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$begingroup$
I have two arrays, say
array1 = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9};
array2 = {13, 4, 6, 20, 21};
I want to compare these two arrays by cycling array1 through all elements in array2. The output should be the position in each array where they match a condition, where the condition array2[[m]] = array1[[n]]-1 is true. That is, the output should be:
output = {6,2}
Because in array2, $5-1 = 4$. (array1[[6]] = 5 and array2[[2]] = 4). So far, my code is
output = {Position[array1, #][[1, 1]],
Position[array2, #][[1, 1]]} & /@ (I have no idea)
Where the code on the left hand side of Map gives me the positions of a true condition. On the right hand side of Map, I'm not sure what to do.
To clarify a little bit, the code I want is similar to
output = {Position[array1, #][[1, 1]],
Position[array2, #][[1, 1]]} & /@ Intersection[array1, array2]
Except I want the position of the element when array2 = array1-1 instead of when array1 = array2.
array conditional
edited 2 hours ago
Roman
2,055715
asked 3 hours ago
MetallicSilenceMetallicSilence
112
New contributor
MetallicSilence is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I have two arrays, say
array1 = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9};
array2 = {13, 4, 6, 20, 21};
I want to compare these two arrays by cycling array1 through all elements in array2. The output should be the position in each array where they match a condition, where the condition array2[[m]] = array1[[n]]-1 is true. That is, the output should be:
output = {6,2}
Because in array2, $5-1 = 4$. (array1[[6]] = 5 and array2[[2]] = 4). So far, my code is
output = {Position[array1, #][[1, 1]],
Position[array2, #][[1, 1]]} & /@ (I have no idea)
Where the code on the left hand side of Map gives me the positions of a true condition. On the right hand side of Map, I'm not sure what to do.
To clarify a little bit, the code I want is similar to
output = {Position[array1, #][[1, 1]],
Position[array2, #][[1, 1]]} & /@ Intersection[array1, array2]
Except I want the position of the element when array2 = array1-1 instead of when array1 = array2.
array conditional
edited 2 hours ago
Roman
2,055715
asked 3 hours ago
MetallicSilenceMetallicSilence
112
New contributor
MetallicSilence is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I have two arrays, say
array1 = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9};
array2 = {13, 4, 6, 20, 21};
I want to compare these two arrays by cycling array1 through all elements in array2. The output should be the position in each array where they match a condition, where the condition array2[[m]] = array1[[n]]-1 is true. That is, the output should be:
output = {6,2}
Because in array2, $5-1 = 4$. (array1[[6]] = 5 and array2[[2]] = 4). So far, my code is
output = {Position[array1, #][[1, 1]],
Position[array2, #][[1, 1]]} & /@ (I have no idea)
Where the code on the left hand side of Map gives me the positions of a true condition. On the right hand side of Map, I'm not sure what to do.
To clarify a little bit, the code I want is similar to
output = {Position[array1, #][[1, 1]],
Position[array2, #][[1, 1]]} & /@ Intersection[array1, array2]
Except I want the position of the element when array2 = array1-1 instead of when array1 = array2.
array conditional
edited 2 hours ago
Roman
2,055715
asked 3 hours ago
MetallicSilenceMetallicSilence
112
New contributor
MetallicSilence is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I have two arrays, say
array1 = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9};
array2 = {13, 4, 6, 20, 21};
I want to compare these two arrays by cycling array1 through all elements in array2. The output should be the position in each array where they match a condition, where the condition array2[[m]] = array1[[n]]-1 is true. That is, the output should be:
output = {6,2}
Because in array2, $5-1 = 4$. (array1[[6]] = 5 and array2[[2]] = 4). So far, my code is
output = {Position[array1, #][[1, 1]],
Position[array2, #][[1, 1]]} & /@ (I have no idea)
Where the code on the left hand side of Map gives me the positions of a true condition. On the right hand side of Map, I'm not sure what to do.
To clarify a little bit, the code I want is similar to
output = {Position[array1, #][[1, 1]],
Position[array2, #][[1, 1]]} & /@ Intersection[array1, array2]
Except I want the position of the element when array2 = array1-1 instead of when array1 = array2.
