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awk + sum all numbers

Is there a way to sum up the size of files listed?How can I quickly sum all numbers in a file?AWK sum column in file specify as a argumentHow to sum many numbers inside 2D array using awkawk + count field separator in csv and print line numberHow to join duplicates and sum their numbers with awkawk to sum the numbers(floating) and group it on a unique keySumming rows in a new column using sed, awk and perl?Check which process is using most memory and summary total used memoryAWK how to count sumSum and count in for loop

2

1

We want to calculate the first numbers that we get from du

du -b /tmp/*

6       /tmp/216c6f99-6671-4865-b8bc-7205f5388752_resources

668669  /tmp/hadoop7887078727316788325.tmp

6       /tmp/hadoop-hdfs

42456   /tmp/hive

32786   /tmp/hsperfdata_hdfs

6       /tmp/hsperfdata_hive

32786   /tmp/hsperfdata_root

262244  /tmp/hsperfdata_yarn

so final sum will be

sum=6+668669+6+42456+32786+6+32786+262244





echo $sum

How we can do it by awk or perl one liners?

linux shell-script awk perl disk-usage

share|improve this question

edited 23 mins ago

PRY

2,57831026

asked 48 mins ago

yaelyael

2,63722571

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  du -bs /tmp would get you the answer too
  
  – roaima
  44 mins ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  See How can I quickly sum all numbers in a file?
  
  – glenn jackman
  42 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  See also Is there a way to sum up the size of files listed?
  
  – Jeff Schaller
  40 mins ago
  
  
  

  
  
  
  

add a comment | 

2

1

We want to calculate the first numbers that we get from du

du -b /tmp/*

6       /tmp/216c6f99-6671-4865-b8bc-7205f5388752_resources

668669  /tmp/hadoop7887078727316788325.tmp

6       /tmp/hadoop-hdfs

42456   /tmp/hive

32786   /tmp/hsperfdata_hdfs

6       /tmp/hsperfdata_hive

32786   /tmp/hsperfdata_root

262244  /tmp/hsperfdata_yarn

so final sum will be

sum=6+668669+6+42456+32786+6+32786+262244





echo $sum

How we can do it by awk or perl one liners?

linux shell-script awk perl disk-usage

share|improve this question

edited 23 mins ago

PRY

2,57831026

asked 48 mins ago

yaelyael

2,63722571

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  du -bs /tmp would get you the answer too
  
  – roaima
  44 mins ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  See How can I quickly sum all numbers in a file?
  
  – glenn jackman
  42 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  See also Is there a way to sum up the size of files listed?
  
  – Jeff Schaller
  40 mins ago
  
  
  

  
  
  
  

add a comment | 

We want to calculate the first numbers that we get from du

du -b /tmp/*

6       /tmp/216c6f99-6671-4865-b8bc-7205f5388752_resources

668669  /tmp/hadoop7887078727316788325.tmp

6       /tmp/hadoop-hdfs

42456   /tmp/hive

32786   /tmp/hsperfdata_hdfs

6       /tmp/hsperfdata_hive

32786   /tmp/hsperfdata_root

262244  /tmp/hsperfdata_yarn

so final sum will be

sum=6+668669+6+42456+32786+6+32786+262244





echo $sum

How we can do it by awk or perl one liners?

linux shell-script awk perl disk-usage

share|improve this question

edited 23 mins ago

PRY

2,57831026

asked 48 mins ago

yaelyael

2,63722571

du -b /tmp/*

6       /tmp/216c6f99-6671-4865-b8bc-7205f5388752_resources

668669  /tmp/hadoop7887078727316788325.tmp

6       /tmp/hadoop-hdfs

42456   /tmp/hive

32786   /tmp/hsperfdata_hdfs

6       /tmp/hsperfdata_hive

32786   /tmp/hsperfdata_root

262244  /tmp/hsperfdata_yarn

sum=6+668669+6+42456+32786+6+32786+262244





echo $sum

share|improve this question

edited 23 mins ago

PRY

2,57831026

asked 48 mins ago

yaelyael

2,63722571

share|improve this question

edited 23 mins ago

PRY

2,57831026

asked 48 mins ago

yaelyael

2,63722571

edited 23 mins ago

PRY

2,57831026

edited 23 mins ago

PRY

2,57831026

asked 48 mins ago

yaelyael

2,63722571

asked 48 mins ago

yaelyael

2,63722571

3 Answers
3

active

oldest

votes

4

In AWK:

