I've read posts mentioning that after selecting your model with the validation set you train with the full set (training + validation set) and test the model on the test set.  However, what if the validation set was used for early stopping during training?  The training part might early stop at 12 epochs but that doesn't mean that it is the optimal amount of epochs for any training set in order to achieve generalization in unseen data.  Even worse, in my case, I am training a neural network where the amount the learning rate decreases in each iteration depends on the maximum number of epochs.  So even if I find that the optimal amount of epochs during training was 12, setting the maximum amount of epochs at 12 in order to retrain will yield a different result than early stopping at 12 when the maximum amount of epochs is 30.  So my question is:   Do I have to somehow retrain the model and include the validation set in the training data? If yes, how?  Or do I just use the model trained with early stopping and test it on the test set without retraining the model using the validation set?
I am a little bit confused by early stopping and in particular by how it can be inserted inside a CV framework. As far as I understand, I can fix the optimal number of epochs (for NN, or number of trees for XGB) by early stopping, that is:   pick a validation set, train with increasing number of epochs until a predefined metric evaluated on the validation set starts worsening that is the optimal number of epochs to prevent overfitting   and that's fine. But then I would like to insert this early stopping framework inside a CV framework: Suppose I have a model with 10 hyperparameters I want to fix via CV. And suppose that there is an eleventh hyperparameter, the number of epochs. My feeling is that one can do like this:   create the K resampled folds, for each of which you have a training and validation set choose a suitable grid for your 10 hyperparameters for each point on the grid train your model in each fold with early stopping, that is use the validation set of the fold to keep track of the preferred metric and stop when it gets worse take the mean of the K validation metric choose the point of the grid (i.e. the set of hyperparameters) that gives the best metric   Questions:   What number of epochs should I choose as optimal? in each of the K folds I have, in general, a different number of stopping epochs.  seems to suggest they take the mean of the K stopping epochs. Is this right? Is it actually a "fair" practice to use the validation metric coming from the early stopping as a proxy of the out-of-sample metric? As  in section 3.4.5, maybe the best thing to do would be:     ...if you want to do early stopping, then in each fold you should take your training set and split it again, holding out a small early-stopping-set to guide the early stopping, and then evaluate the model on the validation set of that fold.   But this seems to me a really intricate process...   Unfortunately, I wasn't able to find references where this issue is presented in a clear and transparent form.
I have a lot of 3D objects of various sizes, shapes in blender 2.76b and I would like to set the origin to the bottom center of the objects bounds.  I cannot select the bottom edge and snap to cursor etc as each shapes edges are in different X/Y positions.  Is there a script or tool that will allow me to set the origin to the lowest edge z position? Disabling Intel Turbo Boost in ubuntu Why are my sauces splitting? Until now, I have been compiling my LaTeX documents with pdfLaTeX. My standard document preamble includes the "usual suspects" for non-English speakers (, , ) and a bunch of other packages. Assuming I were to convert to XeLaTeX, what modifications of my preamble are advisable? I'm aware that the expansion=true option of the  package is not (yet) available for XeLaTeX, and that  is sort of a default package for OpenType fonts. What other packages (and package options) should be removed and, vice versa, incorporated when switching from pdfLaTeX to XeLaTeX? iPhone's Bluetooth - What? Why? How? I googled a lot and read many documents. But I can't understand the concept behind iPhone's Bluetooth technology.  In my knowledge, Bluetooth was used for:   Transferring files between devices Wireless communication for things like video gaming and earphones   Recently I tried to copy some of the photos from my old Android mobile to my iPhone 4S via Bluetooth. It's quite simple in my older phones to connect with other devices by Bluetooth for file sharing. But I can't pair other devices (even another iPhone :( ) with my iPhone 4S.   Copying songs/videos may affect their iTunes as well as their business. But what about images? Why is the usage of Bluetooth restricted and what's the use of it being able to transfer images? Is there any other way to use iPhone's Bluetooth to transfer files? Maybe using an app? But I don't want to jailbreak my device.   