Replacing the Zeros of One List by the (i-1)th Element of Another List
Clash Royale CLAN TAG#URR8PPP
$begingroup$
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 Feb 11 at 17:03
WillWill
3348
$endgroup$
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 Feb 11 at 17:03
WillWill
3348
$endgroup$
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 Feb 11 at 17:03
WillWill
3348
$endgroup$
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 Feb 11 at 17:03
WillWill
3348
asked Feb 11 at 17:03
WillWill
3348
asked Feb 11 at 17:03
WillWill
3348
asked Feb 11 at 17:03
WillWill
3348
asked Feb 11 at 17:03
WillWill
3348
3348
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 Feb 11 at 17:32
RomanRoman
2,624717
$endgroup$
$begingroup$
What a simple way to do this. Thank you so much.
$endgroup$
– Will
Feb 11 at 17:51
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 Feb 11 at 17:12
Henrik SchumacherHenrik Schumacher
55.5k576154
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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 Feb 11 at 17:13
Carl WollCarl Woll
69.3k393179
$endgroup$
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 Feb 11 at 17:16
Mane.andreaMane.andrea
1816
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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 Feb 11 at 17:12
MikeYMikeY
3,258614
$endgroup$
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 Feb 11 at 17:32
RomanRoman
2,624717
$endgroup$
$begingroup$
What a simple way to do this. Thank you so much.
$endgroup$
– Will
Feb 11 at 17:51
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 Feb 11 at 17:32
RomanRoman
2,624717
$endgroup$
$begingroup$
What a simple way to do this. Thank you so much.
$endgroup$
– Will
Feb 11 at 17:51
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 Feb 11 at 17:32
RomanRoman
2,624717
$endgroup$
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 Feb 11 at 17:32
RomanRoman
2,624717
edited Feb 11 at 17:59
answered Feb 11 at 17:32
RomanRoman
2,624717
answered Feb 11 at 17:32
RomanRoman
2,624717
answered Feb 11 at 17:32
RomanRoman
2,624717
2,624717
$begingroup$
What a simple way to do this. Thank you so much.
$endgroup$
– Will
Feb 11 at 17:51
add a comment |
$begingroup$
What a simple way to do this. Thank you so much.
$endgroup$
– Will
Feb 11 at 17:51
$begingroup$
What a simple way to do this. Thank you so much.
$endgroup$
– Will
Feb 11 at 17:51
$begingroup$
What a simple way to do this. Thank you so much.
$endgroup$
– Will
Feb 11 at 17:51
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 Feb 11 at 17:12
Henrik SchumacherHenrik Schumacher
55.5k576154
$endgroup$
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 Feb 11 at 17:12
Henrik SchumacherHenrik Schumacher
55.5k576154
$endgroup$
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 Feb 11 at 17:12
Henrik SchumacherHenrik Schumacher
55.5k576154
$endgroup$
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 Feb 11 at 17:12
Henrik SchumacherHenrik Schumacher
55.5k576154
answered Feb 11 at 17:12
Henrik SchumacherHenrik Schumacher
55.5k576154
answered Feb 11 at 17:12
Henrik SchumacherHenrik Schumacher
55.5k576154
answered Feb 11 at 17:12
Henrik SchumacherHenrik Schumacher
55.5k576154
55.5k576154
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 Feb 11 at 17:13
Carl WollCarl Woll
69.3k393179
$endgroup$
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 Feb 11 at 17:13
Carl WollCarl Woll
69.3k393179
$endgroup$
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 Feb 11 at 17:13
Carl WollCarl Woll
69.3k393179
$endgroup$
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 Feb 11 at 17:13
Carl WollCarl Woll
69.3k393179
answered Feb 11 at 17:13
Carl WollCarl Woll
69.3k393179
answered Feb 11 at 17:13
Carl WollCarl Woll
69.3k393179
answered Feb 11 at 17:13
Carl WollCarl Woll
69.3k393179
69.3k393179
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 Feb 11 at 17:16
Mane.andreaMane.andrea
1816
$endgroup$
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 Feb 11 at 17:16
Mane.andreaMane.andrea
1816
$endgroup$
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 Feb 11 at 17:16
Mane.andreaMane.andrea
1816
$endgroup$
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 Feb 11 at 17:16
Mane.andreaMane.andrea
1816
answered Feb 11 at 17:16
Mane.andreaMane.andrea
1816
answered Feb 11 at 17:16
Mane.andreaMane.andrea
1816
answered Feb 11 at 17:16
Mane.andreaMane.andrea
1816
1816
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 Feb 11 at 17:12
MikeYMikeY
3,258614
$endgroup$
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 Feb 11 at 17:12
MikeYMikeY
3,258614
$endgroup$
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 Feb 11 at 17:12
MikeYMikeY
3,258614
$endgroup$
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 Feb 11 at 17:12
MikeYMikeY
3,258614
answered Feb 11 at 17:12
MikeYMikeY
3,258614
answered Feb 11 at 17:12
MikeYMikeY
3,258614
answered Feb 11 at 17:12
MikeYMikeY
3,258614
3,258614
add a comment |
add a comment |
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