How to transpose "foo" and "bar" in "foo and bar" in emacs with the least number of key strokes?
input:
foo and bar
output:
bar and foo
Another way:
A numeric prefix of 0 to M-t will transpose the word ending after the mark with the word ending after the point.
So, if ^ is mark and | is point:
f^oo and ba|r
will become, after pressing M-0 M-t:
|bar and ^foo
So, in your example, if you are typing foo and bar|, the key sequence can be C-space M-3 M-b M-0 M-t (set mark at end of line, back 3 words to foo, transpose those words).
Here's one way (starting from the beginning of the phrase): M-t M-t M-m M-t.
If there's text on the line before foo, replace M-m with M-b M-b.
Related
My code has a ton of occurrences of something like:
idof(some_object)
I want to replace them with:
some_object["id"]
It sounds simple:
sed -i 's/idof(\([^)]\+\))/\1["id"]/g' source.py
The problem is that some_object might be something like idof(get_some_object()), or idof(my_class().get_some_object()), in which case, instead of getting what I want (get_some_object()["id"] or my_class().get_some_object()["id"]), I get get_some_object(["id"]) or my_class(["id"].get_some_object()).
Is there a way to have sed match closing bracket, so that it internally keeps track of any opening/closing brackets inside my (), and ignores those?
It needs to keep everything that's between those brackets: idof(ANYTHING) becomes ANYTHING["id"].
Using sed
$ sed -E 's/idof\(([[:alpha:][:punct:]]*)\)/\1["id"]/g' input_file
Using ERE, exclude idof and the first opening parenthesis.
As a literal closing parenthesis is also excluded, everything in-between the capture parenthesis including additional parenthesis will be captured.
[[:alpha:]] will match all alphabetic characters including upper and lower case while [[:punct:]] will capture punctuation characters including ().-{} and more.
The g option will make the substitution as many times as the pattern is found.
Theoretically, you can write a regex that will handle all combinations of idof(....) up to some limit of nested () calls inside ..... Such regex would have to list with all possible combinations of calls, like idof(one(two(three))) or idof(one(two(three)four(five)) you can match with an appropriate regex like idof([^()]*([^()]*([^()]*)[^()]*)[^()]*) or idof([^()]*([^()]*([^()]*)[^()]*([^()]*)[^()]*) respectively.
The following regex handles only some cases, but shows the complexity and general path. Writing a regex to handle all possible cases to "eat" everything in front of the trailing ) is left to OP as an exercise why it's better to use something else. Note that handling string literals ")" becomes increasingly complex.
The following Bash code:
sed '
: begin
# No idof? Just print the line!
/^\(.*\)idof(\([^)]*)\)/!n
# Note: regex is greedy - we start from the back!
# Note: using newline as a stack separator.
s//\1\n\2/
# hold the front
{ h ; x ; s/\n.*// ; x ; s/[^\n]*\n// ; }
: handle_brackets
# Eat everything before final ) up to some number of nested ((())) calls.
# Insert more jokes here.
: eat_brackets
/^[^()]*\(([^()]*\(([^()]*\(([^()]*\(([^()]*\(([^()]*\(([^()]*)\)\?[^()]*)\)\?[^()]*)\)\?[^()]*)\)\?[^()]*)\)\?[^()]*)\)/{
s//&\n/
# Hold the front.
{ H ; x ; s/\n\([^\n]*\)\n.*/\1/ ; x ; s/[^\n]*\n// ; }
b eat_brackets
}
/^\([^()]*\))/!{
s/^/ERROR: eating brackets did not work: /
q1
}
# Add the id after trailing ) and remove it.
s//\1["id"]/
# Join with hold space and clear the hold space for next round
{ H ; s/.*// ; x ; s/\n//g ; }
# Restart for another idof if in input.
b begin
' <<EOF
before idof(some_object) after
before idof(get_some_object()) after
before idof(my_class().get_some_object()) after
before idof(one(two(three)four)five) after
before idof(one(two(three)four)five) between idof(one(two(three)four)five) after
before idof( one(two(three)four)five one(two(three)four)five ) after
before idof(one(two(three(four)five)six(seven(eight)nine)ten) between idof(one(two(three(four)five)six(seven(eight)nine)ten) after
EOF
Will output:
before some_object["id"] after
before get_some_object()["id"] after
before my_class().get_some_object()["id"] after
before one(two(three)four)five["id"] after
before one(two(three)four)five["id"] between one(two(three)four)five["id"] after
before one(two(three)four)five one(two(three)four)five ["id"] after
ERROR: eating brackets did not work: one(two(three(four)five)six(seven(eight)nine)ten) after
The last line is not handled correctly, because (()()) case is not correctly handled. One would have to write a regex to match it.
I am trying to parse all the files and verify if any of the file content has strings TESTDIR or TEST_DIR
Files contents might look something like:-
TESTDIR = foo
include $(TESTDIR)/chop.mk
...
