I'm working with OPE IDs. One file has them with two trailing zeros, eg, [998700, 1001900]. The other file has them with one or two leading zeros for a total length of six, eg, [009987, 010019]. I want to convert every OPE ID (in both files) to an eight-digit string with exactly two leading zeros and however many zeros at the end to get it to be eight digits long.
Try this:
a = [ "00123123", "077934", "93422", "1231234", "12333" ]
a.map { |n| n.gsub(/^0*/, '00').ljust(8, '0') }
=> ["00123123", "00779340", "00934220", "001231234", "00123330"]
If you have your data parsed and stored as strings, it could be done like this, for example.
n = ["998700", "1001900", "009987", "0010019"]
puts n.map { |i|
i =~ /^0*([0-9]+?)0*$/
"00" + $1 + "0" * [0, 6 - $1.length].max
}
Output:
00998700
00100190
00998700
00100190
This example on codepad.
I'm note very sure though, that I got the description exactly right. Please check the comments and I correct in case it's not exactly what you were looking for.
With the help of the answers given by #detunized & #nimblegorilla, I came up with:
"998700"[0..-3].rjust(6, '0').to_sym
to make the first format I described (always with two trailing zeros) equal to the second.
Related
I need to check if the last character in a string is a digit, and if so, increment it.
I have a directory structure of /u01/app/oracle/... and that's where it goes off the rails. Sometimes it ends with the version number, sometimes it ends with dbhome_1 (or 2, or 3), and sometimes, I have to assume, it will take some other form. If it ends with dbhome_X, I need to parse that and bump that final digit, if it is a digit.
I use split to split the directory structure on '/', and use include? to check if the final element is something like "dbhome". As long as my directory structure ends with dbhome_X it seems to work. As I was testing, though, I tried a path that ended with dbhome, and found that my check for the last character being a digit didn't work.
db_home = '/u01/app/oracle/product/11.2.0/dbhome'
if db_home.split('/')[-1].include?('dbhome')
homedir=db_home.split('/')[-1]
if homedir[-1].to_i.is_a? Numeric
homedir=homedir[0...-1]+(homedir[-1].to_i+1).to_s
new_path="/"+db_home.split('/')[1...-1].join("/")+"/"+homedir.to_s
end
else
new_path=db_home+"/dbhome_1"
end
puts new_path
I did not expect the output to be /u01/app/oracle/11.2.0/product/dbhom1 - it seems to have fallen into the if block that added 1 to the final character.
If I set the initial path to /u01/app/.../dbhome_1, I get the expected /u01/app/.../dbhome_2 as the output.
You could use a regular expression to make matching a tad bit easier
if !!(db_home[/.*dbhome.*\z]) ..
You could use regex's
/[0-9]$/.match("How3").nil?
I need to check if the last character in a string is a digit, and if
so, increment it.
This is one option:
s = 'string9'
s[-1].then { |last| last.to_i.to_s == last ? [s[0..-2], last.to_i+1].join : s }
#=> "string10"
'/u01/app/11.2.0/dbhome'.sub(/\d\z/) { |s| s.succ }
#=> "/u01/app/11.2.0/dbhome"
'/u01/app/11.2.0/dbhome9'.sub(/\d\z/) { |s| s.succ }
#=> "/u01/app/11.2.0/dbhome10"
This is a starting point if you're running Ruby v2.6+:
fname = 'filename1'
fname[/\d+$/].then { |digits|
fname[/\d+$/] = digits.to_i.next.to_s if digits
}
fname # => "filename2"
And it's safe if the filename doesn't end with a digit:
fname = 'filename'
fname[/\d+$/].then { |digits|
fname[/\d+$/] = digits.to_i.next.to_s if digits
}
fname # => "filename"
I'm not sure if I like doing it that way better than the more traditional way which works with much older Rubies:
digits = fname[/\d+$/]
fname[/\d+$/] = digits.to_i.next.to_s if digits
except for the fact that digits gets stuck into the variable space after only being used once. There's probably worse things that happen in my code though.
This is taking advantage of String's [] and []= methods.
I have to cut the price from strings like that:
s1 = "somefing $ 100"
s2 = "$ 19081 words $"
s3 = "30$"
s4 = "hi $90"
s5 = "wow 150"
Output should be:
s1 = "100"
s2 = "19081"
s3 = "30"
s4 = "90"
s5 = nil
I use the following regex:
price = str[/\$\s*(\d+)|(\d+)\s*\$/, 1]
But it doesn't work for all types of strings.
Your code always returns the result of the first capture group group whereas in the failing case it is the second capture group that you are interested in. I don't think the [] method has a good way of dealing with this (when using numbered capture groups). You could write this like so
price = str =~ /\$\s*(\d+)|(\d+)\s*\$/ && ($1 || $2)
Although this isn't very legible. If instead you use a named capture group, then you can do
price = str[/\$\s*(?<amount>\d+)|(?<amount>\d+)\s*\$/, 'amount']
Duplicate named capture groups won't always do what you want but when they are in separate alternation branches (as they are here) then it should work.
The problem is that you're always getting value from the first regex group and you don't check the second. So, you're not looking the case after | - the one when digit is before $ sign.
If you look at the graphical representation of your regex, by typing 1 as a second parameter in square brackets, you are covering only the upper path (first case), and you never check lower one (second case).
