I'm looking for a DICOM Tag for Dose-Length Product (DLP) Value. If there is no such a tag, how to calculate it based on other tags?
Maybe it is a calculation between (0018,1302) Scan Length and (0018,0050) Slice Thinkness and some others but I am not sure.
DLP = Dose-Length Product [mGy x cm]
I don't believe that there is a standard DICOM tag for DLP. I know it exists as a value in RDSR (Radiation Dose Structured Reports.) You could, however, calculate DLP from CTDIvol and Scan Length, assuming that you have both of those available in the header.
DLP = CTDIvol * ScanLength
Related
Is there any (simple) random generation function that can work without variable assignment? Most functions I read look like this current = next(current). However currently I have a restriction (from SQLite) that I cannot use any variable at all.
Is there a way to generate a number sequence (for example, from 1 to max) with only n (current number index in the sequence) and seed?
Currently I am using this:
cast(((1103515245 * Seed * ROWID + 12345) % 2147483648) / 2147483648.0 * Max as int) + 1
with max being 47, ROWID being n. However for some seed, the repeat rate is too high (3 unique out of 47).
In my requirements, repetition is ok as long as it's not too much (<50%). Is there any better function that meets my need?
The question has sqlite tag but any language/pseudo-code is ok.
P.s: I have tried using Linear congruential generators with some a/c/m triplets and Seed * ROWID as Seed, but it does not work well, it's even worse.
EDIT: I currently use this one, but I do not know where it's from. The rate looks better than mine:
((((Seed * ROWID) % 79) * 53) % "Max") + 1
I am not sure if you still have the same problem but I might have a solution for you.
What you could do is use Pseudo Random M-sequence generators based on shifting registers. Where you just have to take high enough order of you primitive polynomial and you don't need to store any variables really.
For more info you can check the wiki page
What you would need to code is just the primitive polynomial shifting equation and I have checked in an online editor it should be very easy to do. I think the easiest way for you would be to use Binary base and use PRBS sequences and depending on how many elements you will have you can choose your sequence length. For example this is the implementation for length of 2^15 = 32768 (PRBS15), the primitive polynomial I took from the wiki page (There youcan find the primitive polynomials all the way to PRBS31 what would be 2^31=2.1475e+09)
Basically what you need to do is:
SELECT (((ROWID << 1) | (((ROWID >> 14) <> (ROWID >> 13)) & 1)) & 0x7fff)
The beauty of this approach is if you take the sequence of the PRBS with longer period than your ROWID largest value you will have unique random index. Very simple. :)
If you need help with searching for primitive polynomials you can see my github repo which deals exactly with finding primitive polynomials and unique m-sequences. It is currently written in Matlab, but I plan to write it in python in next few days.
Cheers!
What about using good hash function and map result into [1...max] range?
Along the lines (in pseudocode). sha1 was added to SQLite 3.17.
sha1(ROWID) % Max + 1
Or use any external C code for hash (murmur, chacha, ...) as shown here
A linear congruential generator with appropriately-chosen parameters (a, c, and modulus m) will be a full-period generator, such that it cycles pseudorandomly through every integer in its period before repeating. Although you may have tried this idea before, have you considered that m is equivalent to max in your case? For a list of parameter choices for such generators, see L'Ecuyer, P., "Tables of Linear Congruential Generators of Different Sizes and Good Lattice Structure", Mathematics of Computation 68(225), January 1999.
Note that there are some practical issues to implementing this in SQLite, especially if your SQLite version supports only 32-bit integers and 64-bit floating-point numbers (with 52 bits of precision). Namely, there may be a risk of—
overflow if an intermediate multiplication exceeds 32 bits for integers, and
precision loss if an intermediate multiplication results in a greater-than-52-bit number.
Also, consider why you are creating the random number sequence:
Is the sequence intended to be unpredictable? In that case, a linear congruential generator alone is not enough, and you should generate unique identifiers by other means, such as by combining unique numbers with cryptographically random numbers.
Will the numbers generated this way be exposed in any way to end users? If not, there is no need to obfuscate them by "shuffling" them.
Also, depending on the SQLite API you're using (for your programming language), there may be a way to write a custom function to convert the seed and ROWID to a random unique number. The details, however, depend heavily on the specific SQLite API. Another answer shows an example for Perl.
I am using the HEVC reference software, HM Encoder Version [16.15] (including RExt) on a [Mac OS X][GCC 4.2.1][64 bit] and would like to extract at encoder side:
1) the motion vectors for each block
2) the frame partition information, i.e. the size and location of each block in the frame to which a motion vector refers.
