I new in bash scripting and i bug with this:
tab=( "^[A-Z]\{4,\}[0-9]\{4,\}" )
for (( i=0; i<=$(( ${#tab[*]} - 1 )); i++ ))
do
tmp+=" grep -v \"${tab[i]}\" |"
done
# for remove the last |
chaine=`echo $tmp| rev | cut -c2- | rev`
#result anticipe "cat ${oldConfFile[0]} | grep -v "^[A-Z]\{4,\}[0-9]\{4,\}"
cat ${oldConfFile[0]} | echo $chaine
My trouble is there, how use cat and echo at the same time ?
thanks a lot.
You don't need to grep for each pattern, just join your patterns with pipes (|) and grep once. For example, if you want to filter out lines that contain foo, bar and baz from file, use the following:
grep -v 'foo|bar|baz' file
And you can build up the pattern outside of the invocation for better readability like this:
my_pattern='foo'
my_pattern+='|bar'
my_pattern+='|baz'
grep -v "$my_pattern" file
Related
I have 3 different files that I want to compare
words_freq
words_freq_deduped
words_freq_alpha
For each file, I run a command like so, which I iterate on constantly to compare the results.
For example, I would do this:
$ cat words_freq | grep -v '[soe]'
$ cat words_freq_deduped | grep -v '[soe]'
$ cat words_freq_alpha | grep -v '[soe]'
and then review the results, and then do it again, with an additional filter
$ cat words_freq | grep -v '[soe]' | grep a | grep r | head -n20
a
$ cat words_freq_deduped | grep -v '[soe]' | grep a | grep r | head -n20
b
$ cat words_freq_alpha | grep -v '[soe]' | grep a | grep r | head -n20
c
This continues on until I've analyzed my data.
I would like to write a script that could take the piped portions, and pass it to each of these files, as I iterate on the grep/head portions of the command.
e.g. The following would dump the results of running the 3 commands above AND also compare the 3 results, and dump additional calculations on them
$ myScript | grep -v '[soe]' | grep a | grep r | head -n20
the letters were in all 3 runs, and it took 5 seconds
a
b
c
How can I do this using bash/python or zsh for the myScript part?
EDIT: After asking the question, it occurred to me that I could use eval to do it, like so, which I've added as an answer as well
The following approach allows me to process multiple files by using eval, which I know is frowned upon - any other suggestions are greatly appreciated!
$ myScript "grep -v '[soe]' | grep a | grep r | head -n20"
myScript
#!/usr/bin/env bash
function doIt(){
FILE=$1
CMD="cat $1 | $2"
echo processing file "$FILE"
eval "$CMD"
echo
}
doIt words_freq "$#"
doIt words_freq_deduped "$#"
doIt words_freq_alpha "$#"
You can't avoid your shell from running pipes itself, so using it like that isn't very practical - you'd need to either quote everything and then eval it, which would make it hard to pass arguments with spaces, or quote every pipe, which you can then eval, making it so you have to quote every pipe. But yeah, these solutions are kinda hacky.
I'd suggest doing one of these two:
Keep your editor open, and put whatever you want to run inside the doIt function itself before you run it. Then run it in your shell without any arguments:
#!/usr/bin/env bash
doIt() {
# grep -v '[soe]' < "$1"
grep -v '[soe]' < "$1" | grep a | grep r | head -n20
}
doIt words_freq
doIt words_freq_deduped
doIt words_freq_alpha
Or, you could always use a "for" in your shell, which you can use Ctrl+r to find in your history when you want to use:
$ for f in words_freq*; do grep -v '[soe]' < "$f" | grep a | grep r | head -n20; done
But if you really want your approach, I tried to make it accept spaces, but it ended up being even hackier:
#!/usr/bin/env bash
doIt() {
local FILE=$1
shift
echo processing file "$FILE"
local args=()
for n in $(seq 1 $#); do
arg=$1
shift
if [[ $arg == '|' ]]; then
args+=('|')
else
args+=("\"$arg\"")
fi
done
eval "cat '$FILE' | ${args[#]}"
}
doIt words_freq "$#"
doIt words_freq_deduped "$#"
doIt words_freq_alpha "$#"
With this version you can use it like this:
$ ./myScript grep "a a" "|" head -n1
Notice that it need you to quote the |, and that it now handles arguments with spaces.
Not fully understood problem correctly.
I understood you want to write a script without pipes, by including the filtering logic into the script.
And feeding the filtering patterns as arguments.
