I have a csv and I would like to know how to replace newline by -, just in the brothers column, with bash:
name,brothers,age,adress
------------------------
john,"marc
peter
paul
alex",18,street
thomas,mike,20,place
Awk is perfect for this
awk -v RS='^$' -v ORS= '{while ( match($0,/"[^"]+"/,a) ) {gsub(/\n/," ",a[0]); print substr($0,1,RSTART-1) a[0]; $0=substr($0,RSTART+RLENGTH)} print}' your.csv
outputs:
me,brothers,age,adress
------------------------
john,"marc peter paul alex",18,street
thomas,mike,20,place
Ungainly combo of csvtool, sed, & bash:
csvtool pastecol 2 1- \
input.csv
<(csvtool col 2 input.csv | \
sed -n '/"/,/"/{:a;N;$!ba;s/\([^"]\)\n/\1-/g;};p') | \
csvtool trim r -
Output:
name,brothers,age,adress
------------------------
john,marc-peter-paul-alex,18,street
thomas,mike,20,place
Except for the sed part, it's not that bad. csvtool replaces column 2 with an edited copy. At the end it trims an extra comma that csvtool stuck in there.
Related
I have a file with thousands of rows. I want to print the rows which do not contain a period.
awk '{print$2}' file.txt | head
I have used this to print the column I am interested in, column 2 (The file only has two columns).
I have removed the head and then did
awk '{print$2}' file.txt | grep -v "." | head
But I only get blank lines not any actual values which is expected, I think it has included the spaces between the rows but I am not sure.
Is there an alternative command?
As suggested by Jim, I did-
awk '{print$2}' file.txt | grep -v "\." | head
However the number of lines is greater than before, is this expected? Also, my output is a list of numbers but with spaces in between them (Vertical), is this normal?
file.txt example below-
120.4 3
270.3 7.9
400.8 3.9
200.2 4
100.2 8.7
300.2 3.4
102.3 6
49.0 2.3
38.0 1.2
So the expected (and correct) output would be 3 lines, as there is 3 values in column 2 without the period:
$ awk '{print$2}' file.txt | grep -v "\." | head
3
4
6
However, when running the code as above, I instead get 5, which is also counting the spaces between the rows I think:
$ awk '{print$2}' file.txt | grep -v "\." | head
3
4
6
You seldom need to use grep if you're already using awk
This would print the second column on each line where that second column doesn't contain a dot:
awk '$2 !~ /\./ {print $2}'
But you also wanted to skip empty lines, or perhaps ones where the second column is not empty. So just test for that, too:
awk '$2 != "" && $2 !~ /\./ {print $2}'
(A more amusing version would be awk '$2 ~ /./ && $2 !~ /\./ {print $2}' )
As you said, grep -v "." gives you only blank lines. That's because the dot means "any character", and with -v, the only lines printed are those that don't contain, well, any characters.
grep is interpreting the dot as a regex metacharacter (the dot will match any single character). Try escaping it with a backslash:
awk '{print$2}' file.txt | grep -v "\." | head
If I understand well, you can try this sed
sed ':A;N;${s/.*/&\n/};/\n$/!bA;s/\n/ /g;s/\([^ ]*\.[^ ]* \)//g' file.txt
output
3
4
6
I have a text file /tmp/some.txt with below values
JOHN YES 6 6 2345762
SHAUN NO 6 6 2345748
I want to create a csv file with below format (i.e based on rows. NOT based on columns).
JOHN,YES,6,6,2345762
SHAUN,NO,6,6,2345748
i tried below code
for i in `wc -l /tmp/some.txt | awk '{print $1}'`
do
awk 'NR==$i' /tmp/some.txt | awk '{print $1","$2","$3","$4","$5}' >> /tmp/some.csv
done
here wc -l /tmp/some.txt | awk '{print $1}' will get the value as 2 (i.e 2 rows in text file).
and for each row awk 'NR==$i' /tmp/some.txt | awk '{print $1","$2","$3","$4","$5}' will print the 5 fields into some.csvfile which is separated by comma.
when i execute each command separately it will work. but when i make it as a shell script i'm getting empty some.csv file.
#Kart: Could you please try following.
awk '{$1=$1;} 1' OFS=, Input_file > output.csv
I hope this helps you.
I suggest:
sed 's/[[:space:]]\+/,/g' /tmp/some.txt
You almost got it. awk already process the file row by row, so you don't need to iterate with the for loop.
So you just need to run:
awk '{print $1","$2","$3","$4","$5}' /tmp/some.txt >> /tmp/some.csv
With tr, squeezing (-s), and then transliterating space/tab ([:blank:]):
tr -s '[:blank:]' ',' <file.txt
With sed, substituting one or more space/tab with ,:
sed 's/[[:blank:]]\+/,/g' file.txt
With awk, replacing one ore more space/tab with , using gsub() function:
awk 'gsub("[[:blank:]]+", ",", $0)' file.txt
Example
% cat foo.txt
JOHN YES 6 6 2345762
SHAUN NO 6 6 2345748
% tr -s '[:blank:]' ',' <foo.txt
JOHN,YES,6,6,2345762
SHAUN,NO,6,6,2345748
% sed 's/[[:blank:]]\+/,/g' foo.txt
JOHN,YES,6,6,2345762
SHAUN,NO,6,6,2345748
% awk 'gsub("[[:blank:]]+", ",", $0)' foo.txt
JOHN,YES,6,6,2345762
SHAUN,NO,6,6,2345748
Given input such as:
1
1a
1.1b
2.0c
How to extract the integer/decimal number at beginning of each input line, using only Linux/Unix command line utilities?
