I have a csv file like the following:
entity_name,data_field_name,type
Unit,id
Track,id,LONG
The second row is missing a comma. I wonder if there might be some regex or awk like tool in order to append commas to the end of line in case there are missing commas in these rows?
Update
I know the requirements are a little vague. There might be several alternative ways to narrow down the requirements such as:
The header row should define the number of columns (and commas) that is valid for the whole file. The script should read the header row first and find out the correct number of columns.
The number of columns might be passed as an argument to the script.
The number of columns can be hardcoded into the script.
I didn't narrow down the requirements at first because I was ok with any of them. Of course, the first alternative is the best but I wasn't sure if this was easy to implement or not.
Thanks for all the great answers and comments. Next time, I will state acceptable alternative requirements explicitly.
You can use this awk command to fill up all rows starting from 2nd row with the empty cell values based on # of columns in the header row, in order to avoid hard-coding # of columns:
awk 'BEGIN{FS=OFS=","} NR==1{nc=NF} NF{$nc=$nc} 1' file
entity_name,data_field_name,type
Unit,id,
Track,id,LONG
Earlier solution:
awk 'BEGIN{FS=OFS=","} NR==1{nc=NF} {printf "%s", $0;
for (i=NF+1; i<=nc; i++) printf "%s", OFS; print ""}' file
I would use sed,
sed 's/^[^,]*,[^,]*$/&,/' file
Example:
$ echo 'Unit,id' | sed 's/^[^,]*,[^,]*$/&,/'
Unit,id,
$ echo 'Unit,id,bar' | sed 's/^[^,]*,[^,]*$/&,/'
Unit,id,bar
Try this:
$ awk -F , 'NF==2{$2=$2","}1' file
Output:
entity_name,data_field_name,type
Unit,id,
Track,id,LONG
With another awk:
awk -F, 'NF==2{$3=""}1' OFS=, yourfile.csv
to present balance to all the awk solutions, following could be a vim only solution
:v/,.*,/norm A,
rationale
/,.*,/ searches for 2 comma's in a line
:v apply a global command on each line NOT matching the search
norm A, enters normal mode and appends a , to the end of the line
This MIGHT be all you need, depending on the info you haven't shared with us in your question:
$ awk -F, '{print $0 (NF<3?FS:"")}' file
entity_name,data_field_name,type
Unit,id,
Track,id,LONG
Related
I have a list like this
ABC|Hello1
ABC|Hello2
ABC|Hello3
DEF|Test
GHJ|Blabla1
GHJ|Blabla2
And i want it to be this:
ABC|Hello1
DEF|Test
GHJ|Blabla1
so i want to remove the duplicates in each line before the: |
and only let the first one there.
A simple way using awk
$ awk -F"|" '!seen[$1]++ {print $0}' file
ABC|Hello1
DEF|Test
GHJ|Blabla1
The trick here is to set the appropriate field separator "|" in this case after which the individual columns can be accessed column-wise starting with $1. In this answer, am maintaining a unique-value array seen and printing the line only if the value from $1 is not seen previously.
Good day,
Problem
There are two lines, the field separator is comma. Therefore, each line have 6 fields.
abc,def,ghi,jkl,mno,pqr
abc,def,ghi,jkl,,pqr
Target
If field five is empty, don't print that line.
Expected Output
abc,def,ghi,jkl,mno,pqr
So far I have done
awk '{print ($5=="")?:$5}' file
Thank so much in advance for giving me any clue.
awk -F',' '$5!=""' file
.......
It can be much simpler:
awk -F, '$5' file
i.e. print any line which has non-empty $5.
Two more awks,which wont break if field just contains 0.
awk -F, '$5~/./' file
awk -F, 'x!=$5' file
I have a huge .txt file, 300GB to be more precise, and I would like to put all the distinct strings from the first column, that match my pattern into a different .txt file.
awk '{print $1}' file_name | grep -o '/ns/.*' | awk '!seen[$0]++' > test1.txt
This is what I've tried, and as far as I can see it works fine but the problem is that after some time I get the following error:
awk: program limit exceeded: maximum number of fields size=32767
FILENAME="file_name" FNR=117897124 NR=117897124
Any suggestions?
The error message tells you:
line(117897124) has to many fields (>32767).
You'd better check it out:
sed -n '117897124{p;q}' file_name
Use cut to extract 1st column:
cut -d ' ' -f 1 < file_name | ...
Note: You may change ' ' to whatever the field separator is. The default is $'\t'.
