Problem 41. Cell joiner

Created by Cody Team

Solve Later

You are given a cell array of strings and a string delimiter. You need to produce one string which is composed of each string from the cell array separated by the delimiter.

For example, this input

 in_cell = {'Lorem', 'ipsum', 'dolor', 'sit', 'amet', 'consectetur'};
 delim = ' ';

should produce this output:

 out_str = 'Lorem ipsum dolor sit amet consectetur';

Solve

Solution Stats

6307 Solutions
2255 Solvers

Last Solution submitted on Jan 19, 2020

Last 200 Solutions

Problem Comments

7 Comments

7 Comments

Show 4 older comments

hari laxman on 31 May 2013

There has been a function added in 2013a for this!!

Mineralter on 26 Jun 2013

I make it in R2013a,but I don't know the R2012b or other previous versions have this function.

Jean-Marie Sainthillier on 5 Jul 2013

Most of the solutions consider that delim=1char

Gyoung Deuk Kim on 17 Nov 2016

i don't know

jubin soni on 9 Mar 2017

tricky one!!

Arief Anbiya on 19 Apr 2017

Regards.
May I ask.
I have tried in Matlab, all three test cases were validated but I got incorrect from cody.
Here is what inside the function :
%%%%%%%%%%
out_str=char(strjoin(in_cell,delim));
l=out_str==' ';
out_str(find((l(2:end)+l(1:end-1))==2))=[];
%%%%%%%%%
Thanks and regards.

ren yi xin ren on 11 Jan 2020 at 3:24

function out_str=cellstr_joiner(in_cell, delim)
L=strlength(in_cell);
n=length(L);
summ=L(1);
m=length(delim);
for k=2:n-1
L(k)=summ+L(k);
summ=L(k);
end
L(end)=[];
out_str=cell2mat(in_cell);
for i=1:length(L)
out_str=strcat(out_str(1:L(i)),abs(delim),out_str(L(i)+1:end));
L=L+m;
end
end
who can tell me why my code is wrong.thank you

Solution Comments

Solution 1708581

1 Comment

1 Comment

Carlos Castillo on 18 Jan 2019

This would be the replication of a single built-in function.

Solution 1432379

1 Comment

1 Comment

Jiahang Li on 2 Feb 2018

funny！

Solution 1276005

2 Comments

2 Comments

Thomas Rösch on 25 Sep 2017

Assertion fails in Cody even though the assertion passes in MATLAB R2014b. Does any one know why?

David Verrelli on 3 Oct 2017

Hi. The assertion could only have passed outside of Cody if the wrong "correct solution" were specified for comparison, and this would happen if you copy the HTML-rendered text — due to a kind of bug in the way the solution vector is displayed on the Cody webpage (better: copy from the HTML _source code_). -----
The correct solution has delimiters (here shown as ^ for clarity) as in 'this^one^^has^ ^some tricky^stuff', whereas your code produces 'this^one^has^some tricky^stuff', so you don't get the correct result between "one" and "has", nor between "has" and "some".

Solution 1167406

3 Comments

3 Comments

Vishvaksenan A. on 21 Apr 2017

I think that the test 2 must be checked.

I compared the test 2 correct answer with the program output using strcmp, and it gave 1 as output.( This was before introducing the program lines which also take into consideration the single space given in 5th element of in_cell in test 2)

Anyone, who has solved or has an idea how to solve, Please let me know what is the error in my program.

Thanks, By the way, program might be a bit lengthy.

Vishvaksenan A. on 21 Apr 2017

THE TEST CASE 2 OUTPUT YOU HAVE DISPLAYED IS WRONG
CORRECT IT.

David Verrelli on 3 Oct 2017

There are two issues here:
(1) The HTML display in Cody automatically only displays a single space. Therefore, while the solution *vector* _specified_ as the answer to Test 2 in Cody is correct, the *text* that is _displayed_ is not. You can confirm that the correct information was specified by viewing the HTML source. It is unfortunate that this is not displayed correctly. Of course, any results should match the solution, not the incorrect display.
(2) Your candidate solution fails because it is incorrect. Even if it matches the HTML-parsed display, it doesn't match the true specified solution vector. To make this apparent, imagine if delim='^' (no other changes). The correct solution to Test 2 would then be 'this^one^^has^ ^some tricky^stuff', whereas your output would be 'this^one^has^ ^some tricky^stuff' (provided in Solution 1280078). In other words, you needed a double space between "one" and "has".

Solution 1141460

1 Comment

1 Comment

Abdeslam Aannaque on 19 Mar 2017

My solution is:
function out_str = cellstr_joiner(in_cell, delim)
y=in_cell;
y(find(cellfun(@isempty,cellfun(@strtrim,y,'UniformOutput',false))))=[];
out_str = strjoin(y);
out_str(find(out_str==' '))=delim;
end
On Matlab all tests pass but on the cody website the second test fail!

Solution 900974

2 Comments

2 Comments

Venkatapathi on 3 Jun 2016

This code is working as expected in MATLAB command window. can i know why it is failed here

Jan Orwat on 3 Jun 2016

why deblank?

Solution 432260

2 Comments

2 Comments

Lorenzo on 15 Apr 2014

ok, is not really elegant not short, but it's correct. why doesn't it pass the test?

Mattias on 4 Jan 2015

The problem is that your output is a cell string and not a character array. You can get a char array by adding out_str=char(out_str); in the end, or work with char arrays from the beginning by using in_cell{1} and in_cell{i} instead of in_cell(1) and in_cell(i).

Solution 220215