Dropping list elements from nested list after evaluation Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern) Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?How do I delete items at the same position from every sub-list within a list?Using and replacing sequential elements of a listFinding the main parent after sorting in a multiplication processDeleting certain elements from a nested listData selection by comparing elements from different sublists in a nested listSelect from nested list and dropping non matching elementsSelecting elements in nested listRemove elements from deep nested listEliminate empty elements from a list with a specific pattern
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Dropping list elements from nested list after evaluation
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)
Announcing the arrival of Valued Associate #679: Cesar Manara
Unicorn Meta Zoo #1: Why another podcast?How do I delete items at the same position from every sub-list within a list?Using and replacing sequential elements of a listFinding the main parent after sorting in a multiplication processDeleting certain elements from a nested listData selection by comparing elements from different sublists in a nested listSelect from nested list and dropping non matching elementsSelecting elements in nested listRemove elements from deep nested listEliminate empty elements from a list with a specific pattern
$begingroup$
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 Apr 11 at 16:30
Roman
5,73111131
asked Apr 11 at 13:50
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 Apr 11 at 16:30
Roman
5,73111131
asked Apr 11 at 13:50
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 Apr 11 at 16:30
Roman
5,73111131
asked Apr 11 at 13:50
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 Apr 11 at 16:30
Roman
5,73111131
asked Apr 11 at 13:50
morsmors
716
edited Apr 11 at 16:30
Roman
5,73111131
asked Apr 11 at 13:50
morsmors
716
edited Apr 11 at 16:30
Roman
5,73111131
edited Apr 11 at 16:30
Roman
5,73111131
edited Apr 11 at 16:30
Roman
5,73111131
5,73111131
asked Apr 11 at 13:50
morsmors
716
asked Apr 11 at 13:50
morsmors
716
asked Apr 11 at 13:50
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 Apr 11 at 16:29
RomanRoman
5,73111131
$endgroup$
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 Apr 11 at 14:01
Henrik SchumacherHenrik Schumacher
60.6k584170
$endgroup$
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 Apr 11 at 14:16
J42161217J42161217
4,598324
$endgroup$
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 Apr 12 at 0:11
Michael E2Michael E2
151k12203483
$endgroup$
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 Apr 11 at 16:29
RomanRoman
5,73111131
$endgroup$
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 Apr 11 at 16:29
RomanRoman
5,73111131
$endgroup$
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 Apr 11 at 16:29
RomanRoman
5,73111131
$endgroup$
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 Apr 11 at 16:29
RomanRoman
5,73111131
answered Apr 11 at 16:29
RomanRoman
5,73111131
answered Apr 11 at 16:29
RomanRoman
5,73111131
answered Apr 11 at 16:29
RomanRoman
5,73111131
5,73111131
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 Apr 11 at 14:01
Henrik SchumacherHenrik Schumacher
60.6k584170
$endgroup$
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 Apr 11 at 14:01
Henrik SchumacherHenrik Schumacher
60.6k584170
$endgroup$
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 Apr 11 at 14:01
Henrik SchumacherHenrik Schumacher
60.6k584170
$endgroup$
Delete[
list1,
Position[Abs[list1[[All, 3]]], 1]
]
1, 1, x, 2 x, 3 x, 1, 1, -x, -2 x, -3 x
answered Apr 11 at 14:01
Henrik SchumacherHenrik Schumacher
60.6k584170
answered Apr 11 at 14:01
Henrik SchumacherHenrik Schumacher
60.6k584170
answered Apr 11 at 14:01
Henrik SchumacherHenrik Schumacher
60.6k584170
answered Apr 11 at 14:01
Henrik SchumacherHenrik Schumacher
60.6k584170
60.6k584170
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 Apr 11 at 14:16
J42161217J42161217
4,598324
$endgroup$
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 Apr 11 at 14:16
J42161217J42161217
4,598324
$endgroup$
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 Apr 11 at 14:16
J42161217J42161217
4,598324
$endgroup$
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 Apr 11 at 14:16
J42161217J42161217
4,598324
answered Apr 11 at 14:16
J42161217J42161217
4,598324
answered Apr 11 at 14:16
J42161217J42161217
4,598324
answered Apr 11 at 14:16
J42161217J42161217
4,598324
4,598324
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 Apr 12 at 0:11
Michael E2Michael E2
151k12203483
$endgroup$
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 Apr 12 at 0:11
Michael E2Michael E2
151k12203483
$endgroup$
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 Apr 12 at 0:11
Michael E2Michael E2
151k12203483
$endgroup$
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 Apr 12 at 0:11
Michael E2Michael E2
151k12203483
answered Apr 12 at 0:11
Michael E2Michael E2
151k12203483
answered Apr 12 at 0:11
Michael E2Michael E2
151k12203483
answered Apr 12 at 0:11
Michael E2Michael E2
151k12203483
151k12203483
add a comment |
add a comment |
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