How to reverse order a list? (Python 3.x)
Traverse a list in reverse order in Python
I want to see what nodes exist under a certain domain from within the domain. Short of querying every valid IP address within that domain, is there a way to get all records/map the network? Why is the node editor showing different kinds of nodes? I've just upgraded to Windows 10 from Windows 8.1 on my partitioned laptop that has Ubuntu 14.04 installed. Grub no longer appeared on boot so I ran boot-repair from an Ubuntu liveUSB. This didn't seem to have any effect. I've checked and secure boot is still disabled. Log from boot-repair:  So what's next?  Update: I've attempted to use chroot to update grub from my Ubuntu partition. update-grub runs successfully finding both linux and windows boot manager but the process still doesn't have any effect. grub-install complains that it cannot find EFI directory (but I have mounted everything correctly).  Update2: Just discovered that running efibootmgr after chroot into ubuntu shows ubuntu as not being in the boot order at all. I manually changed the order and rebooted to discover that the bootorder was reset. I run in windows: bcdedit /set {bootmgr} path \EFI\ubuntu\shimx64.efi But this has no effect either even after reporting 'Operation Successful'. Can you choose which version of a game to play on Steam? I want to play a game in an older update, but can't see a way to choose a specific version. Is this possible? I want to show for $x\in\mathbb{R}$, the following inequality holds. $$ \left|\sin x\right|+\left|\sin (x+1)\right|+\left|\sin (x+2)\right|&gt;8/5. $$ This trigonometric inequality has been verified by Mathematica using the Plot commend. However, using the sum formula for the sine function seems doesn't work, and I cannot give a rigorous proof of it. Any suggestion, idea, or comment is welcome, thanks! Associated elements in a ring Please help me to find elements $a,b$ in a ring $R$ such that $a\mid b$ and $b\mid a$, but there does not exist any unit $u$ in $R$ such that $a=ub$. Does Earth really have two high-tide bulges on opposite sides? The bit that makes sense – tidal forces My physics teacher explained that most tidal effect is caused by the Moon rotating around the Earth, and some also by the Sun. They said that in the Earth and Moon system, the bodies are in freefall about each other. But that points on the surface of Earth, not being at Earth's centre of gravity, experience a slightly different pulls towards the Moon. The pull is a little greater if they are on the Moon's side, and slightly lesser if they are on the side away from the Moon. Once freefall is removed, on the Moon side this feels like a pull towards the Moon and on the the opposite side it feels like a repulsion from the Moon. This makes sense to me, and is backed up by other questions and answers here, like  and also  Phys.SE question. The bit that doesn't make sense – tidal bulges They also said that there are &quot;tidal bulges&quot; on opposite sides of the Earth caused by these forces. The bulges stay still relative to the Moon and Earth rotating through the bulges explains why we get two tides a day. They drew a picture like this one…  An image search for  finds hundreds of similar examples, and  from a scientist on Twitter. …But, if there is a tidal bulge on both sides of Earth, like a big wave with two peaks going round and around, how can an island, like Great Britain where I live, simultaneously have a high tide on one side and a low tide on the other? For example:   on the West coast  on the East  Two ports with tides 6 hours, or 180º apart. It's high tide at one while low tide at the other. But they are only  by road. Great Britain is much smaller than Earth. It's probably not even as big as the letter &quot;A&quot; in the word &quot;TIDAL&quot; in that picture.  To prove this isn't just Britain being a crazy anomaly, here is another example from New Zealand:     Two ports that are 180º (6 hours) apart, but separated by  through a national park. New Zealand, unlike the UK, is in fairly open ocean. Extracting the function from InterpolatingFunction object Problems with Curve Modifier Why have view counters been modified? Reputation widget failure on Android The title says it all. When I try to place the reputation widget on my home screen, I get this:  When I try again, I get this error, presumably because the Stack Exchange service has stopped.  Is this a bug? If so, where should I report it?  I am running Android 5.1 armeabi on a DigiLand DL718M tablet. What is a Pythonic way to pad a numeric string with zeroes to the left, i.e. so the numeric string has a specific length? Conditional Expectation of Composite Function Preliminaries  Let $(\Omega, \mathcal{G}, \mathbb{P})$ be a complete probability space.  Let $D$ be a complete, separable, metrizable topological space with Borel $\sigma$-algebra $\mathcal{B}(D)$ (such as $D = \mathbb{R}^q$ with $\sigma$-algebra $\mathcal{B}(D) = \mathcal{B}(\mathbb{R}^d)$).  Let $\mathbb{R}$ be equipped with its canonical Borel $\sigma$-algebra $\mathcal{B}(\mathbb{R})$.  Let $g: \Omega \times D \rightarrow \mathbb{R}$ be a bounded $(\mathcal{G} \otimes \mathcal{B}(D) ) / \mathcal{B}(\mathbb{R})$-measurable function.  Let $\Pi: \Omega \rightarrow D$ be a $\mathcal{G}/\mathcal{B}(D)$-measurable random variable.  Let $H : \Omega \rightarrow \mathbb{R}$ be a $\mathcal{G}/\mathcal{B}(\mathbb{R})$-measurable random variable, defined by $$ H(\omega) := g(\omega, \Pi(\omega)).$$ Note, that, since $g$ is bounded, we have $H \in \mathcal{L}^2(\Omega, \mathcal{G}, \mathbb{P})$.  Let $j: D \rightarrow \mathcal{L}^2(\Omega, \mathcal{G}, \mathbb{P}) $ be defined by $$ j(\pi)(\omega) := g(\omega, \pi)  $$  For all $\pi \in D$, let $j(\pi)$ be independent of $\Pi$.  