How to determine the number of grouped numbers in a string in bash - bash

I have a string in bash
string="123 abc 456"
Where numbers that are grouped together are considered 1 number.
"123" and "456" would be considered numbers in this case.
How can i determine the number of grouped together numbers?
so
"123"
is determined to be a string with just one number, and
"123 abc 456"
is determined to be a string with 2 numbers.

egrep -o '[0-9]+' <<<"$string" | wc -l
Explanation
egrep: This performs an extended regular expression match on the lines of a given file (or, in this case, a herestring). It usually returns lines of text within the string that contain at least one chunk of text that matches the supplied pattern. However, the -o flag tells it to return only those matching chunks, one per line of output.
'[0-9]+': This is the regular expression that the string is compared against. Here, we are telling it to match successive runs of 1 or more digits, and no other character.
<<< The herestring operator allows us to pass a string into a command as if were the contents of a file.
| This pipes the output of the previous command (egrep) to become the input for the next command (wc).
wc: This performs a word count, normally returning the number of words in a given argument. However, the -l tells it to do a line count instead.
UPDATE: 2018-08-23
Is there any way to adapt your solution to work with floats?
The regular expression that matches both integer numbers and floating point decimal numbers would be something like this: '[0-9]*\.?[0-9]+'. Inserting this into the command above in place of its predecessor, forms this command chain:
egrep -o '[0-9]*\.?[0-9]+' <<<"$string" | wc -l
Focussing now only on the regular expression, here's how it works:
[0-9]: This matches any single digit from 0 to 9.
*: This is an operator that applies to the expression that comes directly before it, i.e. the [0-9] character class. It tells the search engine to match any number of occurrences of the digits 0 to 9 instead of just one, but no other character. Therefore, it will match "2", "26", "4839583", ... but it will not match "9.99" as a singular entity (but will, of course, match the "9" and the "99" that feature within it). As the * operator matches any number of successive digits, this can include zero occurrences (this will become relevant later).
\.: This matches a singular occurrence of a period (or decimal point), ".". The backslash is a special character that tells the search engine to interpret the period as a literal period, because this character itself has special function in regular expression strings, acting as a wildcard to match any character except a line-break. Without the backslash, that's what it would do, which would potentially match "28s" if it came across it, where the "s" was caught by the wildcard period. However, the backslash removes the wildcard functionality, so it will now only match with an actual period.
?: Another operator, like the *, except this one tells the search engine to match the previous expression either zero or one times, but no more. In other words, it makes the decimal point optional.
[0-9]+: As before, this will match digits 0 to 9, the number of which here is determined by the + operator, which standards for at least one, i.e. one or more digits.
Applying this to the following string:
"The value of pi is approximately 3.14159. The value of e is about 2.71828. The Golden Ratio is approximately 1.61803, which can be expressed as (√5 + 1)/2."
yields the following matches (one per line):
3.14159
2.71828
1.61803
5
1
2
And when this is piped through the wc -l command, returns a count of the lines, which is 6, i.e. the supplied string contains 6 occurrences of number strings, which includes integers and floating point decimals.
If you wanted only the floating point decimals, and to exclude the integers, the regular expression is this:
'[0-9]*\.[0-9]+'
If you look carefully, it's identical to the previous regular expression, except for the missing ? operator. If you recall, the ? made the decimal point an optional feature to match; removing this operator now means the decimal point must be present. Likewise, the + operator is matching at least one instance of a digit following the decimal point. However, the * operator before it matches any number of digits, including zero digits. Therefore, "0.61803" would be a valid match (if it were present in the string, which it isn't), and ".33333" would also be a valid match, since the digits before the decimal point needn't be there thanks to the * operator. However, whilst "1.1111" could be a valid match, "1111." would not be, because + operator dictates that there must be at least one digit following the decimal point.
Putting it into the command chain:
egrep -o '[0-9]*\.[0-9]+' <<<"$string" | wc -l
returns a value of 3, for the three floating point decimals occurring in the string, which, if you remove the | wc -l portion of the command, you will see in the terminal output as:
3.14159
2.71828
1.61803
For reasons I won't go into, matching integers exclusively and excluding floating point decimals is harder to accomplish with Perl-flavoured regular expression matching (which egrep is not). However, since you're really only interested in the number of these occurrences, rather than the matches themselves, we can create a regular expression that doesn't need to worry about accurate matching of integers, as long as it produces the same number of matched items. This expression:
'[^.0-9][0-9]+(\.([^0-9]|$)|[^.])'
seems to be good enough for counting the integers in the string, which includes the 5, 1 and 2 (ignoring, of course, the √ symbol), returning these approximately matches substrings:
√5
1)
/2.
I haven't tested it that thoroughly, however, and only formulated it tonight when I read your comment. But, hopefully, you are beginning to get a rough sense of what's going on.

