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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

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 14 mins ago

Jeff Schaller

42.6k1159136

asked 19 mins ago

yaelyael

2,63022571

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

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 14 mins ago

Jeff Schaller

42.6k1159136

asked 19 mins ago

yaelyael

2,63022571

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  See also Is there a way to sum up the size of files listed?
  
  – Jeff Schaller
  12 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 14 mins ago

Jeff Schaller

42.6k1159136

asked 19 mins ago

yaelyael

2,63022571

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

share|improve this question

edited 14 mins ago

Jeff Schaller

42.6k1159136

asked 19 mins ago

yaelyael

2,63022571

share|improve this question

edited 14 mins ago

Jeff Schaller

42.6k1159136

asked 19 mins ago

yaelyael

2,63022571

edited 14 mins ago

Jeff Schaller

42.6k1159136

edited 14 mins ago

Jeff Schaller

42.6k1159136

asked 19 mins ago

yaelyael

2,63022571

asked 19 mins ago

yaelyael

2,63022571

2 Answers
2

active

oldest

votes

3

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 10 mins ago

answered 15 mins ago

Stephen KittStephen Kitt

174k24397472

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I think output of du also contains entry size ., so we should avoid that.
  
  – PRY
  19 secs ago
  

  
  
  
  

add a comment | 

2

It is simple you can use:

 du  | awk 'BEGIN{i=0} {i=i+$1} END{print i}'

But wait, it will not work because output of du is like:

1 file

2 filee

2 .

So we need to avoid the last entry because it tells about total size of current directory, and the calculated sum is also equals to that, so instead of that use:

du | 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 1 min ago

answered 15 mins ago

PRYPRY

2,57831026

• 
  
  
  

  
  1
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

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3

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 10 mins ago

answered 15 mins ago

Stephen KittStephen Kitt

174k24397472

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I think output of du also contains entry size ., so we should avoid that.
  
  – PRY
  19 secs ago
  

  
  
  
  

add a comment | 

3

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 10 mins ago

answered 15 mins ago

Stephen KittStephen Kitt

174k24397472

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I think output of du also contains entry size ., so we should avoid that.
  
  – PRY
  19 secs ago
  

  
  
  
  

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 10 mins ago

answered 15 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 10 mins ago

answered 15 mins ago

Stephen KittStephen Kitt

174k24397472

answered 15 mins ago

Stephen KittStephen Kitt

174k24397472

answered 15 mins ago

Stephen KittStephen Kitt

174k24397472

2

It is simple you can use:

 du  | awk 'BEGIN{i=0} {i=i+$1} END{print i}'

But wait, it will not work because output of du is like:

1 file

2 filee

2 .

So we need to avoid the last entry because it tells about total size of current directory, and the calculated sum is also equals to that, so instead of that use:

du | 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 1 min ago

answered 15 mins ago

PRYPRY

2,57831026

• 
  
  
  

  
  1
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

2

It is simple you can use:

 du  | awk 'BEGIN{i=0} {i=i+$1} END{print i}'

But wait, it will not work because output of du is like:

1 file

2 filee

2 .

So we need to avoid the last entry because it tells about total size of current directory, and the calculated sum is also equals to that, so instead of that use:

du | 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 1 min ago

answered 15 mins ago

PRYPRY

2,57831026

• 
  
  
  

  
  1
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

It is simple you can use:

 du  | awk 'BEGIN{i=0} {i=i+$1} END{print i}'

But wait, it will not work because output of du is like:

1 file

2 filee

2 .

So we need to avoid the last entry because it tells about total size of current directory, and the calculated sum is also equals to that, so instead of that use:

du | 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 1 min ago

answered 15 mins ago

PRYPRY

2,57831026

 du  | awk 'BEGIN{i=0} {i=i+$1} END{print i}'

1 file

2 filee

2 .

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

du -s directory

share|improve this answer

edited 1 min ago

answered 15 mins ago

PRYPRY

2,57831026

answered 15 mins ago

PRYPRY

2,57831026

answered 15 mins ago

PRYPRY

2,57831026