Question  I am interested in the conditional expectation $$\mathbb{E}[H \mid \Pi] :\Omega \rightarrow \mathbb{R}$$ of $H$ with respect to $\Pi$. More specifically, I suspect that (a $\mathbb{P}$-unique version of) this condititional expectation is given by  $$ \mathbb{E}[H \mid \Pi] (\omega) = \mathbb{E}[j(\Pi(\omega))], \quad (\dagger)  $$ whereby $\mathbb{E}[j(\Pi(\omega))]$ can of course also be written as $$\mathbb{E}[j(\Pi(\omega))] = \int_{\Omega} j(\Pi(\omega))(\tilde{\omega}) d\mathbb{P}(\tilde{\omega}) . $$  How can I prove, that $(\dagger)$ is the case? I have tried, tracking the definition of conditional expectation and using Fubini, but with little success so far.  Thanks for any advice! Better explanation of the common general relativity illustration (stretched sheet of fabric) I've seen many science popularisation documentaries and read few books (obviously not being scientist myself). I am able to process and understand basic ideas behind most of these. However for general relativity there is this one illustration, which is being used over and over (image from Wikipedia):    I always thought that general relativity gives another way how you can describe gravity. However for this illustration to work, there needs to be another force, pulling the object down (referring to a direction in the attached image). If I put two non-moving objects in the image, what force will pull them together?  So where is my understanding incorrect? Or is general relativity not about explaining gravity and just describes how heavy objects bends spacetime (in that case the analogy is being used not correctly in my opinion)?    UPDATE Thank you for the answers and comments. Namely the XKCD comics is a spot on. I understand that the analogy with bent sheet of fabric pretty bad, but it seems that it can be fixed if you don't bent the fabric, but just distort the drawn grid.   Would you be so kind and answer the second part of the question as well - whether general relativity is explaining gravitational force. To me it seems that it is not (bending of spacetime simply can not affect two non-moving objects). However most of the time it is being presented that it does. Houston, we have a porblem Since  in titles, and there's not even a way  (or ) we all know users just get around that filter by typing various permutations of the word "problem".  The filter has . The bottom line there was:     "Hacking around" the filter is in fact very very rare.   I beg to disagree. I tried searching Stack Overflow for the following variations of the title:  problme probelm probeml probmle probmel prolbem prolbme prolebm prolemb prolmbe prolmeb proeblm proebml proelbm proelmb proembl proemlb   These are just a couple of permutations from 5040 possible ones. I also tried searching for variations with left out letters, like proble, or prblm. Here's just a graph of a subset from maybe 20 searches—but in reality, there might be more ways to get around the filter:    Was this ever intended? I know it saves a lot of bad titles from being created, but on the other hand it enforces even worse titles as well. 3% based on the  aren't much, but they're 3% crap the filter actively provokes.  Now, my questions up for discussion are:   What are we going to do about this large number of porblematic posts? It seems almost impossible to fix all those. We now have a really good review system that allows people to dig through low quality posts the entire day, so that'd be a way to address this of course. Are there any plans of getting rid of the filter again? This, at least, would give you a good indicator of poor quality. If a post title contains the word "problem" you instantly know something's off. Why not have it reviewed? I understand that some people will choose a better title after being blocked, but I don't even want to start looking at the number of those who merely did a s/problem/issue, or even worse, just used "problems" (because that's not blocked, just sayin').   In the end I'm just thinking if there was a way to clean up the back log on this, or how to prevent more posts of this sort in the future, even with a title filter in place if it has to be. Is there a way to mix LIKE operator with IN ? Something Like  SELECT Id FROM Account WHERE Name LIKE IN:nameList   Where nameList is list of String? Is this is possible ? (I can make a dynamic query by doing OR with all the elements, but was looking for a more elegant way to do this) Quieting down a Dell PowerEdge r710 I run a Dell PowerEdge r710 server out of my house and I'm moving in with roommates in a couple weeks. Unfortunately I don't think they'll appreciate the ~70dBA it generates and I'm looking for ways to make it quieter.  I looked into replacing the fans with something less noisy, but unfortunately none of the quieter options move nearly enough air. The best one I found has been the , but at less than half the CFM (30.3 vs 67.85 on the stock  fans) it doesn't seem powerful enough.  I've also thought about putting  around the server to quiet it down, but acoustic panels also have the downside of acting like insulation, so heat could turn into an issue. A high school Olympiad problem I am trying to assign an image texture to an Image Texture node in 2.80. Very simple script - I'm just starting to learn how this stuff works.  This is what I have. It creates an image and and IT node, but it doesn't assign the one to other. Can anybody help? I've searched around, but can't find it in another question.  import bpy  #create blank image bpy.ops.image.new(name="TestImg", width=1024, height=1024)  #add image texture to object, to hold baked lighting mat = bpy.context.view_layer.objects.active.active_material image_node = mat.node_tree.nodes.new('ShaderNodeTexImage')  #assign image to image texture node image_node.image = "TestImg"   Thanks in advance. How do I disable overlay scrollbars? It looks like Ubuntu 11.04 will come with  by default. I do like them, but I don't like the current mix of scrollbar types as many applications don't yet use those overlay scrollbars.  Is there a way to disable overlay scrollbars (without removing the overlay-scrollbar package)?