I'm attempting to use awk one liner to print lines of a file in which the substring is less than a defined variable. Also the line must start with the letter E. The E condition is working, but not the result for the simple if 'less than' I'm looking for. What am I doing wrong here?? It is incorporated into a larger bash script. Thanks in advance.
#!/bin/bash
minimum_dpt=50
awk -v depth="$minimum_dpt" '{if (/^E/ && int(substr($0,65,6)<depth)) print "Shot: ",substr($0,21,5)," has depth below minimum. Value: ",substr($0,65,6)'}
Input:
E1985020687 1 1 2942984632.99S 88 354.60E 596044.16185585.10000.9 44 826 9
E1985020687 1 1 2943264732.95S 88 359.24E 595917.26185461.80000.5 44 82727
E1985020687 1 1 2944264741.97S 88 450.86E 594520.36185751.92445.3 44 82846
E1985020687 1 1 2945264741.97S 88 450.86E 594520.36185751.90045.3 44 82846
Output:
Shot: 2942 has depth below minimum. Value: 0000.9
Shot: 2943 has depth below minimum. Value: 0000.5
Shot: 2945 has depth below minimum. Value: 0045.3
You probably intended:
int(substr($0,65,6))<depth
or even just:
(substr($0,65,6)+0)<depth
instead of what you have:
int(substr($0,65,6)<depth)
There's probably a better way to do this but without seeing your input and output idk...
A possible solution for the task like that:
$ cat input
102030405060
102030405060
203050601070
904050308090
104030607040
406080903040
$ awk -v dpt=50 '/^1/ && (int(substr($0, 9, 2)) > int(dpt))' <input
104030607040
(edited according to Ed's comment, thanks ;)
Related
I have a file that looks like this:
18 29 293.434
12 32 9954.343
12 343 .
12 45 9493.545
I want to replace the "." on line 3 with "0" using sed.
I can't figure out how to do this without replacing every decimal place in the file.
Any help would be much appreciated.
With GNU sed:
sed 's/\B\.\B/0/g' file
Output:
18 29 293.434
12 32 9954.343
12 343 0
12 45 9493.545
See: \B: non-word boundary
Using any awk in any shell on every UNIX box and assuming you want to replace . with 0 wherever it occurs alone in any field rather than just if it's in the last field of the 3rd line of input:
awk '{for (i=1; i<=NF; i++) if ($i == ".") $i=0} 1' file
If I understand well, if the dot is at the end, remove spaces and substitute this dot with 0:
sed '/\.$/{s/ //g;s/\.$/0/}' file
# ^
$ is an anchor, it means at the end of the line
18 29 293.434
12 32 9954.343
123430
12 45 9493.545
EDITS: For reference, "stuff" is a general variable, as is "KEEP".
KEEP could be "Hi, my name is Dave" on line 2 and "I love pie" on line 7. The numbers I've put here are for illustration only and DO NOT show up in the data.
I had a file that needed to be parsed, keeping every 4th line, starting at the 3rd line. In other words, it looked like this:
1 stuff
2 stuff
3 KEEP
4
5 stuff
6 stuff
7 KEEP
8 stuff etc...
Great, sed solved that easily with:
sed -n -e 3~4p myfile
giving me
3 KEEP
7 KEEP
11 KEEP
Now I have a different file format and a different take on the pattern:
1 stuff
2 KEEP
3 KEEP
4
5 stuff
6 KEEP
7 KEEP etc...
and I still want the output of
2 KEEP
3 KEEP
6 KEEP
7 KEEP
10 KEEP
11 KEEP
Here's the problem - this is a multi-pattern "pattern" for sed. It's "every 4th line, spit out 2 lines, but start at line 2".
Do I need to have some sort of DO/FOR loop in my sed, or do I need a different command like awk or grep? Thus far, I have tried formats like:
sed -n -e '3~4p;4~4p' myfile
and
awk 'NR % 3 == 0 || NR % 4 ==0' myfile
and
sed -n -e '3~1p;4~4p' myfile
and
awk 'NR % 1 == 0 || NR % 4 ==0' myfile
source: https://superuser.com/questions/396536/how-to-keep-only-every-nth-line-of-a-file
If your intent is to print lines 2,3 then every fourth line after those two, you can do:
$ seq 20 | awk 'BEGIN{e[2];e[3]} (NR%4) in e'
2
3
6
7
10
11
14
15
18
19
You were pretty close with your sed:
$ printf '%s\n' {1..12} | sed -n '2~4p;3~4p'
2
3
6
7
10
11
this is the idiomatic way to write in awk
$ awk 'NR%4==2 || NR%4==3' file
however, this special case can be shortened to
$ awk 'NR%4>1' file
This might work for you (GNU sed):
sed '2~4,+1p;d' file
Use a range, the first parameter is the starting line and modulus (in this case from line 2 modulus 4). The second parameter is how man lines following the start of the range (in this case plus one). Print these lines and delete all others.
