# Tag Info

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The issue here is to get to an equation that parses the observed data to signal and noise. If your data is simple then your regression approach might work. Care should be taken to understand some of the assumptions that they are making with Prophet. You should better understand what Prophet does do, as it doesn't just fit a simple model but attempts to add ...

8

I have not used it, but this is their preprint's abstract (emphasis mine): Forecasting is a common data science task that helps organizations with capacity planning, goal setting, and anomaly detection. Despite its importance, there are serious challenges associated with producing reliable and high quality forecasts — especially when there are a ...

4

Laplace smoothing is a way to move probabilities towards uninformed mean. Suppose you have a multinomial variable with sample counts $c_1, c_2,..,c_d$, where $d$ is the number of dimensions. A Laplace smoothed version of estimated probabilities has the form: $(c_i + \alpha)/(N + d\alpha)$, where $\alpha$ is positive. If $\alpha$ is $0$ then we have non ...

4

MAPE is defined as $$\mathcal{L} = \sum_{i=1}^n \frac{|y_i - \hat{y}_i|}{y_i} \tag{1}$$ Option 1: Weighted Least Squares First, an approximate (but very convenient) solution. By using weighted least squares with sample weights of $w_i = 1/y_i$, we obtain the following loss function: $$\mathcal{L} = \sum_{i=1}^n w_i (y_i - \hat{y}_i)^2 \tag{2}$$ ...

2

Are dropout and other forms of regularization enough to prevent over fitting and make K-fold unnecessary? Dropout and other forms of regularization don't entirely prevent overfitting. You still need to hold out a validation set which isn't seen at training time. Is there a reason for this? Neural network models typically take hours, days, or even ...

2

The one you are refering to is the common PSB (Princeton Segmentation Benchmark) dataset [1]. Other Important datasets used for 3D mesh segmentation are: COSEG [2] URL: http://irc.cs.sdu.edu.cn/~yunhai/public_html/ssl/ssd.htm ShapeNet [3] URL: https://www.shapenet.org/ Other from Princeton [4] URL: http://shape.cs.princeton.edu/vkcorrs/papers_small/...

2

Since my point of view KNN is more less complex in time execution that SVM model. thanks so much. Empirical evaluation can't really determine which of two algorithms has lower asymptotic complexity. In fact i'm pretty sure that would violate Rice's theorem. What is more complex? O(n^3) or O(nd) and Why? Well these aren't comparable, because one is a ...

2

Exponential function is a function of the form $f(x)=ab^x$, not just $e^x$. And the formula for the decayed exponential is: decayed_learning_rate = learning_rate *                         decay_rate ^ (global_step / decay_steps) as given in your link, which is quite similar with the mathematical definition. When staircase is true, they just apply integer ...

2

LSTM uses Gated Recurrent Units (GRU), which is known to be sensitive to gradient vanishing or exploding. The usual solution is to introduce a gradient clipping mechanism. Basically we merely rescale the gradients to prevent them from blowing up. The goal is to keep their norm at most a predefined threshold value. In tensorflow, this is straightforward. ...

2

You question is almost meaningless without showing/explaining the data. The tells you want invarinces you need (phase. offset, amplitude, uniform scaling, warping, complexity etc [a]). But see for example [b] [a] https://www.cs.ucr.edu/~eamonn/Complexity-Invariant%20Distance%20Measure.pdf [b] https://www.cs.ucr.edu/~eamonn/time_series_bitmaps.pdf

2

[Naive] bayes is a generative model, which means we can generate data using it if we wanted. In NB, we estimate $p(\mathbf{x}|y)$, where $\mathbf{x}$ is our feature vector and $y$ is the class variable. For example, we first pick a $y$, indicating the class, and then pick word(s) according to the probability distribution, $p(\mathbf{x}|y)$.

2

The reason behind using the weights this way was to try to keep the model away from instances with very high/very low probability. My interpretation is that the model is fairly confident about those instances and hence should focus elsewhere. This is similar to AdaBoost, giving higher weights (during fitting, updating weights between trees) to misclassified ...

1

DTW does not have a notion of "convergence", as it is not an iterative optimization procedure. When DTW is too slow, and doesn't finish in acceptable time, the obviously next best idea is to use bounded DTW. Full DTW can take O(nm) time of n and m are the lengths of your series (at least if implemented well...). You can bound this to O(nk) for a fixed ...

1

I mean, you could compare feature selected SVM to non-feature selected naïve Bayes if you wanted to, I would just say the comparison wouldn't be truly fair to the naïve Bayes classifier. Same way you could decide to tune on of your algorithms but not another. It's probably not a great idea as you will be losing out on potentially better algorithms but you ...