Here is a gawk script (standard Linux awk).
With one sweep on 3 input files, without piping.
script.awk
BEGIN {
RS="!#!#!#!#!#!#!#";
# set record separator to something unlikely matched, causing each file to be read entirely as a single record
}
$0 !~ excludeRegEx # if file does not match excludeRegEx
&& $0 ~ includeRegEx1 # and match includeRegEx1
&& $0 ~ includeRegEx2 { # and match includeRegEx2
system "head -n20 "FILENAME; # call shell command "head -n20 " on current filename
}
Running script.awk
awk -v excludeRegEx='[soe]' \
-v includeRegEx1='a' \
-v includeRegEx2='r' \
-f script.awk words_freq words_freq_deduped words_freq_alpha
The following approach allows me to process multiple files by using eval, which I know is frowned upon - any other suggestions are greatly appreciated!
$ myScript "grep -v '[soe]' | grep a | grep r | head -n20"
myScript
#!/usr/bin/env bash
function doIt(){
FILE=$1
CMD="cat $1 | $2"
echo processing file "$FILE"
eval "$CMD"
echo
}
doIt words_freq "$#"
doIt words_freq_deduped "$#"
doIt words_freq_alpha "$#"
I'm new to Linux shell. I know there are tools to do this thing, such as awk. But I'm wondering if I could do it using grep or wc or other commands? awk seems intimidating to me. Thanks.
I tried grep and wc, like this:
grep tol test.txt | wc -w
But grep will give me the whole line.
If I tried the following:
grep '^tol$*' test.txt | wc -w
It only counts the line begins with mol.
How can I grep the words starting with tol?
Something like that:
grep -o '\<tol[[:alpha:]]*\>' test.txt | wc -w
< - for beginning of the word,
> - the end of the word.
[[:alpha:]] - to avoid match of combinations like tol123 (You said you need only words).
-o - to show only matches, not the entire line.
You can do the same fairly simply with awk, e.g.
awk '{for(i=1;i<=NF;i++) $i~/^tol/ && n++} END {print n}'
Example
$ echo -e "tolerance topaz tolstoy\nbats toluene toledo" |
> awk '{for(i=1;i<=NF;i++) $i~/^tol/ && n++} END {print n}'
4
Another option is to translate all whitespace characters into linefeeds so that each word starts on a new line, then grep can count them itself:
echo -e "tolerance topaz\ttolstoy\nbats toluene toledo" | tr '[:space:]' '\n' | grep -c "^tol"
4
Or, if using a file called words.txt:
tr '[:space:]' '\n' < words.txt | grep -c "^tol"
wc -l file.txt
outputs number of lines and file name.
I need just the number itself (not the file name).
I can do this
wc -l file.txt | awk '{print $1}'
But maybe there is a better way?
Try this way:
wc -l < file.txt
cat file.txt | wc -l
According to the man page (for the BSD version, I don't have a GNU version to check):
If no files are specified, the standard input is used and no file
name is
displayed. The prompt will accept input until receiving EOF, or [^D] in
most environments.
To do this without the leading space, why not:
wc -l < file.txt | bc
Comparison of Techniques
I had a similar issue attempting to get a character count without the leading whitespace provided by wc, which led me to this page. After trying out the answers here, the following are the results from my personal testing on Mac (BSD Bash). Again, this is for character count; for line count you'd do wc -l. echo -n omits the trailing line break.
FOO="bar"
echo -n "$FOO" | wc -c # " 3" (x)
echo -n "$FOO" | wc -c | bc # "3" (√)
echo -n "$FOO" | wc -c | tr -d ' ' # "3" (√)
echo -n "$FOO" | wc -c | awk '{print $1}' # "3" (√)
echo -n "$FOO" | wc -c | cut -d ' ' -f1 # "" for -f < 8 (x)
echo -n "$FOO" | wc -c | cut -d ' ' -f8 # "3" (√)
echo -n "$FOO" | wc -c | perl -pe 's/^\s+//' # "3" (√)
echo -n "$FOO" | wc -c | grep -ch '^' # "1" (x)
echo $( printf '%s' "$FOO" | wc -c ) # "3" (√)
I wouldn't rely on the cut -f* method in general since it requires that you know the exact number of leading spaces that any given output may have. And the grep one works for counting lines, but not characters.
bc is the most concise, and awk and perl seem a bit overkill, but they should all be relatively fast and portable enough.