Basically, try:
price = str[/\$\s*(\d+)|(\d+)\s*\$/, 1] or str[/\$\s*(\d+)|(\d+)\s*\$/, 2]
P.S. I'm not that experienced in Ruby, there might be some more optimal way to type this, but this should do the trick
try this, its much simpler but it may not be the most efficient.
p1 = s1.gsub(' ','')[/\$(\d+)|(\d+)\$/,1]
Right now I am working on a project that issues IDs consisting of both letters and numbers, for example 345A22. I need this program to be able to tell that for example, 345B22 is greater than 345A22. I can't assume that the letters will be in the same position all the time (ie we do have some id's with 22335Q) but when I compare two numbers the letters will be in the same position.
How do I accomplish this in Ruby?
You can use the String#<=> method to compare strings. See documentation here.
>> "345B22" <=> "345A22"
=> 1
Where the 1 return value means that 345B22 is greater.
If a simple string comparison won't do the trick (e.g. different lengths, etc.), try converting the IDs (assuming they all match ^[0-9A-Z]*$) into integers by treating them as base36-encoded data.
In Ruby strings have the same comparison methods as numbers have.
2 > 1 #=> true
"2" > "1" #=> true
"B" > "A" #=> true
Not sure I understand your question, but I'm guessing that you mentally parse the ids into components (so 345B22 is 345, B, 22) and then are wishing for a numeric sort for things that are numbers (i.e., 12 > 2) and a string sort for things that are strings (AB < B).
If this is what you intend, something like the following would do the trick:
ids.sort_by do |id|
id.scan(/\d+|[a-zA-Z]+/).map {|c| c =~ /\d/ ? c.rjust(20) : c.ljust(20) }.join
end
What this does is extract out all consecutive numbers or letters and then justify them right or left based on their type, concatenates the result and then sorts based on this (expanded and canonicalized) id.
I am trying to convert a hex value to a binary value (each bit in the hex string should have an equivalent four bit binary value). I was advised to use this:
num = "0ff" # (say for eg.)
bin = "%0#{num.size*4}b" % num.hex.to_i
This gives me the correct output 000011111111. I am confused with how this works, especially %0#{num.size*4}b. Could someone help me with this?
You can also do:
num = "0ff"
num.hex.to_s(2).rjust(num.size*4, '0')
You may have already figured out, but, num.size*4 is the number of digits that you want to pad the output up to with 0 because one hexadecimal digit is represented by four (log_2 16 = 4) binary digits.
You'll find the answer in the documentation of Kernel#sprintf (as pointed out by the docs for String#%):
http://www.ruby-doc.org/core/classes/Kernel.html#M001433
This is the most straightforward solution I found to convert from hexadecimal to binary:
['DEADBEEF'].pack('H*').unpack('B*').first # => "11011110101011011011111011101111"
And from binary to hexadecimal:
['11011110101011011011111011101111'].pack('B*').unpack1('H*') # => "deadbeef"
Here you can find more information:
Array#pack: https://ruby-doc.org/core-2.7.1/Array.html#method-i-pack
String#unpack1 (similar to unpack): https://ruby-doc.org/core-2.7.1/String.html#method-i-unpack1
This doesn't answer your original question, but I would assume that a lot of people coming here are, instead of looking to turn hexadecimal to actual "0s and 1s" binary output, to decode hexadecimal to a byte string representation (in the spirit of such utilities as hex2bin). As such, here is a good method for doing exactly that:
def hex_to_bin(hex)
# Prepend a '0' for padding if you don't have an even number of chars
hex = '0' << hex unless (hex.length % 2) == 0
hex.scan(/[A-Fa-f0-9]{2}/).inject('') { |encoded, byte| encoded << [byte].pack('H2') }
end
Getting back to hex again is much easier:
def bin_to_hex(bin)
bin.unpack('H*').first
end
Converting the string of hex digits back to binary is just as easy. Take the hex digits two at a time (since each byte can range from 00 to FF), convert the digits to a character, and join them back together.
def hex_to_bin(s) s.scan(/../).map { |x| x.hex.chr }.join end
I have a string that
contains at least one number
can contain multiple numbers
Some examples are:
https://www.facebook.com/permalink.php?story_fbid=53199604568&id=218700384
https://www.facebook.com/username_13/posts/101505775425651120
https://www.facebook.com/username/posts/101505775425699820
I need a way to extract the longest number from the string. So for the 3 strings above, it would extract
53199604568
101505775425651120
101505775425699820
How can I do this?
#get the lines first
text = <<ENDTEXT
https://www.facebook.com/permalink.php?story_fbid=53199604568&id=218700384
https://www.facebook.com/username_13/posts/101505775425651120
https://www.facebook.com/username/posts/101505775425699820
ENDTEXT
lines = text.split("\n")
#this bit is the actual answer to your question
lines.collect{|line| line.scan(/\d+/).sort_by(&:length).last}
Note that i'm returning the numbers as strings here. You could convert them to numbers with to_i
parse the list (to get an int array), then use the Max function. array.Max for syntax.
s = "https://www.facebook.com/permalink.php?story_fbid=53199604568&id=218700384"
s.scan(/\d+/).max{|a,b| a.length <=> b.length}.to_i