Is there a format string to truncate a number to a specific number of digits?
For example, any number greater than 5 digits i would like to truncate to 3 digits.
132456 -> 132
5000000 -> 500
#Erik : Format specifiers like %2d are specific to a language? I actually want to use it in javascript
Pseudo-Code
Function returning a String, receiving a String representing a Number as a parameter
IF the String has more than 5 characters
RETURN a substring containing the first 3 characters.
ELSE
RETURN the string received as a parameter
END IF
END Function
I assume you refer to printf format strings. I couldn't find anything that will truncate an integer argument (i.e. %d). But you can specify the maximum length of a string by referring to a string format string and specifying lengths via "%<MinLength>.<MaxLength>s".
So in your case you could turn your number arguments into strings and then use "%3.3s".
Related
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'));
I am working with Swift's Decimal type, trying to ensure that an user-entered String is a valid Decimal.
I have two String values, each including a letter, within my Playground file. One of the values contains a letter at the start, while the other contains a letter at the end. I initialize a Decimal using each value, and only one Decimal initialization fails; the Decimal initialized with the value that contains the letter at the beginning.
Why does the Decimal initialized with a value that contains a letter at the end return a valid Decimal? I expect nil to be returned.
Attached is a screenshot from my Playground file.
It works this way because Decimal accepts any number values before the letters. The letters act as a terminator for any numbers that comes after it. So in your example:
12a = 12 ( a is the terminator in position 3 )
a12 = nil ( a is the terminator in position 1 )
If wanting both to be invalid if the string contains a letter then you could use Float instead.
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.
String to Integer (atoi)
This problem is implement atoi to convert a string to an integer.
When test input = " +0 123"
My code return = 123
But why expected answer = 0?
======================
And if test input = " +0123"
My code return = 123
Now expected answer = 123
So is that answer wrong?
I think this is expected result as it said
Requirements for atoi:
The function first discards as many whitespace characters as necessary until the first non-whitespace character is found. Then, starting from this character, takes an optional initial plus or minus sign followed by as many numerical digits as possible, and interprets them as a numerical value.
Your first test case has a space in between two different digit groups, and atoi only consider the first group which is '0' and convert into integer
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