'How to efficiently predict if data is compressible
I want to write a storage backend to store larger chunks of data. The data can be anything, but it is mainly binary files (images, pdfs, jar files) or text files (xml, jsp, js, html, java...). I found most of the data is already compressed. If everything is compressed, about 15% disk space can be saved.
I am looking for the most efficient algorithm that can predict with high probability that a chunk of data (let's say 128 KB) can be compressed or not (lossless compression), without having to look at all the data if possible.
The compression algorithm will be either LZF, Deflate, or something similar (maybe Google Snappy). So predicting if data is compressible should be much faster than compressing the data itself, and use less memory.
Algorithms I already know about:
Try to compress a subset of the data, let's say 128 bytes (this is a bit slow)
Calculate the sum of 128 bytes, and if it's within a certain range then it's likely not compressible (within 10% of 128 * 127) (this is fast, and relatively good, but I'm looking for something more reliable, because the algorithm really only looks at the topmost bits for each byte)
Look at the file headers (relatively reliable, but feels like cheating)
I guess the general idea is that I need an algorithm that can quickly calculate if the probability of each bit in a list of bytes is roughly 0.5.
Update
I have implemented 'ASCII checking', 'entropy calculation', and 'simplified compression', and all give good results. I want to refine the algorithms, and now my idea is to not only predict if data can be compressed, but also how much it can be compressed. Possibly using a combination of algorithms. Now if I could only accept multiple answers... I will accept the answer that gave the best results.
Additional answers (new ideas) are still welcome! If possible, with source code or links :-)
Update 2
A similar method is now implemented in Linux.
Solution 1:[1]
From my experience almost all of the formats that can effectively be compressed are non-binary. So checking if about 70-80% of the characters are within in the [0-127] rage should do the trick.
If you want to to it "properly" (even though I really can't see a reason to do that), you either have to run (parts of) your compression algorithm on the data or calculate the entropy, as tskuzzy already proposed.
Solution 2:[2]
Calculate the entropy of the data. If it has high entropy (~1.0), it is not likely going to be further compressed. If it has low entropy (~0.0), then that means that there isn't a lot of "information" in it and can be further compressed.
It provides a theoretical measure of how compressed a piece of data can get.
Solution 3:[3]
I implemented a few methods to test if data is compressible.
Simplified Compression
This basically checks for duplicate byte pairs:
static boolean isCompressible(byte[] data, int len) {
int result = 0;
// check in blocks of 256 bytes,
// and sum up how compressible each block is
for (int start = 0; start < len; start += 256) {
result += matches(data, start, Math.min(start + 255, len));
}
// the result is proportional to the number of
// bytes that can be saved
// if we can save many bytes, then it is compressible
return ((len - result) * 777) < len * 100;
}
static int matches(byte[] data, int i, int end) {
// bitArray is a bloom filter of seen byte pairs
// match counts duplicate byte pairs
// last is the last seen byte
int bitArray = 0, match = 0, last = 0;
if (i < 0 || end > data.length) {
// this check may allow the JVM to avoid
// array bound checks in the following loop
throw new ArrayIndexOutOfBoundsException();
}
for (; i < end; i++) {
int x = data[i];
// the bloom filter bit to set
int bit = 1 << ((last ^ x) & 31);
// if it was already set, increment match
// (without using a branch, as branches are slow)
match -= (-(bitArray & bit)) >> 31;
bitArray |= bit;
last = x;
}
return match;
}
On my (limited) set of test data, this algorithm is quite accurate. It about 5 times faster than compressing itself if the data is not compressible. For trivial data (all zeroes), it is about half as fast however.
Partial Entropy
This algorithm estimates the entropy of the high nibbles. I wanted to avoid using too many buckets, because they have to be zeroed out each time (which is slow if the blocks to check are small). 63 - numberOfLeadingZeros is the logarithm (I wanted to avoid using floating point numbers). Depending on the data, it is faster or slower than the algorithm above (not sure why). The result isn't quite as accurate as the algorithm above, possibly because of using only 16 buckets, and only integer arithmetic.
static boolean isCompressible(byte[] data, int len) {
// the number of bytes with
// high nibble 0, 1,.., 15
int[] sum = new int[16];
for (int i = 0; i < len; i++) {
int x = (data[i] & 255) >> 4;
sum[x]++;
}
// see wikipedia to understand this formula :-)
int r = 0;
for (int x : sum) {
long v = ((long) x << 32) / len;
r += 63 - Long.numberOfLeadingZeros(v + 1);
}
return len * r < 438 * len;
}
Solution 4:[4]
This problem is interesting alone because with for example zlib compressing uncompressible data takes much longer then compressing compressible data. So doing unsuccessful compression is especially expensive (for details see the links). Nice recent work in this area has been done by Harnik et al. from IBM.
