How can i use a bash command in python for example :  # ifconfig eth0 promisc   in code like :  import socket # the public network interface HOST = socket.gethostbyname(socket.gethostname()) # create a raw socket and bind it to the public interface s = socket.socket(socket.AF_INET, socket.SOCK_RAW, socket.IPPROTO_IP) s.bind((HOST, 0)) # Include IP headers s.setsockopt(socket.IPPROTO_IP, socket.IP_HDRINCL, 1) # receive all packages s.ioctl(socket.SIO_RCVALL, socket.RCVALL_ON) # receive a package print s.recvfrom(65565) # disabled promiscuous mode s.ioctl(socket.SIO_RCVALL, socket.RCVALL_OFF)   instead of s.ioctl because linux doesn't support this command ?
How do you call an external command (as if I'd typed it at the Unix shell or Windows command prompt) from within a Python script?
Is it possible to see a list of all my edits using a query? I can understand the choice of material, silicon 28, but why is it a sphere rather than (say) a cube? Article   I would have thought that a sphere would have been the hardest shape to machine accurately. Don't understand proof that interior of a set is open I've installed ubuntu 17.04 and I want to upgrade it to 18.04. When I tried to do:   sudo do-release-upgrade   I got a message saying An upgrade from 'zesty' to 'bionic' is not supported with this tool. Need inputs to proceed on this.  sudo do-release-upgrade Checking for a new Ubuntu release Your Ubuntu release is not supported anymore. For upgrade information, please visit: http://www.ubuntu.com/releaseendoflife  Get:1 Upgrade tool signature [819 B]                                            Get:2 Upgrade tool [1,257 kB]                                                   Fetched 1,258 kB in 0s (0 B/s)                                                  authenticate 'bionic.tar.gz' against 'bionic.tar.gz.gpg'  extracting 'bionic.tar.gz'  Reading cache  Checking package manager  Can not upgrade   An upgrade from 'zesty' to 'bionic' is not supported with this tool. I have a document I am trying to copy to learn TeX. Here is what I have encountered: How can I have the "or" in the equation. This is what I have right now:  \[ f = ma; \]  But $a$ is the change in velocity,  \[ f = m \frac{dv}{dt};\] \[ f = m  \frac{d^2y}{dt^2};\]   Edit: Trying out Stefan's answer I miss those pins I make in Windows.     How can I do the same in Ubuntu 17.10 with GNOME 3? Should I compile release builds with debug info as "full" or "pdb-only"? Operator Exponential $e^A e^B = e^{A+B}$ This question comes from an exam in my functional analysis class.  Suppose $X$ is a Banach space, and $T \in B(X,X)$ is a bounded linear operator on $X$. For any non-negative integer $n$, let $$S_n=\sum_{k=0}^n \frac{1}{k!} T^k$$ where $T^k$ is the composition of $T$ with itself $k$ times and $T^0=I$.  We can show that for any integer $k&gt;0$, $\Vert T^k \Vert \le \Vert T \Vert ^k$. Then we can show that $S_n \in B(X,X)$ and there is some $S \in B(X,X)$ such that $S_n \to S$. We write $S = e^T$ for this operator.  Finally, We were asked to show that $e^T$ has an inverse, and that it is $e^{-T}$. My thought is to prove the following claim first: if $A,B \in B(X,X)$ and $AB = BA$, then $e^A  e^B = e^{A+B}$. If the claim is true, it follows that $e^T e^{-T}=I$.  The claim can be proven provided that the product series can be computed with the Cauchy rule.  $$e^Ae^B=\sum_{i=0}^{\infty}\frac{A^i}{i!}\sum_{j=0}^{\infty}\frac{B^j}{j!}=\sum_{k=0}^{\infty}\sum_{l=0}^{k}\frac{A^lB^{k-l}}{l!(k-l)!}$$ $$=\sum_{k=0}^{\infty}\frac{1}{k!}\sum_{l=0}^{k}\frac{k!}{l!(k-l)!}A^lB^{k-l}= \sum_{k=0}^{\infty}\frac{1}{k!