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.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;

4

1

I have 2 folders one contains images files and the second contains text files each text file have the same name as the image file and contains information about the image.eg:

 -Labels:
 -1.txt
 -2.txt
 -3.txt
 -6.txt
 -Images:
 -1.jpg
 -2.jpg
 -3.jpg
 -4.jpg
 -5.jpg
 -6.jpg

I want to delete images who has not a text file(in this example:4.jpg,5.jpg), I found a method how to determinate the different files but I can't delete them.

diff <(ls -1 ./Images | sed s/.jpg//g) <( ls -1 ./Labels | sed s/.txt//g)

command-line bash

share|improve this question

edited Aug 5 at 10:35

pa4080

16.3k77 gold badges3434 silver badges8181 bronze badges

asked Aug 2 at 15:22

oussama bousselmioussama bousselmi

4155 bronze badges

add a comment | 

4

1

I have 2 folders one contains images files and the second contains text files each text file have the same name as the image file and contains information about the image.eg:

 -Labels:
 -1.txt
 -2.txt
 -3.txt
 -6.txt
 -Images:
 -1.jpg
 -2.jpg
 -3.jpg
 -4.jpg
 -5.jpg
 -6.jpg

I want to delete images who has not a text file(in this example:4.jpg,5.jpg), I found a method how to determinate the different files but I can't delete them.

diff <(ls -1 ./Images | sed s/.jpg//g) <( ls -1 ./Labels | sed s/.txt//g)

command-line bash

share|improve this question

edited Aug 5 at 10:35

pa4080

16.3k77 gold badges3434 silver badges8181 bronze badges

asked Aug 2 at 15:22

oussama bousselmioussama bousselmi

4155 bronze badges

add a comment | 

I have 2 folders one contains images files and the second contains text files each text file have the same name as the image file and contains information about the image.eg:

 -Labels:
 -1.txt
 -2.txt
 -3.txt
 -6.txt
 -Images:
 -1.jpg
 -2.jpg
 -3.jpg
 -4.jpg
 -5.jpg
 -6.jpg

I want to delete images who has not a text file(in this example:4.jpg,5.jpg), I found a method how to determinate the different files but I can't delete them.

diff <(ls -1 ./Images | sed s/.jpg//g) <( ls -1 ./Labels | sed s/.txt//g)

command-line bash

share|improve this question

edited Aug 5 at 10:35

pa4080

16.3k77 gold badges3434 silver badges8181 bronze badges

asked Aug 2 at 15:22

oussama bousselmioussama bousselmi

4155 bronze badges

 -Labels:
 -1.txt
 -2.txt
 -3.txt
 -6.txt
 -Images:
 -1.jpg
 -2.jpg
 -3.jpg
 -4.jpg
 -5.jpg
 -6.jpg

diff <(ls -1 ./Images | sed s/.jpg//g) <( ls -1 ./Labels | sed s/.txt//g)

share|improve this question

edited Aug 5 at 10:35

pa4080

16.3k77 gold badges3434 silver badges8181 bronze badges

asked Aug 2 at 15:22

oussama bousselmioussama bousselmi

4155 bronze badges

share|improve this question

edited Aug 5 at 10:35

pa4080

16.3k77 gold badges3434 silver badges8181 bronze badges

asked Aug 2 at 15:22

oussama bousselmioussama bousselmi

4155 bronze badges

edited Aug 5 at 10:35

pa4080

16.3k77 gold badges3434 silver badges8181 bronze badges

edited Aug 5 at 10:35

pa4080

16.3k77 gold badges3434 silver badges8181 bronze badges

asked Aug 2 at 15:22

oussama bousselmioussama bousselmi

4155 bronze badges

asked Aug 2 at 15:22

oussama bousselmioussama bousselmi

4155 bronze badges

3 Answers
3

active

oldest

votes

7

Here is a small bash script that can help you to solve this task:

#!/bin/bash
for file in Images/*.jpg
do
 if [[ ! -f "Labels/$(basename $file%.*).txt" ]]
 then
 echo rm "$file"
 fi
done

• Remove echo to do the actual changes.

