Estimate compressibility of file
Is there a quick and dirty way of estimating gzip-compressibility of a file without having to fully compress it with gzip?
I could, in bash, do
bc <<<"scale=2;$(gzip -c file | wc -c)/$(wc -c <file)"
This gives me the compression factor without having to write the gz file to disk; this way I can avoid replacing a file on disk with its gz version if the resultant disk space savings do not justify the trouble. But with this approach the file is indeed fully put through gzip; it's just that the output is piped to wc rather than written to disk.
Is there a way to get a rough compressibility estimate for a file without having gzip work on all its contents?
compression gzip
asked Sep 16 '14 at 16:48
iruvariruvar
11.7k62960
add a comment |
Is there a quick and dirty way of estimating gzip-compressibility of a file without having to fully compress it with gzip?
I could, in bash, do
bc <<<"scale=2;$(gzip -c file | wc -c)/$(wc -c <file)"
This gives me the compression factor without having to write the gz file to disk; this way I can avoid replacing a file on disk with its gz version if the resultant disk space savings do not justify the trouble. But with this approach the file is indeed fully put through gzip; it's just that the output is piped to wc rather than written to disk.
Is there a way to get a rough compressibility estimate for a file without having gzip work on all its contents?
compression gzip
asked Sep 16 '14 at 16:48
iruvariruvar
11.7k62960
add a comment |
Is there a quick and dirty way of estimating gzip-compressibility of a file without having to fully compress it with gzip?
I could, in bash, do
bc <<<"scale=2;$(gzip -c file | wc -c)/$(wc -c <file)"
This gives me the compression factor without having to write the gz file to disk; this way I can avoid replacing a file on disk with its gz version if the resultant disk space savings do not justify the trouble. But with this approach the file is indeed fully put through gzip; it's just that the output is piped to wc rather than written to disk.
Is there a way to get a rough compressibility estimate for a file without having gzip work on all its contents?
compression gzip
asked Sep 16 '14 at 16:48
iruvariruvar
11.7k62960
Is there a quick and dirty way of estimating gzip-compressibility of a file without having to fully compress it with gzip?
I could, in bash, do
bc <<<"scale=2;$(gzip -c file | wc -c)/$(wc -c <file)"
This gives me the compression factor without having to write the gz file to disk; this way I can avoid replacing a file on disk with its gz version if the resultant disk space savings do not justify the trouble. But with this approach the file is indeed fully put through gzip; it's just that the output is piped to wc rather than written to disk.
Is there a way to get a rough compressibility estimate for a file without having gzip work on all its contents?
compression gzip
compression gzip
asked Sep 16 '14 at 16:48
iruvariruvar
11.7k62960
asked Sep 16 '14 at 16:48
iruvariruvar
11.7k62960
asked Sep 16 '14 at 16:48
iruvariruvar
11.7k62960
asked Sep 16 '14 at 16:48
iruvariruvar
11.7k62960
asked Sep 16 '14 at 16:48
iruvariruvar
11.7k62960
11.7k62960
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
You could try compressing one every 10 blocks for instance to get an idea:
perl -MIPC::Open2 -nE 'BEGIN{$/=4096;open2(*I,*O,"gzip|wc -c")}
if ($. % 10 == 1) {print O $_; $l+=length}
END{close O; $c = <I>; say $c/$l}'
(here with 4K blocks).
answered Sep 16 '14 at 18:23
Stéphane ChazelasStéphane Chazelas
300k54564916
add a comment |
Here's a (hopefully equivalent) Python version of Stephane Chazelas's solution
python -c "
import zlib
from itertools import islice
from functools import partial
import sys
with open(sys.argv[1]) as f:
compressor = zlib.compressobj()
t, z = 0, 0.0
for chunk in islice(iter(partial(f.read, 4096), ''), 0, None, 10):
t += len(chunk)
z += len(compressor.compress(chunk))
z += len(compressor.flush())
print z/t
" file
edited Apr 13 '17 at 12:36
Community♦
1
answered Sep 16 '14 at 19:14
iruvariruvar
11.7k62960
I don't know if that's equivalent (as python is not by cup of coffee ;-b)), but that gives slightly different results (even for very large files where the gzip header size overhead can be ignored), possibly because zlib.compressobj uses different settings than gzip (I find that it's closer togzip -3) or maybe because zlib.compressobj compresses each chunk in isolation (as opposed to a stream as a whole). In any case both approaches should be good enough. I find that perl is slightly faster.
