Dropping list elements from nested list after evaluation
$begingroup$
I need to create a new list from a nested list but using the evaluation as criteria to drop the elements. For example let's say that that I have the following list:
list1={{1,1,-(-1)^3,x,2*x},{1,1,(-1)^3,x,2*x},
{1,1,x,2*x,3*x},{1,1,-x,-2*x,-3*x}}
and I need to eliminate the elements of list1 that the absolute value of the third element give $1$, i.d. $-(-1)^3$ and $(-1)^3$, to obtain
list2={{1,1,x,2*x,3*x},{1,1,-x,-2*x,-3*x}}
In this case, list1 was created with the code
For[i = 1, i < 4, i++,
For[j = 1, j < 4, j++,
list1[i, j, p_] = Sort[Eigenvalues[mat[i, j, p]]];
] ]
I have been trying to use Select but until now I am not been able to create list2 to plot it with
list2=ParallelTable[Select[Abs[eigval[i, j, p][[3]]],
Abs[#] != 1 &] , {i, 1, 4}, {j,1,4}]
I am still learning to uses cases in Mathematica so I am not sure how to do it. Do you know if there is wise way to do it? Thanks in advance.
list-manipulation filtering
edited yesterday
Roman
5,20011131
asked yesterday
morsmors
716
$endgroup$
add a comment |
$begingroup$
I need to create a new list from a nested list but using the evaluation as criteria to drop the elements. For example let's say that that I have the following list:
list1={{1,1,-(-1)^3,x,2*x},{1,1,(-1)^3,x,2*x},
{1,1,x,2*x,3*x},{1,1,-x,-2*x,-3*x}}
and I need to eliminate the elements of list1 that the absolute value of the third element give $1$, i.d. $-(-1)^3$ and $(-1)^3$, to obtain
list2={{1,1,x,2*x,3*x},{1,1,-x,-2*x,-3*x}}
In this case, list1 was created with the code
For[i = 1, i < 4, i++,
For[j = 1, j < 4, j++,
list1[i, j, p_] = Sort[Eigenvalues[mat[i, j, p]]];
] ]
I have been trying to use Select but until now I am not been able to create list2 to plot it with
list2=ParallelTable[Select[Abs[eigval[i, j, p][[3]]],
Abs[#] != 1 &] , {i, 1, 4}, {j,1,4}]
I am still learning to uses cases in Mathematica so I am not sure how to do it. Do you know if there is wise way to do it? Thanks in advance.
list-manipulation filtering
edited yesterday
Roman
5,20011131
asked yesterday
morsmors
716
$endgroup$
add a comment |
$begingroup$
I need to create a new list from a nested list but using the evaluation as criteria to drop the elements. For example let's say that that I have the following list:
list1={{1,1,-(-1)^3,x,2*x},{1,1,(-1)^3,x,2*x},
{1,1,x,2*x,3*x},{1,1,-x,-2*x,-3*x}}
and I need to eliminate the elements of list1 that the absolute value of the third element give $1$, i.d. $-(-1)^3$ and $(-1)^3$, to obtain
list2={{1,1,x,2*x,3*x},{1,1,-x,-2*x,-3*x}}
In this case, list1 was created with the code
For[i = 1, i < 4, i++,
For[j = 1, j < 4, j++,
list1[i, j, p_] = Sort[Eigenvalues[mat[i, j, p]]];
] ]
I have been trying to use Select but until now I am not been able to create list2 to plot it with
list2=ParallelTable[Select[Abs[eigval[i, j, p][[3]]],
Abs[#] != 1 &] , {i, 1, 4}, {j,1,4}]
I am still learning to uses cases in Mathematica so I am not sure how to do it. Do you know if there is wise way to do it? Thanks in advance.
list-manipulation filtering
edited yesterday
Roman
5,20011131
asked yesterday
morsmors
716
$endgroup$
I need to create a new list from a nested list but using the evaluation as criteria to drop the elements. For example let's say that that I have the following list:
list1={{1,1,-(-1)^3,x,2*x},{1,1,(-1)^3,x,2*x},
{1,1,x,2*x,3*x},{1,1,-x,-2*x,-3*x}}
and I need to eliminate the elements of list1 that the absolute value of the third element give $1$, i.d. $-(-1)^3$ and $(-1)^3$, to obtain
list2={{1,1,x,2*x,3*x},{1,1,-x,-2*x,-3*x}}
In this case, list1 was created with the code
For[i = 1, i < 4, i++,
For[j = 1, j < 4, j++,
list1[i, j, p_] = Sort[Eigenvalues[mat[i, j, p]]];
] ]
I have been trying to use Select but until now I am not been able to create list2 to plot it with
list2=ParallelTable[Select[Abs[eigval[i, j, p][[3]]],
Abs[#] != 1 &] , {i, 1, 4}, {j,1,4}]
I am still learning to uses cases in Mathematica so I am not sure how to do it. Do you know if there is wise way to do it? Thanks in advance.
