I have a 3 column file. I would like to append a third column which is just one word repeated many times. I tried the following
paste file.tsv <(echo 'new_text') > new_file.tsv
But the text 'new_text' only appears on the first line, not every line.
How can I get 'new_text' to appear on every line.
Thanks
sed '1,$ s/$/;ABC/' infile > outfile
This replaces the line end ("$") with ";ABC".
Related
Can anyone help to format this text file(YYYYMMDD) as a date formatted(YYYY-MM-DD) text file using bash script or in Linux command line? I am not sure how to start editing 23millon lines!!!
I have YYYYMMDD format textfile :-
3515034013|50008|20140601|20240730
and I want to edit like YYYY-MM-DD formatted text file(Only 3rd and 4th fields need to be changed for 23million lines):-
3515034013|50008|2014-06-01|2024-07-30
I Want to convert from YYYYMMDD formatted text file to the YYYY-DD-MM format and I want to get specific lines from the text file based on the time period after this file manipulation which is the end goal.
The end goal is to format the 3rd field and 4th field as YYYY-MM-DD and also want to grep the line by date from that formatted text file:- 03rd field is the start date and the 04th field is the end date Let's say for example I need,
(01). The end date(04th field) before today i.e 2022-08-06 - all the old lines
(02). The end date(04th field) is 2 years from now i.e lines in between 2022-08-06th <-> 2024-08-06th?
Please note:- There are more than a 23million lines to edit and analyze based on the date.
How to approach this problem statement? which method is time efficient awk or sed or Bash line-by-line editing?
$ awk '
BEGIN { FS=OFS="|" }
{
for ( i=3; i<=4; i++ ) {
$i = substr($i,1,4) "-" substr($i,5,2) "-" substr($i,7)
}
print
}
' file
3515034013|50008|2014-06-01|2024-07-30
Here is a way to do it with sed. It has the same restrictions as steffens answer: | as fieldseparator and that all dates have the same format i.e. leading zeros in the month and date part.
sed -E 's/^(.*[|])([0-9]{4})([0-9]{2})([0-9]{2})[|]([0-9]{4})([0-9]{2})([0-9]{2})$/\1\2-\3-\4|\5-\6-\7/g'
Here is what the regular expression does:
^(.*[|]) captures the first part of the string from linestart (^) to a | into \1, this captures the first two columns, because the remaining part of the re matches the remaining part of the line up until lineend!
([0-9]{4})([0-9]{2})([0-9]{2})[|] captures the first date field parts into \2 to \4, notice the [|]
([0-9]{4})([0-9]{2})([0-9]{2})$ does the same for the second date column anchored at lineend ($) and captures the parts into \5 to \7, notice the $
the replacement part \1\2-\3-\4|\5-\6-\7 inserts - at the different places
the capturing into \n happens because of the use of (...) parens in the regular expression.
Here's one way to change the format with awk:
awk '{$3=substr($3,1,4) "-" substr($3,5,2) "-" substr($3,7,2); $4=substr($4,1,4) "-" substr($4,5,2) "-" substr($4,7,2); print}' FS='|' OFS='|'
It should work given that
| is only used for field separation
all dates have the same format
You can pipe the transformed lines to a new file or change it in place. Of course you can do the same with sed or ed. I'd go for awk because you'd be able to extract your specific lines just in the same run to an extra file.
This might work for you (GNU sed):
sed -E 's/^([^|]*\|[^|]*\|....)(..)(..\|....)(..)/\1-\2-\3-\4-/' file
Pattern match and insert - where desired.
Or if the file is only 4 columns:
sed -E 's/(..)(..\|....)(..)(..)$/-\1-\2-\3-\4/' file
I have the following text file:
.txt file
In the left hand column all the values are '0' is there a way to change only the left hand column to replace all the zeros with the value 15. I cant find all and replace as other columns contain '0' which cant be altered, this also cant be done manually as the file contains 10,000 lines. I'm wondering if this is possible from the command line or with a script.
Thanks
Using awk:
awk '$1 == 0 { $1 = 15 } 1' file.txt
Replaces the first column with 15 on each line only if the original value is 0.
I want to remove certain lines from a tab-delimited file and write output to a new file.
a b c 2017-09-20
a b c 2017-09-19
es fda d 2017-09-20
es fda d 2017-09-19
The 4th column is Date, basically I want to keep only lines that has 4th column as "2017-09-19" (keep line 2&4) and write to a new file. The new file should have same format as the raw file.
How to write the linux command for this example?
Note: The search criteria should be on the 4th field as I have other fields in the real data and possibly have same value as 4th field.
With awk:
awk 'BEGIN{OFS="\t"} $4=="2017-09-19"' file
OFS: output field separator, a space by default
Use grep to filter:
cat file.txt | grep '2017-09-19' > filtered_file.txt
This is not perfect, since the string 2017-09-19 is not required to appear in the 4th column, but if your file looks like the example, it'll work.
