I want to find and delete all lines that have a specific string, with a specific length, at a specific location.
One line in my data set looks something like this:
STRING 1234567 1234567 7654321 6543217 5432176
Notes:
Entries have field widths of 8
Identification numbers can be repeated in the same line
Identification numbers can be repeated on a different line, but at a different location - these lines should not be deleted
In this example, I want to find lines containing "1234567" located at column 17 and spanning to column 24 (i.e. the third field) and delete them. How can I do this with sed or awk?
I have used the following, but it deletes lines that I want to keep:
sed -i '/1234567/d' ./file_name.dat
Cheers!
You may use
sed -i '/^.\{17\}1234567/d' ./file_name.dat
Details
^ - start of a line
.{17} - any 17 chars
1234567 - a substring.
See the online sed demo:
s="STRING 1234567 1234567 7654321 6543217 5432176
STRING 1234567 5534567 7654321 6543217 5432176"
sed '/^.\{17\}1234567/d' <<< "$s"
# => STRING 1234567 5534567 7654321 6543217 5432176
with awk, print lines except the substring match.
$ awk 'substr($0,17,7)=="1234567"{next}1' file > output_file
or perhaps inverse logic is easier
$ awk 'substr($0,17,7)!="1234567"' file > output_file
Related
So, I have csv files to use with hledger, and last field of every row is the amount for that line transaction.
Lines are in the following format:
date1, date2, description, amount
With the amount format any length between 4 and 6 digits; now for some reason all amounts are missing the period before the last two digits.
Now: 1000
Should be: 10.00
Now: 25452
Should be: 254.52
How to add a '.' before the last two digits of all lines, preferably with sed/awk?
So the input file is:
16.12.2005,18.12.2005,ATM,2000
17.12.2005,18.12.2005,utility,12523
18.12.2005,20.12.2005,salary,459023
desired output
16.12.2005,18.12.2005,ATM,20.00
17.12.2005,18.12.2005,utility,125.23
18.12.2005,20.12.2005,salary,4590.23
Thanks
You could try:
awk -F , '{printf "%s,%s,%s,%-6.2f\n", $1, $2, $3, $4/100.0}'
You should always add a sample of your input file and of the output you want in your question.
In this input you provide, you will have to define what has to happen when the description field contains a ,, or if it is possible to have amount of less than 100 as input.
In function of your answer, I will need to adapt the code or not.
sed 's/..$/.&/'
......................
You can also use cut utility to get the desired output. In your case, you always want to add '.' before the last two digits. So essentially it can be thought as something like this:
Step 1: Get all the characters from the beginning till the last 2 characters.
Step 2: Get the last 2 characters from the end.
Step 3: Concatenate them with the character that you want ('.' in this case).
The corresponding command for each of the step is the following:
$ a='17.12.2005,18.12.2005,utility,12523'
$ b=`echo $a | rev | cut -c3- | rev`
$ c=`echo $a | rev | cut -c1-2 | rev`
$ echo $b"."$c
This would produce the output
17.12.2005,18.12.2005,utility,125.23
16.12.2005,18.12.2005,ATM,20.00
17.12.2005,18.12.2005,utility,125.23
18.12.2005,20.12.2005,salary,4590.23
awk -F, '{sub(/..$/,".& ")}1' file
I would like to split the below data to the expected output:
Raw Data:
931096|376601|1|ART|AT-2151780724|2151780724|2|102809198|I|CGM44I|MIL3VF03|52576377.3600|PENDING|MO|PEND-INFO|Pend ACS4R|N|N|N|N|N|NULL|NULL|NULL|NULL|NULL|NULL|NULL|NULL|NULL|NULL|N|NULL|NULL|N|system|NULL|NULL|52576377.3600|1317720|system|2020-02-13 02:00:42|0
