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'm trying to replace two consecutive lines based on a pattern match, and would want this to repeat for the entire file. Here is the input file:
c aaaaa bbb
+ 0.1
c xxxx
c yyyy
+ 0.2
* c gggg
m eeeee hhhhh
+ 0.3
The command I tried is:
sed '/^c/{N;s/+/*+/}'
I expected to see a * prepended to each line beginning, but only those lines immediatlely following a c line:
c aaaaa bbb
*+ 0.1
c xxxx
c yyyy
*+ 0.2
* c gggg
m eeeee hhhhh
+ 0.3
what I actually get:
c aaaaa bbb
*+ 0.1
c xxxx
c yyyy
+ 0.2
* c gggg
m eeeee hhhhh
+ 0.3
Here, i see only the first occurrence of + (with previous line beginning with c) is getting replaced with *+. The second occurrence of + in the file is not getting replaced.
What am I doing wrong? How do I get the result I want: replacement happens in multiple consecutive lines in the file?
The problem you run into is that when a line that starts with c comes right after another line that comes with c, the N command in your code consumes it, and it isn't available for checking when you process the line that comes next.
Instead of reading ahead to see if the next line should be changed, I'd remember the last line and look back to see if the current line should be changed:
sed 'x; G; /^c/ s/+/*+/; s/.*\n//' file
This works as follows:
x # Swap pattern space and hold buffer. Because we do this here,
# the previous line will be in the hold buffer for every line
# (except the first, then it is empty)
G # append hold buffer to pattern space. Now the pattern space
# contains the previous line followed by the current line.
/^c/ s/+/*+/ # If the pattern space begins with a c (i.e., if the previous
# line began with a c), replace + with *+
s/.*\n// # Remove the first line (the previous one) from the pattern
# space
# Then drop off the end. The changed current line is printed.
sed -e 'H;$!d' -e 'x' -e ':cycle' -e 's/\(\nc[[:alnum:][:blank:][:punct:]]*\n\)+/\1*+/g;t cycle' -e 's/.//' YourFile
Posix version changing the whoe in max 2 internal cycle
load the file in memory (-e 'H;$!d' -e 'x')
Add the * in front of line starting with a + after a line starting with a c ( s/\(\nc[[:alnum:][:blank:][:punct:]]*\n\)+/\1*+/g)
do the same if occur in previous line ( :cycle and t cycle)
use a trick to insure starting with new line( H append current line to buffer also for first line so an extra new line as heading) (for first line with a c) and remove this at the end ('s/.//)
I have files with entries of the form:
$$
y = x^2
$$
I'm looking for a way (specifically using sed) to convert them to:
\begin{equation}
y = x^2
\end{equation}
The solution should not rely on the form of the equation (which may also span mutiple lines) nor on the text preceding the opening $$ or following the closing $$.
Thanks for the help.
sed '
/^\$\$$/ {
x
s/begin/&/
t use_end_tag
s/^.*$/\\begin{equation}/
h
b
: use_end_tag
s/^.*$/\\end{equation}/
h
}
'
Explanation:
sed maintains two buffers: the pattern space (pspace) and the hold space (hspace). It operates in cycles, where during each cycle it reads a line and executes the script for that line. pspace is usually auto-printed at the end of each cycle (unless the -n option is used), and then deleted before the next cycle. hspace holds its contents between cycles.
The idea of the script is that whenever $$ is seen, hspace is first checked to see if it contains the word "begin". If it does, then substitute the end tag; otherwise substitute the begin tag. In either case, store the substituted tag in the hold space so it can be checked next time.
sed '
/^\$\$$/ { # if line contains only $$
x # exchange pspace and hspace
s/begin/&/ # see if "begin" was in hspace
t use_end_tag # if it was, goto use_end_tag
s/^.*$/\\begin{equation}/ # replace pspace with \begin{equation}
h # set hspace to contents of pspace
b # start next cycle after auto-printing
: use_end_tag
s/^.*$/\\end{equation}/ # replace pspace with \end{equation}
h # set hspace to contents of pspace
}
'
This might work for you (GNU sed):
sed -r '1{x;s/^/\\begin{equation}\n\\end{equation}/;x};/\$\$/{g;P;s/(.*)\n(.*)/\2\n\1/;h;d}' file
Prime the hold space with the required strings. On encountering the marker print the first line and then swap the strings in anticipation of the next marker.
I can not help you with sed, but this awk should do:
awk '/\$\$/ && !f {$0="\\begin{equation}";f=1} /\$\$/ && f {$0="\\end{equation}";f=0}1' file
\begin{equation}
y = x^2
\end{equation}
The f=0is not needed, if its not repeated.
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.
Please answer with the shortest possible source code for a program that converts an arbitrary plaintext to its corresponding ciphertext, following the sample input and output I have given below. Bonus points* for the least CPU time or the least amount of memory used.
Example 1:
Plaintext: The quick brown fox jumps over the lazy dog. Supercalifragilisticexpialidocious!
Ciphertext: eTh kiquc nobrw xfo smjup rvoe eth yalz .odg !uioiapeislgriarpSueclfaiitcxildcos
Example 2:
Plaintext: 123 1234 12345 123456 1234567 12345678 123456789
Ciphertext: 312 4213 53124 642135 7531246 86421357 975312468
Rules:
Punctuation is defined to be included with the word it is closest to.
The center of a word is defined to be ceiling((strlen(word)+1)/2).
Whitespace is ignored (or collapsed).
Odd words move to the right first. Even words move to the left first.
You can think of it as reading every other character backwards (starting from the end of the word), followed by the remaining characters forwards. Corporation => XoXpXrXtXoX => niaorCoprto.
