Automatically generate random compositions by layering source images **without creating duplicates** - image-processing

This is a follow on from this question asked by ign and expertly answered by Marc Setchell
It works perfectly, though I am hoping to find a way to avoid duplicates forming in the randomisation process. I am going to be doing hundreds of variations of a bunch of layers, with subtle differences, so I can't really go in and spot duplicates in a fail-safe way.
Here is the code I'm using, as per above - all credit to Marc Setchell!
#!/bin/bash
# Number of output files - edit freely :-)
NFILES=10
# Build arrays of filenames in each layer, assume directories are "Layer0", "Layer1" etc
IFS=$'\n' L0files=($(find "Layer 0" -name "*.png"))
IFS=$'\n' L1files=($(find "Layer 1" -name "*.png"))
IFS=$'\n' L2files=($(find "Layer 2" -name "*.png"))
IFS=$'\n' L3files=($(find "Layer 3" -name "*.png"))
# Produce NFILES output files
for i in `seq 1 $NFILES`; do
# Choose random index into each array of filenames
index0=$( jot -r 1 0 $((${#L0files[#]} - 1)) )
index1=$( jot -r 1 0 $((${#L1files[#]} - 1)) )
index2=$( jot -r 1 0 $((${#L2files[#]} - 1)) )
index3=$( jot -r 1 0 $((${#L3files[#]} - 1)) )
# Pick up files as specified by the random index
f0=${L0files[index0]}
f1=${L1files[index1]}
f2=${L2files[index2]}
f3=${L3files[index3]}
# Generate output filename, "output-nnn.png"
# ... where nnn starts at 0 and goes up till no clash
i=0
while :; do
out="output-$i.png"
[ ! -f "$out" ] && break
((i++))
done
echo $f0, $f1, $f2, $f3 "=> $out"
convert "$f0" "$f1" -composite "$f2" -composite "$f3" -composite "$out"
done

IMHO nothing to do with IM or PS. Just name your files so that they include the index of the layers you are using to create them, and skip if you see that you are creating the same file again.
out="output-$index0-$index1-$index2-$index3.png"
[[ ! -f "$out" ]] && continue
((i++))
This also requires that you don't use seq 1 $NFILES to iterate your files since you want to increment the counter only when you don't skip(*).
When you are done you rename your files to a sequence, if necessary.
Another method in bash is to declare an associative array and put an element in it for each file that you create:
# declare associative array
declare -A doneFiles
# when creating a file:
combination=$index0-$index1-$index2-$index3
donefiles[$combination]="done" # or any info
# testing:
[[ -n ${doneFiles[$combination]} ]] && continue # already done
(*) By the way using i for both the outer and inner loops is crazy even if you don't use the outer variable explicitly. But then the inner loop is terrible code, since you can use the outer i.

Related

Find the count of a specific keyword in multiple files in a directory

Say I have a directory /home/ and within it I have 3 subdirectories /home/red/ /home/blue/ /home/green/
And each subdirectory contains a file each like
/home/red/file1 /home/blue/file2 /home/green/file3
Now I want to find how many times file1,file2, file3 contains the word "hello" within them.
For example,
/home/red/file1 - 23
/home/blue/file2 - 6
/home/green/file3 - 0
Now, going to the locations of file and running the grep command is actually very inefficient when this problem scales.
I have tried using this grep command from the /home/ directory
grep -rnw '/path/to/somewhere/' -e 'pattern'
But this is just giving the occurrences rather than the count.
Is there any command through which I can get what I am looking for?
If the search term occurs at maximum once per line, you can use grep's -c option to report the count instead of the matching lines. So, the command will be grep -rc 'search' (add other options as needed).
If there can be more than one occurrence per line, I'd recommend using ripgrep. Note that rg recursively searches by default, so you can use something like rg -co 'search' from within the home directory (add other options as needed). Add --hidden if you need to search hidden files as well. Add --include-zero if you want to show files even if they didn't have any match.
Instead of grep you can use this find | gnu-awk solution:
cd /home
find {red/file1,blue/file2,green/file3} -type f -exec awk '
{c += gsub(/pattern/, "&")} ENDFILE {print FILENAME, "-", c; c=0}' {} +

grep the file if it matches delete it and save it in same name [duplicate]