Using awk, you could say:
awk '{print $0+0}'
Awk is available in Linux, BSD, and many other Unix-like operating systems. It helps in this way:
echo "1" | awk '{a+=$0; print a}' # output 1
echo "1a" | awk '{a+=$0; print a}' # output 1
echo "1.1b" | awk '{a+=$0; print a}' # output 1.1
echo "2.0c" | awk '{a+=$0; print a}' # output 2
Some more awk
For extracting only digits
$ awk 'gsub(/[[:alpha:]].*/,x,$1) + 1' << EOF
1
1a
1.1b
2.0c
EOF
1
1
1.1
2.0
For integer
$ awk '{print int($0)}' << EOF
1
1a
1.1b
2.0c
EOF
1
1
1
2
---edit---
If there is any blank line in file, you can avoid printing zero from following
$ awk 'NF{$0+=0}1' << EOF
1
1a
1.1b
2foot4c
2
EOF
1
1
1.1
2
2
Here is a way to do this with sed:
echo "12.3abc" | sed -n 's/^\([0-9.][0-9.]*\).*/\1/p'
Output:
12.3
The block in parentheses matches all numbers or periods '.' that occur at the beginning of the line. Everything after that is match by the '.*'.
The \1 says to replace the entire line with just the portion that was matched in the parentheses.
Assuming your version of grep supports -o:
grep -o '^[0-9.]\+' data.in
NB: This will match any sequence of digits and decimal points at the start of the line.
I have a string
ABCDEFGHIJ
I would like it to print.
A
B
C
D
E
F
G
H
I
J
ie horizontal, no editing between characters to vertical. Bonus points for how to put a number next to each one with a single line. It'd be nice if this were an awk or shell script, but I am open to learning new things. :) Thanks!
If you just want to convert a string to one-char-per-line, you just need to tell awk that each input character is a separate field and that each output field should be separated by a newline and then recompile each record by assigning a field to itself:
awk -v FS= -v OFS='\n' '{$1=$1}1'
e.g.:
$ echo "ABCDEFGHIJ" | awk -v FS= -v OFS='\n' '{$1=$1}1'
A
B
C
D
E
F
G
H
I
J
and if you want field numbers next to each character, see #Kent's solution or pipe to cat -n.
The sed solution you posted is non-portable and will fail with some seds on some OSs, and it will add an undesirable blank line to the end of your sed output which will then become a trailing line number after your pipe to cat -n so it's not a good alternative. You should accept #Kent's answer.
awk one-liner:
awk 'BEGIN{FS=""}{for(i=1;i<=NF;i++)print i,$i}'
test :
kent$ echo "ABCDEF"|awk 'BEGIN{FS=""}{for(i=1;i<=NF;i++)print i,$i}'
1 A
2 B
3 C
4 D
5 E
6 F
So I figured this one out on my own with sed.
sed 's/./&\n/g' horiz.txt > vert.txt
One more awk
echo "ABCDEFGHIJ" | awk '{gsub(/./,"&\n")}1'
A
B
C
D
E
F
G
H
I
J
This might work for you (GNU sed):
sed 's/\B/\n/g' <<<ABCDEFGHIJ
for line numbers:
sed 's/\B/\n/g' <<<ABCDEFGHIJ | sed = | sed 'N;y/\n/ /'
or:
sed 's/\B/\n/g' <<<ABCDEFGHIJ | cat -n
I want to find the last appearance of a string in a text file with linux commands. For example
1 a 1
2 a 2
3 a 3
1 b 1
2 b 2
3 b 3
1 c 1
2 c 2
3 c 3
In such a text file, i want to find the line number of the last appearance of b which is 6.
I can find the first appearance with
awk '/ b / {print NR;exit}' textFile.txt
but I have no idea how to do it for the last occurrence.
cat -n textfile.txt | grep " b " | tail -1 | cut -f 1
cat -n prints the file to STDOUT prepending line numbers.
grep greps out all lines containing "b" (you can use egrep for more advanced patterns or fgrep for faster grep of fixed strings)
tail -1 prints last line of those lines containing "b"
cut -f 1 prints first column, which is line # from cat -n
Or you can use Perl if you wish (It's very similar to what you'd do in awk, but frankly, I personally don't ever use awk if I have Perl handy - Perl supports 100% of what awk can do, by design, as 1-liners - YMMV):
perl -ne '{$n=$. if / b /} END {print "$n\n"}' textfile.txt
This can work:
$ awk '{if ($2~"b") a=NR} END{print a}' your_file
We check every second file being "b" and we record the number of line. It is appended, so by the time we finish reading the file, it will be the last one.
Test:
$ awk '{if ($2~"b") a=NR} END{print a}' your_file
6
Update based on sudo_O advise:
$ awk '{if ($2=="b") a=NR} END{print a}' your_file
to avoid having some abc in 2nd field.
It is also valid this one (shorter, I keep the one above because it is the one I thought :D):
$ awk '$2=="b" {a=NR} END{print a}' your_file
Another approach if $2 is always grouped (may be more efficient then waiting until the end):
awk 'NR==1||$2=="b",$2=="b"{next} {print NR-1; exit}' file
or
awk '$2=="b"{f=1} f==1 && $2!="b" {print NR-1; exit}' file