I have a string of format
"'Year'-'Month'-'Day'T'Hour':'Minute':'Second'Z" for example '2020-11-26T16:56:09.676Z'
(Note the milliseconds are considered part of the second)
I would like to convert it to the format:
t1 = 1x6
2020 11 26 16 56 09.676
Or in other words a 1x6 array.
Note: This is to be completed using MatLab.
You can
Use regexp with the 'match' input flag to detect the numbers as one or more digits (\d+) optionally followed by a decimal point (\.?) and then zero or more digits (\d*). This will give a cell array of strings.
Apply str2double to convert the strings to numbers. This will give a numeric row vector.
s = '2020-11-26T16:56:09.676Z';
result = str2double(regexp(s, '\d+\.?\d*', 'match'));
Related
I have a text file with 1 million decimal digits of "e" number with 80 digits on each line excluding the first and the last line which have 76 and 4 digits and the file has 12501 lines. I want to convert it into a vector in matlab with each digit on each row. I tried num2str function, but the problem is that it gets converted like for example '7.1828e79' (13 characters). What can I do?
P.S.1: The first two lines of the text file (76 and 80 digits) are:
7182818284590452353602874713526624977572470936999595749669676277240766303535 47594571382178525166427427466391932003059921817413596629043572900334295260595630
P.S.2: I used "dlmread" and got a 12501x1 vector, with the first and second row of 7.18281828459045e+75 and 4.75945713821785e+79 and the problem is that when I use num2str for example for the first row value, I get: '7.182818284590453e+75' as a string and not the whole 76 digits. My aim was to do something like this:
e1=dlmread('e.txt');
es1=num2str(e1);
for i=1:12501
for j=1:length(es1(1,:))
a1((i-1)*length(es1(1,:))+j)=es1(i,j);
end
end
e_digits=a1.';
but I get a string like this:
a1='7.182818284590453e+754.759457138217852e+797.381323286279435e+799.244761460668082e+796.133138458300076e+791.416928368190255e+79 5...'
with 262521 characters instead of 1 million digits.
P.S.3: I think the problem might be solved if I can manipulate the text file in a way that I have one digit on each line and simply use dlmread.
Well, this is not hard, there are many ways to do it.
So first you want to load in your file as a Char Array using something simple like (you want a Char Array so that you can easily manipulate it to forget about the lines breaks) :
C = fileread('yourfile.txt'); %loads file as Char Array
D = C(~isspace(C)); %Removes SPACES which are line-breaks
Next, you want to actually append a SPACE between each char (this is because you want to use the num2str transform - and matlab needs to see the space), you can do this using a RESHAPE, a STRTRIM or simply a REGEX:
E = strtrim(regexprep(D, '.{1}', '$0 ')); %Add a SPACE after each Numeric Digit
Now you can transform it using str2num into a Vector:
str2num(E)'; %Converts the Char Array back to Vector
which will give you a single digit each row.
Using your example, I get a vector of 156 x 1, with 1 digit each row as you required.
You can get a digit per row like this
fid = fopen('e.txt','r');
c = textscan(fid,'%s');
c=cat(1,c{:});
c = cellfun(#(x) str2num(reshape(x,[],1)),c,'un',0);
c=cat(1,c{:});
And it is not the only possible way.
Could you please tell what is the final task, how do you plan using the array of e digits?
I'm trying to create a string from hex values in an array, but whenever a hex in the array starts with a zero it disappears in the resulting string as well.
I use String(value:radix:uppercase) to create the string.
An example:
Here's an array: [0x13245678, 0x12345678, 0x12345678, 0x12345678].
Which gives me the string: 12345678123456781234567812345678 (32 characters)
But the following array: [0x02345678, 0x12345678, 0x02345678, 0x12345678] (notice that I replaced two 1's with zeroes).
Gives me the string: 234567812345678234567812345678 (30 characters)
I'm not sure why it removes the zeroes. I know the value is correct; how can I format it to keep the zero if it was there?
The number 0x01234567 is really just 0x1234567. Leading zeros in number literals don't mean anything (unless you are using the leading 0 for octal number literals).
Instead of using String(value:radix:uppercase), use String(format:).
let num = 0x1234567
let str = String(format: "%08X", num)
Explanation of the format:
The 0 means to pad the left end of the string with zeros as needed.
The 8 means you want the result to be 8 characters long
The X means you want the number converted to uppercase hex. Use x if you want lowercase hex.
