I have values such as
B146XYZ,
G638XYZ,
G488xBC
I have to write a bash script where when it sees comma it has to remove the comma and add 7 spaces to it and also if it sees comma and a space or just space(no punctuations) it has to add 7 spaces to make all of them fixed length.
if [[ $row = *’,’* ]]
then
first= “${ row%%,*}”
echo “${first } “
I tried but can’t understand how to add conditions for the remaining criteria specially struggling with single value conditions such as G488xBC
What about just:
sed -E 's/[, ]+/ /g' file
Or something like this will print a padded table, so long as no field is longer than 13 characters:
awk -F '[,[:space:]]+' \
'{
for (i=1; i<NF; i++) {
printf("%-14s", $i)
}
print $NF
}'
Or the same thing in pure bash:
while IFS=$', \t' read -ra vals; do
last=$((${#vals[#]} - 1))
for ((i=0; i<last; i++)); do
printf "%-14s" "${vals[i]}"
done
printf '%s\n' "${vals[last]}"
done
newrow="${row//,/ }"
VALUES=`echo $VALUES | sed 's/,/ /g' | xargs`
The sed command will replace the comma with a single space.
The xargs will consolidate any number of whitespaces into a single space.
With that you now have your values in space separated string instead of comma, separated by unknown number of whitespaces.
From there you can use for i in $VALUES; do printf "$i\t"; done
Using the tab character like above will give you aligned output in case your values may be different in length.
But if your values are always same length then you can make it a bit more simple by doing
VALUES=`echo $VALUES | sed 's/,/ /g' | xargs | sed 's/1 space/7 spaces/g'`
echo $VALUES
Related
I have a text file FILENAME. I want to split the string at - of the first column field and extract the last element from each line. Here "$(echo $line | cut -d, -f1 | cut -d- -f4)"; alone is not giving me the right result.
FILENAME:
TWEH-201902_Pau_EX_21-1195060301,15cef8a046fe449081d6fa061b5b45cb.final.cram
TWEH-201902_Pau_EX_22-1195060302,25037f17ba7143c78e4c5a475ee98e25.final.cram
TWEH-201902_Pau_T-1383-1195060311,267364a6767240afab2b646deec17a34.final.cram
code I tried:
while read line; do \
DNA="$(echo $line | cut -d, -f1 | cut -d- -f4)";
echo $DNA
done < ${FILENAME}
Result I want
1195060301
1195060302
1195060311
Would you please try the following:
while IFS=, read -r f1 _; do # set field separator to ",", assigns f1 to the 1st field and _ to the rest
dna=${f1##*-} # removes everything before the rightmost "-" from "$f1"
echo "$dna"
done < "$FILENAME"
Well, I had to do with the two lines of codes. May be someone has a better approach.
while read line; do \
DNA="$(echo $line| cut -d, -f1| rev)"
DNA="$(echo $DNA| cut -d- -f1 | rev)"
echo $DNA
done < ${FILENAME}
I do not know the constraints on your input file, but if what you are looking for is a 10-digit number, and there is only ever one 10-digit number per line... This should do niceley
grep -Eo '[0-9]{10,}' input.txt
1195060301
1195060302
1195060311
This essentially says: Show me all 10 digit numbers in this file
input.txt
TWEH-201902_Pau_EX_21-1195060301,15cef8a046fe449081d6fa061b5b45cb.final.cram
TWEH-201902_Pau_EX_22-1195060302,25037f17ba7143c78e4c5a475ee98e25.final.cram
TWEH-201902_Pau_T-1383-1195060311,267364a6767240afab2b646deec17a34.final.cram
A sed approach:
sed -nE 's/.*-([[:digit:]]+)\,.*/\1/p' input_file
sed options:
-n: Do not print the whole file back, but only explicit /p.