Also note that some of these can be adapted to trim surrounding whitespace from general strings, as well (along with echo `echo $FOO`, another neat trick).
How about
wc -l file.txt | cut -d' ' -f1
i.e. pipe the output of wc into cut (where delimiters are spaces and pick just the first field)
How about
grep -ch "^" file.txt
Obviously, there are a lot of solutions to this.
Here is another one though:
wc -l somefile | tr -d "[:alpha:][:blank:][:punct:]"
This only outputs the number of lines, but the trailing newline character (\n) is present, if you don't want that either, replace [:blank:] with [:space:].
Another way to strip the leading zeros without invoking an external command is to use Arithmetic expansion $((exp))
echo $(($(wc -l < file.txt)))
Best way would be first of all find all files in directory then use AWK NR (Number of Records Variable)
below is the command :
find <directory path> -type f | awk 'END{print NR}'
example : - find /tmp/ -type f | awk 'END{print NR}'
This works for me using the normal wc -l and sed to strip any char what is not a number.
wc -l big_file.log | sed -E "s/([a-z\-\_\.]|[[:space:]]*)//g"
# 9249133
wc -l file.txt
outputs number of lines and file name.
I need just the number itself (not the file name).
I can do this
wc -l file.txt | awk '{print $1}'
But maybe there is a better way?
Try this way:
wc -l < file.txt
cat file.txt | wc -l
According to the man page (for the BSD version, I don't have a GNU version to check):
If no files are specified, the standard input is used and no file
name is
displayed. The prompt will accept input until receiving EOF, or [^D] in
most environments.
To do this without the leading space, why not:
wc -l < file.txt | bc
Comparison of Techniques
I had a similar issue attempting to get a character count without the leading whitespace provided by wc, which led me to this page. After trying out the answers here, the following are the results from my personal testing on Mac (BSD Bash). Again, this is for character count; for line count you'd do wc -l. echo -n omits the trailing line break.
FOO="bar"
echo -n "$FOO" | wc -c # " 3" (x)
echo -n "$FOO" | wc -c | bc # "3" (√)
echo -n "$FOO" | wc -c | tr -d ' ' # "3" (√)
echo -n "$FOO" | wc -c | awk '{print $1}' # "3" (√)
echo -n "$FOO" | wc -c | cut -d ' ' -f1 # "" for -f < 8 (x)
echo -n "$FOO" | wc -c | cut -d ' ' -f8 # "3" (√)
echo -n "$FOO" | wc -c | perl -pe 's/^\s+//' # "3" (√)
echo -n "$FOO" | wc -c | grep -ch '^' # "1" (x)
echo $( printf '%s' "$FOO" | wc -c ) # "3" (√)
I wouldn't rely on the cut -f* method in general since it requires that you know the exact number of leading spaces that any given output may have. And the grep one works for counting lines, but not characters.
bc is the most concise, and awk and perl seem a bit overkill, but they should all be relatively fast and portable enough.
Also note that some of these can be adapted to trim surrounding whitespace from general strings, as well (along with echo `echo $FOO`, another neat trick).
How about
wc -l file.txt | cut -d' ' -f1
i.e. pipe the output of wc into cut (where delimiters are spaces and pick just the first field)
How about
grep -ch "^" file.txt
Obviously, there are a lot of solutions to this.
Here is another one though:
wc -l somefile | tr -d "[:alpha:][:blank:][:punct:]"
This only outputs the number of lines, but the trailing newline character (\n) is present, if you don't want that either, replace [:blank:] with [:space:].
Another way to strip the leading zeros without invoking an external command is to use Arithmetic expansion $((exp))
echo $(($(wc -l < file.txt)))
Best way would be first of all find all files in directory then use AWK NR (Number of Records Variable)
below is the command :
find <directory path> -type f | awk 'END{print NR}'
example : - find /tmp/ -type f | awk 'END{print NR}'
This works for me using the normal wc -l and sed to strip any char what is not a number.
wc -l big_file.log | sed -E "s/([a-z\-\_\.]|[[:space:]]*)//g"
# 9249133
Suppose I have the string 1:2:3:4:5 and I want to get its last field (5 in this case). How do I do that using Bash? I tried cut, but I don't know how to specify the last field with -f.
You can use string operators:
$ foo=1:2:3:4:5
$ echo ${foo##*:}
5
This trims everything from the front until a ':', greedily.
