I've a formatted text and I need to find a sequence of two characters A separated by two any characters. The point is that I need to look for them only in the second column of the formatted text. I need to use the grep command. I came up with this:
grep -E A\.\.\A data.txt
which works correctly for all the columns, but I need to search only in the second one. Any suggestions?
Thank you
Using grep and assuming that , is your field separator, you can use something like this:
grep -E "^[^,]*,[^,]*A[^,]{2}A" data.txt
Here we
skip the first column:
start of line ^
everything up to the first comma [^,]*
the first comma ,
now that we are behind the first comma, e.g. in the second column, we match
optional characters [^,]* before the A
two characters that are not a comma: [^,]{2}
followed by the second A
But as others have already said: awk is probably the better tool for this task.
Related
I have a file that looks like this:
19.217.179.33,175.176.12.8
253.149.205.57,174.210.221.195
222.118.178.218,255.99.100.202
241.55.199.243,167.98.204.104
38.224.198.117,21.11.184.68
Each line is 2 IP addresses, separated by a comma. So, each line should meet these requirements:
Has 1 comma.
Has 6 periods.
Has ONLY numbers, commas, and periods.
If a line is missing a period, has more/less than one commas, has a letter, is blank, or anything like that - it isn't correct. Basically I just want to use sed or something similar to loop through each line in the file and make sure each of them meets the above requirements.
Is this something that can be done with sed? I know you can use it to delete files that do/don't have matching strings, but I wasn't sure about counting specific characters or verifying that a line only has certain characters.
Any help would be greatly appreciated. Thanks!
I think grep is a better tool for this. You just want to ensure that each line matches a particular regex, so invert the grep with -v and label the input invalid if any line gets output. Something like:
grep -qvE '^([0-9]{1,3}\.){3}[0-9]{1,3},([0-9]{1,3}\.){3}[0-9]{1,3}$' input || echo input is valid
You can simplify that a bit:
IP='([0-9]{1,3}\.){3}[0-9]{1,3}'
grep -qvE "^$IP,$IP$" input || echo input is valid
Or if you are more interested in invalid data:
grep -qvE "^$IP,$IP$" input && echo input is invalid
What I'd do is to think up a regular expression that fits the 'proper' lines, and omits them from printing. Like this:
sed -r '/^([0-9]{1,3}\.){3}[0-9]{1,3},([0-9]{1,3}\.){3}[0-9]{1,3}$/d' file
Everything that remains is a wrong line.
Here's the recipe in more detail:
[0-9]{1,3} between one and three digits
\. literal period (just the period is a wildcard and matches any character)
(...){3} three repetitions of something, so together
([0-9]{1,3}\.){3}[0-9]{1,3} makes up something that looks like an IP address. (Though note that it doesn't enforce the <256 rule, so 999.999.999.999 matches.)
/^ ... $/ the match needs to start at the beginning of the line and run until its end.
'/ ... /d' print everything except lines that match what's inside the two slashes
-r is needed to recognise the {1,3} syntax.
This will find and print the lines that are wrong. If you want to delete the wrong lines, you can easily invert this:
sed -i.bak -n -r '/^([0-9]{1,3}\.){3}[0-9]{1,3},([0-9]{1,3}\.){3}[0-9]{1,3}$/p' file
-i.bak means keep a backup, but overwrite the input file
-n means don't output anything unless expressly directed to output, and
/ ... /p output all the lines that match this regex.
If you would like to display only information about file contents correctness , you can use this command:
sed -n -r '/^([0-9]{1,3}\.){3}[0-9]{1,3},([0-9]{1,3}\.){3}[0-9]{1,3}$/!{a \
FILE IS INCORRECT
;q;};$aFILE IS OK'
It's modified version of #chw21 answer, but displays only information text:
FILE IS INCORRECT, or
FILE IS OK.
I am trying to clean up the next file:
1. 10.160.120.10 ; 140.0.0.40 ;Data-- 1155~00120~xtl~12/01/2016 03:00:24~000BBBBBA4FB~ÍežG5„È&gÈe#Ÿ#•Œ‘„¦åEI²6frÞõ+ã:®*ÓÓÂ"ða5»V$è~
2. ¼?Amµxðïej£„7‹ìËÏð‡.4 --
3. 10.160.120.11 ; 140.10.10.10 ;Data-- 1155~00120~xtl~12/01/2016 03:00:54~2B3BB1EB1BBB~£ˆD]†CÀ,£ÑÉ»In&Ry+/jÑ%A¡ã ÷d_#C÷—NÏÕÞ
3. Ü‚úè"åD\’c\ûñ7x °yFæï --
Note that the numbers are not an actual part of the file. They are just reference for the number of line. The size of the line depends on the encoded message (That is why the 3 is reapeated because it basically one line). There are thousands of records but they follow the same pattern. Each record ends with a (--).