I'm confused about the restrictions and Apple's thinking. Flat quads (or, why is my mesh so funky?) I think my question relates to a fundamental of 3D modeling that I haven't picked up on yet. It's very simple:  Let's say I have a flat quad. A plane on the XY plane inserted into am empty blender file is a great example. Now let's say I start moving each of the 4 vertices around on in any direction, especially the Z axis.   At this point, there would still be one face, but its extremely unlikely that the face would be planar (I just learned this word). So blender does its best to make sense of what this should look like.  The problem I'm having is that it seems like my quads are always like this, but they never look the way I want. There are edges and seams that look like they are dangling in space, and what tends to happen is that my mesh looks great when viewed from most angles, but if you look at it from just the right angle, you notice the dangling edges or sometimes even some kind of impossible geometry. And so I might try to nudge the problematic vertices around, but this often seems to either make it worse, or just hide the bad geometry until you see it from a different angle.  For me, this always seems to happen whenever I'm trying to move vertices to add roundness to something that is originally square.   There is another question on here about how to actually flatten the quads. That's great, and I might use some of the techniques mentioned there. My question is:   Why does this happen? (Can someone explain this better?) What are some tips and tricks to avoid this situation? What are some tips and tricks to mitigate this situation?   EDIT Here is an illustration:  The overall mesh as viewed from a distance looks decent (for my skill level)    But if you get closer and look from a certain angle...    Super close up:    The "edge" that is pointed to by the arrow is not actually an edge, but some kind of artifact created when blender tries to show the quad. The way the vertices are positioned, it shouldn't actually be possible to see the sliver of the face that I marked in red, since this face is supposed to be around the corner from the edges. The edges should form the silhouette on the right side of my figure.     Here is the side view. Maybe it is the angled topology that skews the face in a funny way?  But I'm not looking for help with this one particular issue, but rather a better understanding of why this happens, and ideas to prevent this in future models. For $x∈\mathbb{R}^n$ , let $B(x,r)$ denote the closed ball in $\mathbb{R}^n$(with Euclidean norm) of radius $r$ centered at $x$. Write $B=B(0,1)$.If $f,g:B→\mathbb{R}^n$ are continuous functions such that $f(x)≠g(x)$ for all $ x∈ B$, then which of the followings are true?       $f(B)∩g(B)=\varnothing$  There exist $ϵ&gt;0$ such that $||f(x)-g(x)||&gt; ϵ$ for all $ x∈ B$  There exist $ϵ&gt;0$ such that $ B(f(x), ϵ) ∩ B(g(x), ϵ)=\varnothing$ for all $ x∈ B$ ${\rm int }(f(B)) ∩ {\rm int }(g(B))=\varnothing$ , where ${\rm int}(E)$ denotes the  interior of a set $E$      How can I solve this problem? Can anyone help? Good books for a high schooler self-studying Abstract Algebra? Does anyone have any suggestions for abstract algebra books particularly suited to self-study?     Here is some background and motivation, if it's helpful.  I'm currently a junior in high school, but I have some familiarity with groups, rings, polynomials and fields, as I went through Fraleigh's book. I liked its writing style, but many of the problems were not helpful. I would like to get better at algebra with some other book, particularly one that's aimed at someone who already has a little familiarity with the objects, but is by no means completely comfortable.  One of my longer-term goals is to learn a bit of algebraic number theory, as I've only been studying elementary number theory. That's why I feel it  good to learn more algebra first. (I'd also be interested in suggestions of other good subjects to learn prior to tackling algebraic number theory.)  I was considering using one of Lang's books, since they seem pretty universal, but I've heard rumors that they are very terse and not good for self-learning. What is the use of the Dot Product of two vectors? Any command that starts with $ returns  command not found   How can I fix this?  I am  using 16.10 Ubuntu. What does "very approximate" mean? An approximate answer is one which is close to the correct answer.  Likewise, we can talk of an approximate model, or approximate methods in mathematics.  The etymology is from the Latin ad, "to" and proximus, "closest", see e.g. .  I don't think this can be disputed.  Occasionally proximate is used as a synonym, though that word has a different meaning in legal English.  So if an approximate answer is close, a very approximate answer should be very close.  In practice, "very approximate" or "very approximately" is overwhelmingly used with the opposite meaning of "roughly", indicating that an answer or estimate is not necessarily accurate.  For examples,   Is there any justification for this use? Is there any justification for my suggested use?  