How to get the number in the cell containing the letter and space

3 vues (au cours des 30 derniers jours)
eko supriyadi
eko supriyadi le 1 Juin 2022
Commenté : dpb le 2 Juin 2022
Hi everyone
consider i have cell matrix like this:
a={'AABB 01001';'AACC 01101'; 'AACC 01201'; 'AADD 01301'; 'AAEE 01401'}
so, it will produce:
a =
5×1 cell array
{'AABB 01001' }
{'AACC 01101' }
{'AACC 01201' }
{'AADD 01301' }
{'AAEE 01401'}
Now, I have no idea how to take the numbers only (because contain letter and single-multiple space).. so i want the result is (with leading zeros in a number too):
a = 01001
01101
01201
01301
01401
the a result is double format.
thank you for your help :)
  4 commentaires
dpb
dpb le 1 Juin 2022
>> compose('%05d',str2double(extractAfter(a,' ')))
ans =
5×1 cell array
{'01001'}
{'01101'}
{'01201'}
{'01301'}
{'01401'}
>>
Stephen23
Stephen23 le 2 Juin 2022
"with leading zeros in a number too"
No numeric class on any common computer platform stores leading zeros, so what you ask for is not possible.
You could store the data as text, in which case you can keep the leading zeros.

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Réponse acceptée

Stephen23
Stephen23 le 2 Juin 2022
Modifié(e) : Stephen23 le 2 Juin 2022
No indirect, fragile, round-about conversion to other data classes is required. Here are two direct approaches:
a = {'AABB 01001';'AACC 01101'; 'AACC 01201'; 'AADD 01301'; 'AAEE 01401'}
a = 5×1 cell array
{'AABB 01001' } {'AACC 01101' } {'AACC 01201' } {'AADD 01301' } {'AAEE 01401'}
b = extract(a,digitsPattern)
b = 5×1 cell array
{'01001'} {'01101'} {'01201'} {'01301'} {'01401'}
b = regexp(a,'\d+','match','once')
b = 5×1 cell array
{'01001'} {'01101'} {'01201'} {'01301'} {'01401'}

Plus de réponses (1)

dpb
dpb le 1 Juin 2022
Modifié(e) : dpb le 1 Juin 2022
>> str2double(extractAfter(a,' '))
ans =
1001
1101
1201
1301
1401
>>
ADDENDUM:
>> compose('%05d',str2double(extractAfter(a,' ')))
ans =
5×1 cell array
{'01001'}
{'01101'}
{'01201'}
{'01301'}
{'01401'}
>>
or
>> string(compose('%05d',str2double(extractAfter(a,' '))))
ans =
5×1 string array
"01001"
"01101"
"01201"
"01301"
"01401"
>>
  2 commentaires
eko supriyadi
eko supriyadi le 1 Juin 2022
Modifié(e) : eko supriyadi le 1 Juin 2022
b=str2double(extractAfter(a,' '))
for i=1:length(b)
zeroleading(i,:)=sprintf('%05d',b(i))
end
if i want keep zero, is the loop above correct?
dpb
dpb le 2 Juin 2022
If you don't need the numeric values as numeric, there's no point in converting them to numeric just to return them back to string or cellstr() to have the leading zero to display. In that case just extract the numeric pattern in one form or another -- I used the simplest syntax possible, the pattern or regular expression offers more generality.
If want a char() array instead, then just convert directly --
>> char(strtrim(extractAfter(a,' ')))
ans =
5×5 char array
'01001'
'01101'
'01201'
'01301'
'01401'
>>

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