I want to apply a search and replace regular expression pattern that work only in a given range of line and column on a text file like this :
AAABBBFFFFBBBAAABBB
AAABBBFFFFBBBAAABBB
GGGBBBFFFFBHHAAABBB
For example i want to replace BBB with YYY in line range 1 to 2 and from column 4 to 6, then obtaining this output :
AAAYYYFFFFBBBAAABBB
AAAYYYFFFFBBBAAABBB
GGGBBBFFFFBHHAAABBB
Is there a way to do it with Vim ?
:1,2 s/\%3cBBB/YYY/
\%3c means third column (see :help /\%c or more globally :help pattern)
If this is always the first one you want to replace, simply don't specify /g
:1,2s/BBB/YYY/
would work fine.
Alternatively, if you need to exactly specify which column you want replaced, you can use the \%Nv syntax, where N is the virtual column (column as it looks, so tabs are multiple columns, use c instead of v for actual columns)
Replacing the second set of B's on lines 1 and 2 could be done with:
:1,2s/\%11vBBB/YYY/
Related
I have a file of 1000 lines, with 5 to 8 columns in each line separated by :
1:2:3:4:5:6:7:8
4g10:8s:45:9u5b:a:z1
I want to have all lines in some order 4:3:1:2:5:6:7...
How would I swap only first 4 columns with regex?
I think this would probably be easier to do with another approach, but you could use ex to do it, so be in command mode and enter:
:%s/^\([^:]\+\):\([^:]\+\):\([^:]\+\):\([^:]\+\):/\4:\3:\1:\2:/
which will create capture groups for the first 4 colon delimited fields, then replace them in a different order than they were there originally.
Here is a regex that should do what you are looking for:
newtext = re.sub("([^:]+):([^:]+):([^:]+):([^:]+)(:)?(.*)?",r"\4:\3:\1:\2\5\6",text)
The take away is you'll want to use parans for capturing and then reorder them in the order you want them in the replace. Each capture "group" is just one or more non : separated by : If there is possibility of empty groups change each + to a *
Here is a sample in Python for clarity:
import re
textlist = [
"1:2:3:4:5:6:7:8",
"1:2:3:4:5",
"1:2:3:4",
]
for text in textlist:
newtext = re.sub("([^:]+):([^:]+):([^:]+):([^:]+)(:)?(.*)?",r"\4:\3:\1:\2\5\6",text)
print (newtext)
output:
4:3:1:2:5:6:7:8
4:3:1:2:5
4:3:1:2
I think I need RegEx for this, but it is new to me...
What I have in a text file are 200 rows of data, 100 INSERT INTO rows and 100 corresponding VALUE rows.
So it looks like this:
INSERT INTO DB1.Tbl1 (Col1, Col2, Col3........Col20)
VALUES(123, 'ABC', '201450204 15:37:48'........'DEF')
What I want to do is replace every Date/Timestamp value in Col3 with this: CURRENT_TIMESTAMP. The Date/Timestamps are NOT the same for every row. They differ, but they are all in Column 3.
There are 100 records in this table, some other tables have more, that's why I am looking for a shortcut to do this.
Try this:
search with (INSERT[^,]+,[^,]+,)([^,]+,)([^']+'[^']+'[^']+)('[^']+',) and replace with $1$3 and check mark regular expression in the notepad++
Live demo
With
"VALUES" being right at the beginning of the line,
"Col1" values being all numeric, and
no single quotes inside the values for "Col2"
you can search for
^(VALUES\(\d+, '[^']+', )'(\d{9} \d{2}:\d{2}:\d{2})'
and replace with
\1CURRENT_TIMESTAMP
along RegEx101. (Remember, Notepad++ uses the backslash in the replacement string…)
Personally, I'd consider to go straight to the database, and fix the timestamp there - especially, if you have more data to handle. (See my above comment for the general idea.)
Please comment, if and as further detail / adjustment is required.
I try to replace the german special character "ö" in a dataframe by "oe". The charcter occurs in multiple columns so I would like to be able to do this all in one by not having to specify individual columns.
Here is a small example of the data frame
data <- data.frame(a=c("aö","ab","ac"),b=c("bö","bb","ab"),c=c("öc","öb","acö"))
I tried :
data[data=="ö"]<-"oe"
but this did not work since I would need to work with regular expressions here. However when I try :
data[grepl("ö",data)]<-"oe"
I do not get what I want.
The dataframe at the end should look like:
> data
a b c
1 aoe boe oec
2 ab bb oeb
3 ac ab acoe
>
The file is a csv import that I import by read.csv. However, there seems to be no option to change to fix this with the import statement.
How do I get the desired outcome?
Here's one way to do it:
data <- apply(data,2,function(x) gsub("ö",'oe',x))
Explanation:
Your grepl doesn't work because grepl just returns a boolean matrix (TRUE/FALSE) corresponding to the elements in your data frame for which the regex matches. What the assignment then does is replace not just the character you want replaced but the entire string. To replace part of a string, you need sub (if you want to replace just once in each string) or gsub (if you want all occurrences replaces). To apply that to every column you loop over the columns using apply.
If you want to return a data frame, you can use:
data.frame(lapply(data, gsub, pattern = "ö", replacement = "oe"))
I have a file with long lines and need to see/ copy what the values are in a specic location(s) for the whole file but copy the rest of the line.
If the text width is small enough, ~184 columns, I can use :set colorcolumnnum to highlight the value. However over 184 characters it gets a bit unwieldy scrolling.
I tried :g/\%1237c/y Z, for one of the positions I needed, but that yanked the entire line.
eg for a smaller sample :g/\%49c/y Z will yank all of line 1 and 2 but I want to yank, or copy, the character at that column ie = on line 1 and x on line 2.
vim: filetype=help foldmethod=indent foldclose=all modifiable noreadonly
Table of Contents *sfcontents* *vim* *regex* *sfregex*
*sfsearch* - Search specific commands
|Ampersand-replaces-previous-pattern|
|append-a-global-search-to-a-register|
*sfHelp* Various Help related commands
There are two problems with your :g command:
For each matching line, the cursor is positioned on the first column. So even though you've matched at a particular column, that position is lost.
The \%c atom actually matches byte indices (what Vim somewhat confusingly names "columns"), so your measurement will be off for Tab and non-ASCII characters. Use the virtual column atom \%v instead.
Instead of :global, I would use :substitute with a replace-expression, in the idiom described at how to extract regex matches using vim:
:let t=[] | %s/\%49v./\=add(t, submatch(0))[-1]/g | let ## = join(t, "\n")
Alternatively, if you install my ExtractMatches plugin, I'd be that short command invocation:
:YankMatchesToReg /\%50v./
I'm trying to substitute all non matching characters in a single line between certain columns (after a search).
Example:
The search can be everything
In example below the search = test
The substitute character of non matching characters: empty space.
I want to substitute all characters non part of "test" between columns 10 and 30.
Columns 10 and 30 are indicated with |
before: djd<aj.testjal.kjetestjaja testlala ratesttsuvtesta !<-a-
| |
after: djd<aj.test test testlala ratesttsuvtesta !<-a-
How can I realize this?
Use the following substitution command on that line.
:s/\(test\)\zs\|\%>9v\%<31v./\=submatch(1)!=''?'':' '/g
If the range of columns is specified using visual selection, run
:'<,'>s/\(test\)\zs\|\%V./\=submatch(1)!=''?'':' '/g
One method may be to select the appropiate column range using the Visual mode (control+v)
Once selected, the search and replace can be done using (see this question)
%s/\%Vfoo/bar/g
A regular expression for not test can be found here: Regular expression to match a line that doesn't contain a word?