How to get the count of continuous occurrence of specified name in the table

Hi,

I have the below Table:

Name5
Name2
Name2
Name1
Name3
Name2
Name2
Name2
Name2
Name5
Name3
Name3
Name2
Name2
Name2

I want to get the Name2 continuous count, and accumulated count as explained below:

For example: Name2 run continuously three times (2(row2&row3), 4(row6~9), 3(row13~15)), its max is 4, and its accumulated continuous run is 9(2+4+3). Output1:

2
4
3

output2:

Name5  -1
Name2  1
Name2  2
Name1  -1
Name3  -2
Name2  1
Name2  2
Name2  3
Name2  4
Name5  -1
Name3  -2
Name3  -3
Name2  1
Name2  2
Name2  3

Output2 explanation:

Consider Name2 as 1, and if it is continuous the number increase like 1,2,3 ...(till change from Name2 to other) and except Name2, all other weighting given as -1(it increases like -1,-2,-3,-4 etc till Name2 appear). For instance, row1 is Name5(it is -1), row 2&3 are Name2 (1,2), row4&5 Name1&Name3 (-1,-2), row 6~9 Name2(1,2,3,4) etc. My purpose is when Name2 change to other it is given -1 and till it changes to Name2 again (Name2 given as 1 and increase if it is continuous).

Answer 1

[Azzi Abdelmalek]

a={'Name5'
'Name2'
'Name2'
'Name1'
'Name3'
'Name2'
'Name2'
'Name2'
'Name2'
'Name5'
'Name3'
'Name3'
'Name2'
'Name2'
'Name2'
'Name2'}
n=numel(a);
ii=ismember(a,'Name2')'
ii1=strfind([0 ii],[0 1])
ii2=strfind([ii 0],[1 0])
out1=(ii2-ii1+1)'
jj=1
for k=1:numel(ii1)
  out2(jj:ii1(k)-1)=-(1:(ii1(k)-jj))
  out2(ii1(k):ii2(k))=1:ii2(k)-ii1(k)+1
  jj=ii2(k)+1;
end
out2(end+1:n)=-(1:n-ii2(k))

[Mekala balaji]

Sir,

It giving error if Name2 does not exist. Now I handled for the case like:

a={'Name5'
'Name1'
'Name3'
'Name7'
'Name6'
'Name5'
'Name3'
'Name3'
}
a=[a;{'Name2'}]
and delete last row from out2 (out2(end,:)=[]). It works but is there any smart way.