How to remove everything from string except what's inside square brackets?

I have this cell array:
{'0,11:1.03 SPEED MEASURED 1 [rpm]'}
{'0,12:1.03 SPEED MEASURED 1 [rpm]'}
{'0,13:1.03 SPEED MEASURED 1 [rpm]'}
{'0,14:1.03 SPEED MEASURED 1 [rpm]'}
{'0,11:1.06 MOTOR CURRENT [A]' }
{'0,12:1.06 MOTOR CURRENT [A]' }
{'0,13:1.06 MOTOR CURRENT [A]' }
{'0,14:1.06 MOTOR CURRENT [A]' }
{'0,11:2.13 TORQ USED REF [%]' }
{'0,12:2.13 TORQ USED REF [%]' }
{'0,13:2.13 TORQ USED REF [%]' }
{'0,14:2.13 TORQ USED REF [%]' }
{'0,11:1.08 MOTOR TORQUE [%]' }
{'0,12:1.08 MOTOR TORQUE [%]' }
{'0,13:1.08 MOTOR TORQUE [%]' }
{'0,14:1.08 MOTOR TORQUE [%]' }
{'0,11:2.10 TORQUE REF 3 [%]' }
{'0,12:2.10 TORQUE REF 3 [%]' }
{'0,13:2.10 TORQUE REF 3 [%]' }
{'0,14:2.10 TORQUE REF 3 [%]' }
How can I get the unit from between the brackets in each string, so that I'm left with the following:
{'rpm'}
{'rpm'}
{'rpm'}
{'rpm'}
{'A' }
{'A' }
...
{'%' }
Thanks!

6 Kommentare

regexp this is the saviour.
Thanks! This is the command I ended up using:
regexp(headers{2}, '(?<=\[).*?(?=\])', 'match');
Thats great...this function makes me always confused......
Yeah me too. I always struggle with it, even when I'm trying to achieve something really simple (like in this case) ?
Wow, I never even knew about this function! Very intuitive and seems to work just as well as regexp. Are there any benefits to using extractBetween other than the fact that it's more easy to use for this purpose?

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Antworten (1)

Stephen23
Stephen23 am 12 Jul. 2019
Bearbeitet: Stephen23 am 12 Jul. 2019
Where C is your cell array:
>> D = regexp(C,'\[(.+)\]','tokens','once')
>> D = vertcat(D{:})
D =
'rpm'
'rpm'
'rpm'
'rpm'
'A'
'A'
'A'
'A'
'%'
'%'
'%'
'%'
'%'
'%'
'%'
'%'
'%'
'%'
'%'
'%'

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Gefragt:

am 12 Jul. 2019

Bearbeitet:

am 12 Jul. 2019

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