Well, I do not know much about programming at bash, I'm new at it so I'm struggling to find a code to iterate all the lines in a txt file, and count how many words are different.
Example: If a txt file has "Nory was a Catholic because her mother was a Catholic"
So the result must be 7
$ grep -o '[^[:space:]]*' file | sort -u | wc -l
7
Sure. I assume you are ok with defining "words" as things that are separated by space? In which case, try something like this:
cat filename | sed -r -e "s/[ ]+/ /g" -e "s/ /\n/g" | sort -u | wc -l
This command says:
Dump contents of filename
Replace multiple spaces with a single space
Replace spaces with newline
Sort and "uniquify" the list
Print out the count of lines
Per the comment, you can technically get away without using cat if you'd like, with the following:
sed -r -e "s/[ ]+/ /g" -e "s/ /\n/g" filename | sort -u | wc -l
Further, from another comment, you could optionally use tr (importantly with it's -s flag to handle repeated spaces) instead of sed with something like:
tr -s " " "\n" < filename | sort -u | wc -l
The moral of the story is there are several ways this kind of thing can be accomplished, not to mention the other full answers that are given here :-) My personal favorite answer at this point is Ed Morton's which I've upvoted accordingly.
You could also lowercase the text so words compares regardless of casing.
Also filter words with the [:alnum:] character class, rather than [a-zA-Z0-9_] that is only valid for US-ASCII, and will fail dramatically with Greek or Turkish.
#!/usr/bin/env bash
echo "The uniq words are the words that appears at least once, regardless of casing." |
# Turn text to lowercase
tr '[:upper:]' '[:lower:]' |
# Split alphanumeric with newlines
tr -sc '[:alnum:]' '\n' |
# Sort uniq words
sort -u |
# Count lines of unique words
wc -l
I would do it like so, with comments:
echo "Nory was a Catholic because her mother was a Catholic" |
# tr replace
# -s - squeeze
# -c - complementary
# [a-zA-Z0-9_] - all letters, number and underscore
# but complementary set, so all non letters, not numbers and not underscores.
# replace them by newline
tr -sc '[a-zA-Z0-9_]' '\n' |
# and sort unique and display count
sort -u | wc -l
Tested on repl bash.
Decided to use [a-zA-Z0-9_], because this is how GNU sed \w extension matches a word.
cat yourfile.txt | xargs -n1 | sort | uniq -c > youroutputfile.txt
xargs -n1 = put one word per line
sort = sorts
uniq -c = counts occurrences of distinct values
source
Related
Users who are logged on, in alphabetical order, printed on one line.
What are the minimum amount of changes to get this to work because in a bash script?
this is the given script:
for name in $#
do
who | grep -w "^name" | sed 's/ .*//' | uniq
done | sort | tr '\n' ' '
echo
Single line commands:
who | awk '{print $1}' | sort | uniq | tr '\n' ' '
who: list of logged in users.
awk '{print $1}': keep only the first word or each line, which is the usernames.
sort: put the usernames in alphabetical order.
uniq: remove duplicates.
tr '\n' ' ': remove carriage returns, and replace them with spaces.
Ex
$ who
steve tty7 Mar 5 16:25 (:0)
bernard tty7 Mar 5 16:25 (:0)
sarah tty7 Mar 5 16:25 (:0)
$ who | awk '{print $1}' | sort | uniq | tr '\n' ' '
bernard sara steve
Your code did grep -w "^name", which tells grep to output the lines that start with "name". Not the lines that begin with the value of variable "name". For that you would need to do grep -w "^$name".
