I have a dat file with some data in it seperated by '|' character.
I want to read each line and take out 5th and 6th column data( date format) from here and then validate the date using unix. I am new to unix . Please let me know how to do it.
Try using an awk script (http://www.manpagez.com/man/1/awk/) for parsing files. Something like
awk -F\| '{print $5, $6}' test.dat
This can then be extended to perform date validation, depending on what validation needs are. For example - http://unix.derkeiler.com/Newsgroups/comp.unix.shell/2003-08/1340.html seems to perform a reasonable amount of validation.
Related
We have a requirement where we have a csv file with custom delimiter '||' (double-pipes) . We have 40 columns in the file and the file size is approximately between 400 to 500 MB.
We need to sort the file based on 2 columns, first on column 4 and then by column 17.
We found this command using which we can sort for one column, but not able to find a command which can sort based on both columns.
Since we use a delimiter with 2 characters, we are using awk command for sorting.
Command:
awk -F \|\| '{print $4}' abc.csv | sort > output.csv
Please advise.
If your inputs are not too fancy (no newlines in the middle of a record, for instance), the sort utility can almost do what you want, but it supports only one-character field separators. So || would not work. But wait, if you do not have other | characters in your files, we could just consider | as the field separator and account for the extra empty fields:
sort -t'|' -k7 -k33 foo.csv
We sort by fields 7 (instead of 4) and then 33 (instead of 17) because of these extra empty fields. The formula that gives the new field number is simply 2*N-1 where N is the original field number.
If you do have | characters inside your fields a simple solution is to substitute them all by one unused character, sort, and restore the original ||. Example with tabs:
sed 's/||/\t/g' foo.csv | sort -t$'\t' -k4 -k17 | sed 's/\t/||/g'
If tab is also used in your fields chose any unused character instead. Form feed (\f) or the field separator (ASCII code 28, that is, replace the 3 \t with \x1c) are good candidates.
Using PROCINFO in gnu-awk you can use this solution to sort on multi-character delimiter:
awk -F '\\|\\|' '{a[$2,$17] = $0} END {
PROCINFO["sorted_in"]="#ind_str_asc"; for (i in a) print a[i]}' file.csv
You could try following awk code. Written as per your shown attempts only. Set OFS as |(this is putting | as output field separator in case you want it ,comma etc then change OFS value accordingly in program) and print 17th field also as per your requirement in awk program. In sort use 1st and 2nd fields to sort it(because now 4th and 17th fields have become 1st and 2nd fields respectively for sort).
awk -F'\\|\\|' -v OFS='\\|' '{print $4,$17}' abc.csv | sort -t'|' -k1.1 -k2.1 > output.csv
The sort command works on physical lines, which may or may not be acceptable. CSV files can contain quoted fields which contain newlines, which will throw off sort (and most other Unix line-oriented utilities; it's hard to write a correct Awk script for this scenario, too).
If you need to be able to manipulate arbitrary CSV files, probably look to a dedicated utility, or use a scripting language with proper CSV support. For example, assume you have a file like this:
Title,Number,Arbitrary text
"He said, ""Hello""",2,"There can be
newlines and
stuff"
No problem,1,Simple undramatic single-line CSV
In case it's not obvious, CSV is fundamentally just a text file, with some restrictions on how it can be formatted. To be valid CSV, every record should be comma-separated; any literal commas or newlines in the data needs to be quoted, and any literal quotes need to be doubled. There are many variations; different tools accept slightly different dialects. One common variation is TSV which uses tabs instead of commas as delimiters.
Here is a simple Python script which sorts the above file on the second field.
import csv
import sys
with open("test.csv", "r") as csvfile:
csvdata = csv.reader(csvfile)
lines = [line for line in csvdata]
titles = lines.pop(0) # comment out if you don't have a header
writer = csv.writer(sys.stdout)
writer.writerow(titles) # comment out if you don't have a header
writer.writerows(sorted(lines, key=lambda x: x[1]))
Using sys.stdout for output is slightly unconventional; obviously, adapt to suit your needs. The Python csv library documentation is obviously not designed primarily to be friendly for beginners, but it should not be impossible to figure out, and it's not hard to find examples of working code.
In Python, sorted() returns a copy of a list in sorted order. There is also sort() which sorts a list in-place. Both functions accept an optional keyword parameter to specify a custom sort order. To sort on the 4th and 17th fields, use
sorted(lines, key=lambda x: (x[3], x[16]))
(Python's indexing is zero-based, so [3] is the fourth element.)
To use | as a delimiter, specify delimiter='|' in the csv.reader() and csv.writer() calls. Unfortunately, Python doesn't easily let you use a multi-character delimiter, so you might have to preprocess the data to switch to a single-character delimiter which does not occur in the data, or properly quote the fields which contain the character you selected as your delimiter.
I have a bunch of different files.We have used "|" as delimeter All files contain a column titled CARDNO, but not necessarily in the same location in all of the files. I have a function called data_mask. I want to apply to CARDNO in all of the files to change them into NEWCARDNO.
