Creating index and replacing values

Hi,

I have A = [0 0 1 0 1 0 0], and B = [ "030121", "030221", "030321"]

I want to create C = [ "030121", "030121", "030221", "030221", "030321", "030321", "030321"], by looking at A, and treating each sucessive sequence beginning with 1 as a new day, and replacing with the corresponding date from B. his is a generalization of a large problem.

Any help would be appreciated!

Inna

[Scott MacKenzie]

Note that

B = [ '030121', '030221', '030321']

is the same as

B = '030121030221030321'

and that

C = [  '030121', '030121',  '030221',  '030221', '030321', '030321', '030321']

is the same as

C = '030121030121030221030221030321030321030321'

Did you perhaps intend these to be strings, for example

B = ["030121", "030221", "030321"]

[Inna Pelloso]

Correction made. I intended them to be strings.

Answer 1

[Steven Lord]

A = [0 0 1 0 1 0 0];
B = [ "030121", "030221", "030321"];

I'm going to assume that A does not start with 1, or if it does that indicates the first element of the result should be the second element in B.

C = 1 + cumsum(A)

C = 1×7

1 1 2 2 3 3 3

D = B(C)

D = 1×7 string array
    "030121"    "030121"    "030221"    "030221"    "030321"    "030321"    "030321"

Alternately if you're doing this to build a datetime array:

D2 = datetime(2021, 3, C)

D2 = 1×7 datetime array
   01-Mar-2021   01-Mar-2021   02-Mar-2021   02-Mar-2021   03-Mar-2021   03-Mar-2021   03-Mar-2021

D2.Format = 'MMddyy'

D2 = 1×7 datetime array
   030121   030121   030221   030221   030321   030321   030321

[Inna Pelloso]

Elegant solution!

[Scott MacKenzie]

@Steven Lord Hmm, cumsum. Yes, that's way to do this. Thanks.

Answer 2

[Scott MacKenzie]

Below is a loop solution. There might be a way to vectorize this -- not sure.

A = [0 0 1 0 1 0 0];
B = [ "030121", "030221", "030321"];

M = [];
k = 1;
for i=1:length(A)
    if A(i) == 1 % new day
        k = k + 1;
    end    
    M = [M B(k)];
end

Output:

M =

1×7 string array

"030121" "030121" "030221" "030221" "030321" "030321" "030321"

Obviously, the number of 1s in A can't exceed the number of strings in B (minus 1).

[Inna Pelloso]

appreciate the help! Cumsum is rather elegant!