In Tcl how can I remove all zeroes to the left but the zeroes to the right should remain?

remove trailing zeros in r
stata remove leading zeros
strtrim stata
stata remove characters from string variable
stata remove leading and trailing spaces
stata ltrim
stata remove last character from string
command strtrim is unrecognized

Folks! I ran into a problem that I can't solve by myself.

Since the numbers "08" and "09" cannot be read like the others (01,02,03,04, etc ...) and must be treated separately in the language Tcl.

I can't find a way to remove all [I say ALL because there are more than one on the same line] the leading zeros except the one on the right, which must remain intact.

It may sound simple to those who are already thoroughly familiar with the Tcl / Tk language. But for me, who started out and am looking for more information about Tcl / Tk, I read a lot of material on the internet, including this https: // stackoverflow.com/questions/2110864/handling-numbers-with-leading-zeros-in-tcl#2111822 So nothing to show me how to do this in one sweep eliminating all leading zeros.

I need you to give me a return like this: 2:9:10

I need this to later manipulate the result with the expr [arithmetic expression] command.

In this example it just removes a single leading zero:

set time {02:09:10}
puts [regsub {^0*(.+)} $time {\1}]
# Return: 2:09:10

If anyone can give me that strength friend?! I'm grateful right now.

The group (^|:) matches either the beginning of the string or a colon. 0+ matches one or more zeros. Replace with the group match \1, otherwise the colons get lost. And of course, use -all to do all of the matches in the target string.

% set z 02:09:10
02:09:10
% regsub -all {(^|:)0+} $z {\1} x
2
% puts $x
2:9:10
% 

Edit: As Barmar points out, this will change :00 to an empty string. A better regex might be:

regsub -all {(^|:)0} $z {\1} x

This will only remove a single leading 0.

In Tcl how can I remove all zeroes to the left but the , Which Stata is right for me? Suppose you wish to remove leading or trailing zeros from a string variable (or from a in principle we could say any other character, except leading and trailing blanks should be However, subinstr() can remove characters we want to keep. Here the result of count is left behind in r(N). There can be many ways to solve this problem. Following is a simple and interesting way to solve this problem. Traverse the given array ‘arr’ from left to right. While traversing, maintain count of non-zero elements in array. Let the count be ‘count’.

You're only matching the 0 at the beginning of the string, you need to match after each : as well.

puts [regsub -all {(^|:)0*([^:])} $time {\1\2}]

How do I remove leading or trailing zeros from string variables?, Do not worry: we will keep the problem simpler than it is in its full generality. Invalid numbers (that is, strings we do not want to recognise as numbers but the above pattern: all the parts are optional, that is, each part can match a null Tip: To identify these substrings: just count the opening parentheses from left to right. In decimals, this idea of zeros that don’t add value to a number can be extended to trailing zeros. A trailing zero is any zero that appears to the right of both the decimal point and every digit other than zero. For example: 34.8 34.80 34.8000. All three of these numbers are the same.

In general it is best to use scan $str %d to convert a decimal number with possible leading zeroes to its actual value.

But in your case this will also work (and seems simpler to me than the answers given earlier and doesn't rely on the separator being a colon):

regsub -all {0*(\d+)} $time {\1}

This will remove any number of leading zeroes, but doesn't trim 00 down to an empty string. Also trailing zeroes will not be affected.

regsub -all {0*(\d+)} {0003:000:1000} {\1} => 3:0:1000

More Examples Of Regular Expressions, The command operates by scanning formatString from left to right. The format command must be given enough args to meet the needs of all of the Any of these fields may be omitted except for the conversion character. it specifies a minimum number of digits to print (leading zeroes will be added if necessary). The command operates by scanning formatStringfrom left to right. If the character is a %then it is not copied to the result string. a conversion specifier. The conversion specifier controls the conversion of the next successive argto a particular format and the result is appended to

the scan command is useful here to extract three decimal numbers out of that string:

% set time {02:09:10}
02:09:10
% scan $time {%d:%d:%d} h m s
3
% puts [list $h $m $s]
2 9 10

Tcl Built-In Commands - format manual page, I would like to remove MRN leading zeros. sometimes there are 1 zero 3 or 4.. which function should i use to to remove the leading zeros? any you can specify a character or number to trim from the left (leading) side of a variable. or simply (but I know there are those out there that don't like this way). Specifies that a number should always be printed with a sign, even if positive. space Specifies that a space should be added to the beginning of the number if the first character isn't a sign. 0 Specifies that the number should be padded on the left with zeroes instead of spaces. # Requests an alternate output form.