Basically what I am trying to achive is to just get the IPs side by side.
For example:
10.160.120.10 000BBBBBA4FB
My first step would be to delete everything between the first (;) and the fourth (~) since that pattern is the same for each record.
Which leads me to this.
sed 's/;.*~//'
However this particular command would delete everything untill the last (~) and not the fourth.
If it succesfully removes everything between the first (;) and the fourth (~) it would get me something like this:
0.165.65.113 0008B9A4F3~ÍežG5„È&gÈe#Ÿ#•Œ‘„¦åEI²6frÞõ+ã:®*ÓÓÂ"ða5»V$è~
¼?Amµxðïej£„7‹ìËÏð‡.4 --
And then I guess I could delete everything after the first (~) so I can get the desired output.
Am I following the right procedure? Should I achive this with swd or awk? Any suggestion are appreciated!
Instead of trying to remove stuff, why don't you just keep the stuff you want?
sed -r -n 's/^[^0-9]*(([0-9]{1,3}\.){3}[0-9]{1,3}).*([0-9A-F]{12}).*$/\1 \3/p'
# ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ^^^^^^^^^^^^^^
# IP Address 12 Hex digits
Explanation:
\1 \3 means enter everything that matched the first and the third set of parenthesis of the search term.
^[^0-9]* matches all non-digits from the beginning of the file
([0-9]{1,3}\.){3}[0-9]{1,3} matches an IP address. The whole term is in parentheses because we want to keep it. The inner (...) could be referenced as \2 in the replacement term, but we don't need that.
[0-9A-F]{12} is simply 12 hexadecimal digits (upper case, use `[0-9a-fA-F] if you expect lower cases as well)
Assuming your data struture is the same
use several field separator at once with a class including ";" and "~". Be carefull , not space alone as separator like by default that return a different field 3 (and 6)
awk -F '[[:blank:]*[;~][[:blank:]]*' '/--$/ {print $1 " " $7}' YourFile
Assuming there is only space char and no tab as separator and data line have Data
awk -F ' *[;~] *' '/--$/ {print $1 " " $7}' YourFile
I know how to do it with awk, for example, keep lines, which contains number 3 in second column: $ awk '"$2" == 3'
But how to do the same with only grep?
What about for first column?
Grep is not great for this, awk is better. But assuming your columns are separated by spaces, then you want
grep -E '^[^ ]+ +3( |$)'
Explanation: find something that has a start of line, followed by one or more non-space characters (first column), then one or more space characters (column separator), then the number 3, then either a space (because there's another column) or end of line (if there's no other column).
(Updated to fix syntax after testing.)
Here is the longer explanation for my mysterious command grep -P '^[^\t]*\t3\t' your_file from the comments:
I assumed that the column delimiter is a tab. grep without -P would require some strange things to use it directly (see e.g. see here ) . The -P makes it possible to just write \t without any problems. If for example your delimiter is ; then you could replace the \t with ; and you dont need the -P option.
Having said that, lets explain the idea behind the regular expression: You said, you want to match a 3 in the second column:
^ means: at the beginning of the line
[^\t]* means: zero or more (*) occurences of something not a tab ([^\t] here the ^ means "not a")
followed by tab
followed by 3
followed by tab
Now we have effectively expressed the idea that we need a 3 as the content of the second column (\t3\t) and we are not interested in the precise content of the first column. The ^[^\t]*\t is only necessary to express the idea "what follows is in the second column".
If you want to match something in the fourth column, you could use this to "skip" the first three column and match a 4 in the fourth column:
^([^\t]*\t){3}4. (Note the parenthesis and the {3}).
As you can see many details and awk is much more elegant and easy.
You can read this up in the documentation of grep and then you will need to study something about regular expression, e.g. start here.
