Compute the CR for lightweight temporal compression

Question

I am working on the LTC algorithm to compress the data

int SIZE =256;
int data[SIZE ]={700,...};  // input dataset (original data)

After executed the LTC I get on output (compressed data) like this:

//format: "number:count" --> e.g., 511:4

700:1
812:15
511:4
117:25
905:25
117:25
905:25
117:25

How can I compute the compression ratio for this output (the compressed data)?

 // CR = Original/ compressed 
float CR = ((int)(SIZE*(sizeof(int)))/?);

It's unclear what your output represents. Please [edit] and clarify. — Jabberwocky, May 19 '23 at 09:48
Write output to a file in the form you want and ask the OS how big the file is. What you show could be 54 or 62 or 108 bytes, depending how it is written. — stark, May 19 '23 at 09:57
@Jabberwocky, The output represents the data after compression (compressed data) — lena, May 19 '23 at 10:13
@ayla the just count the output bytes. What specific problem do you have? — Jabberwocky, May 19 '23 at 10:50
Are you actually compressing the data or just getting information about the data? — Ian Abbott, May 19 '23 at 11:14
Please show what compressed data following sequence would give: `2,3,3,5,5,5`. Would it be `2:1, 3:2, 5:3`? — Jabberwocky, May 19 '23 at 11:18
@Jabberwocky, according to this format (format: "number:count" --> e.g., 511:4) the number represents the compressed value, and the count represents the number of values that represented by a single number. — lena, May 19 '23 at 11:49
@ayla please answer the question I asked in my previous comment by "yes" or "no", and if it's "no" explain. — Jabberwocky, May 19 '23 at 11:52
So, my question: When I compute the CR I should depend on just the "number"? without the count, is that right? — lena, May 19 '23 at 11:52
@ayla the compression ration is usually the amount of compressed data divided by the amount of uncompressed data. — Jabberwocky, May 19 '23 at 11:54
@ayla _The output for this sequence 2,3,3,5,5,5 is 2:6_ Huh? I'm giving up. — Jabberwocky, May 19 '23 at 11:55
@ayla Tip: `(int)(SIZE*(sizeof(int)))` replaceable with `sizeof data`. — chux - Reinstate Monica, May 19 '23 at 12:53
@Jabberwocky Yes "2,3,3,5,5,5 is 2:6" is unusual, yet LTC is lossy and "3,3,5,5,5" is part of the trend. — chux - Reinstate Monica, May 19 '23 at 14:56

score 2 · Answer 1 · answered May 19 '23 at 13:05

2

How can I compute the compression ratio

// Size of one "number:count" pair.
size_t size_of_number_count = sizeof size_t + sizeof data[0];

// Number of "number:count" pairs.
size_t compressed_count = 8;

// CR = Original/ compressed 
double CR = 1.0 * sizeof data / (compressed_count * size_of_number_count);

answered May 19 '23 at 13:05

chux - Reinstate Monica

143,097
13
135
256

Is the calculation of the `size_of_number_count` value just an example, or does it match some actual implementation? Are there any implementations that store each number:count pair in a variable-length format (so that `size_of_number_count` would not be constant over the data set)? – Ian Abbott May 19 '23 at 18:41
@IanAbbott A sample. As the goal is the ratio of a _compression_, it is reasonable to assume the compressed data lacks padding or unused elements. I suppose code could use a slightly more correct/pessimistic: `double CR = 1.0 * sizeof data / (compressed_count * size_of_number_count + sizeof compressed_count);`. – chux - Reinstate Monica May 19 '23 at 19:24

Compute the CR for lightweight temporal compression

1 Answers1