Should we avoid the phrase altogether? Best resources to learn bash scripting? Nautilus 3.6 doesn't have a status bar? In 12.04, Nautilus showed the amount of remaining space in the status bar , but as of 13.04 the status bar is gone.  gsettings set org.gnome.nautilus.window-state start-with-status-bar true   turned out "no such key".  Is there any way to bring back the status bar? Ugly 'post date' alignment on user profile when the date wraps into multiple lines You can think this of the following as a 3d QFT where we try to calculate the self-energy of two fields. $I$ is a this external self-energy and let us assume it does not depend on the loop momenta which I denote a $\vec{x}=(x,y)$. Consider the following propagator  $$ P(\vec{r}) = \frac{x-ia}{2i^2[\,(x-ia)^2+(y-ib)^2+(z-ic)^2]\,} \,\, \gamma_i. $$ Then we want to calculate $$ I(x) = \int \frac{\,dx \,dy\, dz}{(2\pi)^3} \text{Tr} \left( \frac{x-ia}{2i^2[\,(x-ia)^2+(y-ib)^2+(z-ic)^2]\,}\gamma_x \gamma_x  \right), $$ where Tr$(\gamma_i \gamma_j)=2\delta_{ij}$. Therefore  I am trying to resolve the following integral where $x,y,z \in \mathbb{R}$ and $a,b,c$ are constants (internal self-energies if you wish) in $\mathbb{R}$.  Then, after simplifying the propagator $P(x,y)$ we end up with the following expression for the $x$ component:  $$ I = \frac{1}{2i^2} \int \frac{dxdy}{(2\pi)^2} (x-ia) \int \frac{dz}{2\pi} \frac{1}{[\,(x-ia)^2+(y-ib)^2+(z-ic)^2]\,} $$ but it can be shown that  $$ \int \frac{dq}{(q+iv)^2+w^2} = \frac{\pi}{|w|}\theta(|w|-|v|).  $$ (e.g. see here ) Thus we get that for $|(y-ib)^2+(x-ia)^2|&gt;|a|$, $\theta=1$ (for this is what we are interested about) and then we are left with $$ I = \frac{1}{4i^2} \int \frac{dxdy}{(2\pi)^2} \frac{(x-ia)}{|(x-ia)^2+(y-ib)^2|}.   $$ The question is wether I can make the transformation $$ (x-ia) = q \cos \theta  $$ $$ (y-ib) = q \sin \theta $$ in polar coordinates. Then $dx dy = qdqd\theta$ and the previous integral becomes, assuming $|q|&gt;0$, $$ I = \int_0^{\infty} dq \int_0^{2\pi} d\theta \frac{q \cos \theta}{|q|} =0. $$ Is this correct? I am confused since one can find complex poles in the $I$. If not, what would the correct way to resolve this be? I have a suspicion that this integral should indeed give zero but how can we find this? Moving tfrom Shimano 9- to Shimano 10- speed I'm thinking of changing the 9-speed Tiagra rear mech on one of my (road) bikes to something, still Shimano, but 10-speed.  The obvious replacements I'll need are cassette, right-shifter and chain. But I reckon because of the narrower chain I'm also looking at a new chainset and rear derailleur. However I think the front derailleur and left-shifter should be safe (still keeping the front mech the same). Ditto the bb, regardless of whether I need a new chainset  Has anyone actually done this conversion and is able to confirm (or put me right)?  Thanks  Update  Decided in the end that going from 9- to 10- was too much to spend all in one go. But I used a service as an excuse to change the cassette (to a 9-speed ultegra) and the chain (dura ace) at quite minimal cost. Can upgrade other parts as and when, although I'm basically looking at Shim's 6500 range here so all quite dated now. How to resolve "iTunes could not back up the iPhone because the backup was corrupt"? When I try to sync my iPhone with my computer, I get the following message:       iTunes could not back up the iPhone "iPhone Name"   because the backup was corrupt or not compatible   with the iPhone.      Delete the backup for this iPhone, then try again.   Yet, here is no option to delete the backup. Where are backups stored / how can I delete the backup?  This is iTunes version 10.5.3 running under  Mac OS X 10.6.8. Suppose there are two face down cards each with a positive real number and with one twice the other. Each card has value equal to its number. You are given one of the cards (with value $x$) and after you have seen it, the dealer offers you an opportunity to swap without anyone having looked at the other card.  If you choose to swap, your expected value should be the same, as you still have a $50\%$ chance of getting the higher card and $50\%$ of getting the lower card.  However, the other card has a $50\%$ chance of being $0.5x$ and a $50\%$ chance of being $2x$. If we keep the card, our expected value is $x$, while if we swap it, then our expected value is: $$0.5(0.5x)+0.5(2x)=1.25x$$ so it seems like it is better to swap. Can anyone explain this apparent contradiction? Correct tracked camera orientation I deleted the "/etc/passwd" file and cannot log in I was trying to change my username and also my home directory (/home/username) and my system started to crash. I deleted the passwd file but I had a backup named passwd_bkp. I tried to rename this passwd_bkp as passwd and it did not work. No commands were being executed... I was in a terminal window.   I restarted my system and now I cannot log in. GRUB gives two options: Linux and recovery mode.   I tried to open a sessions as root but it says the filesystem is corrupted. I cannot access my files.  Did I lose all my files?