String operations on dynamic dataset

klb
2021 2 8
1 个回答
回答已采纳
更新时间：2021 2 13
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0 个投票

Hi everyone,
Number of machines activated and hence the duration timestamp, varies each day so the dataset dimention is dynamic.
I am trying to extract the dataset for each machine. Cant get my extractBetwen () to work.
w=[ "Mac_A"
"1 9:53.3"
"2 9:23.5"
"3 2:16.2"
"4 2:45.6"
"5 12:01.2"
"Mac_B"
"1 23:56.5"
"2 28:12.6"
"3 15:34.1"
"4 19:23.0"
"5 1:38.4"
"Mac_C"
"1 11:35.6"]
% ..
% many more machines and thier datasets
l=length(w(strlength(w)==5)) % counts how many machines
for e=1: length(w)
start = w(strlength(w)==5)
stop = w(strlength(w)<8)
itsdata(data(l),:) = extractBetween(w,start,stop) % cant get this to work.
end
expected output is vectors which are the datasets of the machines.
["1 9:53.3,2 9:23.5,3 2:16.2,4 2:45.6,5 12:01.2"]
["1 23:56.5 , 2 28:12.6,3 15:34.1,4 19:23.0,5 1:38.4"]
["1 11:35.6"]
..
.. %more vectors

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Jan
Jan 2021-2-9
编辑：Jan 2021-2-11

1 个投票

w=[ "Mac_A"
"1 9:53.3"
"2 9:23.5"
"3 2:16.2"
"4 2:45.6"
"5 12:01.2"
"Mac_B"
"1 23:56.5"
"2 28:12.6"
"3 15:34.1"
"4 19:23.0"
"51:38.4"
"Mac_C"
"1 11:35.6"];
Sep = [find(startsWith(w, 'Mac_')); numel(w) + 1];
n = numel(Sep) - 1;
itsdata = strings(n, 1);
for k = 1:n
itsdata(k) = [w{Sep(k)+1:Sep(k+1)-1}];
end
The command extractBetween() cuts out a piece of a string, not an array of strings between indices.

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klb
klb 2021-2-9
编辑：klb 2021-2-9
Hi Jan, returns the last extraction only i.e. For Mac_C. it doesnt store the other datasets. If the loop can output an array (rather than vector as I ask) where each row is the string dataset of each machine, I could work with that too. expected output would look like:
itsdata =
"1 9:53.3,2 9:23.5,3 2:16.2,4 2:45.6,5 12:01.2"
"1 23:56.5 , 2 28:12.6,3 15:34.1,4 19:23.0,5 1:38.4"
"1 11:35.6"
...
Jan
Jan 2021-2-9
I had a typo in my code. See above: "(k)" inserted to assign the k.th string of itsdata instead of overwriting it.
klb
klb 2021-2-11
编辑：klb 2021-2-11
Inside the loop, getting "Unable to perform assignment because the indices on the left side are not compatible with the size of the right side." error.
Jan
Jan 2021-2-11
Please try it again with the current code. I do not get the error message you have posted.
klb
klb 2021-2-13
编辑：klb 2021-2-13
Expected output is with a comma sparator:
"1 9:53.3,2 9:23.5,3 2:16.2,4 2:45.6,5 12:01.2"
the code returns is
1 9:53.32 9:23.53 2:16.24 2:45.65 12:01.2
i.e the first value should be 1 9:53.3, but it has combined the serial no. 2 to it and results in 1 9:53.32.
Hence, modified the original dataset by adding a "," at the end of each row.
Works but crude. Could you suggest another way to obtain the Expected output as requested?
Can you also explain this command : Sep = [find(startsWith(w, 'Mac_')); numel(w) + 1]
w=[ "Mac_A"
"1 9:53.3"
"2 9:23.5"
"3 2:16.2"
"4 2:45.6"
"5 12:01.2"
"Mac_B"
"1 23:56.5"
"2 28:12.6"
"3 15:34.1"
"4 19:23.0"
"5 1:38.4"
"Mac_C"
"1 11:35.6"];
w= w+ "," % i added this.
Sep = [find(startsWith(w, 'Mac_')); numel(w) + 1]
n = numel(Sep) - 1;
itsdata = strings(n, 1);
for k = 1:n
itsdata(k) = [w{Sep(k)+1:Sep(k+1)-1} ];
end
itsdata
Jan
Jan 2021-2-13
w=[ "Mac_A"; "1 9:53.3"; "2 9:23.5"; "3 2:16.2"; ...
"4 2:45.6"; "5 12:01.2"; "Mac_B"; "1 23:56.5"; ...
"2 28:12.6"; "3 15:34.1"; "4 19:23.0"; "5 1:38.4"; ...
"Mac_C"; "1 11:35.6"];
Sep = [find(startsWith(w, 'Mac_')); numel(w) + 1]
n = numel(Sep) - 1;
itsdata = strings(n, 1);
for k = 1:n
itsdata(k) = join(w(Sep(k)+1:Sep(k+1)-1), ',');
end
itsdata
klb
klb 2021-2-13
编辑：klb 2021-2-13
Jan, Thank you so very much!
This expresion : Sep = [find(startsWith(w, 'Mac_')); numel(w) + 1], is complex so broke it into its elements and echoed each output. I understand that logic as well now. Thank you again for your time.

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