I have a very trivial example where I'm trying to filter by matching a String:
A = [0:1:999];
B = A(int2str(A) == '999');
This
A(A > 990);
works
This
int2str(5) == '5'
also works
I just can't figure out why I cannot put the two together. I get an error about nonconformant arguments.
int2str(A) produces a very long char array (of size 1 x 4996) containing the string representations of all those numbers (including spacing) appended together end to end.
int2str(A) == '999'
So, in the statement above, you're trying to compare a matrix of size 1 x 4996 with another of size 1 x 3. This, of course, fails as the two either need to be of the same size, or at least one needs to be a scalar, in which case scalar expansion rules apply.
A(A > 990);
The above works because of logical indexing rules, the result will be the elements from the indices of A for which that condition holds true.
int2str(5) == '5'
This only works because the result of the int2str call is a 1 x 1 matrix ('5') and you're comparing it to another matrix of the same size. Try int2str(555) == '55' and it'll fail with the same error as above.
I'm not sure what result you expected from the original statements, but maybe you're looking for this:
A = [0:1:999];
B = int2str(A(A == 999)) % outputs '999'
I am not sure that the int2str() conversion is what you are looking for. (Also, why do you need to convert numbers to strings and then carry out a char comparison?)
Suppose you have a simpler case:
A = 1:3;
strA = int2str(A)
strA =
1 2 3
Note that this is a 1x7 char array. Thus, comparing it against a scalar char:
strA == '2'
ans =
0 0 0 1 0 0 0
Now, you might wanna transpose A and carry out the comparison:
int2str(A')=='2'
ans =
0
1
0
however, this approach will not work if the number of digits of each number is not the same because lower numbers will be padded with spaces (try creating A = 1:10 and comparing against '2').
Then, create a cell array of string without whitespaces and use strcmp():
csA = arrayfun(#int2str,A','un',0)
csA =
'1'
'2'
'3'
strcmp('2',csA)
Should be much faster, and correct to turn the string into a number, than the other way around. Try
B = A(A == str2double ('999'));
Related
for var=1:length(str)
switch str(var)
case {'A':'C'}
disp('2')
case {'D':'F'}
disp('3')
case {'G':'I'}
disp('4')
case {'J':'L'}
disp('5')
case {'M':'O'}
disp('6')
case {'P' 'R' 'S'}
disp('7')
case {'T':'V'}
disp('8')
case {'W':'Y'}
disp('9')
end
I am using this code in trying to make a statement such as ('1-800-TO-WELLS') become '1-800-86-93557', but the output is "str=1-800-TO-WELLS and 7". Any amount of tips and help is very appreciated. If I haven't provided a valid enough question please leave comments so I can improve.
For strings, the cases in the switch block act like calls to strcmp, so case <caseExpression> is true if any(strcmp(<caseExpression>,<switchExpression>)).
This is important because {'P' 'R' 'S'} and {'P':'S'} do not generate the same output:
>> {'P' 'R' 'S'}
ans =
'P' 'R' 'S'
>> {'P':'S'}
ans =
'PQRS'
The first is a 1-by-3 cell array with individual characters as the elements' contents; the second is a 1-by-1 cell array with a 1-by-4 character array as the element's content. Performing a strcmp on the first one will give true if the <switchExpression> is a letter in the set while the second will only give a true if <switchExpression> is exactly 'PQRS':
>> strcmp({'P' 'R' 'S'},'S')
ans =
0 0 1
>> strcmp({'P':'S'},'S')
ans =
0
>> strcmp({'P':'S'},'P':'S')
ans =
1
So the 7 pops out because its case is the only one that gives true when given a matching character.
As a side note, you may consider using the otherwise statement to echo non-matching characters:
switch str(var)
.
.
.
otherwise
disp(str(var));
end
Troy gives a very clear answer as to why your code isn't working: {'A':'C'} results in a 1-element cell array {'ABC'}, but you want to use a 3-element cell array {'A' 'B' 'C'} since the way the switch compares case expressions will result in a match if one cell element exactly matches a given character.