Yes, the prefix method and byte order-0 entropy (called entropy in the other posts) are good indicators. Other good ways to guess if a file is compressable or not are (from the paper):
- Core-set size – The character set that makes up most of the data
- Symbol-pairs distribution indicator
Solution 5:[5]
A faster and more accurate algorithm for estimating compressibility
- 2 to 4 times faster and more accurate answer than judging by Shannon's entropy. It is based on the Huffman coding approach.
- Time complexity of answering does not depend on numerical values of frequency of symbols, rather depends on the number of unique symbols. Shannon's entropy calculates log(frequency), Hence the more the frequency, the more time it will consume to compute this value. In the current approach, mathematical operations on frequency values are avoided.
- For similar reasons as above, precision is also higher since our dependency on floating point operations is avoided as well as we just rely on sum and multiplication operations and how actual Huffman codes will contribute to total compressed size.
- Same algorithm can be enhanced to generate actual Huffman codes in less time which won't involve complex data structures like trees, heaps or priority queue. For our different requirements we just use the same frequency array of symbols.
Following algorithm specifies how to calculate compressibility of a file whose symbol frequency values are stored in map array.Time comparison chart
int compressed_file_size_in_bits = 0, n=256;
/* We sort the map array in increasing order.
* We will be simulating huffman codes algorithm.
* Insertion Sort is used as its a small array of 256 symbols.
*/
insertionSort(map, 256);
for (j = 0; j < n; j++)
if (map[j] != 0)
break;
for (i = j; i + 1 < n; i++) {
j = i + 1;
/* Following is an important step, as we keep on building more
* and more codes bottom up, their contribution to compressed size
* gets governed by following formula. Copy pen simulation is recommended.
*/
compressed_file_size_in_bits = compressed_file_size_in_bits + map[i] + map[j];
/* Least two elements of the map gets summed up and form a new frequency
* value which gets placed at i+1 th index.
*/
map[i + 1] = map[i] + map[j];
// map [i+2-----] is already sorted. Just fix the first element.
Adjust_first_element(map + i + 1, n - i - 1);
}
printf("Entropy per byte %f ", compressed_file_size_in_bits * (1.0) / file_len);
void insertionSort(long arr[], long n) {
long i, key, j;
for (i = 1; i < n; i++) {
key = arr[i];
j = i - 1;
/* Move elements of arr[0..i-1], that are
greater than key, to one position ahead
of their current position */
while (j >= 0 && arr[j] > key) {
arr[j + 1] = arr[j];
j = j - 1;
}
arr[j + 1] = key;
}
}
// Assumes arr[i+1---] is already sorted. Just first
// element needs to be placed at appropriate place.
void Adjust_first_element(long arr[], long n) {
long i, key, j = 1;
key = arr[0];
while (j < n && arr[j] < key) {
arr[j - 1] = arr[j];
j = j + 1;
}
arr[j - 1] = key;
}
Construction of codes using above algorithm
Construction of codes which makes use of the above algorithm is a string manipulation problem where we start with no code for each symbol. Then we follow the same algorithm as compressed file size/ compressibility calculation. Additionally we just keep on maintaining the history of code evolution. After the iteration through frequency array finishes, our final code which contains the evolution of different Huffman code for each symbol gets stored in the top index of codes array. At this point, a string parsing algorithm can parse this evolution and generate individual codes per symbol. Complete phenomenon involves no trees, heaps or priority queue. Just one iteration through frequency array (size 256 in most cases) would generate the evolution of codes as well as final compressed size value.
/* Generate code for map array of frequencies. Final code gets generated at
* codes[r] which can be provided as input to string parsing algorithm to
* generate code for individual symbols.
*/
void generate_code(long map[], int l, int r) {
int i, j, k, compressed_file_size_in_bits = 0;
insertionSort(map + l, r - l);
for (i = l; i + 1 <= r; i++) {
j = i + 1;
compressed_file_size = compressed_file_size_in_bits + map[i] + map[j];
char code[50] = "(";
/* According to algorithm, two different codes from two different
* nodes are getting combined in a way so that they can be separated by
* by a string parsing algorithm. Left node code, codes[i] gets appended by
* 0 and right node code, codes[j] gets appended by 1. These two codes
get
* separated by a comma.
