Remove array elements but also store the element indices that were not removed

I have a long array e.g. a = ["a", "b", "c", "d", "e" ,"f"]

I want to remove first and 5th element. u = [1,5]

For that I can do a(u) = []

But I also want the element indices that were not removed i.e. I want output as [2 3 4 6]

I tried a(~u) but it is not working.

Answer 1

[Steven Lord]

Do you want the indices or the elements that weren't deleted?

a = ["a", "b", "c", "d", "e" ,"f"];
u = [1 5];
indToKeep = setdiff(1:numel(a), u)

indToKeep = 1×4

2 3 4 6

I'm going to make a copy of a so you can compare the original with the modified copy.

a1 = a;
deletedElements = a1(u) % Extract elements 1 and 5 first

deletedElements = 1×2 string array
    "a"    "e"

a1(u) = [] % Then delete them from the orignnal vector

a1 = 1×4 string array
    "b"    "c"    "d"    "f"

[Mohd Aaquib Khan]

thank you, I will accept this after some time,

Actually I am currently using the same logic using setdiff(). Was just exploring if there were some easier,shorter ways to do it. Especially using '~'

Thanks again

Answer 2

[Image Analyst]

There are several ways. Here are two:

a = ["a", "b", "c", "d", "e" ,"f"]

a = 1×6 string array
    "a"    "b"    "c"    "d"    "e"    "f"

rowsToRemove = [1, 5];
aExtracted = a(rowsToRemove)

aExtracted = 1×2 string array
    "a"    "e"

aKept = setdiff(a, aExtracted)

aKept = 1×4 string array
    "b"    "c"    "d"    "f"

% Another way
aKept2 = a; % Initialize
aKept2(rowsToRemove) = []

aKept2 = 1×4 string array
    "b"    "c"    "d"    "f"

[Image Analyst]

I know you really really wanted to use ~ but I think using [] is easier and preferable. This is about as compact as you can get (two lines of code to give the two vectors) and doesn't use setdiff, though it removes the elements from a, thus changing it. If you want to keep a copy of the original, then make a copy first.

a = ["a", "b", "c", "d", "e" ,"f"];
rowsToRemove = [1, 5];
aExtracted = a(rowsToRemove)

aExtracted = 1×2 string array
    "a"    "e"

a(rowsToRemove) = []

a = 1×4 string array
    "b"    "c"    "d"    "f"

If you really insist on using ~, here is one way is can be done:

a = ["a", "b", "c", "d", "e" ,"f"]'
indexes = 1 : numel(a);
rowsToRemove = [1, 5];
logicalIndexes = ismember(indexes, rowsToRemove)
aExtracted = a(rowsToRemove)
aKeepers = a(~logicalIndexes)

[Mohd Aaquib Khan]

[] is best for deleting but I also need the indices that were not removed as mentioned in the question ([2 3 4 6] is the primary output that I desire). It is important because I want to assemble many matrices to a bigger matrix, thats why I wanted a simpler one line solution but I guess setdiff() is the only easy way which Steven showed and I was already using that.

so basically if

a = [10 20 30 40 50 60 70];
entryIndexToRemove = [1, 5];

then I want ExtractedIndices = [2, 3, 4, 6, 7] as the answer

which is done below using

setdiff( 1:length(a) , entryIndexToRemove)

ans = 1×5

2 3 4 6 7

[Image Analyst]

a = [10 20 30 40 50 60 70];
indexes = 1 : numel(a);
rowsToRemove = [1, 5];
logicalIndexes = ismember(indexes, rowsToRemove)

logicalIndexes = 1×7 logical array
   1   0   0   0   1   0   0

aExtracted = a(rowsToRemove)

aExtracted = 1×2

10 50

aKeepers = a(~logicalIndexes)

aKeepers = 1×5

20 30 40 60 70

% Also log what indexes are kept.
keeperIndexes = indexes(~logicalIndexes)

keeperIndexes = 1×5

2 3 4 6 7