As an alternative to using a for loop and switch statement, you could also do this using the ismember function and some indexing:
% Character/number map and test string:
map = ['ABCDEFGHIJKLMNOPRSTUVWXY'; '222333444555666777888999'];
str = '1-800-TO-WELLS';
% Convert test string:
[isInMap, index] = ismember(str, map(1, :));
str(isInMap) = map(2, index(isInMap));
And the result:
str =
1-800-86-93557
I have a cell array consisting of numbers, strings, and empty arrays. I want to find the position (linear or indexed) of all cells containing a string in which a certain substring of interest appears.
mixedCellArray = {
'adpo' 2134 []
0 [] 'daesad'
'xxxxx' 'dp' 'dpdpd'
}
If the substring of interest is 'dp', then I should get the indices for three cells.
The only solutions I can find work when the cell array contains only strings:
http://www.mathworks.com/matlabcentral/answers/2015-find-index-of-cells-containing-my-string
http://www.mathworks.com/matlabcentral/newsreader/view_thread/255090
One work-around is to find all cells not containing strings, and fill them with '', as hinted by this posting. Unfortunately, my approach requires a variation of that solution, probably something like cellfun('ischar',mixedCellArray). This causes the error:
Error using cellfun
Unknown option.
Thanks for any suggestions on how to figure out the error.
I've posted this to usenet
EDUCATIONAL AFTERNOTE: For those who don't have Matlab at home, and end up bouncing back and forth between Matlab and Octave. I asked above why cellfun doesn't accept 'ischar' as its first argument. The answer turns out to be that the argument must be a function handle in Matlab, so you really need to pass #ischar. There are some functions whose names can be passed as strings, for backward compatibility, but ischar is not one of them.
How about this one-liner:
>> mixedCellArray = {'adpo' 2134 []; 0 [] 'daesad'; 'xxxxx' 'dp' 'dpdpd'};
>> index = cellfun(#(c) ischar(c) && ~isempty(strfind(c, 'dp')), mixedCellArray)
index =
3×3 logical array
1 0 0
0 0 0
0 1 1
You could get by without the ischar(c) && ..., but you will likely want to keep it there since strfind will implicitly convert any numeric values/arrays into their equivalent ASCII characters to do the comparison. That means you could get false positives, as in this example:
>> C = {65, 'A'; 'BAD' [66 65 68]} % Note there's a vector in there
C =
2×2 cell array
[ 65] 'A'
'BAD' [1×3 double]
>> index = cellfun(#(c) ~isempty(strfind(c, 'A')), C) % Removed ischar(c) &&
index =
2×2 logical array
1 1 % They all match!
1 1
Just use a loop, testing with ischar and contains (added in R2016b). The various *funs are basically loops and, in general, do not offer any performance advantage over the explicit loop.
mixedCellArray = {'adpo' 2134 []; 0 [] 'daesad'; 'xxxxx' 'dp' 'dpdpd'};
querystr = 'dp';
test = false(size(mixedCellArray));
for ii = 1:numel(mixedCellArray)
if ischar(mixedCellArray{ii})
test(ii) = contains(mixedCellArray{ii}, querystr);
end
end
Which returns:
test =
3×3 logical array
1 0 0
0 0 0
0 1 1
Edit:
If you don't have a MATLAB version with contains you can substitute a regex:
test(ii) = ~isempty(regexp(mixedCellArray{ii}, querystr, 'once'));
z=cellfun(#(x)strfind(x,'dp'),mixedCellArray,'un',0);
idx=cellfun(#(x)x>0,z,'un',0);
find(~cellfun(#isempty,idx))
Here is a solution from the usenet link in my original post:
>> mixedCellArray = {
'adpo' 2134 []
0 [] 'daesad'
'xxxxx' 'dp' 'dpdpd'
}
mixedCellArray =
'adpo' [2134] []
[ 0] [] 'daesad'
'xxxxx' 'dp' 'dpdpd'
>> ~cellfun( #isempty , ...
cellfun( #(x)strfind(x,'dp') , ...
mixedCellArray , ...
'uniform',0) ...
)
ans =
1 0 0
0 0 0
0 1 1
The inner cellfun is able to apply strfind to even numerical cells because, I presume, Matlab treats numerical arrays and strings the same way. A string is just an array of numbers representing the character codes. The outer cellfun identifies all cells for which the inner cellfun found a match, and the prefix tilde turns that into all cells for which there was NO match.
Thanks to dpb.