TEST_DIR := goldimage
MAKE_TESTDIR = var_make
NEW_TEST_DIR = tesing_var
Actually I am only interested in TESTDIR ,$(TESTDIR),TEST_DIR but in my case last two lines should be ignored. I am new to perl , Can anyone help me out with re-rex.
/\bTEST_?DIR\b/
\b means a "word boundary", i.e. the place between a word character and a non-word character. "Word" here has the Perl meaning: it contains characters, numbers, and underscores.
_? means "nothing or an underscore"
Look at "characterset".
Only (space) surrounding allowed:
/^(.* )?TEST_?DIR /
^ beginning of the line
(.* )? There may be some content .* but if, its must be followed by a space
at the and says that a whitespace must be there. Otherwise use ( .*)?$ at the end.
One of a given characterset is allowed:
Should the be other characters then a space be possible you can use a character class []:
/^(.*[ \t(])?TEST_?DIR[) :=]/
(.*[ \t(])? in front of TEST_?DIR may be a (space) or a \t (tab) or ( or nothing if the line starts with itself.
afterwards there must be one of (space) or : or = or ). Followd by anything (to "anything" belongs the "=" of ":=" ...).
One of a given group is allowed:
So you need groups within () each possible group in there devided by a |:
/^(.*( |\t))?TEST_?DIR( | := | = )/
In this case, at the beginning is no change to [ \t] because each group holds only one character and \t.
At the end, there must be (single space) or := (':=' surrounded by spaces) or = ('=' surrounded by spaces), following by anything...
You can use any combination...
/^(.*[ \t(])?TEST_?DIR([) =:]| :=| =|)/
Test it on Debuggex.com. (Use 'PCRE')
As the title suggests, I need to remove an character between two characters along a string.
E.g. I want to remove the semicolon between two parenthesis
<>word (word ; word) word<>
output desidered:
<>(word word)<>
Your description and the desired output do not match!
To remove the semicolon between two brackets you can use
sed '/([^)]*/s/;//'
example
echo "<>word (word ; word) word<>" | sed '/([^)]*/s/;//'
output
<>word (word word) word<>
I've modified the variable paragraph-start to count lines starting with .*: as a paragraph start:
(setq paragraph-start "\f\\|[ \t]*$\\|[ \t]*[0-9.]\.\\|.*:$\\|" )
However, if I have a buffer:
foo:
bar:
baz: some stuff
more
_
(Where _ indicates point location)
Then the first backward-paragraph skips to the beginning of the line 'bar:', not the line starting with 'baz:' as expected. How do I change this behaviour/why is it behaving this way?
It's because you have $ after ::
"\f\\|[ \t]*$\\|[ \t]*[0-9.]\.\\|.*:$\\|"
$ matches at the end of a line. So the part of your regexp that matches something followed by : also requires that nothing follow the :.
The first line (going backward from point) that ends in a : is the bar: line.
(Note too that you might not want .*:, if you want to exclude the possibility that what precedes the : not include a :, e.g., if you want to exclude a:b:c foo. To exclude :, use [^:]* instead of .*. And to exclude a lone :, use [^:]+.)
I have a data file like the following:
----------------------------
a b c d e .............
A B C D E .............
----------------------------
But I want it to be in the following format:
----------------------------
a A
b B
c C
d D
e E
...
...
----------------------------
What is the quickest way to do the transformation in Vim or Perl?
Basically :.s/ /SpaceCtrl+vEnter/gEnterjma:.s/ /Ctrl+vEnter/gEnterCtrl+v'axgg$p'adG will do the trick. :)
OK, let's break that down:
:.s/ /Ctrl+vEnter/gEnter: On the current line (.), substitute (s) spaces (/ /) with a space followed by a carriage return (SpaceCtrl+vEnter/), in all positions (g). The cursor should now be on the last letter's line (e in the example).
j: Go one line down (to A B C D E).
ma: Set mark a to the current position... because we want to refer to this position later.
:.s/ /Ctrl+vEnter/gEnter: Do the same substitution as above, but without the Space. The cursor should now be on the last letter's line (E in the example).
Ctrl+v'a: Select from the current cursor position (E) to mark a (that we set in step 3 above), using the block select.
x: Cut the selection (into the " register).
gg: Move the cursor to the first line.
$: Move the cursor to the end of the line.
p: Paste the previously cut text after the cursor position.
'a: Move the cursor to the a mark (set in step 3).
dG: Delete everything (the empty lines left at the bottom) from the cursor position to the end of the file.
P.S. I was hoping to learn about a "built-in" solution, but until such time...
Simple re-map of the columns:
use strict;
use warnings;
my #a = map [ split ], <>; # split each line on whitespace and store in array
for (0 .. $#{$a[0]}) { # for each such array element
printf "%s %s\n", $a[0]->[$_], $a[1]->[$_]; # print elements in order
}
Usage:
perl script.pl input.txt
Assuming that the cursor is on the first of the two lines, I would use
the command
:s/ /\r/g|+&&|'[-;1,g/^/''+m.|-j