You could try to use “find and replace” feature, set to consider regular expressions, and replace all no-space (x) char with x followed a space in a current selection.

Save it as a macro and apply in other parts of code.
https://10bestgame.com/instagram-cool-captions/

Hmmm…or just put the spaces in by this process:

• position the caret appropriately
• press the space bar once
• REPEAT above steps until desired effect is achieved

x=(((a+5)/(z-7/b)-sin(30))*Log(6))

to this one: (without double spacing)

x = ( ( ( a + 5 ) / ( z - 7 / b ) - sin ( 30 ) ) * Log ( 6 ) )

find:

(([\w])([^\w|\s]))|(([^\w|\s])([\w]))|(([^\w|\s])([^\w|\s]))

replace:

(?1\2 \3)(?4\5 \6)(?7\8 \9)

Spend less than ten seconds to write your own Macro !

I’m not sure how THAT is helpful in any way…please explain if you can…

@guy038 said:

…the statement “x=(((a+5)/(z-7/b)-sin(30))*Log(6))” would be … “x = ( ( ( a + 5 ) / ( z - 7 / b ) - sin ( 30 ) ) * Log ( 6 ) )” with…too many space characters

It is very subjective (i.e., everybody likes it their own way), but I definitely agree that this is too spacey:

x = ( ( ( a + 5 ) / ( z - 7 / b ) - sin ( 30 ) ) * Log ( 6 ) )

but I like this version, which doesn’t add space adjacent to ( or ) :

x = (((a + 5) / (z - 7 / b) - sin(30)) * Log(6))

The original is just visually painful …ouch! :

x=(((a+5)/(z-7/b)-sin(30))*Log(6))

I think I will make the modification to the regex to avoid the “too spacey” condition and then record it as a Replace All, In Selection macro, for handy use when I get a short snippet of code from (e.g.) stackoverflow and don’t like the original whitespacing of an equation. (Is this what @Gogo-Neatza meant?!)

BTW, @guy038, I like this site for timestamp conversion.

Without additional information, it’s quite hard to plan anything ! Which language are you using ?

Anyway, I tried to find out a correct regex S/R, which :

• Trims any trailing horizontal blank characters, before the End of Line

• Keeps any leading indentation

• Keeps the blank characters in any range "..........", too

• Deletes any range of horizontal blank characters, before a comma

• Normalizes to a single space character, any range of blank characters, even none, PRECEDING, either :

  • Any consecutive range of symbols ! " # $ % & ' * + - / : ; = ? @ \ ^ | ~

  • Any single symbol [ ] ( ) < > { }

  • Any consecutive range of word characters or dot Letter Digit _ .

Of course, there are still few weird results ! For instance, the sentence “On the 10/27 of that year” would become “On the 10 / 27 of that year” but that other sentence “The value x=a*10/27+b” would produce the sentence “The value x = a * 10 / 27 + b” which is quite correct !

For example, the statement “x=(((a+5)/(z-7/b)-sin(30))*Log(6))” would be developed as “x = ( ( ( a + 5 ) / ( z - 7 / b ) - sin ( 30 ) ) * Log ( 6 ) )” with, certainly, too many space characters than necessary ;-))

A last example : here is, below, a simple Python script, regarding Unix Timestamp, with a very, very bad presentation ! You may select the option View > Show Symbol > Show All Characters to visualize all the blank characters

Day        =        28			
Month=3
Year		=			2018
Hours			=15                
Minutes=19
Seconds		=0
		                    
			# This small script get any Unix Timestamp
			#        with the different time variables above
			#        		defined with a correct numeric value

Leap=int(Year    %4==0)-int(     Year%100==0)+int(Year%400==0)                            # = 1 if a is a LEAP year, = 0 if NOT

Leap_Years=(Year      -1)//4-(Year-1)     //100+(Year-1)//400-477                           # = TOTAL of LEAP years, from 1970 to the year (Year-1)
			
Calendar_Day   =   round(30.57*(Month-1)-2*int(Month   >   2))+Day+int(Leap==1)		*		int(Month>2) # = TOTAL of days, from the 01/01 of Year, to Day, included

Total_Days=(Year-1970)*365+Leap_Years+Calendar_Day-1                             # = TOTAL of days, from 01/01/1970, till (Day-1) of Year
                                  
Unix_Time=Total_Days*86400+3600*Hours    +       60*Minutes+Seconds                          #  UNIX time, in seconds


print("And,    here are.... the     results  :  ",Leap       ,Leap_Years,          Calendar_Day,Total_Days,Unix_Time, "<     Good Bye     >")

        # The results are :
		
		# Leap=0
        #		Leap_Year		=		12
        # Calendar_Day = 87
        #            Total_Days = 17618
        # Unix_Time		=					1522250340

• Open the Replace dialog ( Ctrl + H )

• In the Find what: zone, type in the regex :

(?-s)(\h+\R)|(\h*,)|(^)?\h*(".*?"|[]()<>{}[]|[!"#$%&'*+/:;=?@\\^`|~-]+|[\w.]+)

• In the Replace with: zone, type in the regex

(?1\r\n:(?2,:(?3$0:\x20\4)))

• Tick the Wrap around option

• Click, once, on the Replace All button ( or several times on the Replace button

Et voilà ! You should get the following new text :

Day = 28
Month = 3
Year = 2018
Hours = 15
Minutes = 19
Seconds = 0

			# This small script get any Unix Timestamp
			# with the different time variables above
			# defined with a correct numeric value

Leap = int ( Year % 4 == 0 ) - int ( Year % 100 == 0 ) + int ( Year % 400 == 0 ) # = 1 if a is a LEAP year, = 0 if NOT

Leap_Years = ( Year - 1 ) // 4 - ( Year - 1 ) // 100 + ( Year - 1 ) // 400 - 477 # = TOTAL of LEAP years, from 1970 to the year ( Year - 1 )

Calendar_Day = round ( 30.57 * ( Month - 1 ) - 2 * int ( Month > 2 ) ) + Day + int ( Leap == 1 ) * int ( Month > 2 ) # = TOTAL of days, from the 01 / 01 of Year, to Day, included

Total_Days = ( Year - 1970 ) * 365 + Leap_Years + Calendar_Day - 1 # = TOTAL of days, from 01 / 01 / 1970, till ( Day - 1 ) of Year

Unix_Time = Total_Days * 86400 + 3600 * Hours + 60 * Minutes + Seconds # UNIX time, in seconds


print ( "And,    here are.... the     results  :  ", Leap, Leap_Years, Calendar_Day, Total_Days, Unix_Time, "<     Good Bye     >" )

        # The results are :

		# Leap = 0
        # Leap_Year = 12
        # Calendar_Day = 87
        # Total_Days = 17618
        # Unix_Time = 1522250340

Of course, the trailing comments and, generally, the possible lists are not aligned, too :-( Anyway, I just ran that Python3 script, after the S/R modifications, without any trouble !

You may use this Unix Timestamp Converter to verify my calculus !

Cheers,

guy038

P.S. :

• This Python example is quite raw and simple ! My initial program, about conversions Date <–> Unix Timestamp, was written in the old Qbasic language and some months ago, I simply translated the main part, as an Python3 exercise , about mathematics ;-))

• About 00h30, in France ! So, I’m a bit lazy to explain how this regex S/R works ! May be next time, if you offer me… some beer ;-))

ab=ab+1;

But even so, I believe “find and replace” with a better regEx you will find out a solution.

Save it as a macro and apply in other parts of code.