If I have string as "TestDataData" I want to replace "Data" with "NewValue" so it should become "TestNewValueData", how do we achieve this ?
Note: I cant replace data directly otherwise last data will also updated..
If you are able to use a RegEx, you could use the following:
str.replace("Data.+", "NewValue")
The .+ assures at least one more letter follows after "Data".
Alternatively, you could check if the index of your needle is less than the length of the haystack minus the length of the needle.
Related
heres the question:
Remove First and Last Character
It's pretty straightforward. Your goal is to create a function that removes the first and last characters of a string. You're given one parameter, the original string. You don't have to worry with strings with less than two characters.
i have no idea how to do it because of the fact that it wants me to remove first and last letter and not the first and last word
Slicing is the best way to go about that.
test_string = 'This is a test!'
print (test_string [1:-1])
From your response that removing the first and last word would be done by list1.pop() and list1.pop(0) I am assuming that you are given a list of strings and are required to remove the first and last character of each string in the list.
If my assumption is correct you can do a combination of slicing and list comprehension.
list_of_str = ["words", "words", "words", "words"]
list_of_shortened_str = [word[1:-1] for word in list_of_str]
# list_of_shortened_str = ["ord", "ord", "ord", "ord"]
I am trying to split the string 'id#namespace' by #.
There is this special case that the id has a format name#gmail which makes the string I am trying to split look like name#gmail#namespace.
Is there any way to achieve that only split by the last # which will give me name#gmail and namespace?
If it is always the last index. Use lastIndex to find the last index of the character and then use substring
Something like this
int idx = string.lastIndexOf("#");
String[] splitStrings = {string.substring(0, idx), string.substring(idx)};
You could match the last # using a negative lookahead (?! to assert that there are no more # following:
#(?!.*#)
Regex demo
System.out.println("name#gmail#namespace".split("#(?!.*#)")[0]); //name#gmail
I'm trying to find exact matches of strings in Lua including, special characters. I want the example below to return that it is an exact match, but because of the - character it returns nil
index = string.find("test-string", "test-string")
returns nil
index = string.find("test-string", "test-")
returns 1
index = string.find("test-string", "test")
also returns 1
How can I get it to do full matching?
- is a pattern operator in a Lua string pattern, so when you say test-string, you're telling find() to match the string test as few times as possible. So what happens is it looks at test-string, sees test in there, and since - isn't an actual minus sign in this case, it's really looking for teststring.
Do as Mike has said and escape it with the % character.
I found this helpful for better understanding patterns.
You can also ask for a plain substring match that ignores magic characters:
string.find("test-string", "test-string",1,true)
you need to escape special characters in the pattern with the % character.
so in this case you are looking for
local index = string.find('test-string', 'test%-string')
I have a problem with splitting string into two parts on special character.
For example:
12345#data
or
1234567#data
I have 5-7 characters in first part separated with "#" from second part, where are another data (characters,numbers, doesn't matter what)
I need to store two parts on each side of # in two variables:
x = 12345
y = data
without "#" character.
I was looking for some Lua string function like splitOn("#") or substring until character, but I haven't found that.
Use string.match and captures.
Try this:
s = "12345#data"
a,b = s:match("(.+)#(.+)")
print(a,b)
See this documentation:
First of all, although Lua does not have a split function is its standard library, it does have string.gmatch, which can be used instead of a split function in many cases. Unlike a split function, string.gmatch takes a pattern to match the non-delimiter text, instead of the delimiters themselves
It is easily achievable with the help of a negated character class with string.gmatch:
local example = "12345#data"
for i in string.gmatch(example, "[^#]+") do
print(i)
end
See IDEONE demo
The [^#]+ pattern matches one or more characters other than # (so, it "splits" a string with 1 character).
Problem: From TrajCompact, i find all the prefix and the value after prefix, using regexp, with this code:
[digits{1:2}] = ndgrid(0:4);
for k=1:25
matches(:,k)=regexp(TrajCompact(:,1),sprintf('%d%d.*',digits{1}(k),digits{2}(k)),'match','once');
end
I want only the postfix of matches, how can delete the prefix from matches?
Method using regular expressions
You can put the .* section in a group by enclosing it in parenthesis (i.e. (.*)). Matlab has some peculiar 'token' nomenclature for this. In any case, an example of how it works:
[match, group] = regexp('25blah',sprintf('%d%d(.*)',2,5),'match','once','tokens');
Then:
match would be a char array containing '25blah'
group would be a 1x1 cell array containing the string 'blah'.
That is, the variable group would hold what you're looking for.
Hack method
Since your prefix is always two digits, you could also just take everything from the 3rd character of the match onwards:
my_string = match(3:end);
other comments
You may want to require the prefix to occur at the beginning of the string by adding ^ to the beginning of your regular expression. Eg., make the line:
[match, group] = regexp('25blah',sprintf('^%d%d(.*)',2,5),'match','once','tokens');
As it is, your current regular expression would match strings like zzzzzzzzz25stuff. I'm not sure if you want that (assuming it can occur in your data).