Try this Shellcheck-clean code:
for name in "$#"
do
who | sed 's/[[:space:]].*//' | grep -xF -- "$name"
done | sort -u | paste -sd ' '
$# should always have double quotes on it ("$#"). See Accessing bash command line args $# vs $*. Shellcheck correctly complains if double quotes are not used.
sed 's/[[:space:]].*//' removes the first whitespace character, and everything after it, on every input line. Using [[:space:]] instead of a literal space character means that the code will still work if the who output uses tabs as separators. It may be easier to read too. The sed command is run first to ensure that usernames occupy whole lines so it's easier to avoid spurious matches at the next pipeline stage.
grep -xF -- "$name" searches for whole lines in the input that are the "$name" string. The -x option forces matching of whole lines. That prevents, for instance, the username mary matching the username mary.jane (a valid username on at least some Linux systems). The -F option means that regular expression patterns in "$name" are treated as literal strings. That prevents, for instance, the name t.m matching the name tim. The -- prevents a leading hyphen in "$name" being treated as a grep option. No system that I know of allows usernames to have leading hyphens, but there's nothing to stop such an invalid name being provided as a command line argument to the code. The -w option to grep wouldn't be useful here because valid names may contain non-word characters (e.g. t.m).
sort -u takes the output of the for loop (an unsorted list of usernames, one per line, possibly with repetitions) and sorts it. The -u option causes it to remove duplicates (like piping to uniq, but saves a process creation).
paste -sd ' ' puts all the lines in to input on a single line, separated by spaces (specified by the -d option and ' ' (space) option argument), and terminated with a newline character. tr '\n' ' ' would have a similar effect but it produces an unterminated line with a trailing space character.
All you need is:
who | sort -k1,1 -u | awk '{u=u s $1; s=OFS} END{print u}'
That will output a blank-separated list of all logged in users, all on 1 line, with a terminating newline to make it a valid POSIX text file, and without an undesirable trailing blank char.
I'm looking to remove any non-alphabetic (English) characters and make the output lower-case from /usr/share/dict/words. Here's what I have so far:
sed "$(shuf -i "1-$(cat /usr/share/dict/words | wc -l)" -n 1)q;d" /usr/share/dict/words | tr '[:upper:]' '[:lower:]' | sed 's/[^-a-z]//g'
This works fine but is it possible to do it all in the one sed command?
EDIT: The American word file looks like this:
A
A's
AMD
AMD's
AOL
AOL's
AWS
AWS's
Aachen
Aachen's
I'm looking to make this lower-case and remove any non-alphabetic characters (as mentioned in my original question). The solution I have works fine but I'm hoping to reduce the number of commands (maybe just sed?). Output of the above would then be:
a
as
amd
amds
aol
aols
aws
awss
aachen
aachens
You don't need sed and wc -- shuf can shuffle the lines of a file.
tr can remove non-alphas, so again don't need sed
shuf -n1 /usr/share/dict/words | tr -dc '[:alpha:]' | tr '[:upper:]' '[:lower:]'
This single awk command should do the job:
awk '{gsub(/[^[:alpha:]]+/, ""); print tolower($0)}' file
a
as
amd
amds
aol
aols
aws
awss
aachen
aachens
This might work for you (GNU sed and shuf):
shuf -n1 /usr/share/dict/words | sed 's/[^[:alpha:]-]//g;s/.*/\L&/'
Choose a random line, remove any non-alpha (except hyphen) characters and lowercase the result.
I run the following gsutil command:
gsutil ls -d gs://mybucket/v${version}/folder1/*/*.whl |
sort -V |
grep -e "/*.whl"
I get:
gs://mybucket/v1.0.0/folder1/1560924028/file1-cp27-cp27mu-linux_x86_64.whl
gs://mybucket/v1.0.0/folder1/1560926922/file1-cp36-cp36m-linux_x86_64.whl
gs://mybucket/v1.0.0/folder1/1560930522/file1-cp35-cp35m-linux_x86_64.whl
gs://mybucket/v1.0.0/folder1/1561568612/file1-cp37-cp37m-linux_x86_64.whl
gs://mybucket/v1.0.0/folder1/1561595893/file1-cp37-cp37m-linux_x86_64.whl
gs://mybucket/v1.0.0/folder1/1561654308/file1-cp37-cp37m-linux_x86_64.whl
gs://mybucket/v1.0.0/folder1/1563319372/file1-cp27-cp27mu-linux_x86_64.whl
gs://mybucket/v1.0.0/folder1/1563319400/file1-cp36-cp36m-linux_x86_64.whl
gs://mybucket/v1.0.0/folder1/1563329633/file1-cp27-cp27mu-linux_x86_64.whl
gs://mybucket/v1.0.0/folder1/1563411368/file1-cp35-cp35m-linux_x86_64.whl
gs://mybucket/v1.0.0/folder1/1565916833/file1-cp27-cp27mu-linux_x86_64.whl
gs://mybucket/v1.0.0/folder1/1565921265/file1-cp35-cp35m-linux_x86_64.whl
gs://mybucket/v1.0.0/folder1/1566258114/file1-cp27-cp27mu-linux_x86_64.whl
Since some files in different folders have the same names, how can I retrieve unique filenames ignoring the path?