I know that if I pass in the column number of CARDNO I can do this pretty simply, say it's the 3rd column in a 5 column file with something like:
awk -v column=$COLNUMBER '{print $1, $2, FUNCTION($column), $4, $5}' FILE
However, if all of my files have hundreds of columns and it's somewhere arbitrary in each file, this is incredibly tedious. I am looking for a way to do something along the lines of this:
awk -v column=$COLNUMBER '{print #All columns before $column, FUNCTION($column), #All columns after $column}' FILE
My function takes a string as an input and changes it into a new one. It takes the value of the column as an input, not the column number. Please suggest me Unix command which can pass the column value to the function and give the desired output.
Thanks in advance
If I understand your problem correctly, the first row of the file is the header and one of those columns is named CARDNO. If this is the case then you just search for the header in that file and process accordingly.
awk 'BEGIN{FS=OFS="|";c=1}
(NR==1){while($c != "CARDNO" && c<=NF) c++
if(c>NF) exit
$c="NEWCARDNO" }
(NR!=1){$c=FUNCTION($c)}
{print}' <file>
As per comment, if there is no header in the file, but you know per file, which column number it is, then you can simply do:
awk -v c=$column 'BEGIN{FS=OFS="|"}{$c=FUNCTION($c)}1' <file>
I want to mask only the 2nd column of the data.
Input:
First_name,second_name,phone_number
ram,prakash,96174535
hari,pallavi,98888234
anurag,aakash,82783784
Output Expected:
First_name,second_name,phone_number
ram,*******,96174535
hari,*******,98888234
anurag,******,82783784
The sed program will do this just fine:
sed '2,$s/,[^,]*,/,*****,/'
The 2,$ only operates on lines 2 through to the end of the file (to leave the header line alone) and the substitute command s/,[^,]*,/,*****,/ will replace anything between the first and second comma with the mask *****.
Note that I've specifically used a fixed number of asterisks in the replacement string. Whether you're hiding passwords or anonymising data (as seems to be the case here), you don't want to leak any information, including the size of the names being replaced.
If you really want to use the same number of characters as in the original data, and you also want to cater for the possibility of replacing multiple fields, you can use something like:
awk -F, 'BEGIN{OFS=FS}NR==1{print;next}{gsub(/./,"*",$2);gsub(/./,"*",$4);print}'
This will also leave the first line untouched but will anonymise columns two and four (albeit with the information leakage previously mentioned):
echo 'First_name,second_name,phone_number,other
ram,prakash,96174535,abc
hari,pallavi,98888234,def
anurag,aakash,82783784,g
bob,santamaria,124,xyzzy' | awk -F, 'BEGIN{OFS=FS}NR==1{print;next}{gsub(/./,"*",$2);gsub(/./,"*",$4);print}'
First_name,second_name,phone_number,other
ram,*******,96174535,***
hari,*******,98888234,***
anurag,******,82783784,*
bob,**********,124,*****
Doing multiple columns with full anonymising would entail using $2="*****" rather than the gsub (for both columns of course).
Another in awk. Using gsub to replace every char in $2 with an *:
$ awk 'BEGIN{FS=OFS=","}NR>1{gsub(/./,"*",$2)}1' file
First_name,second_name,phone_number
ram,*******,96174535
hari,*******,98888234
anurag,******,82783784
try following too once and let me know if this helps you.
awk -F"," 'NR>1{$2="*******"} 1' OFS=, Input_file
I've been going through an online UNIX course and have come across this question which I'm stuck on. Would appreciate any help!
You are provided with a set of files each one of which contains personal details about an individual. Each file is laid out in the following format, with one file per individual:
name:Niko Tanaka
age:41
occupation:Doctor
I know the answer has to be in the form:
n=$(awk -F: ' / /{print }' filename)
n=$(awk -F: '/name/{print $2}' infile)
Whatever is inside of / / are regular expressions. In this case you just want to match on the line that contains 'name'.
I have a file that is '|' delimited. One of the fields within the file is a time stamp. The field is in the following format: MM-dd-yyyy HH:mm:ss I'd like to be able to print to a file unique dates. I can use the cut command (cut -f1 -d'|' _file_name_ |sort|uniq) to extract unique dates. However, with the time portion of the field, I'm seeing hundreds of results. After I run the cut command, I'd like to take the substring of the first eleven characters to display unique dates. I tried using an awk command such as:
awk ' { print substr($1,1-11) }' | cut -f1 -d'|' _file_name_ |sort|uniq > _output_file_
I'm having no luck. Am I going about this the wrong way? Is there a more simple way of extracting the data I need. Any help would be appreciated.
cut -c1-11 will display characters 1-11 of each input line.
if the date is the first (space separated) field in the file, then the list of unique dates is just:
cut -f1 -d' ' filename | sort -u
Update: in addition to #shellter's correct answer, I'll just present an alternative to demonstrate other awk facilities:
awk '{split($10, a); date[a[1]]++} END {for (d in date) print d}' filename
You're all most there. This is based on the idea that the date time stamp is in field 1.
Edit : changed field to 10, also used -u option to sort instead of sep process with uniq
You don't need the cut, awk will do that for you.
awk -F"|" ' { print substr($10,1,11) }' _file_name_ |sort -u > _output_file_
I hope this helps.
P.S. as you appear to be a new user, if you get an answer that helps you please remember to mark it as accepted, or give it a + (or -) as a useful answer