decomposition

Lossless decomposition vs Dependency Preservation

北城以北 提交于 2019-12-10 09:37:50
问题 Does anyone of them implies the other? My logic is that if all dependencies are preserved, then there is no loss of information and similarly, if decomposition is lossless then no functional dependency must have been violated. So essentially, dependency preservation is a way to ensure that your decomposition is lossless. I am having a hard time accepting/denying it. So do both of these guarantee one another or are there cases where one can be achieved without the other? 回答1: In general these

seasonal_decompose: operands could not be broadcast together with shapes on a series

99封情书 提交于 2019-12-07 03:27:07
问题 I know there are many questions on this topic, but none of them helped me to solve this problem. I'm really stuck on this. With a simple series: 0 2016-01-31 266 2016-02-29 235 2016-03-31 347 2016-04-30 514 2016-05-31 374 2016-06-30 250 2016-07-31 441 2016-08-31 422 2016-09-30 323 2016-10-31 168 2016-11-30 496 2016-12-31 303 import statsmodels.api as sm logdf = np.log(df[0]) decompose = sm.tsa.seasonal_decompose(logdf,freq=12, model='additive') decomplot = decompose.plot() i keep getting:

Eigenvector (Spectral) Decomposition

三世轮回 提交于 2019-12-06 07:14:35
I am trying to find a program in C code that will allow me to compute a eigenvalue (spectral) decomposition for a square matrix. I am specifically trying to find code where the highest eigenvalue (and therefore its associated eigenvalue) are located int the first column. The reason I need the output to be in this order is because I am trying to compute eigenvector centrality and therefore I only really need to calculate the eigenvector associated with the highest eigenvalue. Thanks in advance! In any case I would recommend to use a dedicated linear algebra package like Lapack (Fortran but can

Lossless decomposition vs Dependency Preservation

|▌冷眼眸甩不掉的悲伤 提交于 2019-12-06 03:06:09
Does anyone of them implies the other? My logic is that if all dependencies are preserved, then there is no loss of information and similarly, if decomposition is lossless then no functional dependency must have been violated. So essentially, dependency preservation is a way to ensure that your decomposition is lossless. I am having a hard time accepting/denying it. So do both of these guarantee one another or are there cases where one can be achieved without the other? In general these are two independent things: you can have a lossless decomposition without dependency preservation, as well

PySpark PCA: avoiding NotConvergedException

。_饼干妹妹 提交于 2019-12-05 13:13:05
I'm attempting to reduce a wide dataset (51 features, ~1300 individuals) using PCA through the ml.linalg method as follows: 1) Named my columns as one list: features = indi_prep_df.select([c for c in indi_prep_df.columns if c not in{'indi_nbr','label'}]).columns 2) Imported the necessary libraries from pyspark.ml.feature import PCA as PCAML from pyspark.ml.linalg import Vector from pyspark.ml.feature import VectorAssembler from pyspark.ml.linalg import DenseVector 3) Collapsed the features to a DenseVector indi_feat = indi_prep_df.rdd.map(lambda x: (x[0], x[-1], DenseVector(x[1:-2]))).toDF([

Decompose integers larger than 100 digits [closed]

本小妞迷上赌 提交于 2019-12-05 12:07:38
X and Y are integers larger than 100 digits. Find the integer P which is within the range [ X , Y [ and that guaranties the "best" prime decomposition (i.e. the decomposition with the most unique prime factors). What I've done is just check the primality and decompose each number in the range and find the number that respects the rule. Is there any other way to do this? An example on small integers Edit: In the above example, 123456 is decomposed to 2^6 * 3^1 * 643^1 , that's 2 * 2 * 2 * 2 * 2 * 2 * 3 * 643 but only 3 unique factors. While the answer, 123690, is decomposed to 6 unique factors

Svg matrix decomposition

↘锁芯ラ 提交于 2019-12-04 22:26:49
问题 In svg we have method element.getCTM() which returns a SVGMatrix as: [a c e][b d f][0 0 1] I want to calculate sx , sy and angle of rotation from this matrix. 回答1: There is a lot to read and learn on this subject. I'll give a basic answer, but be aware, if you are trying to do a game or animations this is NOT the way to do it. a == sx and d == sy , so you'll access these like this: var r, ctm, sx, sy, rotation; r = document.querySelector('rect'); // access the first rect element ctm = r

What is the auto Bracketed List Syntax?

左心房为你撑大大i 提交于 2019-12-04 16:48:28
W.F. gave a now-deleted answer to my question here which used the line: auto [x, y] = div_t{ 1, 0 }; From the code in the answer it looks like that's like a tie for the div_t struct. I was hoping that someone could explain what was going on here. The complete function code was as follows: constexpr bool first_quot() { auto [x, y] = std::div_t{1, 0}; (void)y; return x; } In the most most recent draft of the C++17 specification it's called "Decomposition declarations" and is defined in section 8.5 [dcl.decomp]. 来源: https://stackoverflow.com/questions/41569419/what-is-the-auto-bracketed-list

Svg matrix decomposition

☆樱花仙子☆ 提交于 2019-12-03 13:33:07
In svg we have method element.getCTM() which returns a SVGMatrix as: [a c e][b d f][0 0 1] I want to calculate sx , sy and angle of rotation from this matrix. There is a lot to read and learn on this subject. I'll give a basic answer, but be aware, if you are trying to do a game or animations this is NOT the way to do it. a == sx and d == sy , so you'll access these like this: var r, ctm, sx, sy, rotation; r = document.querySelector('rect'); // access the first rect element ctm = r.getCTM(); sx = ctm.a; sy = ctm.d; Now for the rotation a == cos(angle) and b == sin(angle) . Asin and acos can't

Decomposing a relation into BCNF

独自空忆成欢 提交于 2019-12-03 02:10:47
问题 I'm having trouble establishing when a relation is in Boyce-Codd Normal Form and how to decompose it info BCNF if it is not. Given this example: R(A, C, B, D, E) with functional dependencies: A -> B, C -> D How do I go about decomposing it? The steps I've taken are: A+ = AB C+ = CD R1 = A+ = **AB** R2 = ACDE (since elements of C+ still exist, continue decomposing) R3 = C+ = **CD** R4 = ACE (no FD closures reside in this relation) So now I know that ACE will compose the whole relation, but the