I would do it like this:
blabla_your_command | rev | sort -t'/' -u -k1,1 | rev
rev reverses lines. Then I unique sort using / as a separator on the first field. After the line is reversed, the first field will be the filename, so sorting -u on it would return only unique filenames. Then the line needs to be reversed back.
The following command:
cat <<EOF |
gs://mybucket/v1.0.0/folder1/1560924028/file1-cp27-cp27mu-linux_x86_64.whl
gs://mybucket/v1.0.0/folder1/1560926922/file1-cp36-cp36m-linux_x86_64.whl
gs://mybucket/v1.0.0/folder1/1560930522/file1-cp35-cp35m-linux_x86_64.whl
gs://mybucket/v1.0.0/folder1/1561568612/file1-cp37-cp37m-linux_x86_64.whl
gs://mybucket/v1.0.0/folder1/1561595893/file1-cp37-cp37m-linux_x86_64.whl
gs://mybucket/v1.0.0/folder1/1561654308/file1-cp37-cp37m-linux_x86_64.whl
gs://mybucket/v1.0.0/folder1/1563319372/file1-cp27-cp27mu-linux_x86_64.whl
gs://mybucket/v1.0.0/folder1/1563319400/file1-cp36-cp36m-linux_x86_64.whl
gs://mybucket/v1.0.0/folder1/1563329633/file1-cp27-cp27mu-linux_x86_64.whl
gs://mybucket/v1.0.0/folder1/1563411368/file1-cp35-cp35m-linux_x86_64.whl
gs://mybucket/v1.0.0/folder1/1565916833/file1-cp27-cp27mu-linux_x86_64.whl
gs://mybucket/v1.0.0/folder1/1565921265/file1-cp35-cp35m-linux_x86_64.whl
gs://mybucket/v1.0.0/folder1/1566258114/file1-cp27-cp27mu-linux_x86_64.whl
EOF
rev | sort -t'/' -u -k1,1 | rev
outputs:
gs://mybucket/v1.0.0/folder1/1560930522/file1-cp35-cp35m-linux_x86_64.whl
gs://mybucket/v1.0.0/folder1/1560926922/file1-cp36-cp36m-linux_x86_64.whl
gs://mybucket/v1.0.0/folder1/1561568612/file1-cp37-cp37m-linux_x86_64.whl
gs://mybucket/v1.0.0/folder1/1560924028/file1-cp27-cp27mu-linux_x86_64.whl
Please check awk option given below, this will print the last occurrence of delimiter '/', it worked for me
example:
gsutil ls gs://mybucket/v1.0.0/folder1/1560930522 | awk -F/ '{print $(NF)}'
print all the file names under '1560930522'
your_command|awk -F/ '!($NF in a){a[$NF]; print}'
gs://mybucket/v1.0.0/folder1/1560924028/file1-cp27-cp27mu-linux_x86_64.whl
gs://mybucket/v1.0.0/folder1/1560926922/file1-cp36-cp36m-linux_x86_64.whl
gs://mybucket/v1.0.0/folder1/1560930522/file1-cp35-cp35m-linux_x86_64.whl
gs://mybucket/v1.0.0/folder1/1561568612/file1-cp37-cp37m-linux_x86_64.whl
4 different ways of saying the same thing
nawk -F'^.+/' '++_[$NF]<NF'
gawk -F'/' '__[$NF]++<!_'
mawk -F/ '_^__[$NF]++'
mawk2 -F/ '!_[$NF]--'
gs://mybucket/v1.0.0/folder1/1560924028/file1-cp27-cp27mu-linux_x86_64.whl
gs://mybucket/v1.0.0/folder1/1560926922/file1-cp36-cp36m-linux_x86_64.whl
gs://mybucket/v1.0.0/folder1/1560930522/file1-cp35-cp35m-linux_x86_64.whl
gs://mybucket/v1.0.0/folder1/1561568612/file1-cp37-cp37m-linux_x86_64.whl
Here's a simple, straightforward solution:
$ your_gsutil_command | xargs -L 1 basename | sort -u
The easiest way to remove paths is with basename. Unfortunately it accepts only a single filename, which must be on the command line (not from stdin), so we need to take the following steps:
Create the list of files.