1

When you are fitting a regression-type model, you are trying to explain or predict variation in one variable with variation in others. If all values are identical, there is no variation for you to work with! There is no meaningful information in there to use, either as a predictor or a target/dependent variable. So you are completely justified in excluding ...

1

The usual way is to test your success metric across test and train data. The gap between the two generally signals overfitting. Also, accuracy isn't a good metric; check here for a good explanation. Moreover, having such tests with random data has its own problems. If your model overfits the data, it might still easily perform 25 % accuracy on random data. ...

1

RNNs are most commonly used as density models over some space of sequences. To be more precise, if we have some sequence $X = x_1, x_2, \ldots, x_k$ then our model describes the distribution $$P(X) = \prod_{i=1}^k p(x_i | x_{<i})$$ (which is valid by the product rule). More specifically, our RNN models each conditional term $p(x_i | x_{<i})$ as $x_i \... 1 Object detection is completely dominated by deep learning. Essentially it is now down to one-stage detectors (e.g., YOLO) when you want speed and two-stage detectors (e.g., faster-RCNN) when you want accuracy. Check out any recent survey on object detection, like this one or this one. However, these deep neural networks are trained on hundreds of classes ... 1 You could either impute or omit those 4 variables. Whether imputing is a good idea depends on whether the covariance between all variables is likely to vary between source hospitals. Regarding your second question, you could include data source (hospital) as a variable in your analyses. Whether that is a good idea is also likely to depend on your goals and ... 1 I think standard scaling mostly depends on the model being used, and normalizing depend on how the data is originated Most of distance based models e.g. k-means need standard scaling so that large-scaled features don't dominate the variation. Same goes to PCA. About the normalization, it mostly depends on the data. For example, if you have sensor data (... 1 There cannot be a general rule on what to do. Any automatic normalization is usually "wrong". They only happen to usually work better than not weighting features at all, so people commony use them - in particular on data they don't understand. But the right way is to weight and scale features such they have the right balanced amount of influence on the ... 1 Indeed you can't say any more accurate explanation without knowing the details of the complexity. Because the time complexity is asymptotic and it might be some constant factors are neglected in this theoretical complexity. Then, those factors could be large or too small and effect on the CPU running time of the algorithm. Therefore, you could not say ... 1 There is nothing special but there are some observations to be done. It is true that it is a linear combination and applying the activation function gives a regression. However this is not a linear regression unless we have an identity as activation. An indeed using identity we will get a linear regression in the end. If we use logistic sigmoid, then the NN ... 1 A VAE models a distribution as $$\log P(x) = \log \int P(x|z)P(z) dz \geq E_{z \sim q}[\log P(x|z)] - \text{KL}(q(z)||p(z))$$ As you can see, there is no direct relationship between the variance of$q$and the probability the model assigns to a given datapoint$x$. In fact decreasing the variance on$q$could increase the KL term, resulting in lower log ... 1 There is no model that will perform this as far as I know. The only option I see, if you really wanted to test pairs, is to build 3 models, one with A, one with B, and one with A and B. After this you check whether the last model performs better than the other ones. And you would have to perform this for every possible pair. If you have 200 million mutations ... 1 Assuming the columns have unit L2 norm, the$\lambda^{*}$which sets all coefficients to zero is given by$\frac{1}{2} \max_{j} (X_j^T y)^2$(the reasoning in the answer above is correct; but the final answer misses the factor of$\frac{1}{2}$). L0Learn centers and then normalizes the columns before fitting the model. The$\lambda$'s are reported after ... 1 F1 is a suitable measure of models tested with imbalance datasets. But I think F1 is mostly a measure for models, rather than datasets. You could not say that dataset A is better than dataset B. There is no better or worse here; dataset is dataset. 1 These sort of algorithms often predate machine learning and have their roots in the linguistic community. A commonly used such measure of "clarity" is the Gunning Fog Index, which estimates the years of formal education a person needs to understand the text on the first reading. An alternative is the Flesch–Kincaid readability test that quantifies how ... 1 I could see possibly two challenges if one penalize the knots locations. It will be a difficult optimization. For simplicity, I will illustrate my point with cubic spline, instead of natural cubic spline. Suppose I have data points$\{(x_1, y_1), \dots, (x_n, y_n)\}$, and I have 2 knots,$\epsilon_1, \epsilon_2\$, whose location I don't know. If the number ...

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