How to take a common particular part of string

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Mekala balaji
Mekala balaji 2016-5-28
评论： Image Analyst 2016-5-28
Hi,
I have the below cell array. Please some one help me how to do this.
PR.VK0K200E5T00_T123
PR.VBK0K800E3F00_R243
PR.VP124K800E5T00
PR.ZVK0K10E5T00_T123_FF99
My output is below.
200E5
800E3
800E5
10E5
Many thanks in advance.
  1 个评论
the cyclist
the cyclist 2016-5-28
The hard part here is interpreting the exact rule you want to use to parse the input. Azzi's solution below assumes you want the "E", and the numeric characters before and after that. (Seems reasonable, given your input, and that it looks like numbers in exponential notation. However, it will break if you have an input like "PR.ZVK0K10E5T00_T123_FF5E99".)

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采纳的回答

Azzi Abdelmalek
Azzi Abdelmalek 2016-5-28
s={'PR.VK0K200E5T00_T123'
'PR.VBK0K800E3F00_R243'
'PR.VP124K800E5T00'
'PR.ZVK0K10E5T00_T123_FF99'}
out=regexp(s,'\d+E\d+','match')
out=[out{:}]'
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Mekala balaji
Mekala balaji 2016-5-28
Sir,
Can it be possible to further generalize, like as you said "PR.ZVK0K10E5T00_T123_FF5E99", it will also break, but to generalize it, the left and right side should be numeric (not a Aplhabets, or _ or .(dot))., So that my out put will avoid 5E99.
Many thanks,
Azzi Abdelmalek
Azzi Abdelmalek 2016-5-28
编辑：Azzi Abdelmalek 2016-5-28
s={'PR.VK0K200E5T00_T123'
'PR.VBK0K800E3F00_R243'
'PR.VP124K800E5T00'
'PR.ZVK0K10E5T00_T123_FF99'
'PR.ZVK0K10E5T00_T123_FF5E99'}
out=regexp(s,'\d+E\d+','match','once')

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更多回答（1 个）

Image Analyst
Image Analyst 2016-5-28
编辑：Image Analyst 2016-5-28
Here's a non-regexp way that I think should be a lot easier to understand:
% Create data.
s1 = 'PR.VK0K200E5T00_T123'
s2 = 'PR.VBK0K800E3F00_R243'
s3 = 'PR.VP124K800E5T00'
s4 = 'PR.ZVK0K10E5T00_T123_FF99'
% Now find the number
str = s1; % To test, set = s2, s3, and s4
% Find the final E
lastE = find(str == 'E', 1, 'last')
% Find the final K
lastK = find(str(1:lastE-1) == 'K', 1, 'last')
% Extract the string from 1 character after the K
% to one character after the last E.
output = str(lastK+1 : lastE+1)
This works for all the cases that you presented.
  1 个评论
Image Analyst
Image Analyst 2016-5-28
That's why it's important to give all cases in advance. So now I've added a line to handle the case of "PR.ZVK0K10E5T00_T123_FF5E99" which has an extra E in the string:
% Now find the number
str = 'PR.ZVK0K10E5T00_T123_FF5E99';
% Truncate string after first underline
str = str(1:find(str=='_', 1, 'first'));
% Find the final E
lastE = find(str == 'E', 1, 'last')
% Find the final K
lastK = find(str(1:lastE-1) == 'K', 1, 'last')
% Extract the string from 1 character after the K
% to one character after the last E.
output = str(lastK+1 : lastE+1)
Now it gives 10E5. I hope that is what you wanted.

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