i have to use awk to print out 4 different columns in a csv file. The problem is the strings are in a $x,xxx.xx format. When I run the regular awk command.
awk -F, {print $1} testfile.csv
my output `ends up looking like
307.00
$132.34
30.23
What am I doing wrong.
"$141,818.88","$52,831,578.53","$52,788,069.53"
this is roughly the input. The file I have to parse is 90,000 rows and about 40 columns
This is how the input is laid out or at least the parts of it that I have to deal with. Sorry if I made you think this wasn't what I was talking about.
If the input is "$307.00","$132.34","$30.23"
I want the output to be in a
$307.00
$132.34
$30.23
Oddly enough I had to tackle this problem some time ago and I kept the code around to do it. You almost had it, but you need to get a bit tricky with your field separator(s).
awk -F'","|^"|"$' '{print $2}' testfile.csv
Input
# cat testfile.csv
"$141,818.88","$52,831,578.53","$52,788,069.53"
"$2,558.20","$482,619.11","$9,687,142.69"
"$786.48","$8,568,159.41","$159,180,818.00"
Output
# awk -F'","|^"|"$' '{print $2}' testfile.csv
$141,818.88
$2,558.20
$786.48
You'll note that the "first" field is actually $2 because of the field separator ^". Small price to pay for a short 1-liner if you ask me.
I think what you're saying is that you want to split the input into CSV fields while not getting tripped up by the commas inside the double quotes. If so...
First, use "," as the field separator, like this:
awk -F'","' '{print $1}'
But then you'll still end up with a stray double-quote at the beginning of $1 (and at the end of the last field). Handle that by stripping quotes out with gsub, like this:
awk -F'","' '{x=$1; gsub("\"","",x); print x}'
Result:
echo '"abc,def","ghi,xyz"' | awk -F'","' '{x=$1; gsub("\"","",x); print x}'
abc,def
In order to let awk handle quoted fields that contain the field separator, you can use a small script I wrote called csvquote. It temporarily replaces the offending commas with nonprinting characters, and then you restore them at the end of your pipeline. Like this:
csvquote testfile.csv | awk -F, {print $1} | csvquote -u
This would also work with any other UNIX text processing program like cut:
csvquote testfile.csv | cut -d, -f1 | csvquote -u
You can get the csvquote code here: https://github.com/dbro/csvquote
The data file:
$ cat data.txt
"$307.00","$132.34","$30.23"
The AWK script:
$ cat csv.awk
BEGIN { RS = "," }
{ gsub("\"", "", $1);
print $1 }
The execution:
$ awk -f csv.awk data.txt
$307.00
$132.34
$30.23
Related
I have a bunch of big csv I want to prefix every header column with fixed string. There is more than 500 columns in every file.
suppose my header is:
number;date;customer;key;amount
I tried this awk line:
awk -F';' 'NR==1{gsub(/[^a-z_]/,"input_file.")} { print }'
but I get (note fist column is missing prefix and separator is removed):
numberinput_file.dateinput_file.customerinput_file.keyinput_file.amount
expected output:
input_file.number;input_file.date;input_file.customer;input_file.key;input_file.amount
In any awk that'd be:
$ awk 'NR==1{gsub(/^|;/,"&input_file.")} 1' file
input_file.number;input_file.date;input_file.customer;input_file.key;input_file.amount
but sed exists to do simple substitutions like that, e.g. using a sed that has -E to enable EREs (e.g. GNU and BSD sed):
$ sed -E '1s/^|;/&input_file./g' file
input_file.number;input_file.date;input_file.customer;input_file.key;input_file.amount
If you're using GNU tools then you could use either of the above to change all of your CSV files at once with either of these:
awk -i inplace 'NR==1{gsub(/^|;/,"&input_file.")} 1' *.csv
sed -i -E '1s/^|;/&input_file./g' *.csv
Your gsub would brutally replace any nonalphabetic character anywhere in the input with the prefix - including your column separators.
