Truncate strings to maximum length
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I have an array of strings (new strings, not old character arrays) of varying lengths, and I would like to truncate them to below some maximum length. I thought that extractBefore would be helpful, but it fails for strings shorter than the truncation length.
>> str = ["ab", "cdefg"];
>> extractBefore(str, 3) % this works fine
ans =
1×2 string array
"ab" "cd"
>> extractBefore(str, 4) % this fails
Error using extractBefore
Numeric value exceeds the number of characters in element 1.
This is my current solution:
>> arrayfun(@(s)extractBefore(s, min(4, s.strlength())+1), str)
ans =
1×2 string array
"ab" "cdef"
However, this is awkward and difficult to read.
Is there no ready-made functionality for doing this?
1 commentaire
dpb
le 21 Oct 2024
"Is there no ready-made functionality for doing this?"
Amazingly, no...with everything they did add, the equivalent of MID$, LEFT$, RIGHT$ are not available in a directly-callable functional form.
Réponses (2)
Jan
le 16 Fév 2023
Modifié(e) : Jan
le 16 Fév 2023
Omit arrayfun:
str = ["ab", "cdefg"];
extractBefore(str, min(4, str.strlength())+1)
If this still looks to clumsy, write your own function:
limitWidth(str, 4)
function s = limitWidth(s, n)
s = extractBefore(s, min(n, s.strlength()) + 1);
end
0 commentaires
Stephen23
le 16 Fév 2023
Modifié(e) : Stephen23
le 31 Oct 2024 à 0:07
str = ["", "ab", "1234", "cdefgh"];
out = regexprep(str,'^(.{1,4}).*?$','$1') % LEFT
out = regexprep(str,'^.*?(.{1,4})$','$1') % RIGHT
or (but note the behavior of the empty string):
out = regexp(str,'^.{1,4}','match','once') % LEFT
out = regexp(str,'.{1,4}$','match','once') % RIGHT
4 commentaires
dpb
le 31 Oct 2024 à 11:47
Modifié(e) : dpb
le 31 Oct 2024 à 13:21
@Stephen23, by writing the function names with the $ sign, I was presuming it would be clear the intent was to mimic BASIC (and VBA) which originally required the trailing $ on string variables. Current dialects have dropped the $ convention, of course, but the functionality is retained.
The definition of MID$
res=mid(str,start,[len])
to pick up to len characters from str beginning at start position in str. If omitted, the rest of the original string is returned, in which case it mimics RIGHT$.
It doesn't mean to try to split the string midway so there's no dependency on whether the string length is even/odd nor any ambiguity in the result.
Stephen23
le 31 Oct 2024 à 13:28
Perhaps:
str = ["", "ab", "1234", "cdefgh"];
out = regexprep(str,'^.{0,2}(.{0,4}).*$','$1') % MID(START,LEN)
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