The script must be executed in the parent directory, here it is formatted as inline command:

for f in Images/*.jpg; do if [[ ! -f "Labels/$(basename $f%.*).txt" ]]; then echo rm "$f"; fi; done

share|improve this answer

edited Aug 2 at 16:09

answered Aug 2 at 16:02

pa4080pa4080

16.3k77 gold badges3434 silver badges8181 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  thank you it works.
  
  – oussama bousselmi
  Aug 2 at 16:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I am way too fond of sed, your solution is more elegant, thanks!
  
  – Dries
  Aug 2 at 16:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  In zsh you can use $file:r instead of $file%.* to get the filename minus one extension. It's smarter about what that means, which can be an advantage over substring matches when you're dealing with mixed paths that don't all fit a simple pattern. (e.g. file="foo.bar/baz"; echo $file:r will return foo.bar/baz, not foo. Though file="foo.bar/.baz" will produce foo.bar/ by either method, so some care is still required.)
  
  – FeRD
  Aug 3 at 3:06
  
  

  
  
  
  

add a comment | 

5

I think I would do it like this:

for i in Images/*; do file=`echo $i | sed -e 's/jpg/txt/' -e 's/Images/Labels/'`; if [ ! -f "$file" ] ; then rm $i ; fi; done

If you want to make sure it works , before actually using it try this first:

for i in Images/*; do file=`echo $i | sed -e 's/jpg/txt/' -e 's/Images/Labels/'`; if [ ! -f "$file" ] ; then echo rm $i ; fi; done

It will show which commands will be executed.

share|improve this answer

edited Aug 2 at 16:06

pa4080

16.3k77 gold badges3434 silver badges8181 bronze badges

answered Aug 2 at 16:02

DriesDries

11811 silver badge99 bronze badges

add a comment | 

2

cd /path/to/Images
LIST=$(find /path/to/Labels -iname *.txt -printf "%f|" | sed 's/.txt/.jpg/g')
rm -i !($LIST)

1. Generates a list of txt files.


2. Changes their extension to jpg.


3. 
   

   Removes anything that is not on the list.

   
   
   
   • 
     -i is just for safety.
   
   

share|improve this answer

edited Aug 2 at 16:22

answered Aug 2 at 16:15

RavexinaRavexina

35.7k1414 gold badges9797 silver badges126126 bronze badges

add a comment | 

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7

Here is a small bash script that can help you to solve this task:

#!/bin/bash
for file in Images/*.jpg
do
 if [[ ! -f "Labels/$(basename $file%.*).txt" ]]
 then
 echo rm "$file"
 fi
done

• Remove echo to do the actual changes.

The script must be executed in the parent directory, here it is formatted as inline command:

for f in Images/*.jpg; do if [[ ! -f "Labels/$(basename $f%.*).txt" ]]; then echo rm "$f"; fi; done

share|improve this answer

edited Aug 2 at 16:09

answered Aug 2 at 16:02

pa4080pa4080

16.3k77 gold badges3434 silver badges8181 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  thank you it works.
  
  – oussama bousselmi
  Aug 2 at 16:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I am way too fond of sed, your solution is more elegant, thanks!
  
  – Dries
  Aug 2 at 16:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  In zsh you can use $file:r instead of $file%.* to get the filename minus one extension. It's smarter about what that means, which can be an advantage over substring matches when you're dealing with mixed paths that don't all fit a simple pattern. (e.g. file="foo.bar/baz"; echo $file:r will return foo.bar/baz, not foo. Though file="foo.bar/.baz" will produce foo.bar/ by either method, so some care is still required.)
  
  – FeRD
  Aug 3 at 3:06
  
  

  
  
  
  

add a comment | 

7

Here is a small bash script that can help you to solve this task:

#!/bin/bash
for file in Images/*.jpg
do
 if [[ ! -f "Labels/$(basename $file%.*).txt" ]]
 then
 echo rm "$file"
 fi
done

• Remove echo to do the actual changes.

