I installed python3-selenium apt package on Ubuntu 16.04. While installing, got a message:  Suggested packages: chromedriver firefoxdriver The following NEW packages will be installed: python3-selenium   When I try to run the following python code,  #! /usr/bin/python3.5 from selenium import webdriver import time  def get_profile():     profile = webdriver.FirefoxProfile()     profile.set_preference("browser.privatebrowsing.autostart", True)     return profile  def main():     browser = webdriver.Firefox(firefox_profile=getProfile())      #browser shall call the URL     browser.get("http://www.google.com")     time.sleep(5)     browser.quit()  if __name__ == "__main__":     main()   I get the following error:  Traceback (most recent call last):     File "./test.py", line 19, in &lt;module&gt;         main()     File "./test.py", line 11, in main         browser = webdriver.Firefox(firefox_profile=getProfile())     File "/usr/lib/python3/dist-packages/selenium/webdriver/firefox      /webdriver.py", line 77, in __init__ self.binary, timeout),     File "/usr/lib/python3/dist-packages/selenium/webdriver/firefox/extension_connection.py", line 47, in __init__ self.profile.add_extension()     File "/usr/lib/python3/dist-packages/selenium/webdriver/firefox/firefox_profile.py", line 91, in add_extension self._install_extension(extension)     File "/usr/lib/python3/dist-packages/selenium/webdriver/firefox/firefox_profile.py", line 251, in _install_extension compressed_file = zipfile.ZipFile(addon, 'r')    File "/usr/lib/python3.5/zipfile.py", line 1009, in __init__ self.fp = io.open(file, filemode) FileNotFoundError: [Errno 2] No such file or directory: '/usr/lib /firefoxdriver/webdriver.xpi'   I did searching for packages name firefoxdriver in Ubuntu repositories but none exist.  Any help with installing the webdrivers appreciated.
When I install Selenium I get the following error:  Shubham@Shubham-To-be-filled-by-O-E-M:~$ sudo apt-get update     Get:1 http://security.ubuntu.com/ubuntu xenial-security InRelease [107 kB]     Hit:2 https://repo.skype.com/deb stable InRelease                            Hit:3 http://in.archive.ubuntu.com/ubuntu xenial InRelease                          Get:4 http://in.archive.ubuntu.com/ubuntu xenial-updates InRelease [109 kB]     Get:5 http://in.archive.ubuntu.com/ubuntu xenial-backports InRelease [107 kB]       Fetched 323 kB in 8s (38.6 kB/s)                                                    Reading package lists... Done     Shubham@Shubham-To-be-filled-by-O-E-M:~$ sudo pip install selenium     Traceback (most recent call last):       File "/usr/bin/pip", line 9, in &lt;module&gt;         from pip import main     ImportError: cannot import name main   How should I proceed?
80s-90s anime (movie or series) in which an armored girl exchanges age with a dying man on a beach What is the first reference in Sci-Fi to a touch-screen computer interface? I know in TNG that all computer interfaces are now touch-screen, and that got me thinking: when was the first reference to touch-screen computer interfaces in Sci-Fi made? Gapless superconductivity I don't understand what is meant by &quot;gapless superconductors&quot;. As far as I know, the superconducting gap is necessairy for both zero resistance and the Meissner effect, the two hallmarks of superconductivity. So, what does it mean to have gapless superconductivity and how does it work? Thanks in advance, Greetings Mutual set of representatives for left and right cosets: what about infinite groups? Let $G$ be a group and $H$ a subgroup of $G$. If $G$ is finite, then according to Philip Hall's "marriage theorem" there is a left transversal $T$ of $H$ in $G$ (that is, $T$ contains precisely one element from each left coset of $H$) such that $T$ is also a right transversal of $H$.   Does this theorem generalize to infinite groups? The original theorem treats only the case where $|H| &lt; \infty$ and $[G:H] &lt; \infty$.  What can we say when $|H| = \infty$ and $[G:H] = \infty$? What about $|H| &lt; \infty$ and $[G:H] = \infty$? Also $|H| = \infty$ and $[G:H] &lt; \infty$? In MySql's interpreter, it's very easy to dump a table to the screen along with its field names.   There seems to be no simple way to export a table to a tab-delimted or CSV outfile including its column headers.  I'm trying to do this using only SQL or the Linux command line, without writing a program in another language.  Thank you Can "real" be used as an adverb to describe an adjective? Is this correct?     That is a real cool answer.   I learned that that was incorrect, since "real" is an adjective which can describe a noun, e.g. "real answer" but it is not an adverb which can describe an adjective, "real cool". Instead you would have to say:     That is a really cool answer.   Since "really" is an adverb. On this site I see hundreds if not a thousand questions being asked about issues with a Wireless LAN adapter not being detected/recognized by Ubuntu. Some mainstream adapters appear to be utterly broken in Linux support, while others just require some simple steps to get it to work.  