bash method to remove last 4 columns from csv file

问题

Is there a way to use bash to remove the last four columns for some input CSV file? The last four columns can have fields that vary in length from line to line so it is not sufficient to just delete a certain number of characters from the end of each row.

回答1:

Cut can do this if all lines have the same number of fields or awk if you don't.

cut -d, -f1-6 # assuming 10 fields

Will print out the first 6 fields if you want to control the output seperater use --output-delimiter=string

awk -F , -v OFS=, '{ for (i=1;i<=NF-4;i++){ printf $i, }; printf "\n"}'

Loops over fields up to th number of fields -4 and prints them out.

回答2:

cat data.csv | rev | cut -d, -f-5 | rev

rev reverses the lines, so it doesn't matter if all the rows have the same number of columns, it will always remove the last 4. This only works if the last 4 columns don't contain any commas themselves.

回答3:

You can use cut for this if you know the number of columns. For example, if your file has 9 columns, and comma is your delimiter:

cut -d',' -f -5

However, this assumes the data in your csv file does not contain any commas. cut will interpret commas inside of quotes as delimiters also.

回答4:

awk -F, '{NF-=4; OFS=","; print}' file.csv

or alternatively

awk -F, -vOFS=, '{NF-=4;print}' file.csv

will drop the last 4 columns from each line.

回答5:

awk one-liner:

awk -F, '{for(i=0;++i<=NF-5;)printf $i", ";print $(NF-4)}'  file.csv

the advantage of using awk over cut is, you don't have to count how many columns do you have, and how many columns you want to keep. Since what you want is removing last 4 columns.

see the test:

kent$  seq 40|xargs -n10|sed 's/ /, /g'           
1, 2, 3, 4, 5, 6, 7, 8, 9, 10
11, 12, 13, 14, 15, 16, 17, 18, 19, 20
21, 22, 23, 24, 25, 26, 27, 28, 29, 30
31, 32, 33, 34, 35, 36, 37, 38, 39, 40

kent$  seq 40|xargs -n10|sed 's/ /, /g' |awk -F, '{for(i=0;++i<=NF-5;)printf $i", ";print $(NF-4)}'
1,  2,  3,  4,  5,  6
11,  12,  13,  14,  15,  16
21,  22,  23,  24,  25,  26
31,  32,  33,  34,  35,  36

回答6:

This might work for you (GNU sed):

sed -r 's/(,[^,]*){4}$//' file

回答7:

This awk solution in a hacked way

awk -F, 'OFS=","{for(i=NF; i>=NF-4; --i) {$i=""}}{gsub(",,,,,","",$0);print $0}' temp.txt

回答8:

None of the mentioned methods will work properly when having CVS files with quoted fields with a <comma> character. So it is a bit hard to just use the <comma>-character as a field separator.

The following two posts are now very handy:

What's the most robust way to efficiently parse CSV using awk?
[U&L] How to delete the last column of a file in Linux (Note: this is only for GNU awk)

Since you work with GNU awk, you can thus do any of the following two:

$ awk -v FPAT='[^,]*|"[^"]+"' -v OFS="," 'NF{NF-=4}1'

Or with any awk, you could do:

$ awk 'BEGIN{ere="([^,]*|\042[^\042]+\042)"
             ere=","ere","ere","ere","ere"$"
       }
       {sub(ere,"")}1'

来源：https://stackoverflow.com/questions/14418511/bash-method-to-remove-last-4-columns-from-csv-file

标签

bash

csv

sed

awk

cut