list-manipulation filtering
list-manipulation filtering
edited yesterday
Roman
5,20011131
asked yesterday
morsmors
716
edited yesterday
Roman
5,20011131
asked yesterday
morsmors
716
edited yesterday
Roman
5,20011131
edited yesterday
Roman
5,20011131
edited yesterday
Roman
5,20011131
5,20011131
asked yesterday
morsmors
716
asked yesterday
morsmors
716
asked yesterday
morsmors
716
716
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
If you prefer using DeleteCases,
list2 = DeleteCases[list1, _?(Abs[#[[3]]] == 1 &)]
{{1, 1, x, 2 x, 3 x}, {1, 1, -x, -2 x, -3 x}}
answered yesterday
RomanRoman
5,20011131
$endgroup$
add a comment |
$begingroup$
Delete[
list1,
Position[Abs[list1[[All, 3]]], 1]
]
{{1, 1, x, 2 x, 3 x}, {1, 1, -x, -2 x, -3 x}}
answered yesterday
Henrik SchumacherHenrik Schumacher
60k582168
$endgroup$
add a comment |
$begingroup$
if you want to use Select, try this
Select[list1,!NumberQ@#[[3]]||Abs[#[[3]]]!=1&]
{{1, 1, x, 2 x, 3 x}, {1, 1, -x, -2 x, -3 x}}
answered yesterday
J42161217J42161217
4,468324
$endgroup$
add a comment |
$begingroup$
This is pretty efficient on unpacked arrays (the Listable attribute assumes list1[[All, 3]] is a flat list, as it is in the OP's example):
Block[{signal},
SetAttributes[signal, Listable];
signal[1] = 1; signal[_] = 0;
Pick[list1, signal@Abs[list1[[All, 3]]], 0]
]
(* {{1, 1, x, 2 x, 3 x}, {1, 1, -x, -2 x, -3 x}} *)
(For packed arrays, one would probably want to use Unitize[x-1] instead of signal.)
answered yesterday
Michael E2Michael E2
150k12203482
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If you prefer using DeleteCases,
list2 = DeleteCases[list1, _?(Abs[#[[3]]] == 1 &)]
{{1, 1, x, 2 x, 3 x}, {1, 1, -x, -2 x, -3 x}}
answered yesterday
RomanRoman
5,20011131
$endgroup$
add a comment |
$begingroup$
If you prefer using DeleteCases,
list2 = DeleteCases[list1, _?(Abs[#[[3]]] == 1 &)]
{{1, 1, x, 2 x, 3 x}, {1, 1, -x, -2 x, -3 x}}
answered yesterday
RomanRoman
5,20011131
$endgroup$
add a comment |
$begingroup$
If you prefer using DeleteCases,
list2 = DeleteCases[list1, _?(Abs[#[[3]]] == 1 &)]
{{1, 1, x, 2 x, 3 x}, {1, 1, -x, -2 x, -3 x}}
answered yesterday
RomanRoman
5,20011131
$endgroup$
If you prefer using DeleteCases,
list2 = DeleteCases[list1, _?(Abs[#[[3]]] == 1 &)]
{{1, 1, x, 2 x, 3 x}, {1, 1, -x, -2 x, -3 x}}
answered yesterday
RomanRoman
5,20011131
answered yesterday
RomanRoman
5,20011131
answered yesterday
RomanRoman
5,20011131
answered yesterday
RomanRoman
5,20011131
5,20011131
add a comment |
add a comment |
$begingroup$
Delete[
list1,
Position[Abs[list1[[All, 3]]], 1]
]
{{1, 1, x, 2 x, 3 x}, {1, 1, -x, -2 x, -3 x}}
answered yesterday
Henrik SchumacherHenrik Schumacher
60k582168
$endgroup$
add a comment |
$begingroup$
Delete[
list1,
Position[Abs[list1[[All, 3]]], 1]
]
{{1, 1, x, 2 x, 3 x}, {1, 1, -x, -2 x, -3 x}}
answered yesterday
Henrik SchumacherHenrik Schumacher
60k582168
$endgroup$
add a comment |
$begingroup$
Delete[
list1,
Position[Abs[list1[[All, 3]]], 1]
]
{{1, 1, x, 2 x, 3 x}, {1, 1, -x, -2 x, -3 x}}
answered yesterday
Henrik SchumacherHenrik Schumacher
60k582168
$endgroup$
Delete[
list1,
Position[Abs[list1[[All, 3]]], 1]
]
{{1, 1, x, 2 x, 3 x}, {1, 1, -x, -2 x, -3 x}}
answered yesterday
Henrik SchumacherHenrik Schumacher
60k582168
answered yesterday
Henrik SchumacherHenrik Schumacher
60k582168
