When I batch operations such as exportings, I'd like to know when it will be over.   Having a progress bar or indicator and something like "X of Y" (X is number of objects processed; Y is total number of objects) would be useful.  Does Blender have a way to accomplish this?
I'm making an python script which create a lot of object and this could take a lot of time. My question is how to show to the user of my script, the "progression" of the script? For now I print an progress bar in the system console.  But you have to open the system console befor and (apparently) the system console only work with Windows.  Is there an other way to show the progression of my script with python like control the mouse percentage icon (the little icon that replace the mouse while baking) or another idea?
How to wait in bash for several subprocesses to finish and return exit code !=0 when any subprocess ends with code !=0? In fact, I would like to ask more general question -- &quot;what does write permission for a directory allow you to do exactly?&quot; -- but let's approach it with a concrete example. It is a long question, if you are in a hurry read the bold -- it should cover the main part. Different sources (, , ) say something similar to the following on the directory permissions:  r, read -- reading of the directory's content (filenames inside) w, write -- changing the directory's attributes (e.g. modification time) and creating/renaming/removing entries inside x, search -- accessing files inside, you have access to the inode of the file, hence you can reach it's actual content  My problem is with the description of w. What directory's attributes does it give you access to? I cannot create/rename/remove a file inside a directory with only write permission: I make a directory (tdir/) and a file inside (afile), chmod -x-r tdir/, mv tdir/afile tdir/af, rm tdir/afile, touch tdir/newfile -- all fail with permission denial, unless I set x permission to the directory as well. And x alone doesn't give you the permission to create/rename/remove files inside the directory. In order to do that you need both x and w. But touch tdir does change the modification time of the directory with w only. I would rephrase the sources above this way for the compliance with the issue: a directory's r allows you to see the filenames inside, but no access to the actual file (to the inode); x gives you access to the inodes of the files (which means you can see their permissions and, according to it, have access to the contents), but you still cannot change anything in the directory; directory is actually some sort of a file and to change something in it you need the w permission. Thus, when you are changing something in the directory you need w permission. If your change requires inodes of the files in the directory -- you need x as well. It explains why you cannot remove a file inside a directory with w only: when removing a file you need to reduce the link count of the inode by 1 -- you need to know the inode -- thus you need x for the directory. But why do you need x for creating (you could ask the system to create a file without exposing the inode?) and renaming/moving the file (when you move a file you don't change it in any way, you only change the records inside the directories and their inode counts?)? Maybe it is just an implementation thing? I.e. indeed you don't need the inode for renaming/creating files -- you need only filenames and w permission; but inode and filename constitute one record in the directory; thus changing the filenames = changing the records = kind of accessing the inodes. And also what attributes do directories have besides modification time, permissions and files records? What else in the directory can you change with w only? Let $R$ be a commutative ring with unit, $I$ an ideal of $R$ and consider the following three constructions.   The  $R_I$ of $R$ at $I$ (i.e. the localization of $R$ at the multiplicative system $R\setminus I$) gives a morphism $$ f_1:\operatorname{Spec}(R_I)\to \operatorname{Spec}(R)=:X $$ The  $\widehat{R}_I$ of $R$ at $I$ gives a morphism $$ f_2:\operatorname{Spec}(\widehat{R}_I)\to X $$ For the special case $I=(a)$, the localization $R_a$ (i.e. the localization of $R$ at the multiplicative system $\{1,a,a^2,\ldots\}$) gives a morphism $$ f_3:\operatorname{Spec}(R_a)\to X $$   My question is:     What is the geometric meaning of all three constructions and how are they related?   This is what I ''know'' already or describes at least the style of answer I would appreciate. As remarked in the comments, $I$ has to be a prime ideal.   $R\to R_I$ is injective iff $R\setminus I$ contains no zero divisors which is the case if $I$ is a prime ideal. The scheme $\operatorname{Spec}(R_I)$ is the intersection of all the neightbourhoods of $I$ in $X$. This is a little contra-intuitive for me since the last statement sounds like ''$f_1$ is injective'' but the $\operatorname{Spec}$-operator should turn 'injective'' and ''surjective'' around somehow (I know this is literally not true but I only want to get a feeling like ''is contained'', ''is bigger'', ''is smaller'', etc.). $\operatorname{Spec}(R_a)$ is somehow the opposite of $\operatorname{Spec}(R_{(a)})$ (= the intersection of all the neightbourhoods of $(a)$ in $X$) because $\operatorname{Spec}(R_a)$ seems to be something like the union of all the open sets of $X$ not containing the point $(a)$. $R\to\widehat{R}_I$ is injective iff $\cap I^n=(0)$ and this holds very often (e.g. for $R$ noetherian and either an integral domain or a local ring). Hence (this is probably false intuition as remarked before), $f_2$ should be something like a ''projection'' (from something ''big'' into something ''small''). But what is geometrically the difference between localization and completion. I don't have a geometric idea of completion at all.   As remarked by Qiaochu Yuan in the comments below, one should not think of $\operatorname{Spec}$ as injective-surjective switching. Solve Euler Project #9 only mathematically - Pythagorean triplet The "Euler Project" problem 9 () asks to solve:  $a^2$ + $b^2$ = $c^2$ a + b + c = 1000  I find answers solving it with brute-force and programmatically, but is there a way to solve the problem ONLY mathematically? Can someone help, please?  Problem as explained in Project Euler website:  A Pythagorean triplet is a set of three natural numbers, a &lt; b &lt; c, for which,  a^2 + b^2 = c^2 For example, 3^2 + 4^2 = 9 + 16 = 25 = 52.  There exists exactly one Pythagorean triplet for which a + b + c = 1000.  Find the product abc. Side-by-side source and output when documenting a style file Possible Duplicate:          I want to know some example of a continuous function which is continuous at exactly one point. We know that $f(x)=\frac{1}{x}$ is continuous everywhere except at $x=0$. But i think this in reverse manner but i dont get any example. So please help me out! "is a field and used like a type" error Sequence converges iff $\limsup = \liminf$ I want to prove that a sequence of real numbers $\{s_n\}$ converges to $s$ if and only if $\limsup_{n \to \infty} s_n = \liminf_{n \to \infty} s_n = s$.  Here are my definitions:  For any sequence of real numbers $\{s_n\}$, let $E$ be the set of all subsequential limits of $\{s_n\}$, including possibly $+\infty$ and/or $-\infty$ if any subsequence of $\{s_n\}$ diverges to infinity.  Then $\limsup_{n \to \infty} s_n = \sup E$, and $\liminf_{n \to \infty} s_n = \inf E$.  I know the theorem that a sequence converges to a point if and only if every one of its subsequences converge to that same point, so one direction of this proof is easy:  If $\{s_n\}$ converges to some point $s \in \mathbb{R}$,  then every subsequence of $\{s_n\}$ converges to $s$.  So the set $E$ of every subsequential limit of $\{s_n\}$ consists of the single point $s$, so $$\limsup_{n \to \infty} s_n = \sup \{s\} = s = \inf \{s\} = \liminf_{n \to \infty} s_n$$  But the other direction seems more tricky...  If $\limsup_{n \to \infty} s_n = \liminf_{n \to \infty} s_n = s$, then every convergent subsequence converges to the same point $s$.  Also, there can be no subsequences which diverge to infinity (otherwise $\limsup_{n \to \infty} s_n$ would be $+\infty$, or $\liminf_{n \to \infty} s_n$ would be $-\infty$).  But can't there be subsequences which diverge otherwise?  And wouldn't that throw off the convergence of $\{s_n\}$?    EDIT:  I'd also be willing to accept a solution which makes use of the "Pinching Theorem" (if $a_n \leq s_n \leq b_n$ for every $n \in \mathbb{N}$, and if $a_n \to s$ and $b_n \to s$, then $s_n \to s$). Let $x$ and $y$ be nonzero elements of an integral domain $D$. Then $x$ and $y$ are associates if and only if $x = yd$ for some unit $d \in D$.    I am done proving the $\Leftarrow$ part. For $\Rightarrow$ what I did was: If $x$ and $y$ are associates then $x \mid y$ and $y \mid x$, so $x \mid y$ implies that $xs = y$ for some $s \in D$ and $y \mid x$ implies that $yt = x$ for some $t \in D$. That is, $yts=y$ which implies $ts=1$. Therefore $s$ is a unit and $t$ is a unit.    Am I on the right track? Can I say that the proof is complete? What EXACTLY cause Voldemort's soul piece to split off when he killed Lily and tried to kill Harry? Cannot vote to close immediately after edit Create UEFI bootable USB stick to install Windows AND Ubuntu from it I know how to make a UEFI bootable stick to install Ubuntu (explained ) and I know how to make one to install Windows 7 (, sorry). I guess for Win 8 the procedure would be similar, but I haven't looked into it.  