I have a very large text file that I have opened it in VIM and I need to add 1 to numbers matching some criteria. for example:
Comment 1
Comment 2
[p2,0,0,115]Live! ConcertGrand
[p2,2,104,5]Live! PopGrand
[p2,3,104,4]Live! RockPiano
[p2,4,104,3]Live! AmbientPiano
End of file.
and I'd like to transform this (by adding say 1 to the second element of the list) to
Comment 1
Comment 2
[p2,1,0,115]Live! ConcertGrand
[p2,3,104,5]Live! PopGrand
[p2,4,104,4]Live! RockPiano
[p2,5,104,3]Live! AmbientPiano
End of file.
How can I do this in vim, please?
I have tried:
%s/\[p\zs\d\+\(,\d\+\)\=/\=(1+str2nr(submatch(1)))/
But it does not work. Any help would be greatly appreciated. CS
Do you really need to do this with search and replace? Because there is builtin functionality for addition and subtraction.
:h CTRL-A
CTRL-A: Add [count] to the number or alphabetic character at or after the cursor.
{Visual}CTRL-A: Add [count] to the number or alphabetic character in the highlighted text. {not in Vi}
So you basically could use VISUAL-BLOCK selection on your column of numbers and press CTRL-A and it will add 1 to all of them
If it is more complicated you could use macro
Related
I need to make this exercise about regexes and text manipulation in vim.
So I have this file about the most scoring soccer players in history, with 50 entries looking like this:
1 Cristiano Ronaldo Portugal 88 121 0.73 03 Manchester United Real Madrid
The whitespaces between the fields are tabs (\t)
The fields each respond to a differen category: etc...
This last field contains one or more clubs the player has played in. (so not a fixed number of clubs)
The question: replace all tabs with a ';', except for the last field, where the clubs need to be seperated by a ','.
So I thought: I just replace all of them with a comma, and then I replace the first 7 commas with a semicolon. But how do you do that? Everything - from regex to vim commands - is allowed.
The first part is easy: :2,$s/\t/,/g
But the second part, I can't seem to figure out.
Any help would be greatly appreciated.
Thanks, Zeno
This answer is similar to #Amadan's, but it makes use of the ability to provide an expression as the replace string to actually do the difficult bit of changing the first set of tabs to semicolons:
%s/\v(.{-}\t){7}/\=substitute(submatch('0'), '\t', ';', 'g')/|%s/\t/,/g
Broken down this is a set of three substitute commands. The first two are cobbled together with a sub-replace-expression:
%s/\v(.{-}\t){7}/\=substitute(submatch('0'), '\t', ';', 'g')/
What this does is find exactly seven occurrances ({7}) of any character followed by a tab, in a non-greedy way. ((.{-}\t)). Then we replace this entire match (submatch(0)) with the result of the substitute expression (\=substitute(...)). The substitute expression is simple by comparison as it just converts all tabs to semicolons.
The last substitute just changes any other tabs on the line to commas.
See :help sub-replace-expression
Here's one way you could do it:
:let #q=":s/\t/;\<cr>"
:2,$norm 7#q
:2,$s/\t/,/g
Explanation:
First, we define a macro 'q' that will replace one tab with a semicolon. Now, on any line we can simply run this macro n times to replace the first n tabs. To automatically do this to every line, we use the norm command:
:2,$norm 7#q
This is essentially the same thing as literally typing 7#q (e.g. "run macro 'q' seven times") on every line in the specified range. From there, we can simply replace every tab with a comma.
:2,$s/\t/,/g
:2,$s/\t\(.*\t\)\#=/;/g
:2,$s/\t/,
Change any tabs where there is a tab later to ;
Change any remaining tabs to ,
EDIT: Misunderstood. Here is a fixed version:
:2,$s/\(\(\t.*\)\{7}\)\#<=\t/,/g
:2,$s/\t/;/g
Change any tabs where there's seven tabs before it to ,
Change any remaining tabs to ;
My PatternsOnText plugin has (among others) a :SubstituteSelected command that allows to specify the match positions. With this, you can easily replace the first 8 tabs with semicolons, and then use a regular substitute to change the remaining tabs into commas:
:2,$SubstituteSelected/\t/;/g 1-8
:2,$s/\t/,/g
We solved the issue by just capturing the first 8 groups manually ([^\t]*\t)(...)(...) and then separate them with a semicolon (\1;\2;...;) then replacing the remaining tabs with comma's | 2,$s/\t/,/g
Thanks to everyone trying to help!
This is my first post so I hope it is clear enough.
I am having a problem regarding cleaning my CSV files before I can read them into R and have spent the entire day trying to find a solution.
My data is supposed to be in the form of two columns. The first column is a timestamp consisting of 10 digits and the second an ID consisting of 11 or 12 Letters and numbers (the first 6 are always numbers).
