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

351 次查看（过去 30 天）
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'}

请先登录，再进行评论。

请先登录，再回答此问题。

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"

请先登录，再进行评论。

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))

请先登录，再进行评论。

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.

请先登录，再进行评论。

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.

类别

在 Help Center 和 File Exchange 中查找有关 Characters and Strings 的更多信息

标签

产品


版本

R2021b

Community Treasure Hunt

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

Start Hunting!

Translated by