Thank you to those who pointed out the inconsistency in my description. This has lead many of you down the wrong path, which I apologize for. Rule #4 should clear things up.
*Bonus points will only be awarded if Jeff Atwood decides to do so. Since I haven't checked with him, the chances are slim. Sorry.
Python, 50 characters
For input in i:
' '.join(x[::-2]+x[len(x)%2::2]for x in i.split())
Alternate version that handles its own IO:
print ' '.join(x[::-2]+x[len(x)%2::2]for x in raw_input().split())
A total of 66 characters if including whitespace. (Technically, the print could be omitted if running from a command line, since the evaluated value of the code is displayed as output by default.)
Alternate version using reduce:
' '.join(reduce(lambda x,y:y+x[::-1],x) for x in i.split())
59 characters.
Original version (both even and odd go right first) for an input in i:
' '.join(x[::2][::-1]+x[1::2]for x in i.split())
48 characters including whitespace.
Another alternate version which (while slightly longer) is slightly more efficient:
' '.join(x[len(x)%2-2::-2]+x[1::2]for x in i.split())
(53 characters)
J, 58 characters
>,&.>/({~(,~(>:#+:#i.#-#<.,+:#i.#>.)#-:)#<:##)&.><;.2,&' '
Haskell, 64 characters
unwords.map(map snd.sort.zip(zipWith(*)[0..]$cycle[-1,1])).words
Well, okay, 76 if you add in the requisite "import List".
Python - 69 chars
(including whitespace and linebreaks)
This handles all I/O.
for w in raw_input().split():
o=""
for c in w:o=c+o[::-1]
print o,
Perl, 78 characters
For input in $_. If that's not acceptable, add six characters for either $_=<>; or $_=$s; at the beginning. The newline is for readability only.
for(split){$i=length;print substr$_,$i--,1,''while$i-->0;
print"$_ ";}print $/
C, 140 characters
Nicely formatted:
main(c, v)
char **v;
{
for( ; *++v; )
{
char *e = *v + strlen(*v), *x;
for(x = e-1; x >= *v; x -= 2)
putchar(*x);
for(x = *v + (x < *v-1); x < e; x += 2)
putchar(*x);
putchar(' ');
}
}
Compressed:
main(c,v)char**v;{for(;*++v;){char*e=*v+strlen(*v),*x;for(x=e-1;x>=*v;x-=2)putchar(*x);for(x=*v+(x<*v-1);x<e;x+=2)putchar(*x);putchar(32);}}
Lua
130 char function, 147 char functioning program
Lua doesn't get enough love in code golf -- maybe because it's hard to write a short program when you have long keywords like function/end, if/then/end, etc.
First I write the function in a verbose manner with explanations, then I rewrite it as a compressed, standalone function, then I call that function on the single argument specified at the command line.
I had to format the code with <pre></pre> tags because Markdown does a horrible job of formatting Lua.
Technically you could get a smaller running program by inlining the function, but it's more modular this way :)
t = "The quick brown fox jumps over the lazy dog. Supercalifragilisticexpialidocious!"
T = t:gsub("%S+", -- for each word in t...
function(w) -- argument: current word in t
W = "" -- initialize new Word
for i = 1,#w do -- iterate over each character in word
c = w:sub(i,i) -- extract current character
-- determine whether letter goes on right or left end
W = (#w % 2 ~= i % 2) and W .. c or c .. W
end
return W -- swap word in t with inverted Word
end)
-- code-golf unit test
assert(T == "eTh kiquc nobrw xfo smjup rvoe eth yalz .odg !uioiapeislgriarpSueclfaiitcxildcos")
-- need to assign to a variable and return it,
-- because gsub returns a pair and we only want the first element
f=function(s)c=s:gsub("%S+",function(w)W=""for i=1,#w do c=w:sub(i,i)W=(#w%2~=i%2)and W ..c or c ..W end return W end)return c end
-- 1 2 3 4 5 6 7 8 9 10 11 12 13
--34567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890
-- 130 chars, compressed and written as a proper function
print(f(arg[1]))
--34567890123456
-- 16 (+1 whitespace needed) chars to make it a functioning Lua program,
-- operating on command line argument
Output:
$ lua insideout.lua 'The quick brown fox jumps over the lazy dog. Supercalifragilisticexpialidocious!'
eTh kiquc nobrw xfo smjup rvoe eth yalz .odg !uioiapeislgriarpSueclfaiitcxildcos
I'm still pretty new at Lua so I'd like to see a shorter solution if there is one.
For a minimal cipher on all args to stdin, we can do 111 chars:
for _,w in ipairs(arg)do W=""for i=1,#w do c=w:sub(i,i)W=(#w%2~=i%2)and W ..c or c ..W end io.write(W ..' ')end
But this approach does output a trailing space like some of the other solutions.
For an input in s:
f=lambda t,r="":t and f(t[1:],len(t)&1and t[0]+r or r+t[0])or r
" ".join(map(f,s.split()))
Python, 90 characters including whitespace.
TCL
125 characters
set s set f foreach l {}
$f w [gets stdin] {$s r {}
$f c [split $w {}] {$s r $c[string reverse $r]}
$s l "$l $r"}
puts $l
Bash - 133, assuming input is in $w variable
Pretty
for x in $w; do
z="";
for l in `echo $x|sed 's/\(.\)/ \1/g'`; do
if ((${#z}%2)); then
z=$z$l;
else
z=$l$z;
fi;
done;
echo -n "$z ";
done;
echo
Compressed
for x in $w;do z="";for l in `echo $x|sed 's/\(.\)/ \1/g'`;do if ((${#z}%2));then z=$z$l;else z=$l$z;fi;done;echo -n "$z ";done;echo
Ok, so it outputs a trailing space.