Example code:
let interestingNumbers = [
"Prime":[2,3,5,7,11,13],
"Fibonacci":[1,1,2,3,5,8],
"Square":[1,4,9,16,25]`,`
]
Question: after "Square:[1,4,9,16,25]", there is a comma(sample code from Apple Swift reference guide book), when I get rid of it, I didn't get any error messages from Xcode, is this just convention at all ? (I remember there is a nil after array or dictionary in objective-C
This is because a comma after the last element in a dictionary is optional.
Consider the simpler example:
let letters = ["A":1,
"B":2,
"C":3
]
A comma placed after the last element, "C":3, is acceptable, but optional.
I believe that there is no specific convention regarding the final comma - some may prefer it as it allows you to add items on following lines without modifying the above line to add the comma (makes source control review simpler). I often leave commas on the last element in an enum declaration for the same reason.
If you know you are likely to add more elements in the future, then having the comma would simplify the source diff in a code review (one added line instead of one removed line and two added lines). I would use the comma where you know you're going to add elements later, and omit in if the list of items is final.
Related
why is there a comma in this variable declaration:
// RinkebyBootnodes are the enode URLs of the P2P bootstrap nodes running on the
// Rinkeby test network.
var RinkebyBootnodes = []string{
"enode://a24ac7c5484ef4ed0c5eb2d36620ba4e4aa13b8c84684e1b4aab0cebea2ae45cb4d375b77eab56516d34bfbd3c1a833fc51296ff084b770b94fb9028c4d25ccf#52.169.42.101:30303", // IE
"enode://343149e4feefa15d882d9fe4ac7d88f885bd05ebb735e547f12e12080a9fa07c8014ca6fd7f373123488102fe5e34111f8509cf0b7de3f5b44339c9f25e87cb8#52.3.158.184:30303", // INFURA
}
I am talking about comma that goes here:
30303", // INFURA
it is the last character of the string array, and it has to go there, otherwise I get a compile error.
In C language you can't have commas at the end of curly braces {} , but in go you have to. Why ? And what does this comma mean?
Comma is an element splitter. Elements could be.written as in a row so in a column:
{1, 2, 3}
{
1,
2,
3,
}
It’s evident comma is unnecessary after the last element. Some languages require their absense like C, some - presence like Go, some allow both variants like Python. Sometimes it causes bugs like with JavaScript in old Internet Explorer.
Why do gophers decide to keep this comma? One of key principles of Go is visual simplicity and readability. This way all the strings look the same way. It’s easier to read them by eyes. Also suppose you add one more line - you have to edit previous one. And this creates visual noise in diff.
Thinking the same you get to the idea comma should not be in one liners. Because spoils readability as extra symbol.
I need to validate that something is an Excel cell range in Ruby, i.e: "A4:A6". By looking at it, the requirement I am looking for is:
<Alphabetical, Capitalised><Integer>:<Integer><Alphabetical, Capitalised>
I am not sure how to form a RegExp for this.
I would appreciate a small explanation for a solution, as opposed to purely a solution.
A bonus would be to check that the range is restricted to within a row or column. I think this would be out of scope of Regular Expressions though.
I have tried /[A-Z]+[0-9]+:[A-Z]+[0-9]+/ this works but allows extra characters on the ends.
This does not work because it allows extra's to be added on to the beginning or end:
"HELLOAA3:A7".match(/\A[A-Z]+[0-9]+:[A-Z]+[0-9]+\z/) also returns a match, but is more on the right track.
How would I limit the number range to 10000?
How would I limit the number of characters to 3?
This is my solution:
(?:(?:\'?(?:\[(?<wbook>.+)\])?(?<sheet>.+?)\'?!)?(?<colabs>\$)?(?<col>[a-zA-Z]+)(?<rowabs>\$)?(?<row>\d+)(?::(?<col2abs>\$)?(?<col2>[a-zA-Z]+)(?<row2abs>\$)?(?<row2>\d+))?|(?<name>[A-Za-z]+[A-Za-z\d]*))
It includes named ranges, but the R1C1 notation is not supported.
The pattern is written in perl compatible regex dialect (i.e. can also be used with C#), I'm not familiar with Ruby, so I can't tell the difference, but you may want to look here: What is the difference between Regex syntax in Ruby vs Perl?
This will do both: match Excel range and that they must be same row or column. Stub
^([A-Z]+)(\d+):(\1\d+|[A-Z]+\2)$
A4:A6 // ok
A5:B10 // not ok
B5:Z5 // ok
AZ100:B100hello // not ok
The magic here is the back-reference group:
([A-Z]+)(\d+) -- column is in capture group 1, row in group 2
(\1\d+|[A-Z]+\2) -- the first column followed by any number; or
-- the first row preceded by any character
I have come across a problem that I cannot see to solve. I have extracted a line from a web page into a variable. lets say for argument sake this is:
rhyme = "three blind mice Version 6.0"
and I want to be able to first of all locate the version number within this string (6.0) and secondly extract this number into another seperate variable - (I want to specifically extract no more than "6.0")
I hope I have clarified this enough, if not please ask me anything you need to know and I will get back to you asap.