In the generic case, you want to keep lines p to p+q and p+n to p+q+n and p+2n to p+q+2n ... So you can write:
awk '(NR - p) % n <= q'
I have a file which contain numbers, say 1 to 300. But the numbers are not continuous. A sample file looks like this
042
043
044
045
078
198
199
200
201
202
203
212
213
214
215
238
239
240
241
242
256
257
258
Now I need to check the continuity of the number series and accordingly write out the output. For example the first 4 numbers are in series, so the output should be
042-045
Next, 078 is a lone number, so the output should be
078
for convenience it can be made to look like
078-078
Then 198 to 203 are continuous. So, next output should be
198-203
and so on. The final output should be like
042-045
078-078
198-203
212-215
238-242
256-258
I just need to know the first and end member of the continuous series and jump on the next series when discontinuity is encountered; The output can be manipulated. I am inclined to use the if statement and can think of a complicated thing like this
num=`cat file | wc -l`
out1=`head -1 file`
for ((i=2;i<=$num;i++))
do
j=`echo $i-1 | bc`
var1=`cat file | awk 'NR='$j'{print}'`
var2=`cat file | awk 'NR='$i'{print}'`
var3=`echo $var2 - $var1 | bc`
if [ $var3 -gt 1 ]
then
out2=$var1
echo $out1-$out2
out1=$var2
fi
done
which works but too lengthy. I am sure there is definitely a short way of doing this.
I am also open to other straight-forward command (or few commands) in shell, awk or a few lines of fortran code that can do it.
Thanking you in anticipation.
This awk one-liner works for given example:
awk 'p+1!=$1{printf "%s%s--",NR==1?"":p"\n",$1}{p=$1}END{print $1}' file
It gives the output for your data as input:
042--045
078--078
198--203
212--215
238--242
256--258
Here is a simple program in Fortran:
program test
implicit none
integer :: first, last, uFile, i, stat
open( file='numbers.txt', newunit=uFile, action='read', status='old' )
read(uFile,*,iostat=stat) i
if ( stat /= 0 ) stop
first = i ; last = i
do
read(uFile,*,iostat=stat) i
if ( stat /= 0 ) exit
if ( i == last+1 ) then
last = i
else
print *,first,'-',last
write(*,'(i3.3,a,i3.3)') first,'-',last
endif
enddo
write(*,'(i3.3,a,i3.3)') first,'-',last
end program
The output is
042-045
078-078
198-203
212-215
238-242
256-258
I have a question. I have a file with coordinates (TAB separated)
2 10
35 50
90 200
400 10000
...
I would like to substract the first column of the second line from the second column of the fist line , i.e. calculate the distance, i.e. I would like a file with
25
40
200
...
How could I do that using awk???
Thank you very much in advance
here is an awk one-liner may help you:
kent$ awk 'a{print $1-a}{a=$2}' file
25
40
200
Here's a pure bash solution:
{
read _ ps
while read f s; do
echo $((f-ps))
((ps=s))
done
} < input_file
This only works if you have (small) integers, as it uses bash's arithmetic. If you want to deal with arbitrary sized integers or floats, you can use bc (with only one fork):
{
read _ ps
while read f s; do
printf '%s-%s\n' "$f" "$ps"
ps=$s
done
} < input_file | bc
Now I leave the others give an awk answer!
Alright, since nobody wants to upvote my answer, here's a really funny solution that uses bash and bc:
a=( $(<input_file) )
printf -- '-(%s)+(%s);\n' "${a[#]:1:${#a[#]}-2}" | bc
or the same with dc (shorter but doesn't work with negative numbers):
a=( $(<input_file) )
printf '%s %sr-pc' "${a[#]:1:${#a[#]}-2}" | dc
using sed and ksh for evaluation
sed -n "
1x
1!H
$ !b
x
s/^ *[0-9]\{1,\} \(.*\) [0-9]\{1,\} *\n* *$/\1 /
s/\([0-9]\{1,\}\)\(\n\)\([0-9]\{1,\}\) /echo \$((\3 - \1))\2/g
s/\n *$//
w /tmp/Evaluate.me
"
. /tmp/Evaluate.me
rm /tmp/Evaluate.me
I am trying to convert all negative numbers to positive numbers and have so far come up with this
echo "-32 45 -45 -72" | sed -re 's/\-([0-9])([0-9])\ /\1\2/p'
but it is not working as it outputs:
3245 -45 -72
I thought by using \1\2 I would have got the positive number back ?
Where am I going wrong ?
Why not just remove the -'s?
[root#vm ~]# echo "-32 45 -45 -72" | sed 's/-//g'
32 45 45 72
My first thought is not using sed, if you don't have to. awk can understand that they're numbers and convert them thusly:
echo "-32 45 -45 -72" | awk -vRS=" " -vORS=" " '{ print ($1 < 0) ? ($1 * -1) : $1 }'
-vRS sets the "record separator" to a space, and -vORS sets the "output record separator" to a space. Then it simply checks each value, sees if it's less than 0, and multiplies it by -1 if it is, and if it's not, just prints the number.
In my opinion, if you don't have to use sed, this is more "correct," since it treats numbers like numbers.
This might work for you:
echo "-32 45 -45 -72" | sed 's/-\([0-9]\+\)/\1/g'
Reason why your regex is failing is
Your only doing a single substitution (no g)
Your replacement has no space at the end.
The last number has no space following so it will always fail.
This would work too but less elegantly (and only for 2 digit numbers):
echo "-32 45 -45 -72" | sed -rn 's/-([0-9])([0-9])(\s?)/\1\2\3/gp'
Of course for this example only:
echo "-32 45 -45 -72" | tr -d '-'
You are dealing with numbers as with a string of characters. More appropriate would be to store numbers in an array and use built in Shell Parameter Expansion to remove the minus sign:
[~] $ # Creating and array with an arbitrary name:
[~] $ array17=(-32 45 -45 -72)
[~] $ # Calling all elements of the array and removing the first minus sign:
[~] $ echo ${array17[*]/-}
32 45 45 72
[~] $