– Stéphane Chazelas
Sep 17 '14 at 11:55
@StéphaneChazelas, judging by the documentationcompressobjshould be acting on the stream as a whole. I find that both Python and Perl solutions produce approximately the same result on my data files (they differ in the second decimal place, that's good enough for me). Thanks for the great idea
– iruvar
Sep 17 '14 at 12:48
add a comment |
I had a multi-gigabyte file and I wasn't sure if it was compressed, so I test-compressed the first 10M bytes:
head -c 10000000 large_file.bin | gzip | wc -c
It's not prefect but it worked well for me.
answered 26 mins ago
aidanaidan
1063
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
You could try compressing one every 10 blocks for instance to get an idea:
perl -MIPC::Open2 -nE 'BEGIN{$/=4096;open2(*I,*O,"gzip|wc -c")}
if ($. % 10 == 1) {print O $_; $l+=length}
END{close O; $c = <I>; say $c/$l}'
(here with 4K blocks).
answered Sep 16 '14 at 18:23
Stéphane ChazelasStéphane Chazelas
300k54564916
add a comment |
You could try compressing one every 10 blocks for instance to get an idea:
perl -MIPC::Open2 -nE 'BEGIN{$/=4096;open2(*I,*O,"gzip|wc -c")}
if ($. % 10 == 1) {print O $_; $l+=length}
END{close O; $c = <I>; say $c/$l}'
(here with 4K blocks).
answered Sep 16 '14 at 18:23
Stéphane ChazelasStéphane Chazelas
300k54564916
add a comment |
You could try compressing one every 10 blocks for instance to get an idea:
perl -MIPC::Open2 -nE 'BEGIN{$/=4096;open2(*I,*O,"gzip|wc -c")}
if ($. % 10 == 1) {print O $_; $l+=length}
END{close O; $c = <I>; say $c/$l}'
(here with 4K blocks).
answered Sep 16 '14 at 18:23
Stéphane ChazelasStéphane Chazelas
300k54564916
You could try compressing one every 10 blocks for instance to get an idea:
perl -MIPC::Open2 -nE 'BEGIN{$/=4096;open2(*I,*O,"gzip|wc -c")}
if ($. % 10 == 1) {print O $_; $l+=length}
END{close O; $c = <I>; say $c/$l}'
(here with 4K blocks).
answered Sep 16 '14 at 18:23
Stéphane ChazelasStéphane Chazelas
300k54564916
answered Sep 16 '14 at 18:23
Stéphane ChazelasStéphane Chazelas
300k54564916
answered Sep 16 '14 at 18:23
Stéphane ChazelasStéphane Chazelas
300k54564916
answered Sep 16 '14 at 18:23
Stéphane ChazelasStéphane Chazelas
300k54564916
300k54564916
add a comment |
add a comment |
Here's a (hopefully equivalent) Python version of Stephane Chazelas's solution
python -c "
import zlib
from itertools import islice
from functools import partial
import sys
with open(sys.argv[1]) as f:
compressor = zlib.compressobj()
t, z = 0, 0.0
for chunk in islice(iter(partial(f.read, 4096), ''), 0, None, 10):
t += len(chunk)
z += len(compressor.compress(chunk))
z += len(compressor.flush())
print z/t
" file
edited Apr 13 '17 at 12:36
Community♦
1
answered Sep 16 '14 at 19:14
iruvariruvar
11.7k62960
I don't know if that's equivalent (as python is not by cup of coffee ;-b)), but that gives slightly different results (even for very large files where the gzip header size overhead can be ignored), possibly because zlib.compressobj uses different settings than gzip (I find that it's closer togzip -3) or maybe because zlib.compressobj compresses each chunk in isolation (as opposed to a stream as a whole). In any case both approaches should be good enough. I find that perl is slightly faster.