{ sum += $1 }

END { print sum }

So

du -b /tmp/* | awk '{ sum += $1 } END { print sum }'

du -s will also calculate the sum for you (on all subdirectories and files in /tmp, including hidden ones):

du -sb /tmp

share|improve this answer

edited 38 mins ago

answered 44 mins ago

Stephen KittStephen Kitt

174k24397472

add a comment | 

2

It is simple you can use:

 du -b /tmp/* | awk 'BEGIN{i=0} {i=i+$1} END{print i}'

If you are not using wildcard, if you are using directory name like /tmp, then you need to avoid the last entry because output of du -b /tmp is like:

size1 file1

size2 file2

size_total .

So now you should avoid this last entry, so use:

du -b /tmp | awk 'BEGIN{i=0} {if( $2 != "." ){i=i+$1}} END{print i}'

However you can also use -s option, it will calculate the summary for you then you don't need to add the values, just print the last one, i.e.:

du -s directory

share|improve this answer

edited 20 mins ago

answered 43 mins ago

PRYPRY

2,57831026

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  variables initialize to zero, if you'd like to golf some bytes off :)
  
  – Jeff Schaller
  42 mins ago
  

  
  
  
  

add a comment | 

-1

You can also produce a total sum of selected files with du -c. This works even if an argument of du is not a directory, what is not the case of du -s:

$ du -sb file1 file2

17  file1

18  file2



$ du -cb file1 file2

17  file1

18  file2

35  total

BTW, for interactive use I recommend adding -h option instead of -b or any other multiplier of block-size. This will print the size in human readable unit format.

$ du -ch file1 file2

4.0K    file1

4.0K    file2

8.0K    total

share|improve this answer

answered 11 mins ago

jimmijjimmij

31.9k874108

add a comment | 

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4

In AWK:

{ sum += $1 }

END { print sum }

So

du -b /tmp/* | awk '{ sum += $1 } END { print sum }'

du -s will also calculate the sum for you (on all subdirectories and files in /tmp, including hidden ones):

du -sb /tmp

share|improve this answer

edited 38 mins ago

answered 44 mins ago

Stephen KittStephen Kitt

174k24397472

add a comment | 

4

In AWK:

{ sum += $1 }

END { print sum }

So

du -b /tmp/* | awk '{ sum += $1 } END { print sum }'

du -s will also calculate the sum for you (on all subdirectories and files in /tmp, including hidden ones):

du -sb /tmp

share|improve this answer

edited 38 mins ago

answered 44 mins ago

Stephen KittStephen Kitt

174k24397472

add a comment | 

In AWK:

{ sum += $1 }

END { print sum }

So

du -b /tmp/* | awk '{ sum += $1 } END { print sum }'

du -s will also calculate the sum for you (on all subdirectories and files in /tmp, including hidden ones):

du -sb /tmp

share|improve this answer

edited 38 mins ago

answered 44 mins ago

Stephen KittStephen Kitt

174k24397472

{ sum += $1 }

END { print sum }

du -b /tmp/* | awk '{ sum += $1 } END { print sum }'

du -sb /tmp

share|improve this answer

edited 38 mins ago

answered 44 mins ago

Stephen KittStephen Kitt

174k24397472

answered 44 mins ago

Stephen KittStephen Kitt

174k24397472

answered 44 mins ago

Stephen KittStephen Kitt

174k24397472

2

It is simple you can use:

 du -b /tmp/* | awk 'BEGIN{i=0} {i=i+$1} END{print i}'

If you are not using wildcard, if you are using directory name like /tmp, then you need to avoid the last entry because output of du -b /tmp is like:

size1 file1

size2 file2

size_total .