I have a file f1:
line1
line2
line3
line4
..
..
I want to delete all the lines which are in another file f2:
line2
line8
..
..
I tried something with cat and sed, which wasn't even close to what I intended. How can I do this?
grep -v -x -f f2 f1 should do the trick.
Explanation:
-v to select non-matching lines
-x to match whole lines only
-f f2 to get patterns from f2
One can instead use grep -F or fgrep to match fixed strings from f2 rather than patterns (in case you want remove the lines in a "what you see if what you get" manner rather than treating the lines in f2 as regex patterns).
Try comm instead (assuming f1 and f2 are "already sorted")
comm -2 -3 f1 f2
For exclude files that aren't too huge, you can use AWK's associative arrays.
awk 'NR == FNR { list[tolower($0)]=1; next } { if (! list[tolower($0)]) print }' exclude-these.txt from-this.txt
The output will be in the same order as the "from-this.txt" file. The tolower() function makes it case-insensitive, if you need that.
The algorithmic complexity will probably be O(n) (exclude-these.txt size) + O(n) (from-this.txt size)
Similar to Dennis Williamson's answer (mostly syntactic changes, e.g. setting the file number explicitly instead of the NR == FNR trick):
awk '{if (f==1) { r[$0] } else if (! ($0 in r)) { print $0 } } ' f=1 exclude-these.txt f=2 from-this.txt
Accessing r[$0] creates the entry for that line, no need to set a value.
Assuming awk uses a hash table with constant lookup and (on average) constant update time, the time complexity of this will be O(n + m), where n and m are the lengths of the files. In my case, n was ~25 million and m ~14000. The awk solution was much faster than sort, and I also preferred keeping the original order.
if you have Ruby (1.9+)
#!/usr/bin/env ruby
b=File.read("file2").split
open("file1").each do |x|
x.chomp!
puts x if !b.include?(x)
end
Which has O(N^2) complexity. If you want to care about performance, here's another version
b=File.read("file2").split
a=File.read("file1").split
(a-b).each {|x| puts x}
which uses a hash to effect the subtraction, so is complexity O(n) (size of a) + O(n) (size of b)
here's a little benchmark, courtesy of user576875, but with 100K lines, of the above:
$ for i in $(seq 1 100000); do echo "$i"; done|sort --random-sort > file1
$ for i in $(seq 1 2 100000); do echo "$i"; done|sort --random-sort > file2
$ time ruby test.rb > ruby.test
real 0m0.639s
user 0m0.554s
sys 0m0.021s
$time sort file1 file2|uniq -u > sort.test
real 0m2.311s
user 0m1.959s
sys 0m0.040s
$ diff <(sort -n ruby.test) <(sort -n sort.test)
$
diff was used to show there are no differences between the 2 files generated.
Some timing comparisons between various other answers:
$ for n in {1..10000}; do echo $RANDOM; done > f1
$ for n in {1..10000}; do echo $RANDOM; done > f2
$ time comm -23 <(sort f1) <(sort f2) > /dev/null
real 0m0.019s
user 0m0.023s
sys 0m0.012s
$ time ruby -e 'puts File.readlines("f1") - File.readlines("f2")' > /dev/null
real 0m0.026s
user 0m0.018s
sys 0m0.007s
$ time grep -xvf f2 f1 > /dev/null
real 0m43.197s
user 0m43.155s
sys 0m0.040s
sort f1 f2 | uniq -u isn't even a symmetrical difference, because it removes lines that appear multiple times in either file.
comm can also be used with stdin and here strings:
echo $'a\nb' | comm -23 <(sort) <(sort <<< $'c\nb') # a
Seems to be a job suitable for the SQLite shell:
create table file1(line text);
create index if1 on file1(line ASC);
create table file2(line text);
create index if2 on file2(line ASC);
-- comment: if you have | in your files then specify “ .separator ××any_improbable_string×× ”
.import 'file1.txt' file1
.import 'file2.txt' file2
.output result.txt
select * from file2 where line not in (select line from file1);
.q
Did you try this with sed?
sed 's#^#sed -i '"'"'s%#g' f2 > f2.sh
sed -i 's#$#%%g'"'"' f1#g' f2.sh
sed -i '1i#!/bin/bash' f2.sh
sh f2.sh
Not a 'programming' answer but here's a quick and dirty solution: just go to http://www.listdiff.com/compare-2-lists-difference-tool.
Obviously won't work for huge files but it did the trick for me. A few notes:
I'm not affiliated with the website in any way (if you still don't believe me, then you can just search for a different tool online; I used the search term "set difference list online")
The linked website seems to make network calls on every list comparison, so don't feed it any sensitive data
A Python way of filtering one list using another list.
Load files:
>>> f1 = open('f1').readlines()
>>> f2 = open('f2.txt').readlines()
Remove '\n' string at the end of each line:
>>> f1 = [i.replace('\n', '') for i in f1]
>>> f2 = [i.replace('\n', '') for i in f2]
Print only the f1 lines that are also in the f2 file:
>>> [a for a in f1 if all(b not in a for b in f2)]
$ cat values.txt
apple
banana
car
taxi
$ cat source.txt
fruits
mango
king
queen
number
23
43
sentence is long
so what
...
...
I made a small shell scrip to "weed" out the values in source file which are present in values.txt file.
$cat weed_out.sh
from=$1
cp -p $from $from.final
for x in `cat values.txt`;
do
grep -v $x $from.final > $from.final.tmp
mv $from.final.tmp $from.final
done
executing...
$ ./weed_out source.txt
and you get a nicely cleaned up file....