*/
strcat(code, codes[i]);
strcat(code, "0");
strcat(code, ",");
strcat(code, codes[j]);
strcat(code, "1");
strcat(code, ")");
map[i + 1] = map[i] + map[j];
strcpy(codes[i + 1], code);
int n = r - l;
/* Adjust_first_element now takes an additional 3rd argument.
* this argument helps in adjusting codes according to how
* map elements are getting adjusted.
*/
Adjust_first_element(map + i + 1, n - i - 1, i + 1);
}
}
void insertionSort(long arr[], long n) {
long i, key, j;
// if(n>3)
// n=3;
for (i = 1; i < n; i++) {
key = arr[i];
j = i - 1;
/* Move elements of arr[0..i-1], that are
greater than key, to one position ahead
of their current position */
while (j >= 0 && arr[j] > key) {
arr[j + 1] = arr[j];
j = j - 1;
}
arr[j + 1] = key;
}
}
// Assumes arr[i+1---] is already sorted. Just first
// element needs to be placed at appropriate place.
void Adjust_first_element(long arr[], long n, int start) {
long i, key, j = 1;
char temp_arr[250];
key = arr[0];
/* How map elements will change position, codes[] element will follow
* same path
*/
strcpy(temp_arr, codes[start]);
while (j < n && arr[j] < key) {
arr[j - 1] = arr[j];
/* codes should also move according to map values */
strcpy(codes[j - 1 + start], codes[j + start]);
j = j + 1;
}
arr[j - 1] = key;
strcpy(codes[j - 1 + start], temp_arr);
}
Solution 6:[6]
I expect there's no way to check how compressible something is until you try to compress it. You could check for patterns (more patterns, perhaps more compressible), but then a particular compression algorithmn may not use the patterns you checked for - and may do better than you expect. Another trick may be to take the first 128000 bytes of data, push it through Deflate/Java compression, and see if it's less than the original size. If so - chances are it's worthwhile compressing the entire lot.
Solution 7:[7]
Fast compressor such as LZ4 already have built-in checks for data compressibility. They quickly skip the bad segments to concentrate on more interesting ones. To give a proper example, LZ4 on non-compressible data works at almost RAM speed limit (2GB/s on my laptop). So there is little room for a detector to be even faster. You can try it for yourself : http://code.google.com/p/lz4/
Solution 8:[8]
It says on your profile that you're the author of the H2 Database Engine, a database written in Java.
If I am guessing correctly, you are looking to engineer this database engine to automatically compress BLOB data, if possible.
But -- (I am guessing) you have realized that not everything will compress, and speed is important -- so you don't want to waste a microsecond more than is necessary when determining if you should compress data...
My question is engineering in nature -- why do all this? Basically, isn't it second-guessing the intent of the database user / application developer -- at the expense of speed?
Wouldn't you think that an application developer (who is writing data to the blob fields in the first place) would be the best person to make the decision if data should be compressed or not, and if so -- to choose the appropriate compression method?
The only possible place I can see automatic database compression possibly adding some value is in text/varchar fields -- and only if they're beyond a certain length -- but even so, that option might be better decided by the application developer... I might even go so far as to allow the application developer a compression plug-in, if so... That way they can make their own decisions for their own data...
If my assumptions about what you are trying to do were wrong -- then I humbly apologize for saying what I said... (It's just one insignificant user's opinion.)
Solution 9:[9]
Also -- Why not try lzop? I can personally vouch for the fact that it's faster, much faster (compression and decompression) than bzip, gzip, zip, rar...
Using it for disk image compression makes the process disk-IO bound. Using any of the other compressors makes the process CPU-bound (i.e., the other compressors use all available CPU, lzop (on a reasonable CPU) can handle data at the same speed a 7200 RPM stock hard drive can dish it out...)
I'll bet if you tested it with the first X bytes of a 'test compression' string, it would be much faster than most other methods...
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
| Solution | Source |
|---|---|
| Solution 1 | Chronial |
| Solution 2 | tskuzzy |
| Solution 3 | |
| Solution 4 | dmeister |
| Solution 5 | sweetesh |
| Solution 6 | cs94njw |
| Solution 7 | Cyan |
| Solution 8 | Peter Sherman |
| Solution 9 | Peter Sherman |