Given this line of code in C:
printf("%3.0f\t%6.1f\n", fahr, ((5.0/9.0) * (fahr-32)));
Is there a way to delete or yank from the first bold parenthesis to its matching parenthesis? I thought about df), but that only will get you to just after the 9.0.
Is there a similar way to get vim to grab everything between matching braces, regardless of newlines?
What about dib or di(.
It will delete the inner (...) block where the cursor is.
I love text-object motions and selections!
Various Motions: %
The % command jumps to the match of the item under the cursor. Position the cursor on the opening (or closing) paren and use y% for yanking or d% for deleting everything from the cursor to the matching paren.
This works because % is a "motion command", so it can be used anywhere vim expects such a command. From :help y:
["x]y{motion} Yank {motion} text [into register x]. When no
characters are to be yanked (e.g., "y0" in column 1),
this is an error when 'cpoptions' includes the 'E'
flag.
By default, "item" includes brackets, braces, parens, C-style comments and various precompiler statements (#ifdef, etc.).
There is a plugin for "extended % matching" that you can find on the Vim homepage.
You can read the documentation on % and related motion commands by entering :help various-motions in command mode.
object-select
There is another set of motion commands that you can use in Visual mode to select various text objects.
To solve your specific problem you would do the following:
printf("%3.0f\t%6.1f\n", fahr, ((5.0/9.0) * (fahr-32)));
^
Let's say your cursor is positioned at ^. Enter the following sequence to select the part you are looking for:
v2a)
First v enters Visual mode, then you specify that you want to go 2 levels of parens up. Finally the a) selects "a block". After that you can use d or x to delete, etc.
If you don't want to include the outer parens, you can use "inner block" instead:
v2i)
See :help object-select for the complete list of related commands.
To delete all that is inside a pair of parentheses, you can always issue di( and its derivatives.
Note :
As #porglezomb suggested in his comment, you can use a ("along with") instead of i ("inside") to include the parentheses. So, using da( deletes everything inside ( and ) including ( and ).
Deleting text inside the immediate outer pair of parentheses :
So, for this line of code
printf("%3.0f\t%6.1f\n", fahr, ((5.0/9.0) * (fahr-32)));
^ ^
| |
\_______\___---> Cursor range
assuming that your cursor is inside the above mentioned cursor range, you can issue the following commands :
di( --> Deletes '5.0/9.0'
ci( --> Substitutes '5.0/9.0'
yi( --> Yanks '5.0/9.0'
Deleting text inside the n-th outer pair of parentheses :
To grab everything inside the n-th outer pair of parentheses, just add n before the above command. So, with the same cursor position as above,
2di( --> Deletes '(5.0/9.0) * (fahr-32)'
2ci( --> Substitutes '(5.0/9.0) * (fahr-32)'
2yi( --> Yanks '(5.0/9.0) * (fahr-32)'
3di( --> Deletes '"%3.0f\t%6.1f\n", fahr, ((5.0/9.0) * (fahr-32))'
3ci( --> Substitutes '"%3.0f\t%6.1f\n", fahr, ((5.0/9.0) * (fahr-32))'
3yi( --> Yanks '"%3.0f\t%6.1f\n", fahr, ((5.0/9.0) * (fahr-32))'
You can use d% for deleting and y% for yanking.
Place your cursor on the first parenthesis, then press v%y or v%d.
Try ci[block-surrounder]
In your case, place the cursor anywhere between the 2 parenthesis that you highlighed and try the keys: ci(
As answer of David Norman says,
Place your cursor on the first parenthesis, then press v%y or v%d.
Explanation from http://vimdoc.sourceforge.net/htmldoc/vimindex.html:
tag char note action in Normal mode
------------------------------------------------------------------------------
|v| v start characterwise Visual mode
|%| % 1 find the next (curly/square) bracket on
this line and go to its match, or go to
matching comment bracket, or go to matching
|d| ["x]d{motion} 2 delete Nmove text [into buffer x]
This means it will select everything between and including the two brackets (%) while showing the selection to you visually (v) and then yank/copy y or delete/cut d it. (To the default buffer.)
You can put/paste with p.
Made this answer to "teach myself to fish".
Related
I am tryng to get rid of shortcodes inside a Google Sheet column. I have many items such as [spacer type="1" height="20"][spacer] or [FinalTilesGallery id="37"] I just would like to cancel them. Is there any simple way to do it?
Thanks !