Best way to search the path in shell

I've got a small script called "onewhich". Its purpose is to behave like which, except that it will only give the FIRST occurrence of any executables specified as options, as found in the order they'd appear in the path.
So for example, if my path is /opt/bin:/usr/bin:/bin, and I have both /opt/bin/runme and /usr/bin/runme, then the command onewhich runme would return /opt/bin/runme.
But if I also have a /usr/bin/doit, then the command onewhich doit runme would return /usr/bin/doit instead.
The idea is to walk through the path, check for each executable specified, and if it exists, show it and exit.
Here's the script so far.
#!/bin/sh
for what in "$#"; do
for loc in `echo "${PATH}" | awk -vRS=: 1`; do
if [ -f "${loc}/${what}" ]; then
echo "${loc}/${what}"
exit 0
fi
done
done
exit 1
The problem is, I want to be better about PATH directories with special characters. Every second shell question here on StackOverflow talks about how bad it is to parse paths with tools like awk and sed. There's even a bash faq entry about it. (Proviso: I'm not using bash for this, but the recommendation is still valid.)
So I tried rewriting the script to separate paths in a pipe, like this"
#!/bin/sh
for what in "$#"; do
echo "${PATH}" | awk -vRS=: 1 | while read loc ; do
if [ -f "${loc}/${what}" ]; then
echo "${loc}/${what}"
exit 0
fi
done
done
exit 1
I'm not sure if this gives me any real advantage (since $loc is still inside quotes), but it also doesn't work because for some reason, the exit 0 seems to be ignored. Or ... it exits something (the sub-shell with the while loop that terminates the pipe, maybe), but the script exits with a value of 1 every time.
What's a better way to step through directories in ${PATH} without the risk that special characters will confuse things?
Alternately, am I reinventing the wheel? Is there maybe a way to do this that's built in to existing shell tools?
This needs to run in both Linux and FreeBSD, which is why I'm writing it in Bourne instead of bash.
Thanks.
This doesn't directly answer your question, but does eliminate the need to parse PATH at all:
onewhich () {
for what in "$#"; do
which "$what" 2>/dev/null && break
done
}
This just calls which on each command on the input list until it finds a match.
To parse PATH, you can simply set `IFS=':'.
if [ "${IFS:-x}" = "${IFS-x}" ]; then
# Only preserve the value of IFS if it is currently set
OLDIFS=$IFS
fi
IFS=":"
for f in $PATH; do # Do not quote $PATH, to allow word splitting
echo $f
done
if [ "${OLDIFS:-x}" = "${OLDIFS-x}" ]; then
IFS=$OLDIFS
fi
The above will fail if any of the directories in PATH actually contain colons.
Your first method looks to me as if it should work. In practical terms, if it's really the $PATH you'll be searching, it's unlikely you'll have spaces and newlines embedded in directories there. If you do, it's probably time to refactor.
But still, I don't think you're at risk from the possibility of bad names clobbering your loop, since you're wrapping variables in quotes. At worst, I suspect you might miss the odd valid executable, but I can't see how the script would generate errors. (I don't see how the script would miss valid executables, and I haven't tested - I'm just saying I don't see problems at first glance.)
As for your second question, about the loop, I think you've hit the nail on the head. When you run a pipe like this | that | while condition; do things; done, the while loop runs in its own shell at the end of the pipe. Exiting that shell may terminate the actions of the pipe, but that only brings you back to the parent shell, which has its own thread of execution that terminates with exit 1.
As for a better way to do this, I would consider which.
#!/bin/sh
for what in "$#"; do
which "$what"
done | head -1
And if you really want the exit values as well:
#!/bin/sh
for what in "$#"; do
which "$what" && exit 0
done
exit 1
The second might even be fewer resources, as it doesn't have to open a file handle and pipe through head.
You can also split your path using IFS. For example, if you wanted to wrap your loops the other way around, you could do this:
#!/bin/sh
IFS=":"
for loc in $PATH; do
for what in "$#"; do
if [ -x "$loc"/"$what" ]; then
echo "$loc"/"$what"
exit 0
fi
done
done
exit 1
Note that under normal circumstances, you might want to save the old value of $IFS, but you seem to be doing things in a stand-alone script, so the "new" value gets thrown out when the script exits.
All the above code is untested. YMMV.
Another way to get around the need to parse PATH at all is to run the builtin type command in new shell with a stripped environment (i. e. there simply are no functions or aliases to look up; cf. env -i sh -c 'type cmd 2>/dev/null).
# using `cmd` instead of $(cmd) for portability
onewhich() {
ec=0 # exit code
for cmd in "$#"; do
command -p env -i PATH="$PATH" sh -c '
export LC_ALL=C LANG=C
cmd="$1"
path="`type "$cmd" 2>/dev/null`"
if [ X"$path" = "X" ]; then
printf "%s\n" "error: command \"${cmd}\" not found in PATH" 1>&2
exit 1
else
case "$path" in
*\ /*)
path="/${path#*/}"
printf "%s\n" "$path";;
*)
printf "%s\n" "error: no disk file: $path" 1>&2
exit 1;;
esac
exit 0
fi
' _ "$cmd"
[ $? != 0 ] && ec=1
done
[ $ec != 0 ] && return 1
}
onewhich awk ls sed
onewhich builtin
onewhich if
Since which on success returns two full command paths if two commands are specified as arguments, exit 0 in the first onewhich script above aborts the program prematurely. In addition, if two commands are specified as arguments to which, the exit code of which is set to 1 even if only one command lookup failed (cf. which awk sedxyz ls; echo $?). To mimic this behaviour of the which command it is necessary to toggle on/off two variables (cnt and nomatches below).
onewhich() (
IFS=":"
nomatches=0
for cmd in "$#"; do
cnt=0
for loc in $PATH ; do
if [ $cnt = 0 ] && [ -x "$loc"/"$cmd" ]; then
echo "$loc"/"$cmd"
cnt=1
fi
done
[ $cnt = 0 ] && nomatches=1
done
[ $nomatches = 1 ] && exit 1 || exit 0 # exit 1: at least one cmd was not in PATH
)
onewhich awk ls sed
onewhich awk lsxyz sed
onewhich builtin
onewhich if