The script must be executed in the parent directory, here it is formatted as inline command:

for f in Images/*.jpg; do if [[ ! -f "Labels/$(basename $f%.*).txt" ]]; then echo rm "$f"; fi; done

share|improve this answer

edited Aug 2 at 16:09

answered Aug 2 at 16:02

pa4080pa4080

16.3k77 gold badges3434 silver badges8181 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  thank you it works.
  
  – oussama bousselmi
  Aug 2 at 16:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I am way too fond of sed, your solution is more elegant, thanks!
  
  – Dries
  Aug 2 at 16:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  In zsh you can use $file:r instead of $file%.* to get the filename minus one extension. It's smarter about what that means, which can be an advantage over substring matches when you're dealing with mixed paths that don't all fit a simple pattern. (e.g. file="foo.bar/baz"; echo $file:r will return foo.bar/baz, not foo. Though file="foo.bar/.baz" will produce foo.bar/ by either method, so some care is still required.)
  
  – FeRD
  Aug 3 at 3:06
  
  

  
  
  
  

add a comment | 

Here is a small bash script that can help you to solve this task:

#!/bin/bash
for file in Images/*.jpg
do
 if [[ ! -f "Labels/$(basename $file%.*).txt" ]]
 then
 echo rm "$file"
 fi
done

• Remove echo to do the actual changes.

The script must be executed in the parent directory, here it is formatted as inline command:

for f in Images/*.jpg; do if [[ ! -f "Labels/$(basename $f%.*).txt" ]]; then echo rm "$f"; fi; done

share|improve this answer

edited Aug 2 at 16:09

answered Aug 2 at 16:02

pa4080pa4080

16.3k77 gold badges3434 silver badges8181 bronze badges

#!/bin/bash
for file in Images/*.jpg
do
 if [[ ! -f "Labels/$(basename $file%.*).txt" ]]
 then
 echo rm "$file"
 fi
done

for f in Images/*.jpg; do if [[ ! -f "Labels/$(basename $f%.*).txt" ]]; then echo rm "$f"; fi; done

share|improve this answer

edited Aug 2 at 16:09

answered Aug 2 at 16:02

pa4080pa4080

16.3k77 gold badges3434 silver badges8181 bronze badges

answered Aug 2 at 16:02

pa4080pa4080

16.3k77 gold badges3434 silver badges8181 bronze badges

answered Aug 2 at 16:02

pa4080pa4080

16.3k77 gold badges3434 silver badges8181 bronze badges

5

I think I would do it like this:

for i in Images/*; do file=`echo $i | sed -e 's/jpg/txt/' -e 's/Images/Labels/'`; if [ ! -f "$file" ] ; then rm $i ; fi; done

If you want to make sure it works , before actually using it try this first:

for i in Images/*; do file=`echo $i | sed -e 's/jpg/txt/' -e 's/Images/Labels/'`; if [ ! -f "$file" ] ; then echo rm $i ; fi; done

It will show which commands will be executed.

share|improve this answer

edited Aug 2 at 16:06

pa4080

16.3k77 gold badges3434 silver badges8181 bronze badges

answered Aug 2 at 16:02

DriesDries

11811 silver badge99 bronze badges

add a comment | 

5

I think I would do it like this:

for i in Images/*; do file=`echo $i | sed -e 's/jpg/txt/' -e 's/Images/Labels/'`; if [ ! -f "$file" ] ; then rm $i ; fi; done

If you want to make sure it works , before actually using it try this first:

for i in Images/*; do file=`echo $i | sed -e 's/jpg/txt/' -e 's/Images/Labels/'`; if [ ! -f "$file" ] ; then echo rm $i ; fi; done

It will show which commands will be executed.

share|improve this answer

edited Aug 2 at 16:06

pa4080

16.3k77 gold badges3434 silver badges8181 bronze badges

answered Aug 2 at 16:02

DriesDries

11811 silver badge99 bronze badges

add a comment | 

I think I would do it like this:

for i in Images/*; do file=`echo $i | sed -e 's/jpg/txt/' -e 's/Images/Labels/'`; if [ ! -f "$file" ] ; then rm $i ; fi; done

