Change date format in single column - bash

I have a file with a 1000 lines. The first column has a date in then European standard (DD.MM.YYYY). There are about 20 other columns with different data. The columns are separated by a semicolon. Here's a simplified example.
10.12.2020;name;address
18.12.2020;name2;address2
21.12.2020;name3;address3
What I want to do is to change the date format in the first column to YYYY.MM.DD
In this case the final line should read
2020.12.10;name;address
2020.12.18;name2;address2
2020.12.21;name3;address3
I tried to do it with a combination of awk and sed
awk -F';' 'BEGIN {OFS = FS} NR != 0 { sed 's/\([0-9]\{2\}\).\([0-9]\{2\}\).\([0-9]\{4\}\)/\3.\2.\1/g'; print; }'
which results in errors. There is probably a better way to do it with gsub but I wasn't able to understand the syntax.
Can anyone help me achieve this result? Could be with sed, gsub or any other way.

No need to combine sed and awk. Either is sufficient:
sed -E 's/^(..)\.(..)\.(....);/\3.\2.\1;/'
or
awk -F\; -v OFS=\; '{split($1,a,"."); $1=a[3] "." a[2] "." a[1]; print}'

Related

Extract the last three columns from a text file with awk

I have a .txt file like this:
ENST00000000442 64073050 64074640 64073208 64074651 ESRRA
ENST00000000233 127228399 127228552 ARF5
ENST00000003100 91763679 91763844 CYP51A1
I want to get only the last 3 columns of each line.
as you see some times there are some empty lines between 2 lines which must be ignored. here is the output that I want to make:
64073208 64074651 ESRRA
127228399 127228552 ARF5
91763679 91763844 CYP51A1
awk  '/a/ {print $1- "\t" $-2 "\t" $-3}'  file.txt.
it does not return what I want. do you know how to correct the command?
Following awk may help you in same.
awk 'NF{print $(NF-2),$(NF-1),$NF}' OFS="\t" Input_file
Output will be as follows.
64073208 64074651 ESRRA
127228399 127228552 ARF5
91763679 91763844 CYP51A1
EDIT: Adding explanation of command too now.(NOTE this following command is for only explanation purposes one should run above command only to get the results)
awk 'NF ###Checking here condition NF(where NF is a out of the box variable for awk which tells number of fields in a line of a Input_file which is being read).
###So checking here if a line is NOT NULL or having number of fields value, if yes then do following.
{
print $(NF-2),$(NF-1),$NF###Printing values of $(NF-2) which means 3rd last field from current line then $(NF-1) 2nd last field from line and $NF means last field of current line.
}
' OFS="\t" Input_file ###Setting OFS(output field separator) as TAB here and mentioning the Input_file here.
You can use sed too
sed -E '/^$/d;s/.*\t(([^\t]*[\t|$]){2})/\1/' infile
With some piping:
$ cat file | tr -s '\n' | rev | cut -f 1-3 | rev
64073208 64074651 ESRRA
127228399 127228552 ARF5
91763679 91763844 CYP51A1
First, cat the file to tr to squeeze out repeted \ns to get rid of empty lines. Then reverse the lines, cut the first three fields and reverse again. You could replace the useless cat with the first rev.

awk rounding the float numbers

My first line of file a.txt contains following and fields are separated by (,)
ab,b1,c,d,5.986627e738,e,5.986627e738
cd,g2,h,i,7.3423542344,j,7.3423542344
ef,l3,m,n,9.3124234323,o,9.3124234323
when I issue the below command
awk -F"," 'NR>-1{OFS=",";gsub($5,$5+10);OFS=",";print }' a.txt
it is printing
ab,b1,c,d,inf,e,inf
cd,g2,h,i,17.3424,j,17.3424
ef,l3,m,n,19.3124,o,19.3124
Here I have two issues
I asked awk to add 10 to only 5th column but it has added to 7th column as well due to duplicate entries
It is rounding up the numbers, instead, I need decimals to print as it is
How can I fix this?
awk 'BEGIN {FS=OFS=","}{$5=sprintf("%.10f", $5+10)}7' file
in your data, the $5 from line#1 has an e, so it was turned into 10.0000... in output.
you did substitution with gsub, therefore all occurrences will be replaced.
printf/sprintf should be considered to output in certain format.
tested with gawk
If you want to set the format in printf dynamically:
kent$ cat f
ab,b1,c,d,5.9866,e,5.986627e738
cd,g2,h,i,7.34235,j,7.3423542344
ef,l3,m,n,9.312423,o,9.3124234323
kent$ awk 'BEGIN {FS=OFS=","}{split($5,p,".");$5=sprintf("%.*f",length(p[2]), $5+10)}7' f
ab,b1,c,d,15.9866,e,5.986627e738
cd,g2,h,i,17.34235,j,7.3423542344
ef,l3,m,n,19.312423,o,9.3124234323
what you did is replacement on the whole record, what you really want to do is
awk 'BEGIN {FS=OFS=","}
{$5+=10}1' a.txt

How can I replace a character in a specific column? [duplicate]