I also know how to make a legacy bootable stick from which I can install Windows and Ubuntu using YUMI. I have failed so far though at creating a UEFI bootable stick that can do this. When I do a UEFI boot from my YUMI multiboot stick it gives me no option to choose between Ubuntu and Windows (hardly surprising I guess) and instead simply UEFI boots Windows (which I find kind of surprising to be honest but maybe this is due to the fact that I added the Windows image after the Ubuntu image and this simply replaced Ubuntu's EFI boot files).  So how would I proceed to create such a bootable stick? Privacy/Security concers for setting up business email using Gmail app on Nexus devices I need to set up my business email*(non gmail email)* on the mobile devices (Nexus 5 and Nexus7). The business has its own private mail server. The way I can do it is by using Gmail app on these devices. The question I have is does Gmail app on the mobile devices store the IMAP emails(non gmail emails) on the gmail servers? Or Gmail app is just a client and does not bother storing the IMAP emails(non gmail emails) on its servers that are not gmail emails? Thanks for the insight... Flag dialogue popup is stacked behind realtime-post-deleted-notification banners There is a little bug that becomes visible when a page contains realtime-post-deleted-notification banners (that's the CSS classname of these red banners appearing on answers which just got deleted, see screenshot below) and you try to open the flag dialogue above them:    Obviously, the dialogue appears below the banners, which it clearly should not do. It also has no input focus, trying to interact with the dialogue fails and either closes it (clicking outside of the red banners) or reloads the page (clicking on the red banners).  The cause for this is simply the z-index ordering in the site's CSS. Relevant CSS snippets:  .realtime-post-deleted-notification {     position: absolute;     text-align: center;     z-index: 100; }  .popup {     display: none;     padding: 15px;     position: absolute;     z-index: 10; }   So the fix for this minor bug would probably be to raise the z-index of the popup class to something greater than 100. Prove that if $\gcd(a,b)=1$ then $\gcd(ab,c) = \gcd(a,c) \gcd(b,c)$ Well structured URLs vs. URLs optimized for SEO Let’s take an example of a movie site. So a well structured set of URLs for this site can be:     Movie page - example.com/movies/the-matrix Review for movie - example.com/movies/the-matrix/reviews Trailers for movie - example.com/movies/the-matrix/trailers Specific trailer for movie - example.com/movies/the-matrix/trailers/trailer1   where it’s possible to navigate the resource even by directly deleting/appending resource name seamlessly.  For SEO purpose I read somewhere that for two given pages, Google will prefer the page which have search query up in the URL hierarchy among other things. So for the given 2 URLs and the query "the matrix trailer"    example.com/movies/the-matrix/trailers example.org/trailers/the-matrix/    `     Google might prefer the second URL as it has the keyword "trailer" at first level comparing to first URL which has "trailer" at 3rd level (a resource deeper in URL hierarchy may be less important).    If it is correct, one may be tempted to create URL hierarchy somewhat like      Movie page - example.com/movies/the-matrix Review for movie - example.com/reviews/the-matrix/ Trailers for movie - example.com/trailers/the-matrix/ Specific trailer for movie - example.com/trailers/the-matrix/trailer1     which will break the seamless hierarchy.    My questions are:   Is it true that resource deeper in URL hierarchy is tend to be less important when comparing to URLs of other site?   Is sacrificing meaningful, seamless hierarchy worth for this SEO optimization? We can write Bayes' theorem as  $$p(\theta|x) = \frac{f(X|\theta)p(\theta)}{\int_{\theta} f(X|\theta)p(\theta)d\theta}$$  where $p(\theta|x)$ is the posterior, $f(X|\theta)$ is the conditional distribution, and $p(\theta)$ is the prior.  or  $$p(\theta|x) = \frac{L(\theta|x)p(\theta)}{\int_{\theta} L(\theta|x)p(\theta)d\theta}$$  where $p(\theta|x)$ is the posterior, $L(\theta|x)$ is the likelihood function, and $p(\theta)$ is the prior.  My question is   Why is Bayesian analysis done using the likelihood function and not the conditional distribution? Can you say in words what the difference between the likelihood and conditional distribution is?  I know the likelihood is not a probability distribution and $L(\theta|x) \propto f(X|\theta)$. Interface vs Abstract Class (general OO) How to draw a return arrow from node-3 to node-1 How can I view Google Photos that are not in a Google Photos album? Go to  Add a few photos to a photo album Go back to All Photos   See all of the photos there, even ones you copied to the Album. No problem here, per se. At least not yet.  What if you want to delete photos that you did not copy to that album?  Where is the selector or filter that filters those "in an album already" photos from this main view?  Or is there a way to create such a filtered view such that I can then cherry-pick photos for deletion?