If you want to make sure it works , before actually using it try this first:

for i in Images/*; do file=`echo $i | sed -e 's/jpg/txt/' -e 's/Images/Labels/'`; if [ ! -f "$file" ] ; then echo rm $i ; fi; done

It will show which commands will be executed.

share|improve this answer

edited Aug 2 at 16:06

pa4080

16.3k77 gold badges3434 silver badges8181 bronze badges

answered Aug 2 at 16:02

DriesDries

11811 silver badge99 bronze badges

for i in Images/*; do file=`echo $i | sed -e 's/jpg/txt/' -e 's/Images/Labels/'`; if [ ! -f "$file" ] ; then rm $i ; fi; done

for i in Images/*; do file=`echo $i | sed -e 's/jpg/txt/' -e 's/Images/Labels/'`; if [ ! -f "$file" ] ; then echo rm $i ; fi; done

share|improve this answer

edited Aug 2 at 16:06

pa4080

16.3k77 gold badges3434 silver badges8181 bronze badges

answered Aug 2 at 16:02

DriesDries

11811 silver badge99 bronze badges

edited Aug 2 at 16:06

pa4080

16.3k77 gold badges3434 silver badges8181 bronze badges

edited Aug 2 at 16:06

pa4080

16.3k77 gold badges3434 silver badges8181 bronze badges

answered Aug 2 at 16:02

DriesDries

11811 silver badge99 bronze badges

answered Aug 2 at 16:02

DriesDries

11811 silver badge99 bronze badges

2

cd /path/to/Images
LIST=$(find /path/to/Labels -iname *.txt -printf "%f|" | sed 's/.txt/.jpg/g')
rm -i !($LIST)

1. Generates a list of txt files.


2. Changes their extension to jpg.


3. 
   

   Removes anything that is not on the list.

   
   
   
   • 
     -i is just for safety.
   
   

share|improve this answer

edited Aug 2 at 16:22

answered Aug 2 at 16:15

RavexinaRavexina

35.7k1414 gold badges9797 silver badges126126 bronze badges

add a comment | 

2

cd /path/to/Images
LIST=$(find /path/to/Labels -iname *.txt -printf "%f|" | sed 's/.txt/.jpg/g')
rm -i !($LIST)

1. Generates a list of txt files.


2. Changes their extension to jpg.


3. 
   

   Removes anything that is not on the list.

   
   
   
   • 
     -i is just for safety.
   
   

share|improve this answer

edited Aug 2 at 16:22

answered Aug 2 at 16:15

RavexinaRavexina

35.7k1414 gold badges9797 silver badges126126 bronze badges

add a comment | 

cd /path/to/Images
LIST=$(find /path/to/Labels -iname *.txt -printf "%f|" | sed 's/.txt/.jpg/g')
rm -i !($LIST)

1. Generates a list of txt files.


2. Changes their extension to jpg.


3. 
   

   Removes anything that is not on the list.

   
   
   
   • 
     -i is just for safety.
   
   

share|improve this answer

edited Aug 2 at 16:22

answered Aug 2 at 16:15

RavexinaRavexina

35.7k1414 gold badges9797 silver badges126126 bronze badges

cd /path/to/Images
LIST=$(find /path/to/Labels -iname *.txt -printf "%f|" | sed 's/.txt/.jpg/g')
rm -i !($LIST)

share|improve this answer

edited Aug 2 at 16:22

answered Aug 2 at 16:15

RavexinaRavexina

35.7k1414 gold badges9797 silver badges126126 bronze badges

answered Aug 2 at 16:15

RavexinaRavexina

35.7k1414 gold badges9797 silver badges126126 bronze badges

answered Aug 2 at 16:15

RavexinaRavexina

35.7k1414 gold badges9797 silver badges126126 bronze badges