In case you need to know the number of grouped digits in string then following may help you.
string="123 abc 456"
echo "$string" | awk '{print gsub(/[0-9]+/,"")}'
Explanation: Adding explanation too here, following is only for explanation purposes.
string="123 abc 456" ##Creating string named string with value of 123 abc 456.
echo "$string" ##Printing value of string here with echo.
| ##Putting its output as input to awk command.
awk '{ ##Initializing awk command here.
print gsub(/[0-9]+/,"") ##printing value of gsub here(where gsub is for substituting the values of all digits in group with ""(NULL)).
it will globally substitute the digits and give its count(how many substitutions happens will be equal to group of digits present).
}' ##Closing awk command here.

Related

How can I mask everything but the last four characters of a credit card number (PAN) with "#" symbols? [duplicate]

This question already has answers here:
How to mask all but last four characters in a string
(6 answers)
Closed 2 years ago.
I have a credit card number like 1234567891234 and I want to show only the last 4 characters of this string, like #########1234. How can I do this?
string.gsub(/.(?=....)/, '*')
=> "*********1234"
gsub without the ! does not mutate the original object that string points to and can take regex arguements.
. matches with a character that is not a line break and ?= is a positive lookahead, so any character that has four characters beyond it, that are not line breaks, will be replaced with the second gsub parameter, which is *.
string.gsub(/\d(?=[0-9]{4})/, '*')
=> "*********1234"
produces the same output, looking for digits with \d and doing a positive lookahead with [0-9]{4} which matches for four characters between zero and nine.
Masking All Digits Except Last Four
If you're just trying to mask the credit card number, there are a number of ways to do that. However, what makes it potentially tricky is that credit card numbers can have anywhere from 13-19 digits, although 16 is certainly the most common.
One of the easiest ways to work around this expected variation is to use is String#slice! to save the last four digits, and then String#tr to convert the remainder of the digits to your masking character. For example:
def mask_credit_card card_number
credit_card_number = String(card_number).scan(/\d/).join
last_four_digits = credit_card_number.slice! -4..-1
credit_card_number.tr("0-9", "#") << last_four_digits
end
# Test against various lengths & formats.
[
"1234567890123456",
"1234-5678-9012-3456",
1234567890123456,
"1234-567890-12345",
].map { |card_number| mask_credit_card card_number }
#=> ["############3456", "############3456", "############3456", "###########2345"]
Caveats & Considerations
Some cards like Diners Club can start with a zero, making the card number unsuitable for processing as an Integer. Treating the card number as a String can be more reliable, but forces you to think about how you'll handle unexpected characters or spacing.
Extracting digits with #scan is safer than using #delete when invoking the mask on unsanitized input. For example, String(card_number).delete "-\s\t" would normalize the example data above, but might not catch other unexpected characters. Never trust user input!
If you want to preserve spacing, dashes, and so forth in your masking, you run the risk that a malformed string (e.g. "1234-5678-9012-34 5-6") will yield unexpected results like " 5-6" as the last four digits. It's usually better to normalize your inputs, and apply your chosen formatting to your outputs (e.g. with printf or sprintf) instead. Of course, your specific use case may vary.

Need help understanding why this string in grep pulls IP addresses rather than this other string

The following statement is from a homework question which I tested out and answered, but I'm just not understanding how come this line behaves the way it does and I want to understand why. I realize why this expression is flawed to find an IP address but I don't fully understand why it behaves the way it does since it seems as if the question mark doesn't actually behave as 0 or 1 times in like it's supposed to.
"user#machine:~$ grep -E '[01]?[0-9][0-9]?' "
To my understanding "[01]?" should look for any number 0-1 as indicated by the brackets while the question mark tells grep to look for zero or one instance only and similar with "[0-9]?". Thing is this line will print an unlimited number of digits far exceeding 3 digits. I ruled out that it was due to the 3rd bracket that didn't have a proceeding question mark since it would still print an unlimited amount of digits if I piped an echo or used a testing .txt file full of numbers.
This above example made me than wonder how to find IP's with grep the correct way. So I found countless examples like the following expression for IPv4 octets:
\.(25[0-5]\|2[0-4][0-9]\|[01][0-9][0-9]\|[0-9][0-9]).\
Is this telling me to look for any number 2-5 anywhere from 0-5 times? 0-5 is too many digits for an octet. Is it telling me to look for any number 0-5 up to 25 times? Again that's way too many digits for an octet. What does \2[0-4][0-9]\ mean in this case? I'm confused about how this expression finds numbers strictly between 1-255?
Look at it this way: x?[0-9]x? matches anything which contains a digit because both the x:es are optional. You might as well leave them out because they do not constrain the match at all.
25[0-5] looks for 25 followed by a digit in the range 0-5. In other words, the expression matches a number in the range 250-255.
The full expression in your example looks for a number in the range 00-255 by enumerating strings beginning with 25, 20-24, etc; though it's incomplete in that it doesn't permit single-digit numbers.
The expression matches a single octet (incompletely), not an entire IP address. Here is a common way to match an IPv4 address:
([3-9][0-9]?|2([0-4][0-9]?|5[0-9]?|[6-9])?|1([0-9][0-9]?)?)(\.([3-9][0-9]?|2([0-4][0-9]?|5[0-9]?|[6-9])?|1([0-9][0-9]?)?){3}
where the square brackets express character classes which match a single character out of a set, and the final curly braces {3} express a repetition.
Some regex dialects (e.g. POSIX grep) require backslashes before | and \( but I have used the extended notation (a la grep -E and most online regex exploration tools) which doesn't want backslashes.