${foo <-- from variable foo
## <-- greedy front trim
* <-- matches anything
: <-- until the last ':'
}
Another way is to reverse before and after cut:
$ echo ab:cd:ef | rev | cut -d: -f1 | rev
ef
This makes it very easy to get the last but one field, or any range of fields numbered from the end.
It's difficult to get the last field using cut, but here are some solutions in awk and perl
echo 1:2:3:4:5 | awk -F: '{print $NF}'
echo 1:2:3:4:5 | perl -F: -wane 'print $F[-1]'
Assuming fairly simple usage (no escaping of the delimiter, for example), you can use grep:
$ echo "1:2:3:4:5" | grep -oE "[^:]+$"
5
Breakdown - find all the characters not the delimiter ([^:]) at the end of the line ($). -o only prints the matching part.
You could try something like this if you want to use cut:
echo "1:2:3:4:5" | cut -d ":" -f5
You can also use grep try like this :
echo " 1:2:3:4:5" | grep -o '[^:]*$'
One way:
var1="1:2:3:4:5"
var2=${var1##*:}
Another, using an array:
var1="1:2:3:4:5"
saveIFS=$IFS
IFS=":"
var2=($var1)
IFS=$saveIFS
var2=${var2[#]: -1}
Yet another with an array:
var1="1:2:3:4:5"
saveIFS=$IFS
IFS=":"
var2=($var1)
IFS=$saveIFS
count=${#var2[#]}
var2=${var2[$count-1]}
Using Bash (version >= 3.2) regular expressions:
var1="1:2:3:4:5"
[[ $var1 =~ :([^:]*)$ ]]
var2=${BASH_REMATCH[1]}
$ echo "a b c d e" | tr ' ' '\n' | tail -1
e
Simply translate the delimiter into a newline and choose the last entry with tail -1.
Using sed:
$ echo '1:2:3:4:5' | sed 's/.*://' # => 5
$ echo '' | sed 's/.*://' # => (empty)
$ echo ':' | sed 's/.*://' # => (empty)
$ echo ':b' | sed 's/.*://' # => b
$ echo '::c' | sed 's/.*://' # => c
$ echo 'a' | sed 's/.*://' # => a
$ echo 'a:' | sed 's/.*://' # => (empty)
$ echo 'a:b' | sed 's/.*://' # => b
$ echo 'a::c' | sed 's/.*://' # => c
There are many good answers here, but still I want to share this one using basename :
basename $(echo "a:b:c:d:e" | tr ':' '/')
However it will fail if there are already some '/' in your string.
If slash / is your delimiter then you just have to (and should) use basename.
It's not the best answer but it just shows how you can be creative using bash commands.
If your last field is a single character, you could do this:
a="1:2:3:4:5"
echo ${a: -1}
echo ${a:(-1)}
Check string manipulation in bash.
Using Bash.
$ var1="1:2:3:4:0"
$ IFS=":"
$ set -- $var1
$ eval echo \$${#}
0
echo "a:b:c:d:e"|xargs -d : -n1|tail -1
First use xargs split it using ":",-n1 means every line only have one part.Then,pring the last part.
Regex matching in sed is greedy (always goes to the last occurrence), which you can use to your advantage here:
$ foo=1:2:3:4:5
$ echo ${foo} | sed "s/.*://"
5
A solution using the read builtin:
IFS=':' read -a fields <<< "1:2:3:4:5"
echo "${fields[4]}"
Or, to make it more generic:
echo "${fields[-1]}" # prints the last item
for x in `echo $str | tr ";" "\n"`; do echo $x; done
improving from #mateusz-piotrowski and #user3133260 answer,
echo "a:b:c:d::e:: ::" | tr ':' ' ' | xargs | tr ' ' '\n' | tail -1
first, tr ':' ' ' -> replace ':' with whitespace
then, trim with xargs
after that, tr ' ' '\n' -> replace remained whitespace to newline
lastly, tail -1 -> get the last string
For those that comfortable with Python, https://github.com/Russell91/pythonpy is a nice choice to solve this problem.
$ echo "a:b:c:d:e" | py -x 'x.split(":")[-1]'
From the pythonpy help: -x treat each row of stdin as x.
With that tool, it is easy to write python code that gets applied to the input.
Edit (Dec 2020):
Pythonpy is no longer online.
Here is an alternative:
$ echo "a:b:c:d:e" | python -c 'import sys; sys.stdout.write(sys.stdin.read().split(":")[-1])'
it contains more boilerplate code (i.e. sys.stdout.read/write) but requires only std libraries from python.