931097|375789|1|AYT|AT-2151509210|2151509210|7|102614605|A|CTHGMI|OZF19|444006.6400|APPROVED|NULL|APPROVED|Approved|N|N|N|N|N|NULL|NULL|NULL|NULL|NULL|NULL|NULL|NULL|NULL|NULL|N|NULL|NULL|N|kg17718|NULL|NULL|0.0000|1317722|system|2020-02-13 02:00:43|0931098|375979|1|AHT|AT-2151780726|2151780726|2|102809199|I|CGMI|MILaesLF11|26312.0000|PENDING|MO|PEND-INFO|Pend ACRES|N|N|N|N|N|NULL|NULL|NULL|NULL|NULL|NULL|NULL|NULL|NULL|NULL|N|NULL|NULL|N|system|NULL|NULL|26312.0000|1317721|system|2020-02-13 02:00:43|0
931099|376572|1|AT|AT-2151399812|2151399812|5|102673999|I|CG2rMI|WEL44LF15|60991.6956|PENDING|MO|PEND-INFO|Pend ACERS|N|N|N|N|N|NULL|NULL|NULL|NULL|NULL|NULL|NULL|NULL|NULL|NULL|N|NULL|NULL|N|system|NULL|NULL|0.0000|1317723|system|2020-02-13 02:00:45|0
Expected Output:
931096|376601|1|ART|AT-2151780724|2151780724|2|102809198|I|CGM44I|MIL3VF03|52576377.3600|PENDING|MO|PEND-INFO|Pend ACS4R|N|N|N|N|N|NULL|NULL|NULL|NULL|NULL|NULL|NULL|NULL|NULL|NULL|N|NULL|NULL|N|system|NULL|NULL|52576377.3600|1317720|system|2020-02-13 02:00:42|0
931097|375789|1|AYT|AT-2151509210|2151509210|7|102614605|A|CTHGMI|OZF19|444006.6400|APPROVED|NULL|APPROVED|Approved|N|N|N|N|N|NULL|NULL|NULL|NULL|NULL|NULL|NULL|NULL|NULL|NULL|N|NULL|NULL|N|kg17718|NULL|NULL|0.0000|1317722|system|2020-02-13 02:00:43|0
931098|375979|1|AHT|AT-2151780726|2151780726|2|102809199|I|CGMI|MILaesLF11|26312.0000|PENDING|MO|PEND-INFO|Pend ACRES|N|N|N|N|N|NULL|NULL|NULL|NULL|NULL|NULL|NULL|NULL|NULL|NULL|N|NULL|NULL|N|system|NULL|NULL|26312.0000|1317721|system|2020-02-13 02:00:43|0
931099|376572|1|AT|AT-2151399812|2151399812|5|102673999|I|CG2rMI|WEL44LF15|60991.6956|PENDING|MO|PEND-INFO|Pend ACERS|N|N|N|N|N|NULL|NULL|NULL|NULL|NULL|NULL|NULL|NULL|NULL|NULL|N|NULL|NULL|N|system|NULL|NULL|0.0000|1317723|system|2020-02-13 02:00:45|0
Basically the \n character is getting lost sometimes in the data and the lines are getting merged. Sometimes more than 1 line gets merged as well (even the opposite happens but we can get to that later).
The data always has 43 columns | separated. The last but one column(42nd) always is a timestamp and the last column is usually 0 or 1.
Trying for the below approach:
If cols > 43
Split 44th column to add \n and print the remaining.
Repeat process until cols=43
echo "${curr}" | awk -F\| ' { if(NF > 43) {for(i=43;i<NF;i++) "sed '${NR}s/\(^0\)/\1\n/p' $i" }}' filename
less complex
awk 'BEGIN {FS=OFS="|"}
NF>43 {for(i=43;i<=NF;i+=42) {t=$i; $i=substr(t,1,1) ORS substr(t,2)}}1' file
931096|376601|1|ART|AT-2151780724|2151780724|2|102809198|I|CGM44I|MIL3VF03|52576377.3600|PENDING|MO|PEND-INFO|Pend ACS4R|N|N|N|N|N|NULL|NULL|NULL|NULL|NULL|NULL|NULL|NULL|NULL|NULL|N|NULL|NULL|N|system|NULL|NULL|52576377.3600|1317720|system|2020-02-13 02:00:42|0
931097|375789|1|AYT|AT-2151509210|2151509210|7|102614605|A|CTHGMI|OZF19|444006.6400|APPROVED|NULL|APPROVED|Approved|N|N|N|N|N|NULL|NULL|NULL|NULL|NULL|NULL|NULL|NULL|NULL|NULL|N|NULL|NULL|N|kg17718|NULL|NULL|0.0000|1317722|system|2020-02-13 02:00:43|0
931098|375979|1|AHT|AT-2151780726|2151780726|2|102809199|I|CGMI|MILaesLF11|26312.0000|PENDING|MO|PEND-INFO|Pend ACRES|N|N|N|N|N|NULL|NULL|NULL|NULL|NULL|NULL|NULL|NULL|NULL|NULL|N|NULL|NULL|N|system|NULL|NULL|26312.0000|1317721|system|2020-02-13 02:00:43|0
931099|376572|1|AT|AT-2151399812|2151399812|5|102673999|I|CG2rMI|WEL44LF15|60991.6956|PENDING|MO|PEND-INFO|Pend ACERS|N|N|N|N|N|NULL|NULL|NULL|NULL|NULL|NULL|NULL|NULL|NULL|NULL|N|NULL|NULL|N|system|NULL|NULL|0.0000|1317723|system|2020-02-13 02:00:45|0
following your spec
If cols > 43 Split 44th 43th column to add
\n and print the remaining. Repeat process until cols=43 the end.
