I am trying to split a string at the character ":" but cant create two separate strings from the split. If somebody could help me, I would appreciate it.
In RealBasic, the Split method doesn't create two (or more) separate strings but rather a single string array.
Dim s() As String = Split("Zero:One:Two", ":")
's() now contains the substrings like so:
's(0) = "Zero"
's(1) = "One"
's(2) = "Two"
Actually, the code is incorrect. It should be:
Dim s() As String = Split("Zero:One:Two", ":")
If you don't pass in the delimiter it assumes a space which wouldn't work in this case.
The online docs are at http://docs.realsoftware.com/index.php/Split
Split is best for actually splitting the text, but you can also use the string-manipulation methods: Left, Right, Mid and InStr.
Related
Original string '4.0.0-4.0-M-672092'
How to modify the Original string to "4.0-M-672092" using a one line code.
Any Help is highly appreciated .
Thanks and Regards
The 'split' method works in this case
https://apidock.com/ruby/String/split
'4.0.0-4.0-M-672092'.split('-')[1..-1].join('-')
# => "4.0-M-672092"
Just be careful, in this application is fine, but in long texts this might become unoptimized, since it splits all the string and then joins the array all over again
If you need this in wider texts to be more optimized, you can find the "-" index (which is your split) and use the next position to make a substring
text = '4.0.0-4.0-M-672092'
text[(text.index('-') + 1)..-1]
# => "4.0-M-672092"
But you can't do it in one line, and not finding a split character will result in an error, so use a rescue statement if that is possible to happen
Simplest way:
'4.0.0-4.0-M-672092'.split('-', 2).second
"4.0.0-4.0-M-672092"[/(?<=-).*/]
#=> "4.0-M-672092"
The regular expression reads, "Match zero or more characters other than newlines, as many as possible (.*), provided the match is preceded by a hyphen. (?<=-) is a positive lookbehind. See String#[].
Lets say I have several strings I have to inspect. Here is a list of them:
/gigs-services/online-marketing/
/gigs-services/;
/scrapersnbots-files/footer.php';?
/scrapersnbots-files/ratings.js
Lets say the name of the variable that holds each string is strUrls
I need to filter out strings that end with ; and ;? and ';?
Is there a more efficient way to do so than this one below?
if right(strUrls,1) = ";" or right(strUrls,2) = ;? or right(strUrls,3) = ';?
I am wondering how to make something where if X=5 and Y=2, then have it output something like
Hello 2 World 5.
In Java I would do
String a = "Hello " + Y + " World " + X;
System.out.println(a);
So how would I do that in TI-BASIC?
You have two issues to work out, concatenating strings and converting integers to a string representation.
String concatenation is very straightforward and utilizes the + operator. In your example:
"Hello " + "World"
Will yield the string "Hello World'.
Converting numbers to strings is not as easy in TI-BASIC, but a method for doing so compatible with the TI-83+/84+ series is available here. The following code and explanation are quoted from the linked page:
:"?
:For(X,1,1+log(N
:sub("0123456789",ipart(10fpart(N10^(-X)))+1,1)+Ans
:End
:sub(Ans,1,length(Ans)-1?Str1
With our number stored in N, we loop through each digit of N and store
the numeric character to our string that is at the matching position
in our substring. You access the individual digit in the number by
using iPart(10fPart(A/10^(X, and then locate where it is in the string
"0123456789". The reason you need to add 1 is so that it works with
the 0 digit.
In order to construct a string with all of the digits of the number, we first create a dummy string. This is what the "? is used
for. Each time through the For( loop, we concatenate the string from
before (which is still stored in the Ans variable) to the next numeric
character that is found in N. Using Ans allows us to not have to use
another string variable, since Ans can act like a string and it gets
updated accordingly, and Ans is also faster than a string variable.
By the time we are done with the For( loop, all of our numeric characters are put together in Ans. However, because we stored a dummy
character to the string initially, we now need to remove it, which we
do by getting the substring from the first character to the second to
last character of the string. Finally, we store the string to a more
permanent variable (in this case, Str1) for future use.
Once converted to a string, you can simply use the + operator to concatenate your string literals with the converted number strings.
You should also take a look at a similar Stack Overflow question which addresses a similar issue.
For this issue you can use the toString( function which was introduced in version 5.2.0. This function translates a number to a string which you can use to display numbers and strings together easily. It would end up like this:
Disp "Hello "+toString(Y)+" World "+toString(X)
If you know the length of "Hello" and "World," then you can simply use Output() because Disp creates a new line after every statement.
i have a CSV in the below way. "India,Inc" is a company name which is single value which contains , in it
How to Get the Values in LINQ
12321,32432,423423,Kevin O'Brien,"India,Inc",234235,23523452,235235
Assuming that you will always have the columns that you specify and that the only variable is that company name can have commas inside, this UGLY code can help you achieve your goal.
var file = File.ReadLines("test.csv");
var value = from p in file
select new string[]
{ p.Split(',')[0],
p.Split(',')[1],
p.Split(',')[2],
p.Split(',')[3],
p.Split(',').Count() == 7 ? p.Split(',')[4] :
(p.Split(',').Count() > 7 ? String.Join(",",p.Split(',').Skip(4).Take(p.Split(',').Count() - 7).ToArray() ) : ""),
p.Split(',')[p.Split(',').Count() - 3],
p.Split(',')[p.Split(',').Count() - 2],
p.Split(',')[p.Split(',').Count() - 1]
};
A regular expression would work, bit nasty due to the recursive nature but it does achieve your goal.