Unexpected arithmetic result with zero padded numbers

I have a problem in my script wherein I'm reading a file and each line has data which is a representation of an amount. The said field always has a length of 12 and it's always a whole number. So let's say I have an amount of 25,000, the data will look like this 000000025000.
Apparently, I have to get the total amount of these lines but the zero prefixes are disrupting the computation. If I add the above mentioned number to a zero value like this:
echo $(( 0 + 000000025000 ))
Instead of getting 25000, I get 10752 instead. I was thinking of looping through 000000025000 and when I finally get a non-zero value, I'm going to substring the number from that index onwards. However, I'm hoping that there must be a more elegant solution for this.
The number 000000025000 is an octal number as it starts with 0.
If you use bash as your shell, you can use the prefix 10# to force the base number to decimal:
echo $(( 10#000000025000 ))
From the bash man pages:
Constants with a leading 0 are interpreted as octal numbers. A leading 0x or 0X denotes hexadecimal. Otherwise, numbers take the form [base#]n, where the optional base is a decimal number between 2 and 64 representing the arithmetic base, and n is a number in that base.
Using Perl
$ echo "000000025000" | perl -ne ' { printf("%d\n",scalar($_)) } '
25000

Can someone give me an example of regular expressions using {x} and {x,y}?

I just learned from a book about regular expressions in the Ruby language. I did Google it, but still got confused about {x} and {x,y}.
The book says:
{x}→Match x occurrences of the preceding character.
{x,y}→Match at least x occurrences and at most y occurrences.
Can anyone explain this better, or provide some examples?
Sure, look at these examples:
http://rubular.com/r/sARHv0vf72
http://rubular.com/r/730Zo6rIls
/a{4}/
is the short version for:
/aaaa/
It says: Match exact 4 (consecutive) characters of 'a'.
where
/a{2,4}/
says: Match at least 2, and at most 4 characters of 'a'.
it will match
/aa/
/aaa/
/aaaa/
and it won't match
/a/
/aaaaa/
/xxx/
Limiting Repetition good online tutorial for this.
I highly recommend regexbuddy.com and very briefly, the regex below does what you refer to:
[0-9]{3}|\w{3}
The [ ] characters indicate that you must match a number between 0 and 9. It can be anything, but the [ ] is literal match. The { } with a 3 inside means match sets of 3 numbers between 0 and 9. The | is an or statement. The \w, is short hand for any word character and once again the {3} returns only sets of 3.
If you go to RegexPal.com you can enter the code above and test it. I used the following data to test the expression:
909 steve kinzey
and the expression matched the 909, the 'ste', the 'kin' and the 'zey'. It did not match the 've' because it is only 2 word characters long and a word character does not span white space so it could not carry over to the second word.
Interval Expressions
GNU awk refers to these as "interval expressions" in the Regexp Operators section of its manual. It explains the expressions as follows:
{n}
{n,}
{n,m}
One or two numbers inside braces denote an interval expression. If there is one number in the braces, the preceding regexp is repeated n times. If there are two numbers separated by a comma, the preceding regexp is repeated n to m times. If there is one number followed by a comma, then the preceding regexp is repeated at least n times:
The manual also includes these reference examples:
wh{3}y
Matches ‘whhhy’, but not ‘why’ or ‘whhhhy’.
wh{3,5}y
Matches ‘whhhy’, ‘whhhhy’, or ‘whhhhhy’, only.
wh{2,}y
Matches ‘whhy’ or ‘whhhy’, and so on.
See Also
Ruby's Regexp class.
Quantifiers section of Ruby's oniguruma engine.

How does backtracking differ from back-referencing in regular expressions?

How does backtracking differ from back-referencing in regular expressions?
How does back-referencing win limitation having with backtracking or vice-versa?
Backtracking is a way for a state machine to back up and retry other matches for a regular expression. It's something that's pretty much internal to the regex engine.
For example, say you're trying to match the regex [a-z]*a, any number of lower case characters followed by an a.
Given the input abca, a greedy match will assign all of that to the [a-z] portion of the regex but then there's no way to match the final a. Backtracking allows the engine to back up by returning that final a to the input stream and trying again, assigning abc to the [a-z] portion and a to the a portion.
Back-referencing on the other hand, is a means for a user of the regex engine to reference previously captured groups. For example,
s/^([a-z])([a-z])/\1_\2/
\_____/\_____/
| |
| +- capture group 2
+-------- capture group 1
may be a command to insert _ between two consecutive lower case letters at the start of each line. The \N back-reference (where N represents a number) refers back to the groups captured within ().

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