For example:
logger10 |
0821164100 | 010300033ADD
0821164523 | 010300033ADD
0821164531 | 010700EDDA0F0831102744
010700EDDA0F|
would become:
0821164100 | 010300033ADD
0821164523 | 010300033ADD
0821164531 | 010700EDDA0F
0831102744 | 010700EDDA0F
(please excuse the lines in the middle, that was my attempt at separating the columns...).
The csv file seems to occasionally be missing a comma which means that sometimes one row will end up like this:
0923120531,010300033ADD0925075301,010700EDD00A
My hardware also adds the word logger10 (or whichever number logger this is) whenever it restarts which gives a similar problem e.g. logger10logger100831102744.
I think I have managed to solve the logger text problem (see code) but I am sure this could be improved. Also, I really don't want to delete any of the data.
My real trouble is making sure there is a line break in the right place after the ID and, if not, I would like to add one. I thought I could use regex for this but I'm having difficulty understanding it.
Any help would be greatly appreciated!
Here is my attempt:
temp <- list.files(pattern="*.CSV") #list of each csv/logger file
for(i in temp){
#clean each csv
tmp<-readLines(i) #check each line in file
tmp<-gsub("logger([0-9]{2})","",tmp) #remove logger text
pattern <- ("[0-9]{10}\\,[0-9]{6}[A-Z,0-9]{5,6}") #regex pattern ??
if (tmp!= pattern){
#I have no idea where to start here...
}
}
here is some raw data:
logger01
0729131218,020700EE1961
0729131226,020700EE1961
0831103159,0203000316DB
0831103207,0203000316DB0831103253,010700EDE28C
0831103301,010700EDE28C
0831103522,010300029815
0831103636,010300029815
0831103657,020300029815
If you want to do this in a single pass:
(?:logger\d\d )?([\dA-F]{10}),?([\dA-F]{12}) ?
can be replaced with
\1\t\2\n
What this does is look for any of those rogue logger01 entries (including the space after it) optionally: That trailing ? after the group means that it can match 0 or 1 time: if it does match, it will. If it's not there, the match just keeps going anyway.
Following that, you look for (and capture) exactly 10 hex values (either digits or A-F). The ,? means that if a comma exists, it will match, but it can match 0 or 1 time as well (making it optional).
Following that, look for (and capture) exactly 12 hex values. Finally, to get rid of any strange trailing spaces, the ? (a space character followed by ?) will optionally match the trailing space.
Your replacement will replace the first captured group (the 10 hex digits), add in a tab, replace the second captured group (the 12 hex digits), and then a newline.
You can see this in use on regex101 to see the results. You can use code generator on the left side of that page to get some preformatted PHP/Javascript/Python that you can just drop into a script.
If you're doing this from the command line, perl could be used:
perl -pe 's/(?:logger\d\d )?([\dA-F]{10}),?([\dA-F]{12}) ?/\1\t\2\n/g'
If another language, you may need to adapt it slightly to fit your needs.
EDIT
Re-reading the OP and comments, a slightly more rigid regex could be
(?:logger\d\d\ )?([\dA-F]{10}),?(\d{6}[\dA-F]{5,6})\ ?
I updated the regex101 link with the changes.
This still looks for the first 10 hex values, but now looks for exactly 6 digits, followed by 5-6 hex values, so the total number of characters matched is 11 or 12.
The replacement would be the same.
Paste your regex here https://regex101.com/ to see whether it catches all cases. The 5 or 6 letters or digits could pose an issue as it may catch the first digit of the timestamp when the logger misses out a comma. Append an '\n' to the end of the tmp string should work provided the regex catches all cases.
I need a RegEx pattern that will return the first N words using a custom word boundary that is the normal RegEx white space (\s) plus punctuation like .,;:!?-*_
EDIT #1: Thanks for all your comments.
To be clear:
I'd like to set the characters that would be the word delimiters
Lets call this the "Delimiter Set", or strDelimiters
strDelimiters = ".,;:!?-*_"
nNumWordsToFind = 5
A word is defined as any contiguous text that does NOT contain any character in strDelimiters
The RegEx word boundary is any contiguous text that contains one or more of the characters in strDelimiters
I'd like to build the RegEx pattern to get/return the first nNumWordsToFind using the strDelimiters.
EDIT #2: Sat, Aug 8, 2015 at 12:49 AM US CT
#maraca definitely answered my question as originally stated.
But what I actually need is to return the number of words ≤ nNumWordsToFind.
So if the source text has only 3 words, but my RegEx asks for 4 words, I need it to return the 3 words. The answer provided by maraca fails if nNumWordsToFind > number of actual words in the source text.
For example:
one,two;three-four_five.six:seven eight nine! ten
It would see this as 10 words.
If I want the first 5 words, it would return:
one,two;three-four_five.
I have this pattern using the normal \s whitespace, which works, but NOT exactly what I need:
([\w]+\s+){<NumWordsOut>}
where <NumWordsOut> is the number of words to return.
