how to sum file size from ls like output log with Bytes, KiB, MiB, GiB - bash

Question

I have a pre-computed ls like output (it is not from actual ls command) and I cannot modify or recalcuate it. It looks like as follows:
2016-10-14 14:52:09 0 Bytes folder/
2020-04-18 05:19:04 201 Bytes folder/file1.txt
2019-10-16 00:32:44 201 Bytes folder/file2.txt
2019-08-26 06:29:46 201 Bytes folder/file3.txt
2020-07-08 16:13:56 411 Bytes folder/file4.txt
2020-04-18 03:03:34 201 Bytes folder/file5.txt
2019-10-16 08:27:11 1.1 KiB folder/file6.txt
2019-10-16 10:13:52 201 Bytes folder/file7.txt
2019-10-16 08:44:35 920 Bytes folder/file8.txt
2019-02-17 14:43:10 590 Bytes folder/file9.txt
The log may have GiB, MiB, KiB, Bytes at least. Among possibile values are zero values, or values w/wo comma for each of prefixes:
0 Bytes
3.9 KiB
201 Bytes
2.0 KiB
2.7 MiB
1.3 GiB
A similar approach is the following
awk 'BEGIN{ pref[1]="K"; pref[2]="M"; pref[3]="G";} { total = total + $1; x = $1; y = 1; while( x > 1024 ) { x = (x + 1023)/1024; y++; } printf("%g%s\t%s\n",int(x*10)/10,pref[y],$2); } END { y = 1; while( total > 1024 ) { total = (total + 1023)/1024; y++; } printf("Total: %g%s\n",int(total*10)/10,pref[y]); }'
but does not work correctly in my case:
$ head -n 10 files_sizes.log | awk '{print $3,$4}' | sort | awk 'BEGIN{ pref[1]="K"; pref[2]="M"; pref[3]="G";} { total = total + $1; x = $1; y = 1; while( x > 1024 ) { x = (x + 1023)/1024; y++; } printf("%g%s\t%s\n",int(x*10)/10,pref[y],$2); } END { y = 1; while( total > 1024 ) { total = (total + 1023)/1024; y++; } printf("Total: %g%s\n",int(total*10)/10,pref[y]); }'
0K Bytes
1.1K KiB
201K Bytes
201K Bytes
201K Bytes
201K Bytes
201K Bytes
411K Bytes
590K Bytes
920K Bytes
Total: 3.8M
This output wrongly calculate the size. My desidered output is the correct total sum of the input log file:
0 Bytes
201 Bytes
201 Bytes
201 Bytes
411 Bytes
201 Bytes
1.1 KiB
201 Bytes
920 Bytes
590 Bytes
Total: 3.95742 KiB
NOTE
The correct value as the result of the sum of the Bytes is
201 * 5 + 590 + 920 = 2926, so the total adding the KiB is
2.857422 + 1.1 = 3,95742 KiB = 4052.400 Bytes
[UPDATE]
I have update with the comparison of the results from KamilCuk and Ted Lyngmo and Walter A solutions that gives pretty much the same values:
$ head -n 10 files_sizes.log | tr -s ' ' | cut -d' ' -f3,4 | sed 's/ //; s/Bytes//; s/B//' | gnumfmt --from=auto | awk '{s+=$1}END{print s " Bytes"}'
117538 Bytes
$ head -n 1000 files_sizes.log | tr -s ' ' | cut -d' ' -f3,4 | sed 's/ //; s/Bytes//; s/B//' | gnumfmt --from=auto | awk '{s+=$1}END{print s " Bytes"}'
1225857 Bytes
$ head -n 10000 files_sizes.log | tr -s ' ' | cut -d' ' -f3,4 | sed 's/ //; s/Bytes//; s/B//' | gnumfmt --from=auto | awk '{s+=$1}END{print s " Bytes"}'
12087518 Bytes
$ head -n 1000000 files_sizes.log | tr -s ' ' | cut -d' ' -f3,4 | sed 's/ //; s/Bytes//; s/B//' | gnumfmt --from=auto | awk '{s+=$1}END{print s " Bytes"}'
77238840381 Bytes
$ head -n 100000000 files_sizes.log | tr -s ' ' | cut -d' ' -f3,4 | sed 's/ //; s/Bytes//; s/B//' | gnumfmt --from=auto | awk '{s+=$1}END{print s " Bytes"}'
2306569381835 Bytes
and
$ head -n 10 files_sizes.log | ./count_files.sh
3.957422 KiB
$ head -n 1000 files_sizes.log | ./count_files.sh
1.168946 MiB
$ head -n 10000 files_sizes.log | ./count_files.sh
11.526325 MiB
$ head -n 1000000 files_sizes.log | ./count_files.sh
71.934024 GiB
$ head -n 100000000 files_sizes.log | ./count_files.sh
2.097807 TiB
and
(head -n 100000000 files_sizes.log | tr -s ' ' | cut -d' ' -f3,4 | sed 's/Bytes//; s/KiB/* 1024/; s/MiB/* 1024 * 1024/;s/GiB/* 1024 * 1024 * 1024/; s/$/ + /; $s/+ //' | tr -d '\n' ; echo) | bc
2306563692898.8
where
2.097807 TiB = 2.3065631893 TB = 2306569381835 Bytes
Computationally, I have compared all the three solution for speed:
$ time head -n 100000000 files_sizes.log | ./count_files.sh
2.097807 TiB
real 2m7.956s
user 2m10.023s
sys 0m1.696s
$ time head -n 100000000 files_sizes.log | tr -s ' ' | cut -d' ' -f3,4 | sed 's/ //; s/Bytes//; s/B//' | gnumfmt --from=auto | awk '{s+=$1}END{print s " Bytes"}'
2306569381835 Bytes
real 4m12.896s
user 5m45.750s
sys 0m4.026s
$ time (head -n 100000000 files_sizes.log | tr -s ' ' | cut -d' ' -f3,4 | sed 's/Bytes//; s/KiB/* 1024/; s/MiB/* 1024 * 1024/;s/GiB/* 1024 * 1024 * 1024/; s/$/ + /; $s/+ //' | tr -d '\n' ; echo) | bc
2306563692898.8
real 4m31.249s
user 6m40.072s
sys 0m4.252s