List<string> matches = new List<string>();
string subjectString = "12321,32432,423423,Kevin O'Brien,\"India,Inc\",234235,23523452,235235";
Regex regexObj = new Regex(#"(?<="")\b[123456789a-z,']+\b(?="")|[123456789a-z']+", RegexOptions.IgnoreCase);
Match matchResults = regexObj.Match(subjectString);
while (matchResults.Success)
{
matches.Add(matchResults.Value);
// matched text: matchResults.Value
// match start: matchResults.Index
// match length: matchResults.Length
matchResults = matchResults.NextMatch();
}
This should suffice in most cases. It handles quoted strings, strings with double quotes within them, and embedded commas.
var subjectString = "12321,32432,423423,Kevin O'Brien,\"India,Inc\",234235,\"Test End\"\"\",\"\"\"Test Start\",\"Test\"\"Middle\",23523452,235235";
var result=Regex.Split(subjectString,#",(?=(?:[^""]*""[^""]*"")*[^""]*$)")
.Select(x=>x.StartsWith("\"") && x.EndsWith("\"")?x.Substring(1,x.Length-2):x)
.Select(x=>x.Replace("\"\"","\""));
It does however break, if you have a field with a single double quote inside it, and the string itself is not enclosed in double quotes -- this is invalid in most definitions of a CSV file, where any field that contains CR, LF, Comma, or Double quote must be enclosed in double quotes.
You should be able to reuse the same Regex expression to break on lines as well for small CSV files. Larger ones you would want a better implementation. Replace the double quotes with LF, and remove the matching ones (unquoted LF's). Then use the regular expression again replacing the quotes with CR, and split on matching.
Another option is to use CSVHelper and not traying to reinvent the wheel
var csv = new CsvHelper.CsvReader(new StreamReader("test.csv"));
while (csv.Read())
{
Console.WriteLine(csv.GetField<int>(0));
Console.WriteLine(csv.GetField<string>(1));
Console.WriteLine(csv.GetField<string>(2));
Console.WriteLine(csv.GetField<string>(3));
Console.WriteLine(csv.GetField<string>(4));
}
Guide
I would recommend LINQ to CSV, because it is powerful enough to handle special characters including commas, quotes, and decimals. They have really worked a lot of these issues out for you.
It only takes a few minutes to set up and it is really worth the time because you won't run into these types of issues down the road like you would with custom code. Here are the basic steps, but definitely follow the instructions in the link above.
Install the Nuget package
Create a class to represent a line item (name the fields the way they're named in the csv)
Use CsvContext.Read() to read into an IEnumerable which you can easily manipulate with LINQ
Use CsvContext.Write() to write a List or IEnumerable to a CSV
This is very easy to setup, has very little code, and is much more scalable than doing it yourself.
becuase you're only reading values delminated bycommas, the spaces shouldn't cause an issue if you just treat them like any other character.
var values = File.ReadLines(path)
SelectMany(line => line.Split(','));
I would like to convert the following string into an array/nested array:
str = "[[this, is],[a, nested],[array]]"
newarray = # this is what I need help with!
newarray.inspect # => [['this','is'],['a','nested'],['array']]
You'll get what you want with YAML.
But there is a little problem with your string. YAML expects that there's a space behind the comma. So we need this
str = "[[this, is], [a, nested], [array]]"
Code:
require 'yaml'
str = "[[this, is],[a, nested],[array]]"
### transform your string in a valid YAML-String
str.gsub!(/(\,)(\S)/, "\\1 \\2")
YAML::load(str)
# => [["this", "is"], ["a", "nested"], ["array"]]
You could also treat it as almost-JSON. If the strings really are only letters, like in your example, then this will work:
JSON.parse(yourarray.gsub(/([a-z]+)/,'"\1"'))
If they could have arbitrary characters (other than [ ] , ), you'd need a little more:
JSON.parse("[[this, is],[a, nested],[array]]".gsub(/, /,",").gsub(/([^\[\]\,]+)/,'"\1"'))
For a laugh:
ary = eval("[[this, is],[a, nested],[array]]".gsub(/(\w+?)/, "'\\1'") )
=> [["this", "is"], ["a", "nested"], ["array"]]
Disclaimer: You definitely shouldn't do this as eval is a terrible idea, but it is fast and has the useful side effect of throwing an exception if your nested arrays aren't valid
Looks like a basic parsing task. Generally the approach you are going to want to take is to create a recursive function with the following general algorithm
base case (input doesn't begin with '[') return the input
recursive case:
split the input on ',' (you will need to find commas only at this level)
for each sub string call this method again with the sub string
return array containing the results from this recursive method
The only slighlty tricky part here is splitting the input on a single ','. You could write a separate function for this that would scan through the string and keep a count of the openbrackets - closedbrakets seen so far. Then only split on commas when the count is equal to zero.
Make a recursive function that takes the string and an integer offset, and "reads" out an array. That is, have it return an array or string (that it has read) and an integer offset pointing after the array. For example:
s = "[[this, is],[a, nested],[array]]"
yourFunc(s, 1) # returns ['this', 'is'] and 11.
yourFunc(s, 2) # returns 'this' and 6.
Then you can call it with another function that provides an offset of 0, and makes sure that the finishing offset is the length of the string.