Shell: Find Matching Lines Across Many Files

I am trying to use a shell script (well a "one liner") to find any common lines between around 50 files.
Edit: Note I am looking for a line (lines) that appears in all the files
So far i've tried grep grep -v -x -f file1.sp * which just matches that files contents across ALL the other files.
I've also tried grep -v -x -f file1.sp file2.sp | grep -v -x -f - file3.sp | grep -v -x -f - file4.sp | grep -v -x -f - file5.sp etc... but I believe that searches using the files to be searched as STD in not the pattern to match on.
Does anyone know how to do this with grep or another tool?
I don't mind if it takes a while to run, I've got to add a few lines of code to around 500 files and wanted to find a common line in each of them for it to insert 'after' (they were originally just c&p from one file so hopefully there are some common lines!)
Thanks for your time,
When I first read this I thought you were trying to find 'any common lines'. I took this as meaning "find duplicate lines". If this is the case, the following should suffice:
sort *.sp | uniq -d
Upon re-reading your question, it seems that you are actually trying to find lines that 'appear in all the files'. If this is the case, you will need to know the number of files in your directory:
find . -type f -name "*.sp" | wc -l
If this returns the number 50, you can then use awk like this:
WHINY_USERS=1 awk '{ array[$0]++ } END { for (i in array) if (array[i] == 50) print i }' *.sp
You can consolidate this process and write a one-liner like this:
WHINY_USERS=1 awk -v find=$(find . -type f -name "*.sp" | wc -l) '{ array[$0]++ } END { for (i in array) if (array[i] == find) print i }' *.sp
old, bash answer (O(n); opens 2 * n files)
From #mjgpy3 answer, you just have to make a for loop and use comm, like this:
#!/bin/bash
tmp1="/tmp/tmp1$RANDOM"
tmp2="/tmp/tmp2$RANDOM"
cp "$1" "$tmp1"
shift
for file in "$#"
do
comm -1 -2 "$tmp1" "$file" > "$tmp2"
mv "$tmp2" "$tmp1"
done
cat "$tmp1"
rm "$tmp1"
Save in a comm.sh, make it executable, and call
./comm.sh *.sp
assuming all your filenames end with .sp.
Updated answer, python, opens only each file once
Looking at the other answers, I wanted to give one that opens once each file without using any temporary file, and supports duplicated lines. Additionally, let's process the files in parallel.
Here you go (in python3):
#!/bin/env python
import argparse
import sys
import multiprocessing
import os
EOLS = {'native': os.linesep.encode('ascii'), 'unix': b'\n', 'windows': b'\r\n'}
def extract_set(filename):
with open(filename, 'rb') as f:
return set(line.rstrip(b'\r\n') for line in f)
def find_common_lines(filenames):
pool = multiprocessing.Pool()
line_sets = pool.map(extract_set, filenames)
return set.intersection(*line_sets)
if __name__ == '__main__':
# usage info and argument parsing
parser = argparse.ArgumentParser()
parser.add_argument("in_files", nargs='+',
help="find common lines in these files")
parser.add_argument('--out', type=argparse.FileType('wb'),
help="the output file (default stdout)")
parser.add_argument('--eol-style', choices=EOLS.keys(), default='native',
