How do you get up to first n characters in a string?

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Dylan
Dylan 2021-10-26
回答： Geoff 2024-4-12
I have an array of strings of variable length s_arr. I want to limit the strings to only contain the first 3 elements. Some of the string have less than 3 elements, in which case they should be unchanged. I tried
arrayfun(@(s) s(1:3), s_arr)
but this gives an error when it finds a string with fewer than 3 elements. Why does MATLAB give an error and not just return the string unchanged? How can I achieve this goal?
  1 个评论
Stephen23
Stephen23 2024-2-2
Ran in:
You were very close:
s_arr = ["a";"ab";"abc";"abcd";"abcde"]
s_arr = 5×1 string array
"a" "ab" "abc" "abcd" "abcde"
cellfun(@(s) s(1:min(end,3)), s_arr, 'uni',0)
ans = 5×1 cell array
{'a' } {'ab' } {'abc'} {'abc'} {'abc'}

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回答（5 个）

Dyuman Joshi
Dyuman Joshi 2024-2-2
Ran in:
A simple approach using extractBetween -
s_arr = ["a";"ab";"abc";"abcd";"abcde";"abcdef";"xyzxyzxyz"]
s_arr = 7×1 string array
"a" "ab" "abc" "abcd" "abcde" "abcdef" "xyzxyzxyz"
pos = 4; % pos = no. of characters + 1
extractBetween(s_arr, 1, min(pos, strlength(s_arr)))
ans = 7×1 string array
"a" "ab" "abc" "abcd" "abcd" "abcd" "xyzx"
Paul
Paul 2021-10-27
编辑：Paul 2024-2-2
Ran in:
If a string array:
s_arr = ["a";"ab";"abc";"abcd";"abcde"]
s_arr = 5×1 string array
"a" "ab" "abc" "abcd" "abcde"
s_arr(strlength(s_arr)>3) = extractBefore(s_arr(strlength(s_arr)>3),4)
s_arr = 5×1 string array
"a" "ab" "abc" "abc" "abc"
Exact same code works on a cell array
s_arr = cellstr(["a";"ab";"abc";"abcd";"abcde"])
s_arr = 5×1 cell array
{'a' } {'ab' } {'abc' } {'abcd' } {'abcde'}
s_arr(strlength(s_arr)>3) = extractBefore(s_arr(strlength(s_arr)>3),4)
s_arr = 5×1 cell array
{'a' } {'ab' } {'abc'} {'abc'} {'abc'}
  1 个评论
Stephen23
Stephen23 2024-2-2
Ran in:
+1 This gracefully handles the shorter elements. Stylistically I would define the index beforehand:
s_arr = ["a";"ab";"abc";"abcd";"abcde"]
s_arr = 5×1 string array
"a" "ab" "abc" "abcd" "abcde"
N = 3;
X = strlength(s_arr)>N;
s_arr(X) = extractBefore(s_arr(X),1+N)
s_arr = 5×1 string array
"a" "ab" "abc" "abc" "abc"

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Image Analyst
Image Analyst 2021-10-26
% Character array
charArray = 'abcdefgh' % Single quotes
n = 3
out = charArray(1:n) % Another way
% String
str = "abcdefgh" % Double quotes
n = 3
charArray = char(str)
out = charArray(1:n)
charArray =
'abcdefgh'
n =
3
out =
'abc'
str =
"abcdefgh"
n =
3
charArray =
'abcdefgh'
out =
'abc'
  1 个评论
Image Analyst
Image Analyst 2021-10-26
For the case where n is longer than the string, or less than the string (so more general and robust):
% Character array
charArray = 'ab' % Single quotes
n = 30
out = charArray(1:min(length(charArray), n)) % Another way
% String
str = "ab" % Double quotes
charArray = char(str)
out = charArray(1:min(length(charArray), n))

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Ramanan
Ramanan 2024-2-2
You can use the extractBefore function in the array function instead.
pos = 4; % pos = no. of characters + 1
out = arrayfun(@(s) extractBefore(s,pos), s_arr);
  2 个评论
Stephen23
Stephen23 2024-2-2
编辑：Stephen23 2024-2-2
Ran in:
This fails due to the shorter elements:
s_arr = ["a";"ab";"abc";"abcd";"abcde"]
s_arr = 5×1 string array
"a" "ab" "abc" "abcd" "abcde"
pos = 4; % pos = no. of characters + 1
out = arrayfun(@(s) extractBefore(s,pos), s_arr)
Error using extractBefore
Numeric value exceeds the number of characters in element 1.

Error in solution>@(s)extractBefore(s,pos) (line 3)
out = arrayfun(@(s) extractBefore(s,pos), s_arr)
See Paul's answer from three years ago for one approach to handle this.
Ramanan
Ramanan 2024-2-2
Ran in:
You can add the min function and make use of the length of the string inside the arrayfun.
I don't notice the previously answered one.
s_arr = ["a";"ab";"abc";"abcd";"abcde";"abcdef";"xyzxyzxyz"];
pos = 4; % pos = no. of characters + 1
out = arrayfun(@(s) extractBefore(s,min(pos,strlength(s))), s_arr)
out = 7×1 string array
"" "a" "ab" "abc" "abc" "abc" "xyz"
Also I think a more simplified version is submitted by Dyuman Joshi.

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Geoff
Geoff 2024-4-12
Learning about and using regex (regular expressions) has solved many string dilemmas for me. Here's a regex one liner for this question:
regexprep(s_arr,"(.{3}).*","$1")
Explanation:
The function regexprep ("reg ex replace") is just a fancy text search-and-replace. In a nutshell, our strategy is to find the entire string and replace it with just its first {3} characters.
The first input s_arr is your string array. regexprep will operate on each element of the array.
The next two inputs are our search "(.{3}).*" and our replacement "$1". You can read about regex syntax here. In a nutshell:
  • The period . matches any character.
  • And .{3} matches any 3 consecutive characters.
  • And .* matches any number of characters -- i.e. the rest of the string.
So, we've searched for the entire string, represented it in two parts: the first 3 characters (.{3}) and the rest .*. The surrounding parentheses in (.{3}) say we want to capture (i.e. remember) those characters. The $1 in our replacement string is what recalls those captured characters.
So, the entire string has been replaced by the first three characters. If the string is less than 3 characters, regexprep doesn't find any matches in its search, so nothing is replaced.
Pros and cons:
Con: it's cryptic.
Pro: It doesn't require Matlab to compute the length of the strings in the array.

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