What is the difference between %{} and %s? - ruby

In exercise 9 in the Learn Ruby the Hard Way book, I am asked to write the following:
formatter = "%{first} %{second} %{third} %{fourth}"
puts formatter % {first: 1, second: 2, third: 3, fourth: 4}
which just evaluates to
1 2 3 4
When searching around, I noticed that many people have written this instead:
formatter = "%s %s %s %s"
puts formatter % [1, 2, 3, 4]
I believe the latter example is from an older version of the book. Can someone explain to me what the differences are between the two examples?

The quick answer to that is that the %{} syntax allows you to pass in named arguments to be substituted into your string whereas the %s syntax only substitutes items in the order they are given.
You can do more things with the %{} version, for example if you have the same string you need to substitute in multiple times you could write it out like this:
string = "Hello %{name}, nice to meet you %{name}. Now I have said %{name} three times, I remember your name."
string % { :name => "Magnus" }
With the %s syntax, you would have had to write:
string = "Hello %s, nice to meet you %s. Now I have said %s three times, I remember your name."
string % ["Magnus", "Magnus", "Magnus"]
There are many other formats and ways to write substitutions for strings in Ruby. The full explanation can be found here in the Ruby documentation.

formatter = "%{first} %{second} %{third} %{fourth}"
and
formatter = "%s %s %s %s"
are essentially the same in that the formatting method will take values and substitute them into a string, however the first format string uses named placeholders vs. unnamed placeholders in the second.
This affects how you pass the values being substituted. The first accepts a hash of symbols, and uses the keys of that hash to identify which of the fields get that associated value. In the second, you pass an array, and the values are picked positionally from the array when being substituted into the format string.
The second is more common, and has been around for years, so you'll see it more often. The first, because it's newer isn't going to run on old Rubies, but has the advantage of resulting in a bit more readable format strings which is good for maintenance. Also, as #griffin says in his comment, the first also allows you to more easily repeat a value if necessary. You can do it when passing an array into the old-style format, but a hash would be more efficient memory-wise, especially for cases when you've got a lot of variables you're reusing.
Note: You'll see %{foo}f and %<foo>f. There is a difference in how the variable passed in is formatted, so be careful which you use:
'%<foo>f' % {:foo => 1} # => "1.000000"
'%{foo}f' % {:foo => 1} # => "1f"
I think the difference is too subtle and will cause problems for unaware developers. For more information, see the Kernel#sprintf documentation.

Related

How to use gsubstitution with more letters

I've printed the code, wit ruby
string = "hahahah"
pring string.gsub("a","b")
How do I add more letter replacements into gsub?
string.gsub("a","b")("h","l") and string.gsub("a","b";"h","l")
didnt work...
*update I have tried this too but without any success .
letters = {
"a" => "l"
"b" => "n"
...
"z" => "f"
}
string = "hahahah"
print string.gsub(\/w\,letters)
You're overcomplicating. As with most method calls in Ruby, you can simply chain #gsub calls together, one after the other:
str = 'adfh'
print str.gsub("a","b").gsub("h","l") #=> 'bdfl'
What you're doing here is applying the second #gsub to the result of the first one.
Of course, that gets a bit long-winded if you do too many of them. So, when you find yourself stringing too many together, you'll want to look for a regex solution. Rubular is a great place to tinker with them.
The way to use your hash trick with #gsub and a regex expression is to provide a hash for all possible matches. This has the same result as the two #gsub calls:
print str.gsub(/[ah]/, {'a'=>'b', 'h'=>'l'}) #=> 'bdfl'
The regex matches either a or h (/[ah]/), and the hash is saying what to substitute for each of them.
All that said, str.tr('ah', 'bl') is the simplest way to solve your problem as specified, as some commenters have mentioned, so long as you are working with single letters. If you need to work with two or more characters per substitution, you'll need to use #gsub.