Does anybody have hints on what are the variables where this info is stored for each coding unit? Thanks!
All you need is available in the TComDataCU class.
1) For motion information, there is the function getCUMvField() which returns the motion vector. It's not easy to work with it though.
Basically, to access almost any of the PU/CU level syntax elements, you need to be able to work with the absolute index of that PU/CU. This unique index tells you where exactly your PU/CU is located in the CTU by pointing to the up-left 4x4 block of that part.
I rememberthat that most of the times this index is stored in the variable uiAbsPartIdx.
If you get to know how to work with this index, then you will be able to get the block partitioning information in the CTU level. so for 2) my suggestion is that you go to the slice level when you have a loop over CUTs (I think this is done in the compressSlice() function). And after the compressCtu() function is called for each CTU (which means that all RDO decisions have been made and the CTU partitioning is decided), you put a loop over all uiAbsPartIdxs of the CTU and get their width and height. For example if your CTU size is 128, then you will have 32*32=1024 unique 4x4 blocks in your CTU. The function for getting width/height of the CU corresponding to a certain uiAbsPartIdx is pCtu->getWidth(uiAbsPartIdx).
I hope it was clear.
I am using Apriori algorithm to identify the frequent item sets of the customer.Based on the identified frequent item sets I want to prompt suggest items to customer when customer adds a new item to his shopping list.Assume my one identified frequent set is [2,3,5].My question is;
if user has already added item 2 and item 5, i want check the confidence of the rule to suggest item 3. for that;
confidence = support of (2,3,5)/ support (3) ?
OR
confidence = support of (2,3,5)/ support (2,5)?
which equation is correct? please help!!
If the association rule is (2,5) -> (3), than is X = (2,5) and Y = (3). The confidence of an association rule is the support of (X U Y) divided by the support of X. Therefore, the confidence of the association rule is in this case the support of (2,5,3) divided by the support of (2,5).
Suppose A^B -> C then
Confidence = support(A^B->C)
i.e. a number of transactions in which all three items are present / support(A,B)
i.e. a number of transactions in which both A and B are present.
So the answer is confidence= support(2,5,3)/support (2,5)
If you just want the answer without any explanation:
confidence = support of (2,3,5)/ support (2,5) in your question is the answer.
What is your antecedent?
Stop.treating equarions as black boxes you need to lok up. understand them or you will fail.
How does google's sparse hash table handle collisions? i.e. when 2 elements map to the same bucket, how does it decide where to put the new (colliding) element? I'm reading What is the main implementation idea behind sparse hash table? but that answer doesn't cover the collision idea.
Your question is answered in the documentation here, specifically:
2c) If t.sparsetable[i % 32] is assigned, but to a value other than foo, look at t.sparsetable[(i+1) % 32]. If that also fails, try t.sparsetable[(i+3) % 32], then t.sparsetable[(i+6) % 32]. In general, keep trying the next triangular number.
You can read about triangular numbers here.
I want code to render n bits with n + x bits, non-sequentially. I'd Google it but my Google-fu isn't working because I don't know the term for it.
For example, the input value in the first column (2 bits) might be encoded as any of the output values in the comma-delimited second column (4 bits) below:
0 1,2,7,9
1 3,8,12,13
2 0,4,6,11
3 5,10,14,15
My goal is to take a list of integer IDs, and transform them in a way they can still be used for persistent URLs, but that can't be iterated/enumerated sequentially, and where a client cannot determine programmatically if a URL in a search result set has been visited previously without visiting it again.
I would term this process "encoding". You'll see something similar done to permit the use of communications channels that have special symbols that are not permitted in data. Examples: uuencoding and base64 encoding.
That said, you still need to (and appear at first blush to have) ensure that there is only one correct de-code; and accept the increase in size of the output (in the case above, the output will be double the size, bit-for-bit as the input).
I think you'd be better off encrypting the number with a cheap cypher + a constant secret key stored on your server(s), adding a random character or four at the end, and a cheap checksum, and simply reject any responses that don't have a valid checksum.
<encrypt(secret)>
<integer>+<random nonsense>
</encrypt>
+
<checksum()>
<integer>+<random nonsense>
</checksum>
Then decrypt the first part (remember, cheap == fast), validate the ciphertext using the checksum, throw off the random nonsense, and use the integer you stored.
There are probably some cryptographic no-no's here, but let's face it, the cost of this algorithm being broken is a touch on the low side.