– Stéphane Chazelas
Sep 17 '14 at 11:55
@StéphaneChazelas, judging by the documentationcompressobjshould be acting on the stream as a whole. I find that both Python and Perl solutions produce approximately the same result on my data files (they differ in the second decimal place, that's good enough for me). Thanks for the great idea
– iruvar
Sep 17 '14 at 12:48
add a comment |
Here's a (hopefully equivalent) Python version of Stephane Chazelas's solution
python -c "
import zlib
from itertools import islice
from functools import partial
import sys
with open(sys.argv[1]) as f:
compressor = zlib.compressobj()
t, z = 0, 0.0
for chunk in islice(iter(partial(f.read, 4096), ''), 0, None, 10):
t += len(chunk)
z += len(compressor.compress(chunk))
z += len(compressor.flush())
print z/t
" file
edited Apr 13 '17 at 12:36
Community♦
1
answered Sep 16 '14 at 19:14
iruvariruvar
11.7k62960
I don't know if that's equivalent (as python is not by cup of coffee ;-b)), but that gives slightly different results (even for very large files where the gzip header size overhead can be ignored), possibly because zlib.compressobj uses different settings than gzip (I find that it's closer togzip -3) or maybe because zlib.compressobj compresses each chunk in isolation (as opposed to a stream as a whole). In any case both approaches should be good enough. I find that perl is slightly faster.
– Stéphane Chazelas
Sep 17 '14 at 11:55
@StéphaneChazelas, judging by the documentationcompressobjshould be acting on the stream as a whole. I find that both Python and Perl solutions produce approximately the same result on my data files (they differ in the second decimal place, that's good enough for me). Thanks for the great idea
– iruvar
Sep 17 '14 at 12:48
add a comment |
Here's a (hopefully equivalent) Python version of Stephane Chazelas's solution
python -c "
import zlib
from itertools import islice
from functools import partial
import sys
with open(sys.argv[1]) as f:
compressor = zlib.compressobj()
t, z = 0, 0.0
for chunk in islice(iter(partial(f.read, 4096), ''), 0, None, 10):
t += len(chunk)
z += len(compressor.compress(chunk))
z += len(compressor.flush())
print z/t
" file
edited Apr 13 '17 at 12:36
Community♦
1
answered Sep 16 '14 at 19:14
iruvariruvar
11.7k62960
Here's a (hopefully equivalent) Python version of Stephane Chazelas's solution
python -c "
import zlib
from itertools import islice
from functools import partial
import sys
with open(sys.argv[1]) as f:
compressor = zlib.compressobj()
t, z = 0, 0.0
for chunk in islice(iter(partial(f.read, 4096), ''), 0, None, 10):
t += len(chunk)
z += len(compressor.compress(chunk))
z += len(compressor.flush())
print z/t
" file
edited Apr 13 '17 at 12:36
Community♦
1
answered Sep 16 '14 at 19:14
iruvariruvar
11.7k62960
edited Apr 13 '17 at 12:36
Community♦
1
edited Apr 13 '17 at 12:36
Community♦
1
edited Apr 13 '17 at 12:36
Community♦
1
1
answered Sep 16 '14 at 19:14
iruvariruvar
11.7k62960
answered Sep 16 '14 at 19:14
iruvariruvar
11.7k62960
answered Sep 16 '14 at 19:14
iruvariruvar
11.7k62960
11.7k62960
I don't know if that's equivalent (as python is not by cup of coffee ;-b)), but that gives slightly different results (even for very large files where the gzip header size overhead can be ignored), possibly because zlib.compressobj uses different settings than gzip (I find that it's closer togzip -3) or maybe because zlib.compressobj compresses each chunk in isolation (as opposed to a stream as a whole). In any case both approaches should be good enough. I find that perl is slightly faster.