For in-place replacement, the quick option would be to use the Find and Replace dialog (Ctrl + H) with Search Using Regular Expressions turned on, which is more powerful than your standard Find and Replace.
Find: \[.*?\] - Match anything within an open-bracket up to the very next close-bracket. This should work assuming you have no nested brackets, e.g. [[no][no]].
If you do have nested brackets, you'll have to change this to \[[^\[\]]*\]. And continue to Replace All until all the codes are gone.
Replace: Nothing.
Replace All. If you don't want to affect other sheets that may be in your document, make sure you select the right range to work with, too.
This just erases everything within the brackets.
If you want to erase any redundant spaces left by this, simply Find and Replace again (with Regular Expressions) on + (space and plus), which will match 1 or more spaces and replace with (single space).
E.g.:
string [] [] string2 -> string string2 after the shortcode replacement.
After replacing spaces, it will become string string2.
Let's say your original strings are in the range A2:A. Place the following into B2 of an otherwise completely empty Column B (or the second cell of any other empty column):
=ArrayFormula(IF(A2:A="",,TRIM(REGEXREPLACE(A2:A,"\[[^\[\]]+\]",""))))
I can't see your data, so I don't know what kind of information is between these shortcodes. If you find that this leaves you with concatenated pieces of data where there should be spaces between them, replace the above with this version:
=ArrayFormula(IF(A2:A="",,TRIM(REGEXREPLACE(SUBSTITUTE(SUBSTITUTE(A2:A,"["," ["),"]","] "),"\[[^\[\]]+\]",""))))
I can't teach regular expression language here. But I will note that, since square brackets have specific meaning within regex, your literal square brackets must be indicated with the escape character: the backslash.
Here is the regex expression alone:
\[[^\[\]]+\]
The opening \[ and the closing \], then, reference your actual opening and closing bracket sets. If we remove those, we have this left:
[^\[\]]+
Again, you see the escaped opening and closing square brackets, which I'll replace with the word these:
[^these]+
What remains there are opening and closing brackets with regex meaning, i.e., "anything in this group." And the circumflex symbol ^ as the first character within this set of square brackets means "anything except." The + symbol means "in any string length of one or more characters."
So that whole regex expression then reads: "A literal open square bracket, followed by one or more characters that are anything except square brackets, ending with a literal closing square bracket."
And we are REGEXREPLACE-ing any instance of that with "" (i.e., nothing).
Here is a simple lists where I like to change the numbers: the entries are as below and it got over 300 entries like it
tom112
smith113
harry114
linda115
cindy106
samantha147
It need to be changed to
tom212
smith213
harry214
...and so on.
Please assist using notepad++ regular expression.
Thanks.
Assuming it's just a matter of replacing a name followed by a number starting with 1 with the same number but starting with 2 instead:
Ctrl + H for search & replace.
Check Regular expression under Search Mode.
Next to Find what type or copy in ([a-zA-Z]+)1([0-9]+).
Next to Replace with type or copy in \12\2.
Click Replace All and that should do it.
Add any other characters that might appear in the name before the number inside the first set of brackets with a-zA-Z.
I need to remove all \text generated by TeXForm in Mathematica.
What I am doing now is this:
MyTeXForm[a_]:=StringReplace[ToString[TeXForm[a]], "\\text" -> ""]
But the result keeps the braces, for example:
for a=fx,
the result of TeXForm[a] is \text{fx}
the result of MyTeXForm[a] is {fx}
But what I would like is it to be just fx
You should be able to use string patterns. Based on http://reference.wolfram.com/mathematica/tutorial/StringPatterns.html, something like the following should work:
MyTeXForm[a_]:=StringReplace[ToString[TeXForm[a]], "\\text{"~~s___~~"}"->s]
I don't have Mathematica handy right now, but this should say 'Match "\text{" followed by zero or more characters that are stored in the variable s, followed by "}", then replace all of that with whatever is stored in s.'
UPDATE:
The above works in the simplest case of there being a single "\text{...}" element, but the pattern s___ is greedy, so on input a+bb+xx+y, which Mathematica's TeXForm renders as "a+\text{bb}+\text{xx}+y", it matches everything between the first "\text{" and last "}" --- so, "bb}+\text{xx" --- leading to the output
In[1]:= MyTeXForm[a+bb+xx+y]
Out[1]= a+bb}+\text{xx+y
A fix for this is to wrap the pattern with Shortest[], leading to a second definition
In[2]:= MyTeXForm2[a_] := StringReplace[
ToString[TeXForm[a]],
Shortest["\\text{" ~~ s___ ~~ "}"] -> s
]
which yields the output
In[3]:= MyTeXForm2[a+bb+xx+y]
Out[3]= a+bb+xx+y
as desired.
