3, 1, 0}
answered 11 hours ago
Carl WollCarl Woll
67.9k389176
$endgroup$
Here's another possibility:
Module[{tmp=list1},
With[{i = Pick[Range[Length[list1]-1], Rest @ list1, 0]},
tmp[[i+1]]=list2[[i]]
];
tmp
]
{0, 1, 1, 4, 7, 1, 1, 10, 1, 11, 3, 1, 0}
answered 11 hours ago
Carl WollCarl Woll
67.9k389176
answered 11 hours ago
Carl WollCarl Woll
67.9k389176
answered 11 hours ago
Carl WollCarl Woll
67.9k389176
answered 11 hours ago
Carl WollCarl Woll
67.9k389176
67.9k389176
add a comment |
add a comment |
$begingroup$
Also MapIndexed works fine
(# /. List -> 0) & /@ MapIndexed[Replace[#1, 0 -> list2[[#2[[1]] - 1]]] &, list1]
The first piece just replaces the special case when a zero element is in the first entry (in Mathematica the 0th element is the Head, which in this case is List). Then the function MapIndexed does its job.
answered 11 hours ago
Mane.andreaMane.andrea
1366
$endgroup$
add a comment |
$begingroup$
Also MapIndexed works fine
(# /. List -> 0) & /@ MapIndexed[Replace[#1, 0 -> list2[[#2[[1]] - 1]]] &, list1]
The first piece just replaces the special case when a zero element is in the first entry (in Mathematica the 0th element is the Head, which in this case is List). Then the function MapIndexed does its job.
answered 11 hours ago
Mane.andreaMane.andrea
1366
$endgroup$
add a comment |
$begingroup$
Also MapIndexed works fine
(# /. List -> 0) & /@ MapIndexed[Replace[#1, 0 -> list2[[#2[[1]] - 1]]] &, list1]
The first piece just replaces the special case when a zero element is in the first entry (in Mathematica the 0th element is the Head, which in this case is List). Then the function MapIndexed does its job.
answered 11 hours ago
Mane.andreaMane.andrea
1366
$endgroup$
Also MapIndexed works fine
(# /. List -> 0) & /@ MapIndexed[Replace[#1, 0 -> list2[[#2[[1]] - 1]]] &, list1]
The first piece just replaces the special case when a zero element is in the first entry (in Mathematica the 0th element is the Head, which in this case is List). Then the function MapIndexed does its job.
answered 11 hours ago
Mane.andreaMane.andrea
1366
answered 11 hours ago
Mane.andreaMane.andrea
1366
answered 11 hours ago
Mane.andreaMane.andrea
1366
answered 11 hours ago
Mane.andreaMane.andrea
1366
1366
add a comment |
add a comment |
$begingroup$
MapThread[If[#1 == 0, #2, #1] &, {Join[list1, {0}], Join[{0}, list2]}] // Most
{0, 1, 1, 4, 7, 1, 1, 10, 1, 11, 3, 1, 0}
Your answer from above
{0, 1, 1, 4, 7, 1, 1, 10, 1, 11, 3, 1, 0}
answered 11 hours ago
MikeYMikeY
2,697412
$endgroup$
add a comment |
$begingroup$
MapThread[If[#1 == 0, #2, #1] &, {Join[list1, {0}], Join[{0}, list2]}] // Most
{0, 1, 1, 4, 7, 1, 1, 10, 1, 11, 3, 1, 0}
Your answer from above
{0, 1, 1, 4, 7, 1, 1, 10, 1, 11, 3, 1, 0}
answered 11 hours ago
MikeYMikeY
2,697412
$endgroup$
add a comment |
$begingroup$
MapThread[If[#1 == 0, #2, #1] &, {Join[list1, {0}], Join[{0}, list2]}] // Most
{0, 1, 1, 4, 7, 1, 1, 10, 1, 11, 3, 1, 0}
Your answer from above
{0, 1, 1, 4, 7, 1, 1, 10, 1, 11, 3, 1, 0}
answered 11 hours ago
MikeYMikeY
2,697412
$endgroup$
MapThread[If[#1 == 0, #2, #1] &, {Join[list1, {0}], Join[{0}, list2]}] // Most
{0, 1, 1, 4, 7, 1, 1, 10, 1, 11, 3, 1, 0}
Your answer from above
{0, 1, 1, 4, 7, 1, 1, 10, 1, 11, 3, 1, 0}
answered 11 hours ago
MikeYMikeY
2,697412
answered 11 hours ago
MikeYMikeY
2,697412
answered 11 hours ago
MikeYMikeY
2,697412
answered 11 hours ago
MikeYMikeY
2,697412
2,697412
add a comment |
add a comment |
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