The print can be abbreviated to the common idiom 1 at the very end of your script; this simply means "this condition is true; perform the default action for every line (i.e. print it all)" though this is just a stylistic change.
awk -F';' 'NR==1{
sub(/^/, "input_file."); gsub(/;/, ";input_file."); }
1' filename
If you want to perform this on multiple files, probably put a shell loop around it. If you only want to concatenate everything to standard output, you can give all the files to Awk in one go (in which case you probably don't want to print the header line for any file after the first; maybe change the 1 to NR==1 || FNR != 1).
I would use GNU AWK following way, let file.txt content be
number;date;customer;key;amount
1;2;3;4;5
6;7;8;9;10
then
awk 'BEGIN{FS=";";OFS=";input_file."}NR==1{$1="input_file." $1}{print}' file.txt
output
input_file.number;input_file.date;input_file.customer;input_file.key;input_file.amount
1;2;3;4;5
6;7;8;9;10
Explanation: I set OFS to ; followed by prefix. Then in first line I add prefix to first column, which trigger string rebuilding. No modification is done in any other line, thus they are printed as is.
(tested in GNU Awk 5.0.1)
Also with awk using for loop and printf:
awk 'BEGIN{FS=OFS=";"} NR==1{for (i=1; i<=NF; i++) printf "%s%s", "input_file." $i, (i<NF ? OFS : ORS)}' file
input_file.number;input_file.date;input_file.customer;input_file.key;input_file.amount
I have a sizeable dataset of about 7 million lines and I am trying to find the number of rows in column $2 that contain "/2020" in the date ($2 is all dates in the format mm/dd/yyyy). However, all of the awk commands I'm trying are either giving me 0 or aren't printing anything at all, and I'm not sure why.
awk -F',' '$2 == "/2020" { count++ } END { print count }' file.csv
prints nothing
awk -v variable="2020" '$2 ~ variable' file.csv | wc -l
prints 0
awk ' BEGIN {count=0;} { if ($2 =="2020") count += 1} END {print count}' file.csv
prints 0
I'd appreciate some help. thanks!
The syntax to use is:
awk -F, '$2 ~ /\/2020/{cnt++} END {print cnt}' file.csv
== would mean that second field will be exactly like the pattern, while ~ means that it is matching the pattern, just a part of the field can be like the pattern.
See also the related part of the GNU awk manual
Also, your second attempt would have worked if you have added the field separator, note that here you match only the year without the slash.
awk -F, -v variable="2020" '$2 ~ variable' file.csv | wc -l
Note: Assuming that there are no separators (commas) nested into quotes fields in your file, at least for the first two fields. If there are, a more complex pattern should be used as the field separator.
Combination of the best parts of your trials is:
$ awk -F, -v variable=2020 '$2~variable{c++}END{print c}' file
2
Since $2 is all dates in the format mm/dd/yyyy no need to put the / in the query (avoid an escaping), 2020 is enough - when using earthbound calendars...
But without a proper sample this is still all guessing.
Could you please try following if you want to use variable.
awk -v variable="2020" 'BEGIN{FS=","} $2 ~ ("/"variable){cnt++} END{print cnt}' file.csv
I have a csv file with thousands of lines in it. I'd like to be able to find values that only appear once in this file.
For instance
dog
dog
cat
dog
bird
I'd like to get as my result:
cat
bird
I tried using the following awk command but it returned one of each value in the file:
awk -F"," '{print $1}' test.csv|sort|uniq
Returns:
dog
cat
bird
Thank you for your help!
Just with awk:
awk -F, '{count[$1]++} END {for (key in count) if (count[key] == 1) print key}' test.csv
Close. Try:
awk -F"," '{print $1}' test.csv |sort | uniq -c | awk '{if ($1 == 1) print $2}'
the -c flag on uniq will give you counts. Next awk will look for any items with the count of 1 (first field) and print the value of the second field ($2)
Only caveat is that this will return bird before cat due to it being previously sroted. you could pipe once more to sort -r to reverse the sort direction. This would be identical to the expected answer you asked for, but it is not the original sort order.
Cutting to first field, then sorting and displaying only uniques:
cut -d ',' -f 1 test.csv | sort | uniq -u
That is, if you append -u to your command, it'd work. This is just using cut instead of awk.