Regex to leave desired string remaining and others removed

In Ruby, what regex will strip out all but a desired string if present in the containing string? I know about /[^abc]/ for characters, but what about strings?
Say I have the string "group=4&type_ids[]=2&type_ids[]=7&saved=1" and want to retain the pattern group=\d, if it is present in the string using only a regex?
Currently, I am splitting on & and then doing a select with matching condition =~ /group=\d/ on the resulting enumerable collection. It works fine, but I'd like to know the regex to do this more directly.
Simply:
part = str[/group=\d+/]
If you want only the numbers, then:
group_str = str[/group=(\d+)/,1]
If you want only the numbers as an integer, then:
group_num = str[/group=(\d+)/,1].to_i
Warning: String#[] will return nil if no match occurs, and blindly calling nil.to_i always returns 0.
You can try:
$str =~ s/.*(group=\d+).*/\1/;
Typically I wouldn't really worry too much about a complex regex. Simply break the string down into smaller parts and it becomes easier:
asdf = "group=4&type_ids[]=2&type_ids[]=7&saved=1"
asdf.split('&').select{ |q| q['group'] } # => ["group=4"]
Otherwise, you can use regex a bunch of different ways. Here's two ways I tend to use:
asdf.scan(/group=\d+/) # => ["group=4"]
asdf[/(group=\d+)/, 1] # => "group=4"
Try:
str.match(/group=\d+/)[0]

How should I parse a fixed length record file in Ruby?

I was wondering if anyone had any advice on parsing a file with fixed length records in Ruby. The file has several sections, each section has a header, n data elements and a footer. For example (This is total nonsense - but has roughly similar content)
1923 000-230SomeHeader 0303030
209231-231992395 MoreData
293894-329899834 SomeData
298342-323423409 OtherData
3 3423942Footer record 9832422
Headers, Footers and Data rows each begin with a specific number (1,2 & 3) in this example.
I have looked at http://rubyforge.org/projects/file-formatter/ and it looks good - except that the documentation is light and I can't see how to have n data elements.
Cheers,
Dan
There are a number of ways to do this. The unpack method of string could be used to define a pattern of fields as follows :-
"209231-231992395 MoreData".unpack('aa5A1A9a4Z*')
This returns an array as follows :-
["2", "09231", "-", "231992395", " ", "MoreData"]
See the documentation for a description of the pack/unpack format.
Several options exist as usual.
If you want to do it manually I would suggest something like this:
very pseudo-code:
Read file
while lines in file
handle_line(line)
end
def handle_line
type=first_char
parse_line(type)
end
def parse_line
split into elements and do_whatever_to_them
end
Splitting the line into elements of fixed with can be done with for instance unpack()
irb(main):001:0> line="1923 000-230SomeHeader 0303030"
=> "1923 000-230SomeHeader 0303030"
irb(main):002:0* list=line.unpack("A1A5A7a15A10")
=> ["1", "923", "000-230", "SomeHeader ", "0303030"]
irb(main):003:0>
The pattern used for unpack() will vary with field lengths on the different kinds of records and the code will depend on wether you want trailing spaces and such. See unpack reference for details.

How can I output leading zeros in Ruby?

I'm outputting a set of numbered files from a Ruby script. The numbers come from incrementing a counter, but to make them sort nicely in the directory, I'd like to use leading zeros in the filenames. In other words
file_001...
instead of
file_1
Is there a simple way to add leading zeros when converting a number to a string? (I know I can do "if less than 10.... if less than 100").
Use the % operator with a string:
irb(main):001:0> "%03d" % 5
=> "005"
The left-hand-side is a printf format string, and the right-hand side can be a list of values, so you could do something like:
irb(main):002:0> filename = "%s/%s.%04d.txt" % ["dirname", "filename", 23]
=> "dirname/filename.0023.txt"
Here's a printf format cheat sheet you might find useful in forming your format string. The printf format is originally from the C function printf, but similar formating functions are available in perl, ruby, python, java, php, etc.
If the maximum number of digits in the counter is known (e.g., n = 3 for counters 1..876), you can do
str = "file_" + i.to_s.rjust(n, "0")
Can't you just use string formatting of the value before you concat the filename?
"%03d" % number
Use String#next as the counter.
>> n = "000"
>> 3.times { puts "file_#{n.next!}" }
file_001
file_002
file_003
next is relatively 'clever', meaning you can even go for
>> n = "file_000"
>> 3.times { puts n.next! }
file_001
file_002
file_003
As stated by the other answers, "%03d" % number works pretty well, but it goes against the rubocop ruby style guide:
Favor the use of sprintf and its alias format over the fairly
cryptic String#% method
We can obtain the same result in a more readable way using the following:
format('%03d', number)
filenames = '000'.upto('100').map { |index| "file_#{index}" }
Outputs
[file_000, file_001, file_002, file_003, ..., file_098, file_099, file_100]

How do I convert a Ruby string with brackets to an array?

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.

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