Removing certain elements from a list of lists

Question

I have this list as an example:

L1 = {{1, 7, 23, 31, 19}, {1, 19, 17, 35, 9}, {1, 9, 25, 37, 21}, {1, 21, 15, 33, 7}, {2, 8, 24, 34, 20}, {2, 20, 18, 38, 10}, {2, 10, 26, 36, 22}, {2, 22, 16, 32, 8}, {3, 11, 28, 34, 24}, {3, 24, 8, 32, 12}, {3, 12, 27, 31, 23}, {3, 23, 7, 33, 11}, {4, 13, 30, 36, 26}, {4, 26, 10, 38, 14}, {4, 14, 29, 37, 25}, {4, 25, 9, 35, 13}, {5, 16, 22, 36, 30}, {5, 30, 13, 35, 17}, {5, 17, 19, 31, 27}, {5, 27, 12, 32, 16}, {6, 15, 21, 37, 29}, {6, 29, 14, 38, 18}, {6, 18, 20, 34, 28}, {6, 28, 11, 33, 15}}

I want to remove all elements {1, 2, 3, 4, 5, 6} from this list to give the following list as an output:

L2 = {{7, 23, 31, 19}, {19, 17, 35, 9}, {9, 25, 37, 21}, {21, 15, 33, 7}, {8, 24, 34, 20}, {20, 18, 38, 10}, {10, 26, 36,22}, {22, 16, 32, 8}, {11, 28, 34, 24}, {24, 8, 32, 12}, {12, 27, 31, 23}, {23, 7, 33, 11}, {13, 30, 36, 26}, {26, 10, 38, 14}, {14, 29, 37, 25}, {25, 9, 35, 13}, {16, 22, 36, 30}, {30, 13, 35, 17}, {17, 19, 31, 27}, {27, 12, 32, 16}, {15, 21, 37, 29}, {29, 14, 38, 18}, {18, 20, 34, 28}, {28, 11, 33, 15}}

In this case each time the first element of a of the list {a,b,c,d,e} is removed, however this might not always be the case for other examples.

My question is what is the best way to do this. I was thinking of two for-loops, but I am sure there is a nicer way to do this.

Thank you in advance.

Answer 1

DeleteCases[L1, Alternatives @@ Range@6, {2}]

Similar to @bmf but using the levelspec argument instead of mapping.

Answer 2

I was thinking of two for-loops,

That would probably be the Fortran or C way. In Mathematica there is no need to do this. You just need to find the builtin function instead and use that.

There are at least 10 ways to do this in Mathematica. One way could be

L1 = {{1, 7, 23, 31, 19}, {1, 19, 17, 35, 9}, {1, 9, 25, 37, 21}, {1, 
   21, 15, 33, 7}, {2, 8, 24, 34, 20}, {2, 20, 18, 38, 10}, {2, 10, 
   26, 36, 22}, {2, 22, 16, 32, 8}, {3, 11, 28, 34, 24}, {3, 24, 8, 
   32, 12}, {3, 12, 27, 31, 23}, {3, 23, 7, 33, 11}, {4, 13, 30, 36, 
   26}, {4, 26, 10, 38, 14}, {4, 14, 29, 37, 25}, {4, 25, 9, 35, 
   13}, {5, 16, 22, 36, 30}, {5, 30, 13, 35, 17}, {5, 17, 19, 31, 
   27}, {5, 27, 12, 32, 16}, {6, 15, 21, 37, 29}, {6, 29, 14, 38, 
   18}, {6, 18, 20, 34, 28}, {6, 28, 11, 33, 15}}

del={1,2,3,4,5,6}
Map[DeleteElements[#,del]&,L1]

see https://reference.wolfram.com/language/ref/DeleteElements.html

Answer 3

Just joining the party hoping that the following

DeleteCases[1 | 2 | 3 | 4 | 5 | 6] /@ L1;

counts as a distinct enough solution.

The check

LinearAlgebra`Private`ZeroArrayQ@Flatten[% - L2]

gives True.

Edit: there's also this that came to mind

Pick[L1, L1, Except[1 | 2 | 3 | 4 | 5 | 6, _Integer]];
LinearAlgebra`Private`ZeroArrayQ@Flatten[% - L2]

Answer 4

You can ReplaceAll such elements with Nothing, which gets removed from lists.