Unfortunately this still won't work when the text itself contains a closing brace. For example, the input f["a}b","c}d"] (for some reason...) would give
In[4]:= MyTeXForm2[f["a}b","c}d"]]
Out[4]= f(a$\$b},c$\$d})
instead of "f(a$\}$b,c$\}$d)", which would be the proper processing of the TeXForm output "f(\text{a$\}$b},\text{c$\}$d})".
This is what I did (works fine for me):
MyTeXForm[a_] := ToString[ToExpression[StringReplace[ToString[TeXForm[a]], "\\text" -> ""]][[1]]]
This is a really late reply, but I just came up against the same issue and discovered a simple solution. Put a space between the variables in the Mathematica expression that you wish to convert using TexForm.
For the original poster's example, the following code works great:
a=f x
TeXForm[a]
The output is as desired: f x
Since LaTeX will ignore that space in math mode, things will format correctly.
(As an aside, I was having the same issue with subscripted expressions that have two side-by-side variables in the subscript. Inserting a space between them solved the issue.)
title says everything plus:
- development language Lua
- code revision control system - Perforce (integrated with IntelliJ IDE)
Posting for posterity - This was the top Google entry when searching "intellij count lines of code" (without quotes)
.
If you're like me and didn't want to install anything else, you can hack it via the native, global search:
Ctrl + Shift + F (to open global search)
Use regex mode (check "regex" checkbox)
In the searchbox, enter only a caret "^" (without the quotes)
You may want to limit the search to a specific directory, via the "directory" tab
Hit the "Open in Find Window" button on the bottom-right
If it asks whether you want to continue, press "Continue"
.
Notes:
In regex, the caret (^) denotes the start of a line, except when inside square brackets, in which case it denotes negation
If you wanted to count non-empty lines, you could instead use "^.*\S" (without quotes), which signifies "The start of a line (^), followed by any number of characters (except newline) (.*), followed by a non-whitespace character (\S)"
You can either turn on the display of lines of code for a single file by right clicking in the left gutter and highlighting "display lines of code". Or you can do it for your entire project by downloading the Statistic plug-in. It's very nice indeed, because it shows LOC and other metrics for your entire project.
I am using TeXnicCenter to edit a LaTeX document.
I now want to remove a certain tag (say, emph{blabla}} which occurs multiple times in my document , but not tag's content (so in this example, I want to remove all emphasization).
What is the easiest way to do so?
May also be using another program easily available on Windows 7.
Edit: In response to regex suggestions, it is important that it can deal with nested tags.
Edit 2: I really want to remove the tag from the text file, not just disable it.
Using a regular expression do something like s/\\emph\{([^\}]*)\}/\1/g. If you are not familiar with regular expressions this says:
s -- replace
/ -- begin match section
\\emph\{ -- match \emph{
( -- begin capture
[^\}]* -- match any characters except (meaning up until) a close brace because:
[] a group of characters
^ means not or "everything except"
\} -- the close brace
and * means 0 or more times
) -- end capture, because this is the first (in this case only) capture, it is number 1
\} -- match end brace
/ -- begin replace section
\1 -- replace with captured section number 1
/ -- end regular expression, begin extra flags
g -- global flag, meaning do this every time the match is found not just the first time
This is with Perl syntax, as that is what I am familiar with. The following perl "one-liners" will accomplish two tasks
perl -pe 's/\\emph\{([^\}]*)\}/\1/g' filename will "test" printing the file to the command line
perl -pi -e 's/\\emph\{([^\}]*)\}/\1/g' filename will change the file in place.
Similar commands may be available in your editor, but if not this will (should) work.
Crowley should have added this as an answer, but I will do that for him, if you replace all \emph{ with { you should be able to do this without disturbing the other content. It will still be in braces, but unless you have done some odd stuff it shouldn't matter.
The regex would be a simple s/\\emph\{/\{/g but the search and replace in your editor will do that one too.
Edit: Sorry, used the wrong brace in the regex, fixed now.
\renewcommand{\emph}[1]{#1}
any reasonably advanced editor should let you do a search/replace using regular expressions, replacing emph{bla} by bla etc.