How can I load a Python module given its full path? Note that the file can be anywhere in the file system, as it is a configuration option. i am using python 3.5 in ubutnu , my code so far is :  Create python module test.py  import sys sys.path.append("test/lib/") from tes1 import Client1 from tes2 import Client2 import tes3
How can I load a Python module given its full path? Note that the file can be anywhere in the filesystem, as it is a configuration option.
node.js conflicts: /sbin/node vs /usr/bin/node I have two Ubuntu 12.10 machines: machine A is a VMWare VM and machine B is and old Acer laptop. On both machines, I installed node using apt-get. However machine A installs the main binary as /usr/bin/node and machine B as /usr/bin/nodejs. Here are some behaviors I observed:   Both machine returns /usr/bin/node in response to the which node command. However, there is no such file on machine B. Issuing the command node works for machine A, not B. On B, I have to use nodejs instead. On B, there is a binary called /sbin/node, A does not have it.   These behaviors causes inconsistency problem when developing on both machines. For now, I worked around by renaming the binaries on B as such:  sudo mv /sbin/node /sbin/node-sbin sudo ln -s /usr/bin/nodejs /usr/bin/node   Is there a way to really fix this, so that node will be installed as /usr/bin/node instead of /usr/bin/nodejs? Does String theory say that spacetime is not fundamental but should be considered an emergent phenomenon? Does String theory say that spacetime is not fundamental but should be considered an emergent phenomenon?  If so, can quantum mechanics describe the universe at high energies where there is no spacetime?  Schrodinger equation assumes the presence of time dimension. So how can such equations be valid at extremely high energy?  And is it meaningful to do physics if there is no time dimension?  Would the concepts of Hilbert space, wavefunction collapse etc be useful in that case? How do I prove that a graph is a topological manifold I'm having problems with that:  Prove that M is a topological manifold.  $f:U \to \mathbb R^k$, U $\subset \mathbb R^{n}$ open, f continuous  M = $\{\left(x,y\right)\in \mathbb R^{n+k} \mid x \in U, y=f(x) \}$  Can anyone help me please? It is quite easy to calculate the total number of functions from a set $X$ with $m$ elements to a set $Y$ with $n$ elements ($n^{m}$), and also the total number of injective functions ($n^{\underline{m}}$, denoting the falling factorial). But I am thinking about how to calculate the total number of surjective functions $f\colon X \twoheadrightarrow Y $.  The way I thought of doing this is as follows: firstly, since all $n$ elements of the codomain $Y$ need to be mapped to, you choose any $n$ elements from the $m$ elements of the set $X$ to be mapped one-to-one with the $n$ elements of $Y$. This results in $n!$ possible pairings. But the number of ways of choosing $n$ elements from $m$ elements is $\frac{m!}{(m-n)!\,n!}$, so the total number of ways of matching $n$ elements in $X$ to be one-to-one with the $n$ elements of $Y$ is $\frac{m!}{(m-n)!\,n!} \times n! = \frac{m!}{(m-n)!}$.  Now we have 'covered' the codomain $Y$ with $n$ elements from $X$, the remaining unpaired $m-n$ elements from $X$ can be mapped to any of the elements of $Y$, so there are $n^{m-n}$ ways of doing this. Therefore I think that the total number of surjective functions should be $\frac{m!}{(m-n)!} \, n^{m-n}$.  Is this anything like correct or have I made a major mistake here? Using congruences, show $\frac{1}{5}n^5 + \frac{1}{3}n^3 + \frac{7}{15}n$ is integer for every $n$ I often see "replace with" and "replace by" used interchangeably, but this doesn't sound right to me:     I replaced that component by this one.   I would use "with" in such a sentence. "By" only seems reasonable in passive, although "with" sounds like it would there work too:     That component was replaced by this one.   That component was replaced with this one.   In my native language, the equivalent of "replace by" can only be used in passive, and even then it's a bit weird unless a person is the object replacing something – perhaps this affects my judgment?  Web searches haven't come up with anything conclusive; the results are contradictory and speculative at best.  Are "replace with" and "replace by" interchangeable in active context? What about passive? Are there stylistic reasons to prefer one over the other? I'm trying to delete some folder but this error is preventing from doing it:     Cannot delete [FileName]: It is being used by another person or program.      Close any programs that might be using the file and try again.   How can I identify what application is using the file?   I want to end the application using a Task Manager, but I didn't know what application is using it.  The folder is not shared and my user account is an administrator. Why does an conditionally convergent infinite series, when deranged, have a different limit? $\sum u_n = 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\dots$ conditionally converges to $\log 2$; and, when deranged to $\sum u'_n = 1-\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{6}-\frac{1}{8}+\dots,$ converges to $\frac{1}{2}\log 2$. Why does a deranged conditionally convergent series have a different limit? First Course in Math Anal, p.93, by Burkill, explains that the sums $s_n, t_n$ of the first n terms of the two series are different functions of $n$. This reasoning is incomplete, because different functions of $n$ can converge to the same limit; for example, let $r_n$ be the first $2n$ terms of $\sum u_n$, then $s_n$ and $r_n$ are different functions; besides, by differently defining terms, i.e. clustering pieces of a series, e.g. \begin{align*} 1-\tfrac{1}{2}-\tfrac{1}{4}+\tfrac{1}{3}-\tfrac{1}{6}-\tfrac{1}{8}+\dots = \left(1-\tfrac{1}{2}-\tfrac{1}{4}\right)+\left(\tfrac{1}{3}-\tfrac{1}{6}-\tfrac{1}{8}\right)+\dots \\ = \left(1-\tfrac{1}{2}-\tfrac{1}{4}+\tfrac{1}{3}-\tfrac{1}{6}-\tfrac{1}{8}\right)+\dots,\\ 1-1+1-1+\dots = (1-1)+(1-1)+\dots \\ =(1-1+1-1)+\dots,\\ 1+1+1+1+\dots = (1+1)+(1+1)+\dots \\ = (1+1+1+1)+\dots, \end{align*} one gets the sum of first n terms as different functions, but the series's limit remains. One could view a changed limit from the perspective of the changed number of terms, Denote 'countably many', the size of natural number, as $\lvert\mathbb{N}\rvert$, then $\lvert\mathbb{N}\rvert = \lvert k\mathbb{N}\rvert = \lvert\mathbb{N}\times\mathbb{N}\rvert$. Then two series with different numbers (e.g. $\lvert \mathbb{N}\rvert$, $\lvert k\mathbb{N}\rvert$) of terms– and so having different limits– paradoxically have same number of terms–and so are possibly formed by the same terms. The paradox could be the fundamental reason. Are there other explanations? Is the use here of 'countably many' is proper? Recursively delete empty directories in Windows I have a directory on my Windows 7 machine that has hundreds if not thousands of sub-directories.  