– Stéphane Chazelas
Sep 17 '14 at 11:55
@StéphaneChazelas, judging by the documentationcompressobjshould be acting on the stream as a whole. I find that both Python and Perl solutions produce approximately the same result on my data files (they differ in the second decimal place, that's good enough for me). Thanks for the great idea
– iruvar
Sep 17 '14 at 12:48
add a comment |
I don't know if that's equivalent (as python is not by cup of coffee ;-b)), but that gives slightly different results (even for very large files where the gzip header size overhead can be ignored), possibly because zlib.compressobj uses different settings than gzip (I find that it's closer togzip -3) or maybe because zlib.compressobj compresses each chunk in isolation (as opposed to a stream as a whole). In any case both approaches should be good enough. I find that perl is slightly faster.
– Stéphane Chazelas
Sep 17 '14 at 11:55
@StéphaneChazelas, judging by the documentationcompressobjshould be acting on the stream as a whole. I find that both Python and Perl solutions produce approximately the same result on my data files (they differ in the second decimal place, that's good enough for me). Thanks for the great idea
– iruvar
Sep 17 '14 at 12:48
I don't know if that's equivalent (as python is not by cup of coffee ;-b)), but that gives slightly different results (even for very large files where the gzip header size overhead can be ignored), possibly because zlib.compressobj uses different settings than gzip (I find that it's closer to
gzip -3) or maybe because zlib.compressobj compresses each chunk in isolation (as opposed to a stream as a whole). In any case both approaches should be good enough. I find that perl is slightly faster.– Stéphane Chazelas
Sep 17 '14 at 11:55
I don't know if that's equivalent (as python is not by cup of coffee ;-b)), but that gives slightly different results (even for very large files where the gzip header size overhead can be ignored), possibly because zlib.compressobj uses different settings than gzip (I find that it's closer to
gzip -3) or maybe because zlib.compressobj compresses each chunk in isolation (as opposed to a stream as a whole). In any case both approaches should be good enough. I find that perl is slightly faster.– Stéphane Chazelas
Sep 17 '14 at 11:55
@StéphaneChazelas, judging by the documentation
compressobj should be acting on the stream as a whole. I find that both Python and Perl solutions produce approximately the same result on my data files (they differ in the second decimal place, that's good enough for me). Thanks for the great idea– iruvar
Sep 17 '14 at 12:48
@StéphaneChazelas, judging by the documentation
compressobj should be acting on the stream as a whole. I find that both Python and Perl solutions produce approximately the same result on my data files (they differ in the second decimal place, that's good enough for me). Thanks for the great idea– iruvar
Sep 17 '14 at 12:48
add a comment |
I had a multi-gigabyte file and I wasn't sure if it was compressed, so I test-compressed the first 10M bytes:
head -c 10000000 large_file.bin | gzip | wc -c
It's not prefect but it worked well for me.
answered 26 mins ago
aidanaidan
1063
add a comment |
I had a multi-gigabyte file and I wasn't sure if it was compressed, so I test-compressed the first 10M bytes:
head -c 10000000 large_file.bin | gzip | wc -c
It's not prefect but it worked well for me.
answered 26 mins ago
aidanaidan
1063
add a comment |
I had a multi-gigabyte file and I wasn't sure if it was compressed, so I test-compressed the first 10M bytes:
head -c 10000000 large_file.bin | gzip | wc -c
It's not prefect but it worked well for me.
answered 26 mins ago
aidanaidan
1063
I had a multi-gigabyte file and I wasn't sure if it was compressed, so I test-compressed the first 10M bytes:
head -c 10000000 large_file.bin | gzip | wc -c
It's not prefect but it worked well for me.
answered 26 mins ago
aidanaidan
1063
answered 26 mins ago
aidanaidan
1063
answered 26 mins ago
aidanaidan
1063
answered 26 mins ago
aidanaidan
1063
1063
add a comment |
add a comment |
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