If Perl is an option, this code is similar to #glenn jackman's:
perl -F, -lane '$c{$F[0]}++; END{for $k (sort keys %c){print $k if $c{$k} == 1}}' test.csv
These command-line options are used:
-n loop around each line of the input file
-l removes newlines before processing, and adds them back in afterwards
-a autosplit mode – split input lines into the #F array. Defaults to splitting on whitespace.
-e execute the perl code
-F autosplit modifier, in this case splits on ,
#F is the array of words in each line, indexed starting with $F[0]
I have a file that has many columns and I only need two of those columns. I am getting the columns I need using
cut -f 2-3 -d, file1.csv > file2.csv
The issue I am having is that the first column is ID and once it gets past 999 it becomes 1,000 and so it is treated as an extra column now. I cant get rid of all commas because I need them to separate the data. Is there a way to use sed to remove commas that only show up between 0-9?
I'd use a real CSV parser, and count backwards from the end of the line:
ruby -rcsv -ne '
row = $_.parse_csv
puts row[-5..-4].to_csv :force_quotes => true
' <<END
999,"someone#example.com","Doe, John","Doe","555-1212","address"
1,234,"email#email.com","name","lastname","phone","address"
END
"someone#example.com","Doe, John"
"email#email.com","name"
This works for the example in the comments:
awk -F'"?,"' '{print $2, $3}' file
The field separator is zero or one " followed by ,". This means that the comma in the first number doesn't count.
To separate the two fields with a comma instead of a space, you can change the OFS variable like this:
awk -F'"?,"' -v OFS=',' '{print $2, $3}' file
Or like this:
awk -F'"?,"' 'BEGIN{OFS=","}{print $2, $3}' file
Alternatively, if you want the quotes as well, you can use printf:
awk -F'"?,"' '{printf "\"%s\",\"%s\"\n", $2, $3}' file
From your comments, it sounds like there is a comma and a space (', ') pattern between tokens.
If this is the case, you can do this easily with sed. The strategy is to first replace all occurrences of , with some unique character sequence (like maybe ||).
's:, :||:g'
From there you can remove all commas:
's:,::g'
Finally, replace the double pipes with comma-space again.
's:||:, :g'
Putting it into one statement:
sed -i -e 's:, :||:g;s:,::g;s:||:, :g' your_odd_file.csv
And a command-line example to try before you buy:
bash$ sed -e 's:, :||:g;s:,::g;s:||:, :g' <<< "1,200,000, hello world, 123,456"
1200000, hello world, 123456
If you are in the unfortunate situation where there is not a space between fields in the CSV - you can attempt to 'fake it' by detecting changes in data type - like where there is a numeric field followed by a text field.
's:,\([^0-9]\):, \1:g' # numeric followed by non-numeric
's:\([^0-9]\),:\1, :g' # non-numeric field followed by something (anything)
You can put this all together into one statement, but you are venturing into dangerous waters here - this will definitely be a one-off solution and should be taken with a large grain of salt.
sed -e 's:,\([^0-9]\):, \1:g;s:\([^0-9]\),:\1, :g' \
-e 's:, :||:g;s:,::g;s:||:, :g' file1.csv > file2.csv
And another example:
bash$ sed -e 's:,\([^0-9]\):, \1:g;s:\([^0-9]\),:\1, :g' \
-e 's:, :||:g;s:,::g;s:||:, :g' <<< "1,200,000,hello world,123,456"
1200000, hello world, 123456
I have a file that is delimited by comma ",", but some rows have only one column, and some rows have multiple columns separated by ",". For example:
NM_001066
NM_015378,NM_018156
NM_001006624,NM_001006625,NM_006474,NM_198389
As you can see above, the third line has 4 columns delimited by ",", but I only need to get the first column in every line.
I tried to use awk: cat fileName.txt | awk '{print $1}', but it does not work. I am looking for help with this. Thank you!
I guess you're looking for this:
awk -F, '{print $1}' file.txt
-F, tells awk to use comma as the field separator.
In this simple case, the same thing is simpler with cut:
cut -f1 -d, file.txt
you are close:
awk -F, '{print $1}' file
or
awk -F, '$0=$1' file