-E: Use Extend Regex without need to escape its grammar.
sed Extended REgex:
's/.*-([[:digit:]]+)\,.*/\1/p': Search, capture one or more digit in group 1, preceded by anything and a dash, followed by a comma and anything, and print only the captured group.
Using awk:
awk -F[,] '{ split($1,arr,"-");print arr[length(arr)] }' FILENAME
Using , as a separator, take the first delimited "piece" of data and further split it into an arr using - as the delimiter and awk's split function. We then print the last index of arr.
I have a file consisting of multiple rows like this
10|EQU000000001|12345678|3456||EOMCO042|EOMCO042|31DEC2018|16:51:17|31DEC2018|SHOP NO.5,6,7 RUNWAL GRCHEMBUR MHIN|0000000010000.00|6761857316|508998|6011|GL
I have to split and replace the column 11 into 4 different columns using the count of character.
This is the 11th column containing extra spaces also.
SHOP NO.5,6,7 RUNWAL GRCHEMBUR MHIN
This is I have done
ls *.txt *.TXT| while read line
do
subName="$(cut -d'.' -f1 <<<"$line")"
awk -F"|" '{ "echo -n "$11" | cut -c1-23" | getline ton;
"echo -n "$11" | cut -c24-36" | getline city;
"echo -n "$11" | cut -c37-38" | getline state;
"echo -n "$11" | cut -c39-40" | getline country;
$11=ton"|"city"|"state"|"country; print $0
}' OFS="|" $line > $subName$output
done
But while doing echo of 11th column, its trimming the extra spaces which leads to mismatch in count of character. Is there any way to echo without trimming spaces ?
Actual output
10|EQU000000001|12345678|3456||EOMCO042|EOMCO042|31DEC2018|16:51:17|31DEC2018|SHOP NO.5,6,7 RUNWAL GR|CHEMBUR MHIN|||0000000010000.00|6761857316|508998|6011|GL
Expected Output
10|EQU000000001|12345678|3456||EOMCO042|EOMCO042|31DEC2018|16:51:17|31DEC2018|SHOP NO.5,6,7 RUNWAL GR|CHEMBUR|MH|IN|0000000010000.00|6761857316|508998|6011|GL
The least annoying way to code this that I've found so far is:
perl -F'\|' -lane '$F[10] = join "|", unpack "a23 A13 a2 a2", $F[10]; print join "|", #F'
It's fairly straightforward:
Iterate over lines of input; split each line on | and put the fields in #F.
For the 11th field ($F[10]), split it into fixed-width subfields using unpack (and trim trailing spaces from the second field (A instead of a)).
Reassemble subfields by joining with |.
Reassemble the whole line by joining with | and printing it.
I haven't benchmarked it in any way, but it's likely much faster than the original code that spawns multiple shell and cut processes per input line because it's all done in one process.
A complete solution would wrap it in a shell loop:
for file in *.txt *.TXT; do
outfile="${file%.*}$output"
perl -F'\|' -lane '...' "$file" > "$outfile"
done
Or if you don't need to trim the .txt part (and you don't have too many files to fit on the command line):
perl -i.out -F'\|' -lane '...' *.txt *.TXT
This simply places the output for each input file foo.txt in foo.txt.out.
A pure-bash implementation of all this logic
#!/usr/bin/env bash
shopt -s nocaseglob extglob
for f in *.txt; do
subName=${f%.*}
while IFS='|' read -r -a fields; do
location=${fields[10]}
ton=${location:0:23}; ton=${ton%%+([[:space:]])}
city=${location:23:12}; city=${city%%+([[:space:]])}
state=${location:36:2}
country=${location:38:2}
fields[10]="$ton|$city|$state|$country"
printf -v out '%s|' "${fields[#]}"
printf '%s\n' "${out:0:$(( ${#out} - 1 ))}"
done <"$f" >"$subName.out"
done
It's slower (if I did this well, by about a factor of 10) than pure awk would be, but much faster than the awk/shell combination proposed in the question.