I have an cell-array of dimensions 1x6 like this:
A = {'25_2.mat','25_3.mat','25_4.mat','25_5.mat','25_6.mat','25_7.mat'};
I want to read for example from the A{1} , the number after the '_' i.e 2 for my example
Using cellfun, strfind and str2double
out = cellfun(#(x) str2double(x(strfind(x,'_')+1:strfind(x,'.')-1)),A)
How does it work?
This code simply finds the index of character one number after the occurrence of '_'. Lets call it as start_index. Then finds the character one number lesser than the index of occurrence of '.' character. Lets call it as end_index. Then retrieves all the characters between start_index and end_index. Finally converts those characters to numbers using str2double.
Sample Input:
A = {'2545_23.mat','2_3.mat','250_4.mat','25_51.mat','25_6.mat','25_7.mat'};
Output:
>> out
out =
23 3 4 51 6 7
You can access the contents of the cell by using the curly braces{...}. Once you have access to the contents, you can use indexes to access the elements of the string as you would do with a normal array. For example:
test = {'25_2.mat', '25_3.mat', '25_4.mat', '25_5.mat', '25_6.mat', '25_7.mat'}
character = test{1}(4);
If your string length is variable, you can use strfind to find the index of the character you want.
Assuming the numbers are non-negative integers after the _ sign: use a regular expression with lookbehind, and then convert from string to number:
numbers = cellfun(#(x) str2num(x{1}), regexp(A, '(?<=\_)\d+', 'match'));
I have a string and I need two characters to be returned.
I tried with strsplit but the delimiter must be a string and I don't have any delimiters in my string. Instead, I always want to get the second number in my string. The number is always 2 digits.
Example: 001a02.jpg I use the fileparts function to delete the extension of the image (jpg), so I get this string: 001a02
The expected return value is 02
Another example: 001A43a . Return values: 43
Another one: 002A12. Return values: 12
All the filenames are in a matrix 1002x1. Maybe I can use textscan but in the second example, it gives "43a" as a result.
(Just so this question doesn't remain unanswered, here's a possible approach: )
One way to go about this uses splitting with regular expressions (MATLAB's strsplit which you mentioned):
str = '001a02.jpg';
C = strsplit(str,'[a-zA-Z.]','DelimiterType','RegularExpression');
Results in:
C =
'001' '02' ''
In older versions of MATLAB, before strsplit was introduced, similar functionality was achieved using regexp(...,'split').
If you want to learn more about regular expressions (abbreviated as "regex" or "regexp"), there are many online resources (JGI..)
In your case, if you only need to take the 5th and 6th characters from the string you could use:
D = str(5:6);
... and if you want to convert those into numbers you could use:
E = str2double(str(5:6));
If your number is always at a certain position in the string, you can simply index this position.
In the examples you gave, the number is always the 5th and 6th characters in the string.
filename = '002A12';
num = str2num(filename(5:6));
Otherwise, if the formating is more complex, you may want to use a regular expression. There is a similar question matlab - extracting numbers from (odd) string. Modifying the code found there you can do the following
all_num = regexp(filename, '\d+', 'match'); %Find all numbers in the filename
num = str2num(all_num{2}) %Convert second number from str
i donot know why there is error in this coding:
hex_str1 = '5'
bin_str1 = dec2bin(hex2dec(hex_str1))
hex_str2 = '4'
bin_str2 = dec2bin(hex2dec(hex_str2))
c=xor(bin_str1,bin_str2)
the value of c is not correct when i transform the hex to binary by using the xor function.but when i used the array the value of c is correct.the coding is
e=[1 1 1 0];
f=[1 0 1 0];
g=xor(e,f)
what are the mistake in my first coding to xor of hec to binary value??anyone can help me find the solution...
Your mistake is applying xor on two strings instead of actual numerical arrays.
For the xor command, logical "0"s are represented by actual zero elements. Any non-zero elements are interpreted as logical "1"s.
When you apply xor on two strings, the numerical value of each character (element) is its ASCII value. From xor's point of view, the zeroes in your string are not really zeroes, but simply non-zero values (being equal to the ASCII value of the character '0'), which are interpreted as logical "1"s. The bottom line is that in your example you're xor-ing 111b and 111b, and so the result is 0.
The solution is to convert your strings to logical arrays:
num1 = (bin_str1 == '1');
num2 = (bin_str2 == '1');
c = xor(num1, num2);
To convert the result back into a string (of a binary number), use this:
bin_str3 = sprintf('%d', c);
... and to a hexadecimal string, add this:
hex_str3 = dec2hex(bin2dec(bin_str3));
it is really helpful, and give me the correct conversion while forming HMAC value in matlab...
but in matlab you can not convert string of length more than 52 character using bin2dec() function and similarly hex2dec() can not take hexadecimal character string more than 13 length.