Answer 1

Use numfmt to convert those numbers.
cat <<EOF |
2016-10-14 14:52:09 0 Bytes folder/
2020-04-18 05:19:04 201 Bytes folder/file1.txt
2019-10-16 00:32:44 201 Bytes folder/file2.txt
2019-08-26 06:29:46 201 Bytes folder/file3.txt
2020-07-08 16:13:56 411 Bytes folder/file4.txt
2020-04-18 03:03:34 201 Bytes folder/file5.txt
2019-10-16 08:27:11 1.1 KiB folder/file6.txt
2019-10-16 10:13:52 201 Bytes folder/file7.txt
2019-10-16 08:44:35 920 Bytes folder/file8.txt
2019-02-17 14:43:10 590 Bytes folder/file9.txt
2019-02-17 14:43:10 3.9 KiB folder/file9.txt
2019-02-17 14:43:10 2.7 MiB folder/file9.txt
2019-02-17 14:43:10 1.3 GiB folder/file9.txt
EOF
# extract 3rd and 4th column
tr -s ' ' | cut -d' ' -f3,4 |
# Remove space, remove "Bytes", remove "B"
sed 's/ //; s/Bytes//; s/B//' |
# convert to numbers
numfmt --from=auto |
# sum
awk '{s+=$1}END{print s}'
outputs the sum.

