I have a long string which contains only decimal numbers with two signs after comma
str = "123,457568,22321,5484123,77"
The numbers in string only decimals with two signs after comma. How I can separate them in different numbers like that
arr = ["123,45" , "7568,22" , "321,54" , "84123,77"]
You could try a regex split here:
str = "123,457568,22321,5484123,77"
nums = str.split(/(?<=,\d{2})/)
print nums
This prints:
123,45
7568,22
321,54
84123,77
The logic above says to split at every point where a comma followed by two digits precedes.
Scan String for Commas Followed by Two Digits
This is a case where you really need to know your data. If you always have floats with two decimal places, and commas are decimals in your locale, then you can use String#scan as follows:
str.scan /\d+,\d{2}/
#=> ["123,45", "7568,22", "321,54", "84123,77"]
Since your input data isn't consistent (which can be assumed by the lack of a reliable separator between items), you may not be able to guarantee that each item has a fractional component at all, or that the component has exactly two digits. If that's the case, you'll need to find a common pattern that is reliable for your given inputs or make changes to the way you assign data from your data source into str.
Related
I have strings that consist of two lots of numbers. I am trying to split them into two substrings so that there are two decimal places after each decimal.
I have this code:
if homeodds.length == 10 then
homeoddsp = homeodds[0,5].to_f
bookieh = homeodds[5,5].to_f
else
homeoddsp = homeodds[0,4].to_f
bookieh = homeodds[4,4].to_f
end
It handles an eight figure string fine:
"1.211.90" translates to "1.21" and "1.90".
and a ten figure string also works:
"12.2113.00" translates to "12.21" and "13.00".
When I have a nine figure string like "9.1110.00" or "10.119.55", I need to find out where the first decimal is, and take two digits after that so that the remainder becomes the second substring. These cases, I can't consistently do.
You could use String#scan for this particular problem
def decimal_splits(string)
string.scan(/\d+\.\d{2}/)
end
To break down the regular expression:
\d+ 1 or more number of digits
\. the decimal point
\d{2} two digits
The result will be an array of matches for the expression.
decimal_splits("9.1110.00")
#=> ["9.11", "10.00"]
decimal_splits("12.2113.00")
#=> ["12.21", "13.00"]
decimal_splits("1.211.90")
#=> ["1.21", "1.90"]
decimal_splits("10.119.55")
#=> ["10.11", "9.55"]
I have a string:
'my_array1: ["1445","374","1449","378"], my_array2: ["1445","374", "1449","378"]'
I need to match all sets of digits from my_array2: [...] and count how many of them there.
I need to do something like this with regex and ruby MatchData
string = 'my_array1: ["1445","374", "1449","378"], my_array2: ["1445","374", "1449","378"]'
matches = string.match(/my_array2\:\s[\[,]\"(\d+)\"/)
count_matches = matches.size
Expected result should be 4.
What is the correct way of doing it?
If you are guaranteed that the content of my_array2 is always numeric you could simply use split twice. First you splitby my_array2: [" and then split by ,. This should give you the amount of items you are after.
If you are not guaranteed that, you could still split by my_array2 and instead of splitting again, you use a pattern such as "\d+" (or "\d+(\.\d+)? if you have floating point values) and count.
An example of the expression is available here.
I have a data structure that I'd like to convert back and forth from hex to binary in Ruby. The simplest approach for a binary to hex is '0010'.to_i(2).to_s(16) - unfortunately this does not preserve leading zeroes (due to the to_i call), as one may need with data structures like cryptographic keys (which also vary with the number of leading zeroes).
Is there an easy built in way to do this?
I think you should have a firm idea of how many bits are in your cryptographic key. That should be stored in some constant or variable in your program, not inside individual strings representing the key:
KEY_BITS = 16
The most natural way to represent a key is as an integer, so if you receive a key in a hex format you can convert it like this (leading zeros in the string do not matter):
key = 'a0a0'.to_i(16)
If you receive a key in a (ASCII) binary format, you can convert it like this (leading zeros in the string do not matter):
key = '101011'.to_i(2)
If you need to output a key in hex with the right number of leading zeros:
key.to_s(16).rjust((KEY_BITS+3)/4, '0')
If you need to output a key in binary with the right number of leading zeros:
key.to_s(2).rjust(KEY_BITS, '0')
If you really do want to figure out how many bits might be in a key based on a (ASCII) binary or hex string, you can do:
key_bits = binary_str.length
key_bits = hex_str.length * 4
The truth is, leading zeros are not part of the integer value. I mean, it's a little detail related to representation of this value, not the value itself. So if you want to preserve properties of representation, it may be best not to get to underlying values at all.
Luckily, hex<->binary conversion has one neat property: each hexadecimal digit exactly corresponds to 4 binary digits. So assuming you only get binary numbers that have number of digits divisible by 4 you can just construct two dictionaries for constructing back and forth:
# Hexadecimal part is easy
hex = [*'0'..'9', *'A'..'F']
# Binary... not much longer, but a bit trickier
bin = (0..15).map { |i| '%04b' % i }
Note the use of String#% operator, that formats the given value interpreting the string as printf-style format string.
Okay, so these are lists of "digits", 16 each. Now for the dictionaries:
hex2bin = hex.zip(bin).to_h
bin2hex = bin.zip(hex).to_h
Converting hex to bin with these is straightforward:
"DEADBEEF".each_char.map { |d| hex2bin[d] }.join
Converting back is not that trivial. I assume we have a "good number" that can be split into groups of 4 binary digits each. I haven't found a cleaner way than using String#scan with a "match every 4 characters" regex:
"10111110".scan(/.{4}/).map { |d| bin2hex[d] }.join
The procedure is mostly similar.
Bonus task: implement the same conversion disregarding my assumption of having only "good binary numbers", i. e. "110101".
"I-should-have-read-the-docs" remark: there is Hash#invert that returns a hash with all key-value pairs inverted.
This is the most straightforward solution I found that preserves leading zeros. 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
Dim str as String
str = "30 40 50 60"
I want to count the number of substrings.
Expected Output: 4
(because there are 4 total values: 30, 40, 50, 60)
How can I accomplish this in VB6?
You could try this:
arrStr = Split(str, " ")
strCnt = UBound(arrStr) + 1
msgBox strCnt
Of course, if you've got Option Explicit set (which you should..) then declare the variables above first..
Your request doesn't make any sense. A string is a sequence of text. The fact that that sequence of text contains numbers separated by spaces is quite irrelevant. Your string looks like this:
30 40 50 60
There are not 4 separate values, there is only one value, shown above—a single string.
You could also view the string as containing 11 individual characters, so it could be argued that the "count" of the string would be 11, but this doesn't get you any further towards your goal.
In order to get the result that you expect, you need to split the string into multiple strings at each space, producing 4 separate strings, each containing a 2-digit numeric value.
Of course, the real question is why you're storing this value in a string in the first place. If they're numeric values, you should store them in an array (for example, an array of Integers). Then you can easily obtain the number of elements in the array using the LBound() and UBound() functions.
I agree with everything Cody stated.
If you really wanted to you could loop through the string character by character and count the number of times you find your delimiter. In your example, it is space delimited, so you would simply count the number of spaces and add 1, but as Cody stated, those are not separate values..
Are you trying to parse text here or what? Regardless, I think what you really need to do is store your data into an array. Make your life easier, not more difficult.
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