The 'number of fields' is the number of 'columns' in the input file, so if one of the lines is really long, then that could potentially cause this error.
I suspect that the awk and grep steps could be combined into one:
sed -n 's/\(^pattern...\).*/\1/p' some_file | awk '!seen[$0]++' > test1.txt
That might evade the awk problem entirely (that sed command substitutes any leading text which matches the pattern, in place of the entire line, and if it matches, prints out the line).
Seems to me that your awk implementation has an upper limit for the number of records it can read in one go of 117,897,124. The limits can vary according to your implementation, and your OS.
Maybe a sane way to approach this problem is to program a custom script that uses split to split the large file into smaller ones, with no more than 100,000,000 records each.
Just in case that you don't want to split the file, then maybe you could look for the limits file correspondent to your awk implementation. Maybe you can define unlimited as the Number of Records value, although I believe that is not a good idea, as you might end up using a lot of resources...
If you have enough free space on disk (because creates a temp .swp file) I suggest to use Vim, vim regex has small difference but you can convert from standard regex to vim regex with this tool http://thewebminer.com/regex-to-vim
The error message says your input file contains too many fields for your awk implementation. Just change the field separator to be the same as the record separator and you'll only have 1 field per line and so avoid that problem, then merge the rest of the commands into one:
awk 'BEGIN{FS=RS} {sub(/[[:space:]].*/,"")} /\/ns\// && !seen[$0]++' file_name
If that's a problem then try:
awk 'BEGIN{FS=RS} {sub(/[[:space:]].*/,"")} /\/ns\//' file_name | sort -u
There may be an even simpler solution but since you haven't posted any sample input and expected output, we're just guessing.
I have a large log file from which I need to extract file names.
The file looks like this:
/path/to/loremIpsumDolor.sit /more/text/here/notAlways/theSame/here
/path/to/anotherFile.ext /more/text/here/differentText/here
.... about 10 million times
I need to extract the file names like this:
loremIpsumDolor.sit
anotherFile.ext
I figure my first strategy is to find/replace all /path/to/ with ''. But I'm stuck how to remove all characters after the space.
Can you help?
sed 's/ .*//' file
It doesn't take any more. The transformed output appears on standard output, of course.
In theory, you could also use awk to grab the filename from each line as:
awk '{ print $1 }' input_file.log
That, of course, assumes that there are no spaces in any of the filenames. awk defaults to looking for whitespace as the field delimiters, so the above snippet would take the first "field" from your log file (your filename) for each line, and output it.
Pass it to cut:
cut '-d ' -f1 yourfile
a bash-only solution:
while read path otherstuff; do
echo ${path##*/}
done < filename
I am trying to clean up some data, and I would eventually like to put it in CSV form.
I have used some regular expressions to clean it up, but I'm stuck on one step.
I would like to replace all but every third newline (\n) with a comma.
The data looks like this:
field1
field2
field3
field1
field2
field3
etc..
I need it in
field1,field2,field3
field1,field2,field3
Anyone have a simple way to do this using sed or awk? I could write a program and use a loop with a mod counter to erase every 1st and 2nd newline char, but I'd rather do it from the command line if possible.
With awk:
awk '{n2=n1;n1=n;n=$0;if(NR%3==0){printf"%s,%s,%s\n",n2,n1,n}}' yourData.txt
This script saves the last three lines and print them at every third line. Unfortunately, this works only with files having a multiple of 3 lines.
A more general script is:
awk '{l=l$0;if(NR%3==0){print l;l=""}else{l=l","}}END{if(l!=""){print substr(l,1,length(l)-1)}}' yourData.txt
In this case, the last three lines are concatenated in a single string, with the comma separator inserted whenever the line number is not a multiple of 3. At the end of the file, the string is printed if it is not empty with the trailing comma removed.
Awk version:
awk '{if (NR%3==0){print $0;}else{printf "%s,", $0;}}'
A Perl solution that's a little shorter and that handles files that don't have a multiple of 3 lines:
perl -pe 's/\n/,/ if(++$i%3&&! eof)' yourData.txt
cat file | perl -ne 'chomp(); print $_, !(++$i%3) ? "\n" : ",";'
Use nawk or /usr/xpg4/bin/awk on Solaris:
awk 'ORS=NR%3?OFS:RS' OFS=, infile
This might work for you:
paste -sd',,\n' file
or this:
sed '$!N;$!N;y/\n/,/' file
vim version:
:1,$s/\n\(.*\)\n\(.*\)\n/,\1,\2\r/g
awk '{ORS=NR%3?",":"\n";print}' urdata.txt