Filter lists by discarding sublists that contain integers different from 0 or 1

Question

Imagine I have a list of the type

Lst = {{1,0,1,1},{4,0,1,1}, {1,1,0,0},{2,1,1,1},{0,0,0,1}}

I want to select only those sublists that only contain elements 0 or 1. For example for the above list the reduced list which I want is

IWantThis = {{1,0,1,1},{1,1,0,0},{0,0,0,1}}

Motivation for this is the given situation describes spinless fermions or spin 1/2 objects.

If the order doesn't matter then BitAnd[Lst, 1] ⋂ Lst does it — Coolwater, Feb 27, 2019 at 12:21

JEM_Mosig · Accepted Answer · 2019-02-27 11:20:35Z

4

To filter a list

list = {{1, 0, 1, 1}, {4, 0, 1, 1}, {1, 1, 0, 0}, {2, 1, 1, 1}, {0, 0, 0, 1}};

you can use Select like this:

Select[list, ContainsOnly[{0, 1}]]
(* {{1, 0, 1, 1}, {1, 1, 0, 0}, {0, 0, 0, 1}} *)

This should work with Mathematica 10.2+.

answered Feb 27, 2019 at 11:20

JEM_Mosig

2,95315 silver badges28 bronze badges

$\begingroup$ OK. Thanks a lot $\endgroup$
– Quasar Supernova
Feb 27, 2019 at 11:50
1

$\begingroup$ In earlier versions, use Cases[] instead: Cases[list, {(0 | 1) ..}]. $\endgroup$
– J. M.'s eventual burnout ♦
Mar 1, 2019 at 9:30

Add a comment |

Anjan Kumar · Accepted Answer · 2019-02-27 11:29:51Z

2

Lst = {{1,0,1,1},{4,0,1,1}, {1,1,0,0},{2,1,1,1},{0,0,0,1}};
Select[Lst, Union[#] == {0, 1} &]

{{1,0,1,1},{1,1,0,0},{0,0,0,1}}

I have used Unionspecifically, because it sorts out the list after removing duplicates unlike DeleteDuplicates.

answered Feb 27, 2019 at 11:21

Anjan Kumar

4,9191 gold badge15 silver badges28 bronze badges

$\begingroup$ Ok thanks a lot $\endgroup$
– Quasar Supernova
Feb 27, 2019 at 11:45

Add a comment |

2 Answers 2