The usual way with sed: write a regex that matches 43 | characters with anything in between and a digit. Then insert a newline after the matched string.
sed 's/[0-9]\{6\}\(|[^|]*\)\{41\}|[0-9]/&\n/g ; s/\n$//'
# ^^^^^^^ - remove the leftover newline
# ^ - the matched string
# ^^^^^ - trailing digit
# ^ - 42th pipe character
# ^^^^^^^^^^^^^^^^ - 41 fields with anything in between
# ^^^^^^^^^^ - leading 6 digits
tested on repl
Or maybe match 42 pipes with anything in front and a digit::
sed 's/\([^|]*|\)\{42\}[0-9]/&\n/g ; s/\n$//'
Or match a character after 42 pipes and a digit and insert a newline in between:
sed 's/\(\([^|]*|\)\{42\}[0-9]\)\(.\)/\1\n\3/g'
Could you please try following, written and tested with shown samples. This solution will take care of inserting new lines even if you have more than 1 occurrences present in your single line too.
awk '
match($0,/[0-9]{4}-[0-9]{2}-[0-9]{2} [0-9]{2}:[0-9]{2}:[0-9]{2}\|0/){
val=substr($0,RSTART+RLENGTH)
if(val){
num=gsub(/[0-9]{4}-[0-9]{2}-[0-9]{2} [0-9]{2}:[0-9]{2}:[0-9]{2}\|0/,"&")
while(++count<num){
sub(/[0-9]{4}-[0-9]{2}-[0-9]{2} [0-9]{2}:[0-9]{2}:[0-9]{2}\|0/,"&\n")
}
}
val=count=num=""
}
1
' Input_file
You don't trust the source of the data. Maybe it will add another | and the number of columns is wrong.
Another approach is guessing that you can trust the timestamp field.
So try to split the line when the field after the timestamp has more dan one character (and split after the first).
sed -E 's/([0-9]{4}-[0-9]{2}-[0-9]{2} [0-9]{2}:[0-9]{2}:[0-9]{2}\|.)(.)/\1\n\2/g' file
This might work for you (GNU sed):
sed 's/[^|]*/\n&/44;s/\(|.\)\([^|]*|\)\n/\1\n\2/;P;D' file
If there is a 44th field, insert a newline before it. Then remove that newline and insert it following the first character of the 43rd field. Print the first line, delete the first line and repeat.
apologies if this really basic stuff but i just started with awk
so i have an input file im piping into awk like below. format never changes (like below)
name: Jim
gender: male
age: 40
name: Joe
gender: female
age: 36
name: frank
gender: Male
age: 40
I'm trying to list all names where age is 40
I can find them like so
awk '$2 == "40" {print $2 }'
but cant figure out how to print the name
Could you please try following(I am driving as of now so couldn't test it).
awk '/^age/{if($NF==40){print val};val="";next} /^name/{val=$0}' Input_file
Explanation: 1st condition checking ^name if a line starts from it then store that line value in variable val. Then in other condition checking if a line starts from age; then checking uf that line's 2nd field is greater than 40 then print value if variable val and nullify it too.
Using gnu awk and set Record Selector to nothing makes it works with blocks.
awk -v RS="" '/age: 40/ {print $2}' file
Jim
frank
Some shorter awk versions of suspectus and RavinderSingh13 post
awk '/^name/{n=$2} /^age/ && $NF==40 {print n}' file
awk '/^name/{n=$2} /^age: 40/ {print n}' file
Jim
frank
If line starts with name, store the name in n
IF line starts with age and age is 40 print n
Awk knows the concept records and fields.
Files are split in records where consecutive records are split by the record separator RS. Each record is split in fields, where consecutive fields are split by the field separator FS.
By default, the record separator RS is set to be the <newline> character (\n) and thus each record is a line. The record separator has the following definition:
RS:
The first character of the string value of RS shall be the input record separator; a <newline> by default. If RS contains more than one character, the results are unspecified. If RS is null, then records are separated by sequences consisting of a <newline> plus one or more blank lines, leading or trailing blank lines shall not result in empty records at the beginning or end of the input, and a <newline> shall always be a field separator, no matter what the value of FS is.
So with the file format you give, we can define the records based on RS="".