Going into the constructs used:
All the ${varname%...} and related constructs are parameter expansion. The specific ${varname%pattern} construct removes the shortest possible match for pattern from the value in varname, or the longest match if % is replaced with %%.
Using extglob enables extended globbing syntax, such as +([[:space:]]), which is equivalent to the regex syntax [[:space:]]+.
I'm reading filenames from a textfile line by line in a bash script. However the the lines look like this:
/path/to/myfile1.txt 1
/path/to/myfile2.txt 2
/path/to/myfile3.txt 3
...
/path/to/myfile20.txt 20
So there is a second column containing an integer number speparated by space. I only need the part of the string before the space.
I found only solutions using a "for-loop". But I need a function that explicitly looks for the " "-character (space) in my string and splits it at that point.
In principle I need the equivalent to Matlabs "strsplit(str,delimiter)"
If you are already reading the file with something like
while read -r line; do
(and you should be), then pass two arguments to read instead:
while read -r filename somenumber; do
read will split the line on whitespace and assign the first field to filename and any remaining field(s) to somenumber.
Three (of many) solutions:
# Using awk
echo "$string" | awk '{ print $1 }'
# Using cut
echo "$string" | cut -d' ' -f1
# Using sed
echo "$string" | sed 's/\s.*$//g'
If you need to iterate trough each line of the file anyways, you can cut off everything behind the space with bash:
while read -r line ; do
# bash string manipulation removes the space at the end
# and everything which follows it
echo ${line// *}
done < file
This should work too:
line="${line% *}"
This cuts the string at it's last occurrence (from left) of a space. So it will work even if the path contains spaces (as long as it follows by a space at end).
while read -r line
do
{ rev | cut -d' ' -f2- | rev >> result.txt; } <<< $line
done < input.txt
This solution will work even if you have spaces in your filenames.
I'd like to remove any word which contains a non alpha char from a text file. e.g
"ok 0bad ba1d bad3 4bad4 5bad5bad5"
should become
"ok"
I've tried using
echo "ok 0bad ba1d bad3 4bad4 5bad5bad5" | sed 's/\b[a-zA-Z]*[^a-zA-Z]\+[a-zA-Z]*\b/ /g'
The following sed command does the job:
sed 's/[[:space:]]*[[:alpha:]]*[^[:space:][:alpha:]][^[:space:]]*//g'
It removes all words containing at least one non-alphabetic character. It is better to use POSIX character classes like [:alpha:], because for instance they won't consider the French name "François" as being faulty (i.e. containing a non-alphabetic character).
Explanation
We remove all patterns starting with an arbitrary number of spaces followed by an arbitrary (possibly nil) number of alphabetic characters, followed by at least one non-space and non-alphabetic character, and then glob to the end of the word (i.e. until the next space). Please note that you may want to swap [:space:] for [:blank:], see this page for a detailed explanation of the difference between these two POSIX classes.
Test
$ echo "ok 0bad ba1d bad3 4bad4 5bad5bad5" | sed 's/[[:space:]]*[[:alpha:]]*[^[:space:][:alpha:]][^[:space:]]*//g'
ok
Using awk:
s="ok 0bad ba1d bad3 4bad4 5bad5bad5"
awk '{ofs=""; for (i=1; i<=NF; i++) if ($i ~ /^[[:alpha:]]+$/)
{printf "%s%s", ofs, $i; ofs=OFS} print ""}' <<< "$s"
ok
This awk command loops through all words and if word matches the regex /^[[:alpha:]]+$/ then it writes to standard out. (i<NF)?OFS:RS is a short cut to add OFS if current field no is less than NF otherwise it writes RS.
Using grep + tr together:
s="ok 0bad ba1d bad3 4bad4 5bad5bad5"
r=$(grep -o '[^ ]\+' <<< "$s"|grep '^[[:alpha:]]\+$'|tr '\n' ' ')
echo "$r"
ok
First grep -o breaks the string into individual words. 2nd grep only searches for words with alphabets only. ANd finally tr translates \n to space.