}(A+B)^k= e^{A+B}$$  But why can the product series be summed in the Cauchy way? I know for real-number series, by Cauchy's theorem, if $\sum_{n=1}^{\infty} a_n$ and $\sum_{n=1}^{\infty} b_n$ are absolutely convergent to $A$ and $B$, respectively, then we can add $a_i b_j$ in any way, and the resulting series will converge to $AB$. Does this proposition still hold for commutable operators? (It would be greatly appreciated if ideas of proof or reference is suggested.) I am sure there is something wrong i am doing but i just want someone to point it to me.  Why can't we say that $e^{\frac {\pi i}{3}} = \left(e^{\pi i}\right)^{\frac {1}{3}} = (-1)^{\frac {1}{3}} = -1$ Why do we have to calculate it by the formula as it is (which produces different value)  Another question is why can't we say that $ (-1)^{\frac {1}{2}} = \left(\left (-1\right)^{2}\right)^{\frac {1}{4}} = 1^{\frac {1}{4}} = 1$  Again i have to say that i know this is all wrong i just want to know why it is wrong What is the Linux equivalent to Windows' Program Files? Can the Flash do things like build muscle faster than normal (do his muscles grow right after a work-out or meal)? Can he heal faster than normal? Does he have a faster immune system to stop all illness or at least make the flu last 2 minutes or something along those lines? Call java program from Node.js application $ echo $(( 255 )) 255 $ echo $(( 33 )) 33 $ echo $(( ~33 )) -34 $ echo $(( ~255 )) -256 $    and my kernel is:  $ uname -a Linux HOSTNAME 3.2.0-40-generic-pae #64-Ubuntu SMP Mon Mar 25 21:44:41 UTC 2013 i686 i686 i386 GNU/Linux     QUESTION: ~ is for negating the number AFAIK. But why does ~33 produce -34 and why does ~255 produce -256? I found this odd relationship, $x^2 = \sum_\limits{k = 0}^{x-1} (2k + 1)$. vertical align middle in Scenario:   Create SymLink: C:\Data\workspaces --&gt; C:\Users\****\Documents\workspaces Navigate to C:\Data\Workspaces in explorer Click and drag the the path from the address bar onto "Quick Access" in the navigation pane.   Expected Result:   Clicking on the shortcut takes you to C:\Data\workspaces   Actual Result:   Clicking on the shortcut takes you to C:\Users\****\Documents\workspaces   How do you make it so that the shortcut points to the correct path. (the SymLink)  Command Used to create the SymLink:  mklink /D C:\Data\workspaces C:\Users\****\Documents\workspaces   Updated with pictures:  Please note the workspaces symlink inside C:\Data:    When I double-click on the symlink it works correctly. (note the address bar.    When i pin it to the Quick access bar, however, it uses the address that the symlink points to and NOT the symlink address. What is vertex and edge shading? Determining a matrix from its characteristic polynomial Let $A\in\mathcal{M}_{n}(K)$, where $K$ is a field. Then, we can obtain the characteristic polynomial of $A$ by simply taking $p(\lambda)=\det(A-\lambda I_n)$, which give us something like $$p(\lambda) = (-1)^n\lambda^n + (-1)^{n-1}(\text{tr } A)\lambda^{n-1} +  \cdots + \det A$$ Now, how can we obtain the matrix $A$ knowing the characteristic polynomial? Let $E=l^p=\left\{x=\left(x_n\right): x_n\in\mathbb{R}, \sum_{1}^{\infty}\left|x_n\right|^p&lt;\infty\right\}$ with $1\le p&lt;\infty$, $\left\|x\right\|_E=\left(\sum_{1}^{\infty}\left|x_n\right|^p\right)^{1/p}.$ Let $\left(\lambda_n\right)$ is a bounded sequence in $\mathbb{R}$ and consider the operator $T$ defined by $$T(x)=\left(\lambda_1x_x,\ldots,\lambda_nx_n,\ldots\right)$$ where $x\in l^p$. Prove that $T$ is a compact operator from $E$ to $E$ iff $\lambda_n\rightarrow 0$.  Can anyone give me some hints to solve this problem? Thank you! Let's assume a very simplified situation. There is only one seller asking for 10$ and there is only one buyer bidding 20$. At what price would the transaction go?   Some answers say it won't, because the ask price has to match the bid price. () Other answers say, that the time ordering is important. () Which answer is correct? What is the final price in the presented situation?