finding Missing elements in a vector

I have a vector P =

     1
     2
     3
     4
     5
     6
     7
     8
     9
    10
    11
    12
    13
    14
    15
    16
    17
     2
    19
    20
    21
     3
    23
    24
     6
    26
    27
    28
    29
    30
    31
    32
As per the ascending order (1 2...15..32) of vector P, the elements 18,25,28 are missing and their position is occupied by the elements 2,3,6. now i want a vector which indicates this numbers like n=[2 3 6]. if elements are in order perfectly (1:10), I don't want new vector n.

Answer 1

[Luca Amerio]

This will do the trick

P(~(P==1:length(P)))

just for clarification

   P==1:length(P)

returns a logical array of the position occupied by the right number.

   ~(P==1:length(P))

is the logical array of the position occupied by the WRONG number

   P(~(P==1:length(P)))

get the values in those positions.

[Stephen23]

Or a little simpler:

P(P~=1:length(P))

E.g.:

>> P = [1,2,9,4,5,6,1,8,9];
>> P(P~=1:length(P))
ans =
   9     1

[Raghavendra Reddy P]

Thank you sir, but for P =

     1
     2
     3
     4
     5
     6
     7
     8
     9
    10
    11
    12
    13
    14
    15
    16
    17
     2
    19
    20
    21
     3
    23
    24
     6
    26
    27
    28
    29
    30
    31
    32

your code giving error like >> P(P~=1:length(P)) Error using ~= Matrix dimensions must agree.

[Raghavendra Reddy P]

i have checked that your code P(~(P==1:length(P))) Works if P is a row vector.

what changes i should incorporate in above code to work if P is column vector. Am not interested to convert row to column and vice-versa.

[Andrei Bobrov]

P(P(:)'~=1:length(P))

[Raghavendra Reddy P]

Thank you sir

Answer 2

[Andrei Bobrov]

out = P([1;diff(P)] < 0);

[Raghavendra Reddy P]

Thank you but for >> P = [1,2,9,4,5,6,1,8,9]; Am getting error like >> out = P([1;diff(P)] < 0) Error using vertcat Dimensions of matrices being concatenated are not consistent.

[Thorsten]

You have to convert P to a column vector

P = P(:);
out = P([1;diff(P)] < 0);

or use "," instead of ";", but this works only if P is a row vector

out = P([1,diff(P)] < 0);

[Raghavendra Reddy P]

for P = [1,2,9,4,5,6,1,8,9] if i use code out = P([1,diff(P)] < 0) am getting out = [ 4 1] instead of [9 1]

[Andrei Bobrov]

P(strfind([P(1);diff(P(:))]' ~= 1,[1 1]))

[Raghavendra Reddy P]

Thank you