So based on this, we can immediately list all records who have the line age: 40
$ awk 'BEGIN{RS="";ORS="\n\n"}/age: 40/
There are a couple of problems with the above line:
What if we have a person that is 400 yr old, he will be listed because the line /age: 400/ contains that the requested line.
What if we have a record with a typo stating age:40 or age : 40
What if our record has a line stating wage: 40 USD/min
To solve most of these problems, it is easier to work with well-defined fields in the record and build the key-value-pairs per record:
key value
---------------
name => Jim
gender => male
age => 40
and then, we can use this to select the requested information:
$ awk 'BEGIN{RS="";FS="\n"}
# build the record
{ delete rec;
for(i=1;i<=NF;++i) {
# find the first ":" and select key and value as substrings
j=index($i,":"); key=substr($i,1,j-1); value=substr($i,j+1)
# remove potential spaces from front and back
gsub(/(^[[:blank:]]*|[[:blank:]]$)/,key)
gsub(/(^[[:blank:]]*|[[:blank:]]$)/,value)
# store key-value pair
rec[key] = value
}
}
# select requested information and print
(rec["age"] == 40) { print rec["name"] }' file
This is not a one-liner, but it is robust. Furthermore, this method is fairly flexible and adaptable to make selections based on a more complex logic.
If you are not averse to using grep and the format is always the same:
cat filename | grep -B2 "age: 40" | grep -oP "(?<=name: ).*"
Jim
frank
awk -F':' '/^name/{name=$2} \
/^age/{if ($NF==40)print name}' input_file
I want to extract the value in each row of a file that comes after AA. I can do this like so:
awk -F'[;=|]' '{for(i=1;i<=NF;i++)if($i=="AA"){print toupper($(i+1));next}}'
This gives me the exact information I need and converts to uppercase, which is exactly what I want to do. How can I do this and then print the entire row with this altered value in its previous position? I am essentially trying to do a find and replace where the value is changed to uppercase.
EDIT:
Here is a sample input line:
11 128196 rs576393503 A G 100 PASS AC=453;AF=0.0904553;AN=5008;NS=2504;DP=5057;EAS_AF=0.0159;AMR_AF=0.0259;AFR_AF=0.3071;EUR_AF=0.006;SAS_AF=0.0072;AA=g|||;VT=SNP
and here is a how I would like the output to look:
11 128196 rs576393503 A G 100 PASS AC=453;AF=0.0904553;AN=5008;NS=2504;DP=5057;EAS_AF=0.0159;AMR_AF=0.0259;AFR_AF=0.3071;EUR_AF=0.006;SAS_AF=0.0072;AA=G|||;VT=SNP
All that has changed is the g after AA= is changed to uppercase.
Following awk may help you on same.
awk '
{
match($0,/AA=[^|]*/);
print substr($0,1,RSTART+2) toupper(substr($0,RSTART+3,RLENGTH-3)) substr($0,RSTART+RLENGTH)
}
' Input_file
With GNU sed and perl, using word boundaries
$ echo 'SAS_AF=0.0072;AA=g|||;VT=SNP' | sed 's/\bAA=[^;=|]*\b/\U&/'
SAS_AF=0.0072;AA=G|||;VT=SNP
$ echo 'SAS_AF=0.0072;AA=g|||;VT=SNP' | perl -pe 's/\bAA=[^;=|]*\b/\U$&/'
SAS_AF=0.0072;AA=G|||;VT=SNP
\U will uppercase string following it until end or \E or another case-modifier
use g modifier if there can be more than one match per line
I would like to search a file, using awk, to output rows that have a value commencing at a specific column number. e.g.
I looking for 979719 starting at column number 10:
moobaaraa**979719**
moobaaraa123456
moo**979719**123456
moobaaraa**979719**
moobaaraa123456
As you can see, there are no delimiters. It is a raw data text file. I would like to output rows 1 and 4. Not row 3 which does contain the pattern but not at the desired column number.
awk '/979719$/' file
moobaaraa979719
moobaaraa979719
An simple sed approach.
$ cat file
moobaaraa979719
moobaaraa123456
moo979719123456
moobaaraa979719
moobaaraa123456
Just search for a pattern, that end's up with 979719 and print the line:
$ sed -n '/^.*979719$/p' file
moobaaraa979719
moobaaraa979719
This code works:
awk 'length($1) == 9' FS="979719" raw-text-file
This code sets 979719 as the field separator, and checks whether the first field has a length of 9 characters. Then prints the line (as default action).
awk 'substr($0,10,6) == 979719' file
You can drop the ,6 if you want to search from the 10th char to the end of each line.