I have a csv file I am looking at in bash that I am trying to manipulate. There are several things that I have/am trying to edit. Structure is like so where the first row are the column(field) headers
cat,dog,hippopotamus,zebra
1,,3,2
three species, five species,only one,multiple
at,home, at, home, wild, wild
How can I edit the field (column) names in the csv?
head -1 test.csv
shows what the field (column) names are, but it still has the commas in it as well and this doesn't allow for field name changing at all.
The other part about this is that I want to only edit titles that are greater than 8 characters in length, in which case I will just take the first 8 characters. I'm guessing I would use some sort of loop based on string length but since I don't know how to even edit the field name of just one column I'm not sure how to do this. In scenario above, changing hippopotamus to hippopot.
How can I replace empty cells in the csv to NA or NULL?
sed -i 's/ /NULL/g'
Thought would work but doesn't.
Some of the cells have commas within them, messing with the , delimiter. I used the code below and it seems to work, but is there a better/safer way to do this?
sed -i "s/, /_/g"
Or in a similar situation, if multiple columns contain strings sometimes with spaces within a string but I only want to remove the space in one of the columns while leaving the other columns alone, how can I achieve this?
sed -i 's/ //g' test.csv
Sed will allow a line number as a command prefix, to only work on a single line (or a range of numbers, to work on lines in that range). Try something like this.
sed -e '1s/cat/Feline/' test.csv > test2.csv
CSV files will store an empty field as either a comma at start of line, a comma at end of line, or a comma followed by another comma:
Field1,Field2,Field3
,"<-- empty field1",field3
field1,,"<-- empty field2"
field1,"empty field3-->",
You can use the following sed commands to fix these:
sed -e 's/^,/NA,/;s/,$/,NA/' -e ':loop' -e 's/,,/,NA,/g;tloop' test.csv
Your solution appears good. Be aware, however, that CSV should have quotes around any string containing a comma. And that's legit. It's also the point where sed stops being a good tool for manipulating CSV files. ;-) One suggestion would be to replace "interior" commas with "%2C", which is the HTML encoding for a comma. That's pretty distinctive, and at least somewhat standard.
sed numbers groups starting from the left-most paren. If your groups match multiple times, you can only get the last match contents, but if an outer group contains the multi-match, the outer group is still valid. (I assume here that you have already replaced the "interior" commas with something else.)
sed -e ':loop' -e '^\(\([^,]*,\)\{3\}\)\([^ ,]*\) /\1\3/;tloop'
This will remove the first space in column 4, then loop. It will stop when it finds the comma that ends the column, or end-of-line.
Note that the first part, called \1, is general. You can replace the 3 with whatever field, minus one, and that will get you to the start of the field. The actual work is in the second part, \3, where you can do what you like. (Note that \2 is included within \1, and not particularly useful.)
Maybe a silly question but I have a text file that needs to display everything upto the first pattern match which is a '/'. (all lines contain no blank spaces)
Example.txt:
somename/for/example/
something/as/another/example
thisfile/dir/dir/example
Preferred output:
somename
something
thisfile
I know this grep code will display everything after a matching pattern:
grep -o '/[^\n]*' '/my/file.txt'
So is there any way to do the complete opposite, maybe rm everything after matching pattern or invert to display my preferred output?
Thanks.
If you're calling an external command like grep, you can get the same results your require with the sed command, i.e.
echo "something/as/another/example" | sed 's:/.*::'
something
Instead of focusing on what you want to keep, think about what you want to remove, in this case everything after the first '/' char. This is what this sed command does.
The leading s means substitute, the :/.*: is the pattern to match, with /.* meaning match the first /' char and all characters after that. The 2nd half of thesedcommand is the replacement. With::`, this means replace with nothing.
The traditional idom for sed is to use s/str/rep/, using / chars to delimit the search from the replacement, but you can use any character you want after the initial s (substitute) command.
Some seds expect the / char, and want a special indication that the following character is the sub/replace delimiter. So if s:/.*:: doesn't work, then s\:/.*:: should work.
IHTH.
Yu can use a much simpler reg exp:
/[^/]*/
The forward slash after the carat is what you're matching to.
jsFiddle
Assuming filename as "file.txt"
cat file.txt | cut -d "/" -f 1
Here, we are cutting the input line with "/" as the delimiter (-d "/"). Then we select the first field (-f 1).
You just need to include starting anchor ^ and also the / in a negated character class.
grep -o '^[^/]*' file