Azzera filtri
Azzera filtri

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

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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 Commenti
Heidi Mäkitalo
Heidi Mäkitalo il 12 Lug 2019
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?

Accedi per commentare.

Risposte (1)

Stephen23
Stephen23 il 12 Lug 2019
Modificato: Stephen23 il 12 Lug 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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