Sed solution:
sed -nr "/^([^\t]*\t){3}2017-09-19/p" input.txt >output.txt
this is:
-n - don't output every line
-r - extended regular expresion
/regexp/p - print line that contains regular expression regexp
^ - begin of line
(regexp){3} - repeat regexp 3 times
[^\t] - any character except tab
\t - tab character
* - repeat characters multiple times
2017-09-19 - search text
That is, skip 3 columns separated by a tab from the beginning of the line, and then check that the value of column 4 coincides with the required value.
awk '/2017-09-19/' file >newfile
cat newfile
a b c 2017-09-19
es fda d 2017-09-19
I have data such as below:
1493992429103289,207.55,207.5
1493992429103559,207.55,207.5
1493992429104353,207.55,207.5
1493992429104491,207.6,207.55
1493992429110551,207.55,207.5
Due to the nature of the last two columns, their values change throughout the day and their values are repeated regularly. By grouping the way outlined in my desired output (below), I am able to view each time there was a change in their values (with the enoch time in the first column). Is there a way to achieve the desired output shown below:
1493992429103289,207.55,207.5
1493992429104491,207.6,207.55
1493992429110551,207.55,207.5
So I consolidate the data by the second two columns. However, the consolidation is not completely unique (as can be seen by 207.55, 207.5 being repeated)
I have tried:
uniq -f 1
However the output gives only the first line and does not go on through the list
The awk solution below does not allow the occurrence which happened previously to be outputted again and so gives the output (below the awk code):
awk '!x[$2 $3]++'
1493992429103289,207.55,207.5
1493992429104491,207.6,207.55
I do not wish to sort the data by the second two columns. However, since the first is epoch time, it may be sorted by the first column.
You can't set delimiters with uniq, it has to be white space. With the help of tr you can
tr ',' ' ' <file | uniq -f1 | tr ' ' ','
1493992429103289,207.55,207.5
1493992429104491,207.6,207.55
1493992429110551,207.55,207.5
You can use an Awk statement as below,
awk 'BEGIN{FS=OFS=","} s != $2 && t != $3 {print} {s=$2;t=$3}' file
which produces the output as you need.
1493992429103289,207.55,207.5
1493992429104491,207.6,207.55
1493992429110551,207.55,207.5
The idea is to store the second and third column values in variables s and t respectively and print the line contents only if the current line is unique.
I found an answer which is not as elegant as Inian but satisfies my purpose.
Since my first column is always enoch time in microseconds and does not increase or decrease in characters, I can use the following uniq command:
uniq -s 17
You can try to manually (with a loop) compare current line with previous line.
previous_line=""
# start at first line
i=1
# suppress first column, that don't need to compare
sed 's#^[0-9][0-9]*,##' ./data_file > ./transform_data_file
# for all line within file without first column
for current_line in $(cat ./transform_data_file)
do
# if previous record line are same than current line
if [ "x$prev_line" == "x$current_line" ]
then
# record line number to supress after
echo $i >> ./line_to_be_suppress
fi
# record current line as previous line
prev_line=$current_line
# increment current number line
i=$(( i + 1 ))
done
# suppress lines
for line_to_suppress in $(tac ./line_to_be_suppress) ; do sed -i $line_to_suppress'd' ./data_file ; done
rm line_to_be_suppress
rm transform_data_file
Since your first field seems to have a fixed length of 18 characters (including the , delimiter), you could use the -s option of uniq, which would be more optimal for larger files:
uniq -s 18 file
Gives this output:
1493992429103289,207.55,207.5
1493992429104491,207.6,207.55
1493992429110551,207.55,207.5
From man uniq:
-f num
Ignore the first num fields in each input line when doing comparisons.
A field is a string of non-blank characters separated from adjacent fields by blanks.
Field numbers are one based, i.e., the first field is field one.
-s chars
Ignore the first chars characters in each input line when doing comparisons.
If specified in conjunction with the -f option, the first chars characters after
the first num fields will be ignored. Character numbers are one based,
i.e., the first character is character one.
I have a text file that has about 500 rows of information.
I am adding a few strings to the beginning of each line separated by a comma (Excel recognizes it as another column).
I have this code so far:
sed -e "2,$s#^# =HYPERLINK(B2,C2), https://otrs.city.pittsburgh.pa.us/index.pl?Action=AgentTicketZoom;TicketID=#"** C:\Users\hd\Desktop\newaction.txt > C:\Users\hd\Desktop\test.txt
I have a columns want. Once column is adding on a link to a previous column (easy enough)
Which will be a formula(string) in the first column is =HYPERLINK(B2,C2) and I want to increment the 2's to 3's,4's and so on.
Example:
=HYPERLINK(B2,C2)
=HYPERLINK(B3,C3)
=HYPERLINK(B4,C4)
=HYPERLINK(B5,C5)
=HYPERLINK(B6,C6)
It is my second day coding with sed and awk.
Is there any way I can make this happen using awk and sed?
This Perl one-liner:
perl -pe "BEGIN{$i = 2} s#^#=HYPERLINK(B${i},C${i})#; $i++" "input.txt"
will add =HYPERLINK(B2,C2) to the front of each line and increment the numbers each time.