First you need to decide what the pattern for a version number should be. One possibility would be \d+(\.\d+)*$ (a number followed by zero or more (dot followed by a number) at the end of the string).
Then you can use String#[] to get the substring that matches the pattern:
rhyme[ /\d+(\.\d+)*$/ ] #=> "6.0"
You need to use regular expressions. I would use rhyme.scan(/(\d+\.\d+)/) since it can return an array if multiple matches occur. It can also take a block so that you can add range checks or other checks to ensure the right one is captured.
version = "0.0"
rhyme = "three blind mice Version 6.0"
rhyme.scan(/(\d+\.\d+)/){|x| version = x[0] if x[0].to_f < 99}
p version
If the input can be trusted to yield only one match or if you always are going to use the first match you can just use the solution in this answer.
Edit: So after our discussion just go with that answer.
if rhyme =~ /(\d\.\d)/
version = $1
end
The regexp matches a digit, followed by a period, followed by another digit. The parenthesis captures its contents. Since it is the first pair of parenthesis, it is mapped to $1.
I have been looking at regular expressions to try and do this, but the most I can do is find the start of a line with ^, but not replace it.
I can then find the first characters on a line to replace, but can not do it in such a way with keeping it intact.
Unfortunately I don´t have access to a tool like cut since I am on a windows machine...so is there any way to do what I want with just regexp?
Use notepad++. It offers a way to record an sequence of actions which then can be repeated for all lines in the file.
Did you try replacing the regular expression ^ with the text you want to put at the start of each line? Also you should use the multiline option (also called m in some regex dialects) if you want ^ to match the start of every line in your input rather than just the first.
string s = "test test\ntest2 test2";
s = Regex.Replace(s, "^", "foo", RegexOptions.Multiline);
Console.WriteLine(s);
Result:
footest test
footest2 test2
I used to program on the mainframe and got used to SPF panels. I was thrilled to find a Windows version of the same editor at Command Technology. Makes problems like this drop-dead simple. You can use expressions to exclude or include lines, then apply transforms on just the excluded or included lines and do so inside of column boundaries. You can even take the contents of one set of lines and overlay the contents of another set of lines entirely or within column boundaries which makes it very easy to generate mass assignments of values to variables and similar tasks. I use Notepad++ for most stuff but keep a copy of SPFSE around for special-purpose editing like this. It's not cheap but once you figure out how to use it, it pays for itself in time saved.
I am getting completely different reults from string.scan and several regex testers...
I am just trying to grab the domain from the string, it is the last word.
The regex in question:
/([a-zA-Z0-9\-]*\.)*\w{1,4}$/
The string (1 single line, verified in Ruby's runtime btw)
str = 'Show more results from software.informer.com'
Work fine, but in ruby....
irb(main):050:0> str.scan /([a-zA-Z0-9\-]*\.)*\w{1,4}$/
=> [["informer."]]
I would think that I would get a match on software.informer.com ,which is my goal.
Your regex is correct, the result has to do with the way String#scan behaves. From the official documentation:
"If the pattern contains groups, each individual result is itself an array containing one entry per group."
Basically, if you put parentheses around the whole regex, the first element of each array in your results will be what you expect.
It does not look as if you expect more than one result (especially as the regex is anchored). In that case there is no reason to use scan.
'Show more results from software.informer.com'[ /([a-zA-Z0-9\-]*\.)*\w{1,4}$/ ]
#=> "software.informer.com"
If you do need to use scan (in which case you obviously need to remove the anchor), you can use (?:) to create non-capturing groups.
'foo.bar.baz lala software.informer.com'.scan( /(?:[a-zA-Z0-9\-]*\.)*\w{1,4}/ )
#=> ["foo.bar.baz", "lala", "software.informer.com"]
You are getting a match on software.informer.com. Check the value of $&. The return of scan is an array of the captured groups. Add capturing parentheses around the suffix, and you'll get the .com as part of the return value from scan as well.
The regex testers and Ruby are not disagreeing about the fundamental issue (the regex itself). Rather, their interfaces are differing in what they are emphasizing. When you run scan in irb, the first thing you'll see is the return value from scan (an Array of the captured subpatterns), which is not the same thing as the matched text. Regex testers are most likely oriented toward displaying the matched text.
How about doing this :
/([a-zA-Z0-9\-]*\.*\w{1,4})$/
This returns
informer.com
On your test string.
http://rubular.com/regexes/13670