answered yesterday
Henrik SchumacherHenrik Schumacher
60k582168
answered yesterday
Henrik SchumacherHenrik Schumacher
60k582168
60k582168
add a comment |
add a comment |
$begingroup$
if you want to use Select, try this
Select[list1,!NumberQ@#[[3]]||Abs[#[[3]]]!=1&]
{{1, 1, x, 2 x, 3 x}, {1, 1, -x, -2 x, -3 x}}
answered yesterday
J42161217J42161217
4,468324
$endgroup$
add a comment |
$begingroup$
if you want to use Select, try this
Select[list1,!NumberQ@#[[3]]||Abs[#[[3]]]!=1&]
{{1, 1, x, 2 x, 3 x}, {1, 1, -x, -2 x, -3 x}}
answered yesterday
J42161217J42161217
4,468324
$endgroup$
add a comment |
$begingroup$
if you want to use Select, try this
Select[list1,!NumberQ@#[[3]]||Abs[#[[3]]]!=1&]
{{1, 1, x, 2 x, 3 x}, {1, 1, -x, -2 x, -3 x}}
answered yesterday
J42161217J42161217
4,468324
$endgroup$
if you want to use Select, try this
Select[list1,!NumberQ@#[[3]]||Abs[#[[3]]]!=1&]
{{1, 1, x, 2 x, 3 x}, {1, 1, -x, -2 x, -3 x}}
answered yesterday
J42161217J42161217
4,468324
answered yesterday
J42161217J42161217
4,468324
answered yesterday
J42161217J42161217
4,468324
answered yesterday
J42161217J42161217
4,468324
4,468324
add a comment |
add a comment |
$begingroup$
This is pretty efficient on unpacked arrays (the Listable attribute assumes list1[[All, 3]] is a flat list, as it is in the OP's example):
Block[{signal},
SetAttributes[signal, Listable];
signal[1] = 1; signal[_] = 0;
Pick[list1, signal@Abs[list1[[All, 3]]], 0]
]
(* {{1, 1, x, 2 x, 3 x}, {1, 1, -x, -2 x, -3 x}} *)
(For packed arrays, one would probably want to use Unitize[x-1] instead of signal.)
answered yesterday
Michael E2Michael E2
150k12203482
$endgroup$
add a comment |
$begingroup$
This is pretty efficient on unpacked arrays (the Listable attribute assumes list1[[All, 3]] is a flat list, as it is in the OP's example):
Block[{signal},
SetAttributes[signal, Listable];
signal[1] = 1; signal[_] = 0;
Pick[list1, signal@Abs[list1[[All, 3]]], 0]
]
(* {{1, 1, x, 2 x, 3 x}, {1, 1, -x, -2 x, -3 x}} *)
(For packed arrays, one would probably want to use Unitize[x-1] instead of signal.)
answered yesterday
Michael E2Michael E2
150k12203482
$endgroup$
add a comment |
$begingroup$
This is pretty efficient on unpacked arrays (the Listable attribute assumes list1[[All, 3]] is a flat list, as it is in the OP's example):
Block[{signal},
SetAttributes[signal, Listable];
signal[1] = 1; signal[_] = 0;
Pick[list1, signal@Abs[list1[[All, 3]]], 0]
]
(* {{1, 1, x, 2 x, 3 x}, {1, 1, -x, -2 x, -3 x}} *)
(For packed arrays, one would probably want to use Unitize[x-1] instead of signal.)
answered yesterday
Michael E2Michael E2
150k12203482
$endgroup$
This is pretty efficient on unpacked arrays (the Listable attribute assumes list1[[All, 3]] is a flat list, as it is in the OP's example):
Block[{signal},
SetAttributes[signal, Listable];
signal[1] = 1; signal[_] = 0;
Pick[list1, signal@Abs[list1[[All, 3]]], 0]
]
(* {{1, 1, x, 2 x, 3 x}, {1, 1, -x, -2 x, -3 x}} *)
(For packed arrays, one would probably want to use Unitize[x-1] instead of signal.)
answered yesterday
Michael E2Michael E2
150k12203482
answered yesterday
Michael E2Michael E2
150k12203482
answered yesterday
Michael E2Michael E2
150k12203482
answered yesterday
Michael E2Michael E2
150k12203482
150k12203482
add a comment |
add a comment |
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