array conditional
array conditional
edited 2 hours ago
Roman
2,055715
asked 3 hours ago
MetallicSilenceMetallicSilence
112
New contributor
MetallicSilence is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 hours ago
Roman
2,055715
asked 3 hours ago
MetallicSilenceMetallicSilence
112
New contributor
MetallicSilence is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 hours ago
Roman
2,055715
edited 2 hours ago
Roman
2,055715
edited 2 hours ago
Roman
2,055715
2,055715
asked 3 hours ago
MetallicSilenceMetallicSilence
112
New contributor
MetallicSilence is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
MetallicSilenceMetallicSilence
112
asked 3 hours ago
MetallicSilenceMetallicSilence
112
112
New contributor
MetallicSilence is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
MetallicSilence is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
MetallicSilence is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You can use Tuples to construct all pairs of positions and use Select to pick the pairs that satisfy your condition:
pairs = Tuples[{ Range[Length[array1]], Range[Length[array2]]}];
Select[pairs, array1[[#[[1]]]] - 1 == array2[[#[[2]]]] &]
{{6, 2}, {8, 3}}
You can also use Outer as follows:
Join @@ Outer[If[array1[[#]] - 1 == array2[[#2]], {##}, Nothing] &,
Range[Length[array1]], Range[Length[array2]]]
{{6, 2}, {8, 3}}
Yet other ways: variations on Roman's method using Position and Outer combination:
Position[1] @ Outer[Subtract, array1, array2]
Position[True] @ Outer[Equal, array1 - 1, array2]
Position[{i_, i_}]@Outer[List, array1 - 1, array2]
{{6, 2}, {8, 3}}
answered 3 hours ago
kglrkglr
186k10203422
$endgroup$
$begingroup$
I like yourOuter[Subtract, ...way!
$endgroup$
– Roman
2 hours ago
$begingroup$
thank you @Roman. Didn't think of usingPositionuntil i saw your answer.
$endgroup$
– kglr
2 hours ago
add a comment |
$begingroup$
This is a general solution for any condition between the elements of array1 and array2:
Position[Outer[List, array1, array2], {i_, j_} /; j == i - 1]
{{6, 2}, {8, 3}}
@kglr's solutions are simpler than this when the condition is a difference as in the given problem.
answered 2 hours ago
RomanRoman
2,055715
$endgroup$
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You can use Tuples to construct all pairs of positions and use Select to pick the pairs that satisfy your condition:
pairs = Tuples[{ Range[Length[array1]], Range[Length[array2]]}];
Select[pairs, array1[[#[[1]]]] - 1 == array2[[#[[2]]]] &]
{{6, 2}, {8, 3}}
You can also use Outer as follows:
Join @@ Outer[If[array1[[#]] - 1 == array2[[#2]], {##}, Nothing] &,
Range[Length[array1]], Range[Length[array2]]]
{{6, 2}, {8, 3}}
Yet other ways: variations on Roman's method using Position and Outer combination:
Position[1] @ Outer[Subtract, array1, array2]
Position[True] @ Outer[Equal, array1 - 1, array2]
Position[{i_, i_}]@Outer[List, array1 - 1, array2]
{{6, 2}, {8, 3}}
answered 3 hours ago
kglrkglr
186k10203422
$endgroup$
$begingroup$
I like yourOuter[Subtract, ...way!
$endgroup$
– Roman
2 hours ago
$begingroup$
thank you @Roman. Didn't think of usingPositionuntil i saw your answer.
$endgroup$
– kglr
2 hours ago
add a comment |
$begingroup$
You can use Tuples to construct all pairs of positions and use Select to pick the pairs that satisfy your condition:
pairs = Tuples[{ Range[Length[array1]], Range[Length[array2]]}];
Select[pairs, array1[[#[[1]]]] - 1 == array2[[#[[2]]]] &]
{{6, 2}, {8, 3}}
You can also use Outer as follows:
Join @@ Outer[If[array1[[#]] - 1 == array2[[#2]], {##}, Nothing] &,
Range[Length[array1]], Range[Length[array2]]]
{{6, 2}, {8, 3}}
Yet other ways: variations on Roman's method using Position and Outer combination:
Position[1] @ Outer[Subtract, array1, array2]
Position[True] @ Outer[Equal, array1 - 1, array2]
Position[{i_, i_}]@Outer[List, array1 - 1, array2]
{{6, 2}, {8, 3}}
answered 3 hours ago
kglrkglr
186k10203422
$endgroup$
$begingroup$
I like yourOuter[Subtract, ...way!
$endgroup$
– Roman
2 hours ago
$begingroup$
thank you @Roman. Didn't think of usingPositionuntil i saw your answer.