So now you should avoid this last entry, so use:

du -b /tmp | awk 'BEGIN{i=0} {if( $2 != "." ){i=i+$1}} END{print i}'

However you can also use -s option, it will calculate the summary for you then you don't need to add the values, just print the last one, i.e.:

du -s directory

share|improve this answer

edited 20 mins ago

answered 43 mins ago

PRYPRY

2,57831026

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  variables initialize to zero, if you'd like to golf some bytes off :)
  
  – Jeff Schaller
  42 mins ago
  

  
  
  
  

add a comment | 

2

It is simple you can use:

 du -b /tmp/* | awk 'BEGIN{i=0} {i=i+$1} END{print i}'

If you are not using wildcard, if you are using directory name like /tmp, then you need to avoid the last entry because output of du -b /tmp is like:

size1 file1

size2 file2

size_total .

So now you should avoid this last entry, so use:

du -b /tmp | awk 'BEGIN{i=0} {if( $2 != "." ){i=i+$1}} END{print i}'

However you can also use -s option, it will calculate the summary for you then you don't need to add the values, just print the last one, i.e.:

du -s directory

share|improve this answer

edited 20 mins ago

answered 43 mins ago

PRYPRY

2,57831026

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  variables initialize to zero, if you'd like to golf some bytes off :)
  
  – Jeff Schaller
  42 mins ago
  

  
  
  
  

add a comment | 

It is simple you can use:

 du -b /tmp/* | awk 'BEGIN{i=0} {i=i+$1} END{print i}'

If you are not using wildcard, if you are using directory name like /tmp, then you need to avoid the last entry because output of du -b /tmp is like:

size1 file1

size2 file2

size_total .

So now you should avoid this last entry, so use:

du -b /tmp | awk 'BEGIN{i=0} {if( $2 != "." ){i=i+$1}} END{print i}'

However you can also use -s option, it will calculate the summary for you then you don't need to add the values, just print the last one, i.e.:

du -s directory

share|improve this answer

edited 20 mins ago

answered 43 mins ago

PRYPRY

2,57831026

 du -b /tmp/* | awk 'BEGIN{i=0} {i=i+$1} END{print i}'

size1 file1

size2 file2

size_total .

du -b /tmp | awk 'BEGIN{i=0} {if( $2 != "." ){i=i+$1}} END{print i}'

du -s directory

share|improve this answer

edited 20 mins ago

answered 43 mins ago

PRYPRY

2,57831026

answered 43 mins ago

PRYPRY

2,57831026

answered 43 mins ago

PRYPRY

2,57831026

-1

You can also produce a total sum of selected files with du -c. This works even if an argument of du is not a directory, what is not the case of du -s:

$ du -sb file1 file2

17  file1

18  file2



$ du -cb file1 file2

17  file1

18  file2

35  total

BTW, for interactive use I recommend adding -h option instead of -b or any other multiplier of block-size. This will print the size in human readable unit format.

$ du -ch file1 file2

4.0K    file1

4.0K    file2

8.0K    total

share|improve this answer

answered 11 mins ago

jimmijjimmij

31.9k874108

add a comment | 

-1

You can also produce a total sum of selected files with du -c. This works even if an argument of du is not a directory, what is not the case of du -s:

$ du -sb file1 file2

17  file1

18  file2



$ du -cb file1 file2

17  file1

18  file2

35  total

BTW, for interactive use I recommend adding -h option instead of -b or any other multiplier of block-size. This will print the size in human readable unit format.

$ du -ch file1 file2

4.0K    file1

4.0K    file2

8.0K    total

share|improve this answer

answered 11 mins ago

jimmijjimmij

31.9k874108

add a comment | 

You can also produce a total sum of selected files with du -c. This works even if an argument of du is not a directory, what is not the case of du -s:

$ du -sb file1 file2

17  file1

18  file2



$ du -cb file1 file2

17  file1

18  file2

35  total

BTW, for interactive use I recommend adding -h option instead of -b or any other multiplier of block-size. This will print the size in human readable unit format.

$ du -ch file1 file2

4.0K    file1

4.0K    file2

8.0K    total

share|improve this answer

answered 11 mins ago

jimmijjimmij

31.9k874108

$ du -sb file1 file2

17  file1

18  file2



$ du -cb file1 file2

17  file1

18  file2

35  total

$ du -ch file1 file2

4.0K    file1

4.0K    file2

8.0K    total

share|improve this answer

answered 11 mins ago

jimmijjimmij

31.9k874108

answered 11 mins ago

jimmijjimmij

31.9k874108

answered 11 mins ago

jimmijjimmij

31.9k874108