deleting concret data from csv - bash

Question

I have a long csv file with 5 columns. But 3 lines have 6 columns. One begin with "tomasluck", another "peterblack" and the last one "susanpeeters". I need to delete, in this 3 lines, the fourth element (column) and get only 5 columns.
I put a short example, my file is long and is created automatically.
petergreat, 15, 11-03-2015, 10, 10
tomasluck, 15, 10-03-2015, tl, 10, 10
anaperez, 14, 11-03-2015, 10, 11
and I need
petergreat, 15, 11-03-2015, 10, 10
tomasluck, 15, 10-03-2015, 10, 10
anaperez, 14, 11-03-2015, 10, 11
Exactly, I was thinking in a code, that select the lines that begin with tomasluck, peterblack and susanpeeters, and then delete the 4rht field or colum.

Answer 1

The tricky thing about this is to keep the formatting intact. The simplest way, I think, is to treat the input as plain text and use sed:
sed '/^tomasluck,/ s/,[^,]*//3' file.csv
This removes, in a line that begins with tomasluck,, the third occurrence of a comma followed by a field (non-comma characters). The filter regex can be amended to include other first fields, such as
sed '/^\(tomasluck\|petergreat\|anaperez\),/ s/,[^,]*//3' file.csv
...but in your input data, those lines don't appear to have a sixth field.
Further ideas that may or may not pertain to your use case:
Removing the fourth field on the basis of the number of fields is a little trickier in sed, largely because sed does not have arithmetic functionality and identifying the lines is a bit tedious:
sed 'h; s/[^,]//g; /.\{5\}/ { x; s/,[^,]*//3; x; }; x' file.csv
That is:
h # copy the line to the hold buffer
s/[^,]//g # remove all non-comma characters
/.\{5\}/ { # if five characters remain (if the line has six or more
# fields)
x # exchange pattern space and hold buffer
s/,[^,]*//3 # remove field
x # swap back again
}
x # finally, swap in the actual data before printing.
The x dance is typical of sed scripts that use the hold buffer; the goal is to make sure that regardless of whether the substitution takes place, in the end the line (and not the isolated commas) are printed.
Mind you, if you want the selection condition to be that a line has six or more fields, it is worth considering to use awk, where the condition is easier to formulate but the replacement of the field is more tedious:
awk -F , 'BEGIN { OFS = FS } NF > 5 { for(i = 5; i <= NF; ++i) { $(i - 1) = $i }; --NF; $1 = $1 } 1' file.csv
That is: Split line at commas (-F ,), then
BEGIN { OFS = FS } # output field separator is input FS
NF > 5 { # if there are more than five fields
for(i = 5; i <= NF; ++i) { # shift them back one, starting at the fifth
$(i - 1) = $i
}
--NF # let awk know that there is one less field
$1 = $1 # for BSD awk: force rebuilding of the line
}
1 # whether or not a transformation happened, print.
This should work for most awks; I have tested it with gawk and mawk. However, because nothing is ever easy to do portably, I am told that there is at least one awk out there (on old Solaris, I believe) that doesn't understand the --NF trick. It would be possible to hack something together with sprintf for that, but it's enough of a corner case that I don't expect it to bite you.

Answer 2

a more generic solution is to check, whether we have 5 or 6 fields:
awk -F', ' '{if(NF==6) print $1", "$2", "$3", "$5", "$6; else print $0}' file.csv

Answer 3

You could do this through sed which uses capturing group a capturing group based regex.
$ sed 's/^\(\(tomasluck\|peterblack\|susanpeeters\),[^,]*,[^,]*\),[^,]*/\1/' file
petergreat, 15, 11-03-2015, 10, 10
tomasluck, 15, 10-03-2015, 10, 10
anaperez, 14, 11-03-2015, 10, 11
This captures all the characters upto the third column and matches the fourth column. Replacing the matched characters with the chars inside group 1 will give you the desired output.