$endgroup$
– kglr
2 hours ago
add a comment |
$begingroup$
You can use Tuples to construct all pairs of positions and use Select to pick the pairs that satisfy your condition:
pairs = Tuples[{ Range[Length[array1]], Range[Length[array2]]}];
Select[pairs, array1[[#[[1]]]] - 1 == array2[[#[[2]]]] &]
{{6, 2}, {8, 3}}
You can also use Outer as follows:
Join @@ Outer[If[array1[[#]] - 1 == array2[[#2]], {##}, Nothing] &,
Range[Length[array1]], Range[Length[array2]]]
{{6, 2}, {8, 3}}
Yet other ways: variations on Roman's method using Position and Outer combination:
Position[1] @ Outer[Subtract, array1, array2]
Position[True] @ Outer[Equal, array1 - 1, array2]
Position[{i_, i_}]@Outer[List, array1 - 1, array2]
{{6, 2}, {8, 3}}
answered 3 hours ago
kglrkglr
186k10203422
$endgroup$
You can use Tuples to construct all pairs of positions and use Select to pick the pairs that satisfy your condition:
pairs = Tuples[{ Range[Length[array1]], Range[Length[array2]]}];
Select[pairs, array1[[#[[1]]]] - 1 == array2[[#[[2]]]] &]
{{6, 2}, {8, 3}}
You can also use Outer as follows:
Join @@ Outer[If[array1[[#]] - 1 == array2[[#2]], {##}, Nothing] &,
Range[Length[array1]], Range[Length[array2]]]
{{6, 2}, {8, 3}}
Yet other ways: variations on Roman's method using Position and Outer combination:
Position[1] @ Outer[Subtract, array1, array2]
Position[True] @ Outer[Equal, array1 - 1, array2]
Position[{i_, i_}]@Outer[List, array1 - 1, array2]
{{6, 2}, {8, 3}}
answered 3 hours ago
kglrkglr
186k10203422
edited 2 hours ago
answered 3 hours ago
kglrkglr
186k10203422
answered 3 hours ago
kglrkglr
186k10203422
answered 3 hours ago
kglrkglr
186k10203422
186k10203422
$begingroup$
I like yourOuter[Subtract, ...way!
$endgroup$
– Roman
2 hours ago
$begingroup$
thank you @Roman. Didn't think of usingPositionuntil i saw your answer.
$endgroup$
– kglr
2 hours ago
add a comment |
$begingroup$
I like yourOuter[Subtract, ...way!
$endgroup$
– Roman
2 hours ago
$begingroup$
thank you @Roman. Didn't think of usingPositionuntil i saw your answer.
$endgroup$
– kglr
2 hours ago
$begingroup$
I like your
Outer[Subtract, ... way!$endgroup$
– Roman
2 hours ago
$begingroup$
I like your
Outer[Subtract, ... way!$endgroup$
– Roman
2 hours ago
$begingroup$
thank you @Roman. Didn't think of using
Position until i saw your answer.$endgroup$
– kglr
2 hours ago
$begingroup$
thank you @Roman. Didn't think of using
Position until i saw your answer.$endgroup$
– kglr
2 hours ago
add a comment |
$begingroup$
This is a general solution for any condition between the elements of array1 and array2:
Position[Outer[List, array1, array2], {i_, j_} /; j == i - 1]
{{6, 2}, {8, 3}}
@kglr's solutions are simpler than this when the condition is a difference as in the given problem.
answered 2 hours ago
RomanRoman
2,055715
$endgroup$
add a comment |
$begingroup$
This is a general solution for any condition between the elements of array1 and array2:
Position[Outer[List, array1, array2], {i_, j_} /; j == i - 1]
{{6, 2}, {8, 3}}
@kglr's solutions are simpler than this when the condition is a difference as in the given problem.
answered 2 hours ago
RomanRoman
2,055715
$endgroup$
add a comment |
$begingroup$
This is a general solution for any condition between the elements of array1 and array2:
Position[Outer[List, array1, array2], {i_, j_} /; j == i - 1]
{{6, 2}, {8, 3}}
@kglr's solutions are simpler than this when the condition is a difference as in the given problem.
answered 2 hours ago
RomanRoman
2,055715
$endgroup$
This is a general solution for any condition between the elements of array1 and array2:
Position[Outer[List, array1, array2], {i_, j_} /; j == i - 1]
{{6, 2}, {8, 3}}
@kglr's solutions are simpler than this when the condition is a difference as in the given problem.
answered 2 hours ago
RomanRoman
2,055715
edited 2 hours ago
answered 2 hours ago
RomanRoman
2,055715
answered 2 hours ago
RomanRoman
2,055715
answered 2 hours ago
RomanRoman
2,055715
2,055715
add a comment |
add a comment |
MetallicSilence is a new contributor. Be nice, and check out our Code of Conduct.
MetallicSilence is a new contributor. Be nice, and check out our Code of Conduct.
MetallicSilence is a new contributor. Be nice, and check out our Code of Conduct.
MetallicSilence is a new contributor. Be nice, and check out our Code of Conduct.
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