I have a text file and I'm trying to replace a specific character (.) in the first column to another character (-). Every field is delimited by comma. Some of the lines have the last 3 columns empty, so they have 3 commas at the end.
Example of text file:
abc.def.ghi,123.4561.789,ABC,DEF,GHI
abc.def.ghq,124.4562.789,ABC,DEF,GHI
abc.def.ghw,125.4563.789,ABC,DEF,GHI
abc.def.ghe,126.4564.789,,,
abc.def.ghr,127.4565.789,,,
What I tried was using awk to replace '.' in the first column with '-', then print out the contents.
ETA: Tried out sarnold's suggestion and got the output I want.
ETA2: I could have a longer first column. Is there a way to change ONLY the first 3 '.' in the first column to '-', so I get the output
abc-def-ghi-qqq.www,123.4561.789,ABC,DEF,GHI
abc-def-ghq-qqq.www,124.4562.789,ABC,DEF,GHI
abc-def-ghw-qqq.www,125.4563.789,ABC,DEF,GHI
abc-def-ghe-qqq.www,126.4564.789,,,
abc-def-ghr-qqq.www,127.4565.789,,,
. is regexp notation for "any character". Escape it with \ and it means .:
$ awk -F, '{gsub(/\./,"-",$1); print}' textfile.csv
abc-def-ghi 123.4561.789 ABC DEF GHI
abc-def-ghq 124.4562.789 ABC DEF GHI
abc-def-ghw 125.4563.789 ABC DEF GHI
abc-def-ghe 126.4564.789
abc-def-ghr 127.4565.789
$
The output field separator is a space, by default. Set OFS = "," to set that:
$ awk -F, 'BEGIN {OFS=","} {gsub(/\./,"-",$1); print}' textfile.csv
abc-def-ghi,123.4561.789,ABC,DEF,GHI
abc-def-ghq,124.4562.789,ABC,DEF,GHI
abc-def-ghw,125.4563.789,ABC,DEF,GHI
abc-def-ghe,126.4564.789,,,
abc-def-ghr,127.4565.789,,,
This still allows changing multiple fields:
$ awk -F, 'BEGIN {OFS=","} {gsub(/\./,"-",$1); gsub("1", "#",$2); print}' textfile.csv
abc-def-ghi,#23.456#.789,ABC,DEF,GHI
abc-def-ghq,#24.4562.789,ABC,DEF,GHI
abc-def-ghw,#25.4563.789,ABC,DEF,GHI
abc-def-ghe,#26.4564.789,,,
abc-def-ghr,#27.4565.789,,,
I don't know what -OFS, does, but it isn't a supported command line option; using it to set the output field separator was a mistake on my part. Setting OFS within the awk program works well.
This might work for you:
awk -F, -vOFS=, '{for(n=1;n<=3;n++)sub(/\./,"-",$1)}1' file
abc-def-ghi-qqq.www,123.4561.789,ABC,DEF,GHI
abc-def-ghq-qqq.www,124.4562.789,ABC,DEF,GHI
abc-def-ghw-qqq.www,125.4563.789,ABC,DEF,GHI
abc-def-ghe-qqq.www,126.4564.789,,,
abc-def-ghr-qqq.www,127.4565.789,,,

How to print the csv file excluding first column till end using awk

I have a csv file with dynamic columns.
I've tried to use awk -F , 'NF>1' resul1.txt but it still prints all columns.
Since it has dynamic columns.
Its quite difficult to print using print $1 till end.
Try this awk command:
awk -F, '{$1=""}1' input.txt | awk -vOFS=, '{$1=$1}1' > output.txt
Make the 1st field empty
Print out entire line again
try substr function :
substr(string, start [, length ])
Return a length-character-long substring of string, starting at character number start. The first character of a string is character
number one.For example, substr("washington", 5, 3) returns "ing".*
awk -F, '{print substr($0,length($1)+1+length(FS))}' file
You can use cut:
cut -d',' -f2- yourfile.csv > output.csv
Explanation:
-d - setting delimiter to ,
-f - fields to print
2- - from 2 field to end of line
With awk:
awk -F, '{sub(/[^,]+,/,"",$0);}1' OFS=, yourfile.csv > output.csv
With sed:
sed -i.bak 's/^[^,]\+,//g' yourfile.csv
-i - in-place edit

Cut and replace bash

I have to process a file with data organized like this
AAAAA:BB:CCC:EEEE:DDDD
FF:III:JJJ:KK:LLL
MMMM:NN:OOO:PP
etc
Columns can have different length but lines always have the same number of columns.
I want to be able to cut a specific column of a given line and change it to the value I want.
For example I'd apply my command and change the file to
AAAAA:BB:XXXX:EEEE:DDDD
FF:III:JJJ:KK:LLL
MMMM:NN:OOO:PP
I know how to select a specific line with sed and then cut the field but I have no idea on how to replace the field with the value I have.
Thanks
Here's a way to do it with awk:
Going with your example, if you wanted to replace the 3rd field of the 1st line:
awk 'BEGIN{FS=OFS=":"} {if (NR==1) {$3 = "XXXX"}; print}' input_file
Input:
AAAAA:BB:CCC:EEEE:DDDD
FF:III:JJJ:KK:LLL
MMMM:NN:OOO:PP
Output:
AAAAA:BB:XXXX:EEEE:DDDD
FF:III:JJJ:KK:LLL
MMMM:NN:OOO:PP
Explanation:
awk: invoke the awk command
'...': everything enclosed by single-quotes are instructions to awk
BEGIN{FS=OFS=":"}: Use : as delimiters for both input and output. FS stands for Field Separator. OFS stands for Output Field Separator.
if (NR==1) {$3 = "XXXX"};: If Number of Records (NR) read so far is 1, then set the 3rd field ($3) to "XXXX".
print: print the current line
input_file: name of your input file.
If instead what you are trying to accomplish is simply replace all occurrences of CCC with XXXX in your file, simply do:
sed -i 's/CCC/XXXX/g` input_file
Note that this will also replace partial matches, such as ABCCCDD -> ABXXXXDD
This might work for you (GNU sed):
sed -r 's/^(([^:]*:?){2})CCC/\1XXXX/' file
or
awk -F: -vOFS=: '$3=="CCC"{$3="XXXX"};1' file

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