If you're not concerned about losing different numbers of spaces between each word, you could use something like this in Perl:
perl -ane 'print join(" ", grep { !/[^[:alpha:]]/ } #F), "\n"
the -a switch enables auto-split mode, which splits the text on any number of spaces and stores the fields in the array #F. grep filters out the elements of that array that contain any non-alphabetical characters. The resulting array is joined on a single space.
This might work for you (GNU sed):
sed -r 's/\b([[:alpha:]]+\b ?)|\S+\b ?/\1/g;s/ $//' file
This uses a back reference within alternation to save the required string.
st="ok 0bad ba1d bad3 4bad4 5bad5bad5"
for word in $st;
do
if [[ $word =~ ^[a-zA-Z]+$ ]];
then
echo $word;
fi;
done
I am wrinting a shell script and have a variable like this: something-that-is-hyphenated.
I need to use it in various points in the script as:
something-that-is-hyphenated, somethingthatishyphenated, SomethingThatIsHyphenated
I have managed to change it to somethingthatishyphenated by stripping out - using sed "s/-//g".
I am sure there is a simpler way, and also, need to know how to get the camel cased version.
Edit: Working function derived from #Michał's answer
function hyphenToCamel {
tr '-' '\n' | awk '{printf "%s%s", toupper(substr($0,1,1)), substr($0,2)}'
}
CAMEL=$(echo something-that-is-hyphenated | hyphenToCamel)
echo $CAMEL
Edit: Finally, a sed one liner thanks to #glenn
echo a-hyphenated-string | sed -E "s/(^|-)([a-z])/\u\2/g"
a GNU sed one-liner
echo something-that-is-hyphenated |
sed -e 's/-\([a-z]\)/\u\1/g' -e 's/^[a-z]/\u&/'
\u in the replacement string is documented in the sed manual.
Pure bashism:
var0=something-that-is-hyphenated
var1=(${var0//-/ })
var2=${var1[*]^}
var3=${var2// /}
echo $var3
SomethingThatIsHyphenated
Line 1 is trivial.
Line 2 is the bashism for replaceAll or 's/-/ /g', wrapped in parens, to build an array.
Line 3 uses ${foo^}, which means uppercase (while ${foo,} would mean 'lowercase' [note, how ^ points up while , points down]) but to operate on every first letter of a word, we address the whole array with ${foo[*]} (or ${foo[#]}, if you would prefer that).
Line 4 is again a replace-all: blank with nothing.
Line 5 is trivial again.
You can define a function:
hypenToCamel() {
tr '-' '\n' | awk '{printf "%s%s", toupper(substr($0,0,1)), substr($0,2)}'
}
CAMEL=$(echo something-that-is-hyphenated | hypenToCamel)
echo $CAMEL
In the shell you are stuck with being messy:
aa="aaa-aaa-bbb-bbb"
echo " $aa" | sed -e 's/--*/ /g' -e 's/ a/A/g' -e 's/ b/B/g' ... -e 's/ *//g'
Note the carefully placed space in the echo and the double space in the last -e.
I leave it as an exercise to complete the code.
In perl it is a bit easier as a one-line shell command:
perl -e 'print map{ $a = ucfirst; $a =~ s/ +//g; $a} split( /-+/, $ARGV[0] ), "\n"' $aa
For the records, here's a pure Bash safe method (that is not subject to pathname expansion)—using Bash≥4:
var0=something-that-is-hyphenated
IFS=- read -r -d '' -a var1 < <(printf '%s\0' "${var0,,}")
printf '%s' "${var1[#]^}"
This (safely) splits the lowercase expansion of var0 at the hyphens, with each split part in array var1. Then we use the ^ parameter expansion to uppercase the first character of the fields of this array, and concatenate them.
If your variable may also contain spaces and you want to act on them too, change IFS=- into IFS='- '.