There are a few tricky edge cases here. Specifically, the string 02:09:10:1001:00 covers the key ones (including middle zeroes, only zeroes). We can use a single substitution command to do the work:

regsub -all {\m0+(?=\d)} $str {}

(This uses a word start anchor and lookahead constraint.)

However, I would be more inclined to use other tools for this sort of thing. For times, for example, parsing them is better done with scan:

set time "02:09:10"
scan $time "%d:%d:%d" h m s

Or, depending on what is going on, clock scan (which handles dates as well, making it more useful in some cases and less in others).

Newbie Questions: How to remove the leading zeros in tcl , If matchVar is specified, its value will be only the part of the string that was matched by the exp. by parenthesized bits in the exp, counting open parentheses from left to right. LV: the first regular expression should be seen like this: { . If you want to keep carriage returns or newlines by themselves, but not when they are� We fill the current position by 0 right away,so that unlike the previous solution, we don't need to come back here in next iteration. In other words, the code will maintain the following invariant: All elements before the slow pointer (lastNonZeroFoundAt) are non-zeroes. All elements between the current and slow pointer are zeroes.

Regular Expression Examples - the Tcler's Wiki!, Code Here is a method to print HEX digits with leading zeros that uses shifts of leading zeros to cells, one help you remove same number of string from left of the cells. But i am getting 6671. format() method to add zero padding to a number. In Java, some claimed it should declare as “int”, and use DecimalFormat to� Use the pound sign to prevent extra zeroes. The symbol # is another placeholder character in custom formats. This will prevent all leading zeroes if used at the front of the number, and prevent all trailing zeroes if used after the decimal point.

How to print leading zero in java, Replace method will remove the regex match from the string, and insert the replacement word character to the left of it, and a non-word character to the right of it. zeros in the hexadecimal number between the curly braces. In . NET and XML, you must omit the underscores but keep the hyphens in the block names. Hello ! We have a field that is a numerical value with many digits. AFAIK QV supports up to 11 or 12 digits for a field to be considered numeric. I wanna load this field as String (since it can't be numeric cause it has too many digits) but REMOVE LEADING ZEROS while loading it. This is a sample of

[PDF] Regular Expressions: The Complete Tutorial, You must have all rights to the file to be updated to use CRTTSTDTA. Decimal fields are right adjusted with leading zeros. The characters will be left adjusted and padded with blanks. The high range may be less than the full field length, but may not exceed Any fields not requested for changes will remain as is. The command operates by scanning formatStringfrom left to right. If the character is a %then it is not copied to the result string. a conversion specifier. The conversion specifier controls the conversion of the next successive argto a particular format and the result is appended to

Comments
  • ^ matches the beginning of the string, but you need to match after each : as well.
  • @Barmar Tai the problem is that I know how to make that combination. If you can help me with any answers, I appreciate it.
  • This is a time string (hh:mm:ss), right? Why not treating it like this: clock format [clock scan $time] -format %k:%N. Only bummer, there is no one- xor two-digit format specifier for the seconds of the minute, right?
  • @mrcalvin I did the test with your suggestion, and the feedback was this: time value too large/small to represent.
  • As for "I need this to later manipulate the result with the expr [arithmetic expression] command": If this is still the case (manipulating a time value), then you might want to use the clock facilities for this purpose, and not run the basic arithmetic yourself. Then you can also sidestep the cleansing step for the leading zeros. See clock arithmetic.
  • This will turn 00 into an empty string. E.g. 02:00:00 will become 2::
  • @Brad Lanam Even so, your answer is valid. It will make no difference if you return an empty string. I just need to get the result and calculate. So if you have "00" or WITHOUT both, it's the same.
  • It needs the -all flag to do multiple replacements
  • @DonalFellows: Regular expressions will match the earliest possible substring. So, this one will match at the first digit of a number. It will match as many zeroes as possible, as long as there's at least one digit left. Then it will match and capture as many remaining digits as possible. For 1001, it will match 0 zeroes and capture 4 more digits. The -all switch will not find overlapping substrings, so the zeroes in 1001 are safe: regsub -all {0*(\d+)} 02:09:10:1001:00 {\1} => 2:9:10:1001:0
  • This works with the colon separator shown in the example, but falls apart when an underscore is used as separator.
  • @SchelteBron Of course. But given that we're apparently really talking about parsing times, underscore is really unlikely.
  • Underscore may be unlikely. But your solution is more complex than the one I provided. And mine doesn't have this issue, nor the issue you claimed in your comment.