L1 = {{1, 7, 23, 31, 19}, {1, 19, 17, 35, 9}, {1, 9, 25, 37, 21}, {1, 
    21, 15, 33, 7}, {2, 8, 24, 34, 20}, {2, 20, 18, 38, 10}, {2, 10, 
    26, 36, 22}, {2, 22, 16, 32, 8}, {3, 11, 28, 34, 24}, {3, 24, 8, 
    32, 12}, {3, 12, 27, 31, 23}, {3, 23, 7, 33, 11}, {4, 13, 30, 36, 
    26}, {4, 26, 10, 38, 14}, {4, 14, 29, 37, 25}, {4, 25, 9, 35, 
    13}, {5, 16, 22, 36, 30}, {5, 30, 13, 35, 17}, {5, 17, 19, 31, 
    27}, {5, 27, 12, 32, 16}, {6, 15, 21, 37, 29}, {6, 29, 14, 38, 
    18}, {6, 18, 20, 34, 28}, {6, 28, 11, 33, 15}};
L2 = {{7, 23, 31, 19}, {19, 17, 35, 9}, {9, 25, 37, 21}, {21, 15, 33, 
    7}, {8, 24, 34, 20}, {20, 18, 38, 10}, {10, 26, 36, 22}, {22, 16, 
    32, 8}, {11, 28, 34, 24}, {24, 8, 32, 12}, {12, 27, 31, 23}, {23, 
    7, 33, 11}, {13, 30, 36, 26}, {26, 10, 38, 14}, {14, 29, 37, 
    25}, {25, 9, 35, 13}, {16, 22, 36, 30}, {30, 13, 35, 17}, {17, 19,
     31, 27}, {27, 12, 32, 16}, {15, 21, 37, 29}, {29, 14, 38, 
    18}, {18, 20, 34, 28}, {28, 11, 33, 15}};
L3 = L1 /. Alternatives @@ {1, 2, 3, 4, 5, 6} -> Nothing
L4 = L1 /. 1 | 2 | 3 | 4 | 5 | 6 -> Nothing (* equivalent *)
L2 == L3 == L4

Answer 5

Clear["Global`*"];
L1 = {{1, 7, 23, 31, 19}, {1, 19, 17, 35, 9}, {1, 9, 25, 37, 21}, {1, 
    21, 15, 33, 7}, {2, 8, 24, 34, 20}, {2, 20, 18, 38, 10}, {2, 10, 
    26, 36, 22}, {2, 22, 16, 32, 8}, {3, 11, 28, 34, 24}, {3, 24, 8, 
    32, 12}, {3, 12, 27, 31, 23}, {3, 23, 7, 33, 11}, {4, 13, 30, 36, 
    26}, {4, 26, 10, 38, 14}, {4, 14, 29, 37, 25}, {4, 25, 9, 35, 
    13}, {5, 16, 22, 36, 30}, {5, 30, 13, 35, 17}, {5, 17, 19, 31, 
    27}, {5, 27, 12, 32, 16}, {6, 15, 21, 37, 29}, {6, 29, 14, 38, 
    18}, {6, 18, 20, 34, 28}, {6, 28, 11, 33, 15}};

alt = Alternatives @@ Range[6];
Delete[L1, Position[L1, alt]]

or

Pick[L1, Map[FreeQ[alt], L1, {2}]]

Answer 6

Select[! Between[#, {1, 6}] &] /@ L1

{{7, 23, 31, 19}, {19, 17, 35, 9}, {9, 25, 37, 21}, {21, 15, 33, 7}, {8, 24, 34, 20}, {20, 18, 38, 10}, {10, 26, 36, 22}, {22, 16, 32, 8}, {11, 28, 34, 24}, {24, 8, 32, 12}, {12, 27, 31, 23}, {23, 7, 33, 11}, {13, 30, 36, 26}, {26, 10, 38, 14}, {14, 29, 37, 25}, {25, 9, 35, 13}, {16, 22, 36, 30}, {30, 13, 35, 17}, {17, 19, 31, 27}, {27, 12, 32, 16}, {15, 21, 37, 29}, {29, 14, 38, 18}, {18, 20, 34, 28}, {28, 11, 33, 15}}

Answer 7

Select[Not@*Between[{1, 6}]] /@ L1

A point-free version of @eldo's solution.

If we can assume that the values are all positive integers, we could do this instead:

Select[GreaterThan[6]] /@ L1

Answer 8

Another way using Position and ReplacePart:

ReplacePart[L1, Position[L1, Alternatives @@ Range[6]] -> Nothing] === L2

(*True*)

Or using Drop for this particular case:

Drop[#, {1}] & /@ L1 === L2

(*True*)

Or using Cases for this particular case:

Cases[L1, x : {___} :> Rest@x] === L2

(*True*)

Answer 9

Some more positional solutions

p = Position[L1, Alternatives @@ Range[6]];

Using Delete

r1 = Delete[p] @ L1;

Using ReplaceAt (new in 13.1)

r2 = ReplaceAt[_ :> Nothing, p] @ L1;

Using MapAt

r3 = MapAt[Nothing, p] @ L1;

r1 == r2 == r3 == L2

True