I have a string called indicators, that the original developer of this application used to store single characters to indicate certain components of a model. I need to change the 7th character in the string, which I tried to do with the following code:
indicators[6] = "R"
The problem, I discovered quickly, was that the string is not always 7 characters long. For example, I have one set of values with U 2, that I need to convert to U 2 R (adding an additional space after the 2). Is there an easy way to force character count with Ruby?
use String.ljust(integer, padstr=' ')
If integer is greater than the length of [the receiver], returns a new String of
length integer with [the return value] left justified and padded with padstr;
otherwise, returns [an unmodified version of the receiver].
indicators = indicators.ljust(7)
indicators[6] = "R"
Related
I have a Textbox field that takes in a string with a character limit of 10. I would like to implement a short hand version because there are a lot of zeros in the string that have to be entered. so an example of the string is T000028999. but id like to key in T28999 and have the zeros padded between the "T" and the "28999" and show up as the T000028999 string in the Textbox field.
Is this even possible?
I've tried searching examples on google and have only found ways to pad the beginning and end of the string.
You want to keep the first character, so you can use oldString.Chars(0) to get that.
You want the remainder of the string: oldString.Substring(1), and you can pad it to the width you require with a character of your choice with PadLeft, like this:
Dim newString = oldString.Chars(0) & oldString.Substring(1).PadLeft(9, "0"c)
It would be a good idea to check that oldString is at least 1 character long before doing that otherwise the .Chars(0) will give an error.
Alternatively you could insert a new string of the required quantity of "0"s:
Dim newString = oldString.Insert(1, New String("0"c, 10 - oldString.Length))
A good place to perform the formatting would be in the control's Validating event handler. (The TextChanged event handler would not be a good place because it would interfere with the user's typing.)
Refs:
String.Chars[Int32] Property
String.Substring Method
String.PadLeft Method
String.Insert(Int32, String) Method
String Constructors
When you use .charAt method to convert a particular character in a string of string class to a character of char type, a code unit.
For example:
String greeting = "Hello";
Char character = greeting.charAt(0);
Character of char type is set to 'H'.
But you use a single index number for a supplementary character, only the first or second half is returned.
So for example, invoke the same method on $\mathcal {P} (\mathbb{Z})$ (set of integer), only the first half is returned.
All these are what is said in the book. But what does it meant by first half? First half of the two unit?
The language is java
Let me clarify:
(The latex code for set of integer seems to not be working)
String greeting = "(set of integer) is important";
char ch = greeting.charAt(0)
Index zero refers to the symbol for set of integer in math but b/c it is a supplementary(encoded by "\uD835\uDD6B"). Only the first half of this character is returned and stored in ch.
But what is first half?
I am working on a watermarking project that embeds binary values (i.e 1s and 0s) in the image, for which I have to take the input from the user, and check certain conditions such as
1) no empty string
2) no other character or special character
3) no number other than 0 and 1
is entered.
The following code just checks the first condition. Is there any default function in Matlab to check whether entered string is binary
int_state = get(handles.edit1,'String'); %edit1 is the Tag of edit box
if isempty(int_state)`
fprintf('Error: Enter Text first\n');
else
%computation code
end
There is no such standard function, but the check can be easily implemented.
Use this error condition:
isempty(int_state) || any(~ismember(int_state, '01'))
It returns false (no error) if the string is non-empty and composed of '0's and '1's only.
The function ismember returns a boolean array that indicates for every character in int_state whether it is contained in the second argument, '01'. The advantage is that this can be generalized to arbitrary sets of allowed characters.
I think the 2nd and 3rd can be combined together as 1 condition: your input string can only be a combination of 0 and 1? If it is so, then a small trick with findstr can do that:
if length(findstr(input_str, '1')) + length(findstr(input_str, '0')) == length(input_str)
condition_satisfied;
end
tf = isnumeric(A) returns true if A is a numeric array and false otherwise.
A numeric array is any of the numeric types and any subclasses of those types.
isnumeric(A)
ans =
1 (when A is numeric).
I have MATLAB set to record three webcams at the same time. I want to capture and save each feed to a file and automatically increment it the file name, it will be replaced by experiment_0001.avi, followed by experiment_0002.avi, etc.
My code looks like this at the moment
set(vid1,'LoggingMode','disk');
set(vid2,'LoggingMode','disk');
avi1 = VideoWriter('X:\ABC\Data Collection\Presentations\Correct\ExperimentA_002.AVI');
avi2 = VideoWriter('X:\ABC\Data Collection\Presentations\Correct\ExperimentB_002.AVI');
set(vid1,'DiskLogger',avi1);
set(vid2,'DiskLogger',avi2);
and I am incrementing the 002 each time.
Any thoughts on how to implement this efficiently?
Thanks.
dont forget matlab has some roots to C programming language. That means things like sprintf will work
so since you are printing out an integer value zero padded to 3 spaces you would need something like this sprintf('%03d',n) then % means there is a value to print that isn't text. 0 means zero pad on the left, 3 means pad to 3 digits, d means the number itself is an integer
just use sprintf in place of a string. the s means String print formatted. so it will output a string. here is an idea of what you might do
set(vid1,'LoggingMode','disk');
set(vid2,'LoggingMode','disk');
for (n=1:2:max_num_captures)
avi1 = VideoWriter(sprintf('X:\ABC\Data Collection\Presentations\Correct\ExperimentA_%03d.AVI',n));
avi2 = VideoWriter(sprintf('X:\ABC\Data Collection\Presentations\Correct\ExperimentB_002.AVI',n));
set(vid1,'DiskLogger',avi1);
set(vid2,'DiskLogger',avi2);
end
Dim str as String
str = "30 40 50 60"
I want to count the number of substrings.
Expected Output: 4
(because there are 4 total values: 30, 40, 50, 60)
How can I accomplish this in VB6?
You could try this:
arrStr = Split(str, " ")
strCnt = UBound(arrStr) + 1
msgBox strCnt
Of course, if you've got Option Explicit set (which you should..) then declare the variables above first..
Your request doesn't make any sense. A string is a sequence of text. The fact that that sequence of text contains numbers separated by spaces is quite irrelevant. Your string looks like this:
30 40 50 60
There are not 4 separate values, there is only one value, shown above—a single string.
You could also view the string as containing 11 individual characters, so it could be argued that the "count" of the string would be 11, but this doesn't get you any further towards your goal.
In order to get the result that you expect, you need to split the string into multiple strings at each space, producing 4 separate strings, each containing a 2-digit numeric value.
Of course, the real question is why you're storing this value in a string in the first place. If they're numeric values, you should store them in an array (for example, an array of Integers). Then you can easily obtain the number of elements in the array using the LBound() and UBound() functions.
I agree with everything Cody stated.
If you really wanted to you could loop through the string character by character and count the number of times you find your delimiter. In your example, it is space delimited, so you would simply count the number of spaces and add 1, but as Cody stated, those are not separate values..
Are you trying to parse text here or what? Regardless, I think what you really need to do is store your data into an array. Make your life easier, not more difficult.