I want to check the length of string got more than 20 characters, if more than 20 then will return 1 else return 0 in matrix form [n x 1]. But now, I get the answer of [1x1]. How do I modify my code in if-else statement to get the ans?
str = {'http://www.mathworks.com/matlabcentral/newsreader/view_thread/324182',
'http://jitkomut.lecturer.eng.chula.ac.th/matlab/text.html',
'http://www.ee.ic.ac.uk/pcheung/teaching/ee2_signals/Introduction%20to%20Matlab2.pdf'};
a = cellfun(#length,str)
if a > 20
'1'
else
'0'
end
Output:
a =
68
57
83
ans =
1
I want the output of, lets say
ans =
1
1
1
In Matlab, you can simply use (no if statement is needed):
a = cellfun(#length,str)
(a>20)'
This will give you:
a =
68 57 83
ans =
1
1
1
As #herohuyongtao mentions, you don't actually need an if, the if will only consider the first element of the matrix that it returns, hence giving you only a single value.
But you could actually do this all in your cellfun by using an anonymous function:
cellfun(#(x)(length(x) > 20), str)
And get the result in one shot.
As there is no equivalent of the c ternary operator (?:) in matlab, you can use the following two statements to replace your if then else statement, and achieve what you ask for:
b(a==a)='0'
b(a>20)='1'
The first line initializes the result array, where all value b defaults to the value of the else branch, i.e. '0',
the second line changes the elements for which the conditional > 20 holds to the value in the then branch, i.e. '1'.
If the output values are boolean, you can simply do:
(a>20)
as #herohuyongtao suggested or use #Dan's answer.
Say I typed x = 'BODD' into the command prompt of MATLAB and then said x(1) it would return B. What I want is x(1) to return the empty String ('') or nothing etc. and x(2) to return B and so forth up until x(5) returning the final D?
I assume that you mean that you really do want the empty zero-length string, ''. There have been some Answers to this question that assume that you meant that you wanted the one-character string that contains a space, ASCII value 32.
If that's the case, I'm afraid you can't to that - MATLAB arrays (including character arrays, which is all that a MATLAB "string" is) don't work that way. There are two ways to look at it...
You asked for x(1). Now, the indexing expression that you used, 1, has size 1x1. Therefore, you are guaranteed to get either a 1x1 return value, OR an error. That means that there's no way to get a 0x1 or 0x0 (the true "empty string"). This is similar to the way that, if you had asked for x(2:4), you would be guaranteed to get a 1x3 array of characters back. In that case, 2:4 is a 1x3 array.
There's no way to "meaningfully" prepend a zero-length string to the beginning of another string. If a = 'WXYZ';, then running b = ['' a] just returns 'WXYZ' back. It didn't somehow stick a magical placeholder for an empty string at the front of the original string.
You can't concatenate '' at end or beginning
However, you can have blank/space, like this :-
>> x=BODD;
>> x=[' ' x]; % Use normal matrix concatenation
>> x(1)
ans =
>> x(2)
ans =
B
Try following concatenation
x = [' ' x];
If you want the string itself to still be 'BODD', you could try writing a custom function:
function [char] = emptyConcat(string, index)
if (index == 1)
char = '';
else
char = string(index - 1);
I want to remove a (*) asterisk from my matrix and write out that matrix to a text file and the remaining elements will be concatenated to each other without a space or any kind of delimiters. I wrote this code
for b = 1 : length(out7num_r7_nt_back)
if ~(out7num_r7_nt_back(b) == '*')
out7num_r7_back(b) = '';
end
end
disp(out7num_r7_nt_back);
dlmwrite('my_data.txt',out7num_r7_nt_back, '');
I got this error message:
??? Index of element to remove exceeds matrix dimensions.
You can use a vectorized boolean index, replacing the loop as follows:
out7num_r7_nt_back = out7num_r7_nt_back(out7num_r7_nt_back(b) ~= '*');
That should be much faster as well.
Value of upper bound of for loop (length(out7num_r7_nt_back)) gets evaluated only once!
Say you have '*ab' in a variable. Loop will count to 3 (length of variable). Inside the loop when program erases '*', you'll get 'ab' which is of length 2. Since loop is iterating to 3, program will try to access out7num_r7_nt_back(3) which is out of bounds.
You could remove characters while going backwards:
...
for b = length(out7num_r7_nt_back) : -1 : 1
...
But you should prefer vectorized approach because it's faster and cleaner to write.