We do this with your_gsutil_command, but you can use any command that generates a list of files.
Send each one to basename to remove its path.
The xargs command does this for us by reading its stdin and invoking basename repeatedly, passing the data as command-line arguments. But xargs efficiently tries to reduce the number of invocations by passing multiple filenames on each command line, and that breaks basename. We prevent that with -L 1, limiting it to only one line (that is, one filename) at a time.
Remove duplicates.
The sort -u command does this.
Using your example data:
$ gsutil ls -d gs://mybucket/v${version}/folder1/*/*.whl |
xargs -L 1 basename | sort -u
file1-cp27-cp27mu-linux_x86_64.whl
file1-cp35-cp35m-linux_x86_64.whl
file1-cp36-cp36m-linux_x86_64.whl
file1-cp37-cp37m-linux_x86_64.whl
Caveat: Spaces break everything. 😡
So far we've assumed the filenames and folders do not contain spaces. Spaces break basename because needs exactly one filename, and it would interpret spaces as separators between multiple filenames. We can get around this in two ways:
ls -Q: If you're deduplicating local filenames, you can use the (non-gsutil) ls command with the -Q flag to put the filenames in quotes, so basename will interpret spaces as part of the filenames rather than separators.
gsutil: The -Q flag is unfortunately not supported, so we'll need to escape the spaces manually:
$ your_gsutil_command | sed 's/ /\\ /g' | xargs -L 1 basename | sort -u
Here we use the sed command to escape each space by inserting a backslash before it. (That is, we replace with \ . Note that we also need to escape the backslash in the sed command, which is why we use \\ and not just \.)
I have two files containing list of files. I need to check what files are missing in the list of second file. Problem is that I do not have to match full name, but only need to match last 19 Characters of the file names.
E.g
MyFile12343220150510230000.xlsx
and
MyFile99999620150510230000.xlsx
are same files.
This is a unique problem and I don't know how to start. Kindly help.
awk based solution:
$ awk '
{start=length($0) - 18;}
NR==FNR{a[substr($0, start)]++; next;} #save last 19 characters for every line in file2
{if(!a[substr($0, start)]) print $0;} #If that is not present in file1, print that line.
' file2.list file.list
First you can use comm to match the exact file names and obtain a list of files not matchig. Then you can use agrep. I've never used it, but you might find it useful.
Or, as last option, you can do a brute force and for every line in the first file search into the second:
#!/bin/bash
# Iterate through the first file
while read LINE; do
# Find the section of the filename that has to match in the other file
CHECK_SECTION="$(echo "$LINE" | sed -nre 's/^.*([0-9]{14})\.(.*)$/\1.\2/p')"
# Create a regex to match the filenames in the second file
SEARCH_REGEX="^.*$CHECK_SECTION$"