Answer 2

For input like the described:
2016-10-14 14:52:09 0 Bytes folder/
2020-04-18 05:19:04 201 Bytes folder/file1.txt
2019-10-16 00:32:44 201 Bytes folder/file2.txt
2019-08-26 06:29:46 201 Bytes folder/file3.txt
2020-07-08 16:13:56 411 Bytes folder/file4.txt
2020-04-18 03:03:34 201 Bytes folder/file5.txt
2019-10-16 08:27:11 1.1 KiB folder/file6.txt
2019-10-16 10:13:52 201 Bytes folder/file7.txt
2019-10-16 08:44:35 920 Bytes folder/file8.txt
2019-02-17 14:43:10 590 Bytes folder/file9.txt
You could use a table of units that you'd like to be able to decode:
BEGIN {
unit["Bytes"] = 1;
unit["kB"] = 10**3;
unit["MB"] = 10**6;
unit["GB"] = 10**9;
unit["TB"] = 10**12;
unit["PB"] = 10**15;
unit["EB"] = 10**18;
unit["ZB"] = 10**21;
unit["YB"] = 10**24;
unit["KB"] = 1024;
unit["KiB"] = 1024**1;
unit["MiB"] = 1024**2;
unit["GiB"] = 1024**3;
unit["TiB"] = 1024**4;
unit["PiB"] = 1024**5;
unit["EiB"] = 1024**6;
unit["ZiB"] = 1024**7;
unit["YiB"] = 1024**8;
}
Then just sum it up in the main loop:
{
if($4 in unit) total += $3 * unit[$4];
else printf("ERROR: Can't decode unit at line %d: %s\n", NR, $0);
}
And print the result at the end:
END {
binaryunits[0] = "Bytes";
binaryunits[1] = "KiB";
binaryunits[2] = "MiB";
binaryunits[3] = "GiB";
binaryunits[4] = "TiB";
binaryunits[5] = "PiB";
binaryunits[6] = "EiB";
binaryunits[7] = "ZiB";
binaryunits[8] = "YiB";
for(i = 8;; --i) {
if(total >= 1024**i || i == 0) {
printf("%.3f %s\n", total/(1024**i), binaryunits[i]);
break;
}
}
}
Output:
3.957 KiB
Note that you can add a she-bang to beginning of the awk-script to make it possible to run it on its own so that you won't have to put it in a bash script:
#!/usr/bin/awk -f

Answer 3

You can parse the input before sending them to bc:
echo "0 Bytes
3.9 KiB
201 Bytes
2.0 KiB
2.7 MiB
1.3 GiB" |
sed 's/Bytes//; s/KiB/* 1024/; s/MiB/* 1024 * 1024/;
s/GiB/* 1024 * 1024 * 1024/; s/$/ + /' |
tr -d '\n' |
sed 's/+ $/\n/' |
bc
When your sed doesn't support \n, you can try replacing the '\n' with a real newline like
sed 's/+ $/
/'
or add an echo after parsing (and move part of the last sed into the first command for removing the last +)
(echo "0 Bytes
3.9 KiB
201 Bytes
2.0 KiB
2.7 MiB
1.3 GiB" | sed 's/Bytes//; s/KiB/* 1024/; s/MiB/* 1024 * 1024/;
s/GiB/* 1024 * 1024 * 1024/; s/$/ + /; $s/+ //' | tr -d '\n' ; echo) | bc

Answer 4

Very good idea from #KamilCuk to make use of numfmt. Based on his answer, here is an alternative command which uses a single awk call wrapping numfmt with a two-way pipe. It requires a recent version of GNU awk (ok with 5.0.1, unstable with 4.1.4, not tested in between).
LC_NUMERIC=C gawk '
BEGIN {
conv = "numfmt --from=auto"
PROCINFO[conv, "pty"] = 1
}
{
sub(/B.*/, "", $4)
print $3 $4 |& conv
conv |& getline val
sum += val
}
END { print sum }
' input
Notes
LC_NUMERIC=C (bash/ksh/zsh) is for portability on systems using a non-english locale.
PROCINFO[conv, "pty"] = 1 lets have the output of numfmt flushed on each line (to avoid a dealock).

Answer 5

Let me give you a better way to work with ls: don't use it as a command, but as a find switch:
find . -maxdepth 1 -ls
This returns file sizes in a uniform unity, as explained in find's manpage, which makes it far easier to do calculations on.
Good luck