help="(default: native)")
args = parser.parse_args()
# actual stuff
common_lines = find_common_lines(args.in_files)
# write results to output
to_print = EOLS[args.eol_style].join(common_lines)
if args.out is None:
# find out stdout's encoding, utf-8 if absent
encoding = sys.stdout.encoding or 'utf-8'
sys.stdout.write(to_print.decode(encoding))
else:
args.out.write(to_print)
Save it into a find_common_lines.py, and call
python ./find_common_lines.py *.sp
More usage info with the --help option.
Combining this two answers (ans1 and ans2) I think you can get the result you are needing without sorting the files:
#!/bin/bash
ans="matching_lines"
for file1 in *
do
for file2 in *
do
if [ "$file1" != "$ans" ] && [ "$file2" != "$ans" ] && [ "$file1" != "$file2" ] ; then
echo "Comparing: $file1 $file2 ..." >> $ans
perl -ne 'print if ($seen{$_} .= #ARGV) =~ /10$/' $file1 $file2 >> $ans
fi
done
done
Simply save it, give it execution rights (chmod +x compareFiles.sh) and run it. It will take all the files present in the current working directory and will make an all-vs-all comparison leaving in the "matching_lines" file the result.
Things to be improved:
Skip directories
Avoid comparing all the files two times (file1 vs file2 and file2 vs file1).
Maybe add the line number next to the matching string
Hope this helps.
Best,
Alan Karpovsky
See this answer. I originally though a diff sounded like what you were asking for, but this answer seems much more appropriate.

Extracting n rows of text from a large csv file

I have a CSV file (foo.csv) with 200,000 rows. I need to break it into four files (foo1.csv, foo2.csv... etc.) with 50,000 rows each.
I already tried simple ctrl-v/-c using gui text editors, but the my computer slows to a halt.
What unix command(s) could I use to accomplish this task?
I don't have a terminal handy to try it out, but it should be just split -d -l 50000 foo.csv.
Hopefully the naming isn't terribly important because with the -d option, the output files will be named foo.csv00 .. foo.csv03. You can add the -a 1 option so that the suffixes are 0-3, but there's no simple way to get the suffix to be injected into the middle of the filename.
you should use head and tail.
head -n 50000 myfile > part1.csv
head -n 100000 myfile | tail -n 50000 > part2.csv
head -n 150000 myfile | tail -n 50000 > part3.csv
etc ...
Else, but with no control on file names, you can use unix command split.
sed -n 2000,4000p somefile.txt
will print from lines 2000 to 4000 to stdout.
split -l50000 foo.csv
You can use sed
I wrote this little shell script for this topic very similar at yours.
This shell script + awk works fine for me:
#!/bin/bash
awk -v initial_line=$1 -v end_line=$2 '{
if (NR >= initial_line && NR <= end_line)
print $0
}' $3
Used with this sample file (file.txt):
one
two
three
four
five
six
The command (it will extract from second to fourth line in the file):
edu#debian5:~$./script.sh 2 4 file.txt
Output of this command:
two
three
four
Of course, you can improve it, for example by testing that all argument values are the expected :-)

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