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

10 vues (au cours des 30 derniers jours)
Heidi Mäkitalo
Heidi Mäkitalo le 12 Juil 2019
Modifié(e) : Stephen23 le 12 Juil 2019
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 commentaires
Heidi Mäkitalo
Heidi Mäkitalo le 12 Juil 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?

Connectez-vous pour commenter.

Réponses (1)

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

Tags

Aucun tag saisi pour le moment.

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by