Some of them have files, some do not.  I want to delete all the empty directories.  Looking at the del and rmdir DOS command, it does not look like you can recursively do this without deleting all the files.  Is there a way to do this from the command line? Disable App Update Notifications on OSX? Theory behind partial least squares regression Can anyone recommend a good exposition of the theory behind partial least squares regression (available online) for someone who understands SVD and PCA?  I have looked at many sources online and have not found anything that had the right combination of rigor and accessibility.    I have looked into The Elements of Statistical Learning, which was suggested in a comment on a question asked on Cross Validated, , but I don't think that this reference does the topic justice (it's too brief to do so, and doesn't provide much theory on the subject).  From what I've read, PLS exploits linear combinations of the predictor variables, $z_i=X \varphi_i$ that maximize the covariance $ y^Tz_i $ subject to the constraints $\|\varphi_i\|=1$ and $z_i^Tz_j=0$ if $i \neq j$, where the $\varphi_i$ are chosen iteratively, in the order in which they maximize the covariance.  But even after all I've read, I'm still uncertain whether that is true, and if so, how the method is executed. Reference template variable within Jinja expression Where does Drupal store its content? I have been currently learning Drupal while trying to duplicate a Drupal application into CakePHP.  I have been looking at the  database tables to know where data is being stored so I can safely migrate any data or study how it is working.  I have researched and have come across questions like:     So far, I have only seen the title in my Drupal (7) installation.  What I have seen in my node revisions table for a recipe content type, which includes fields like, field_ingredients, field_instructions, field_description, field_preptime, field_cookingtime, just the title of the recipe. This is a sample record on my node_revision table.  nid   |  vid  |  title         |  log   | timestamp  | status | comment | promote  | sticky ------------------------------------------------------------------------------------------- 1     |  1    |  Awesome Sauce |        | 1369998600 | 1      | 2       | 1        | 0   As you can see I can only see the title. Where are the other field contents? Presheaf which is not a sheaf -- holomorphic functions which admit a holomorphic square root I'm thinking about a problem in Ravi Vakil's algebraic geometry notes  (Exercise 3.2B) and I'm having trouble with the second part of the exercise as my understanding of complex analysis is quite rudimentary.   If to every open set $U\subset \mathbb{C}$ we associate the ring of holomorphic functions which admit a holomorphic square root, this defines a presheaf on $\mathbb{C}$ (I can see this).   Apparently however this is not a sheaf -- supposedly it fails to satisfy the gluing property.   To prove this, we need to take an open cover $\{U_i\}$ of some open $U\subseteq \mathbb{C}$, a holomorphic function $f_i:U_i\to \mathbb{C}$ for each $i$ which has a holomorphic square root (i.e. there exists a holomorphic function $g_i$ for each $i$ such that $f_i(z)=g_i(z)^2$ for all $z\in U_i$), which all agree nicely on the intersections of the various open sets, but such that it is impossible to find a holomorphic function $f:U\to \mathbb{C}$ with a holomorphic function which restricts to $f_i$ on each $U_i$.   Any help would be great -- my problem is in showing that such a function cannot exist. I have a Dell N5110 laptop with integrated Nvidia GeForce GT 525M and an Intel HD Graphics 3000 card; My problem is the default card is the Intel one, and it provides a very bad performance with Windows 8. So can I force to use the Nvidia one all the time including "Desktop Windows Manager" ? Why does my object not show up? Are ores replenished over time? What is the lowest level of mathematics required in order to understand how encryption algorithms work? What mathematical fields of knowledge would be required in order to get a good understanding of encryption algorithms?   Is it basic algebra, or is there a "higher education" mathematical field dedicated to encryption? I know there is the cryptography field, but what is the subset of knowledge required for cryptographers? Can you win a game of Splendor if you still have cards in reserve? In R, what does qqnorm actually do? qqnorm(x) plot(qnorm(seq(80)/81),sort(x))   Finding that the plots produced by the commands above are slightly different from each other, I tried this:  qqnorm(qnorm(seq(80)/81))   I got a slightly less than perfect line.  I'd have tried regressing the result of qnorm(seq(80)/81) on the variable that was plotted on the $x$-axis and plotting residuals against the predictor, expecting to see some graceful curve, but for the fact that I don't know what to use as the predictor.  Possibly such a residual plot would reveal more than just the graceful S-shaped thing I'd anticipate.  So my question is this: if the thing on the $x$-axis in the plot produced by qqnorm is not what I get from qnorm(seq(80)/81) (and what I did shows that indeed it is not), then what is it?