# Search...
egrep "$SEARCH_REGEX" inputFile_2.txt
done < inputFile_1.txt
Here I assumed the filenames end with 14 digits that must match in the other file and a file extension that can be different from file to file but that has to match too:
MyFile12343220150510230000.xlsx
| variable | 14digits |.ext
So, if the first file is FILE1 and the second file is FILE2 then if the intention is only to identify the files in FILE2 that don't exist in FILE1, the following should do:
tmp1=$(mktemp)
tmp2=$(mktemp)
cat $FILE1 | rev | cut -c -19 | sort | uniq > ${tmp1}
cat $FILE2 | rev | cut -c -19 | sort | uniq > ${tmp2}
diff ${tmp1} ${tmp2} | rev
rm ${tmp1} ${tmp2}
In a nutshell, this reverses the characters on each line, and extracts the part you're interested in, saving to a temporary file, for each list of files. The reversal of characters is done since you haven't said whether or not the length of filenames is guaranteed to be constant---the only thing we can rely on here is that the last 19 characters are of a fixed format (in this case, although the format is easily inferred, it isn't really relevant). The sort is important in order for the diff to show you what's not in the second file that is in the first.
If you're certain that there will only ever be files missing from FILE2 and not the other way around (that is, files in FILE2 that don't exist in FILE1), then you can clean things up by removing the cruft introduced by diff, so the last line becomes:
diff ${tmp1} ${tmp2} | rev | grep -i xlsx | sed 's/[[:space:]]\+.*//'
The grep limits the output to those lines with xlsx filenames, and the sed removes everything on a line from the first space encountered onwards.
Of course, technically this only tells you what time-stamped-grouped groups of files exist in FILE1 but not FILE2--as I understand it, this is what you're looking for (my understanding of your problem description is that MyFile12343220150510230000.xlsx and MyFile99999620150510230000.xlsx would have identical content). If the file names are always the same length (as you subsequently affirmed), then there's no need for the rev's and the cut commands can just be amended to refer to fixed character positions.
In any case, to get the final list of files, you'll have to use the "cleaned up" output to filter the content of FILE1; so, modifying the script above so that it includes the "cleanup" command, we can filter the files that you need using a grep--the whole script then becomes:
tmp1=$(mktemp)
tmp2=$(mktemp)
missing=$(mktemp)
cat $FILE1 | rev | cut -c -19 | sort | uniq > ${tmp1}
cat $FILE2 | rev | cut -c -19 | sort | uniq > ${tmp2}
diff ${tmp1} ${tmp2} | rev | grep -i xlsx | sed 's/[[:space:]]\+.*//' > ${missing}
grep -E "("`echo $(<${missing}) | sed 's/[[:space:]]/|/g'`")" ${tmp1}
rm ${tmp1} ${tmp2} ${missing}
The extended grep command (-E) just builds up an "or" regular expression for each timestamp-plus-extension and applies it to the first file. Of course, this is all assuming that there will never be timestamp-groups that exist in FILE2 and not in FILE1--if this is the case, then the "diff output processing" bit needs to be a little more clever.
Or you could use your standard coreutil tools:
for i in $(cat file1 file2 | sort | uniq -u); do
grep -q "$i" f1.txt && \
echo "f2 missing '$i'" || \
echo "f1 missing '$i'"
done
It will identify which non-common entries are missing from which file. You can also manipulate the non-common filenames in any way you like, e.g. parameter expansion/substring extraction, substring removal, or character indexes.
If i use
wc -m filename
it will generate the number of characters
and
wc -w filename
will generate number of words
if i used this info by dividing number of characters/number of words
it will give me misleading result as number of character will include spaces and punctuation
any advice ?
the solution that I came up with without writing a script was to pipe it through a couple of commands like this.
<filename tr -d " \t\n\r\.\?\!" | wc -m
This works to remove all of the spacing, like new line, tabs and normal spacing. A more rigorous tr command that included any sort of other punctuation like a colon can just be added to the list for example \:
Hope That Helps
Subtract out characters you do not want
chars=$(tr -dc '[:alnum:]' < filename | wc -c)
words=$(cat filename | wc -c)
Now do you calculation. I piped into wc to avoid the extra "filename" in output
printf "%.2f" $(echo "$chars/$words" | bc -l)
Edit: thanks BMW