I have also found this word boundary pattern, but I don't know how to use it:
a "real word boundary" that detects the edge between an ASCII letter
and a non-letter.
(?i)(?<=^|[^a-z])(?=[a-z])|(?<=[a-z])(?=$|[^a-z])
However, I would want my words to allow numbers as well.
IAC, I have not been able how to use the above custom word boundary pattern to return the first N words of my text.
BTW, I will be using this in a Keyboard Maestro macro.
Can anyone help?
TIA.
All you have to do is to adapt your pattern ([\w]+\s+){<NumWordsOut>} to, including some special cases:
^[\s.,;:!?*_-]*([^\s.,;:!?*_-]+([\s.,;:!?*_-]+|$)){<NumWordsOut>}
1. 2. 3. 4. 5.
Match any amount of delimiters before the first word
Match a word (= at least one non-delimiter)
The word has to be followed by at least one delimiter
Or it can be at the end of the string (in case no delimiter follows at the end)
Repeat 2. to 4. <NumWordsOut> times
Note how I changed the order of the -, it has to be at the start or end, otherwise it needs to be escaped: \-.
Thanks to #maraca for providing the complete answer to my question.
I just wanted to post the Keyboard Maestro macro that I have built using #maraca's RegEx pattern for anyone interested in the complete solution.
See KM Forum Macro: Get a Max of N Words in String Using RegEx
I am a noob at regex and i've been trying to select 6 numbers from within a file and then replace those 6 numbers with the same numbers plus , new line (making a CSV obviously).
Anyway sample data is simply nonsense like this:
fafksadjlkgtjafglkj210000adsfaklgjadklgjag3600001skfjaklaj093i393593390000002sadfljafkjgakjgasafksadjlkgtjafglkj£94.00 489438adsfaklgjadklgjag7700001skfjaklaj093i393593390000002ssafksa djlkgtjafglkj000000adsfaklgjadklgjag0000001skfj aklaj093i393593£39.00900002ssafksadjlk gtjafglkj000000adsfaklgjadklgjag0000001skfjaklaj093i3935£933.90000002s
Note some of the numbers are attached to currency values as well (and some are next to it but contain a space before hand) but the end will always be 6 numbers (consider them to be random as I can't see a pattern).
So I basically need to select strings matching numerics that are six digits long or longer, if longer then it just uses the last 6 digits.
Then I will replace it with itself and a comma and new line.
I hope that makes sense, i've tried a few things without success..
Thanks, edit the closest I have is:
(\d)\d{6}(?!\d)
In the Find what: text field, type in (\d{6})(\D). In the Replace with: text field, type in $1\r\n$2. Make sure that the regular expression radio button is selected. For your input, that should yield this:
fafksadjlkgtjafglkj210000
adsfaklgjadklgjag3600001
skfjaklaj093i393593390000002
sadfljafkjgakjgasafksadjlkgtjafglkj£94.00 489438
adsfaklgjadklgjag7700001
skfjaklaj093i393593390000002
ssafksa djlkgtjafglkj000000
adsfaklgjadklgjag0000001
skfj aklaj093i393593
£39.00900002
ssafksadjlk gtjafglkj000000
adsfaklgjadklgjag0000001
skfjaklaj093i3935£933.90000002
s
You want
\d{6}(?=\D*$)
Read more about anchors here.
i've been trying to select 6 numbers from within a file and then replace those 6 numbers with the same numbers plus , new line
So you're basically trying to do this, right?:
Find:
(\d{6})(\D)
Replace:
\1\n\2
[Online example]
How about:
Find what: (\d{6,})(?:\D*)$
Replace with: $1,\n
I am trying to extract a specific text from an Outlook subject line. This is required to calculate turn around time for each order entered in SAP. I have a subject line as below
SO# 3032641559 FW: Attached new PO 4500958640- 13563 TYCO LJ
My final output should be like this: 3032641559
I have been able to do this in MS excel with the formulas like this
=IFERROR(INT(MID([#[Normalized_Subject]],SEARCH(30,[#[Normalized_Subject]]),10)),"Not Found")
in the above formula [#[Normalized_Subject]] is the name of column in which the SO number exists. I have asked to do this in oracle but I am very new to this. Your help on this would be greatly appreciated.
Note: in the above subject line the number 30 is common in every subject line.
The last parameter of REGEXP_SUBSTR() indicates the sub-expression you want to pick. In this case you can't just match 30 then some more numbers as the second set of digits might have a 30. So, it's safer to match the following, where x are more digits.
SO# 30xxxxxx
As a regular expression this becomes:
SO#\s30\d+
where \s indicates a space \d indicates a numeric character and the + that you want to match as many as there are. But, we can use the sub-expression substringing available; in order to do that you need to have sub-expressions; i.e. create groups where you want to split the string:
(SO#\s)(30\d+)
Put this in the function call and you have it:
regexp_substr(str, '(SO#\s)(30\d+)', 1, 1, 'i', 2)
SQL Fiddle