The options to get it to work appear to be infinitely huge and I have no clue where to get started!  What is considered 'best practice'? What order of things to try is most efficient in terms of time, effort and imposes minimal risk to my Ubuntu installation? Please help me save time and guide me what to do, step by step. Are upvotes on competing posts counted per question or per answer for Sportsmanship? The definition of the :     Up voted 100 competing answers   doesn't specify whether each upvote on competing answers to the same question counts towards this criteria, or if it's limited to one "point" per question no matter how many additional competing answers you upvote. Can someone explain how it works? Consider square matrices over a field $K$. I don't think additional assumptions about $K$ like algebraically closed or characteristic $0$ are pertinent, but feel free to make them for comfort. For any such matrix $A$, the set $K[A]$ of polynomials in $A$ is a commutative subalgebra of $M_n(K)$; the question is whether for any pair of commuting matrices $X,Y$ at least one such commutative subalgebra can be found that contains both $X$ and $Y$.  I was asking myself this in connection with frequently recurring requests to completely characterise commuting pairs of matrices, like . While providing a useful characterisation seems impossible, a positive anwer to the current question would at least provide some answer.  Note that in many rather likely situations one can in fact take $A$ to be one of the matrices $X,Y$, for instance when one of the matrices , or more generally if its  (so coincides with the characteristic polynomial). However this is not always possible, as can be easily seen for instance for diagonal matrices $X=\operatorname{diag}(0,0,1)$ and $Y=\operatorname{diag}(0,1,1)$. However in that case both will be polynomials in $A=\operatorname{diag}(x,y,z)$ for any distinct values $x,y,z$ (then $K[A]$ consists of all diagonal matrices); although in the example in  the matrices are not both diagonalisable, an appropriate $A$ can be found there as well.   I thought for some time that any maximal commutative subalgebra of $M_n(K)$ was of the form $K[A]$ (which would imply a positive answer) for some $A$ with minimal polynomial of degree$~n$, and that a positive answer to my question was in fact instrumental in proving this. However I was wrong on both counts: there exist (for $n\geq 4$) commutative subalgebras of dimension${}&gt;n$ (whereas $\dim_KK[A]\leq n$ for all $A\in M_n(K)$) as shown in , and I was forced to correct  in the light of this; however it seems (at least in the cases I looked at) that many (all?) pairs of matrices $X,Y$ in such a subalgebra still admit a matrix $A$ (which in general is not in the subalgebra) such that $X,Y\in K[A]$. This indicates that a positive answer to my question would not contradict the existence of such large commutative subalgebras: it would just mean that to obtain a maximal dimensional subalgebra containing $X,Y$ one should in general avoid throwing in an $A$ with $X,Y\in K[A]$. I do think these large subalgebras easily show that my question but for three commuting matrices has a negative answer.  Finally I note that  to the cited MO question mentions a result by Gerstenhaber that the dimension of the subalgebra generated two commuting matrices in $M_n(K)$ cannot exceed$~n$. This unfortunately does not settle my question (if $X,Y$ would generate a subalgebra of dimension${}&gt;n$, it would have proved a negative answer); it just might be that the mentioned result is true because of the existence of $A$ (I don't have access to a proof right now, but given the formulation it seems unlikely that it was done this way).  OK, I've tried to build up the suspense. Honesty demands that I say that I do know the answer to my question, since a colleague of mine provided a convincing one. I will however not give this answer right away, but post it once there has been some time for gathering answers here; who knows somebody will prove a different answer than the one I have (heaven forbid), or at least give the same answer with a different justification. Analogue of the Cantor-Bernstein-Schroeder theorem for general algebraic structures The Cantor-Bernstein-Schroeder theorem states that if there are two sets $A$ and $B$ such that there exist injective (alternatively, surjective, assuming choice I think) maps $A \to B$ and $B \to A$, then $A$ and $B$ are in bijection.  It then seems natural to try to strengthen the result to other structures.  If $A$ and $B$ are, say, groups (or rings, modules, etc) such that there are injective homomorphisms from each to the other, then $A$ and $B$ are isomorphic.  I'm wondering if there are any results on this, or if there are known counterexamples.  The theorem for sets alone is nontrivial, so I feel like any results about how it holds for other structures would be quite interesting. This is a sentence in a book by Rudy Giuliani.     I could ensure my deputies and commissioners were working off the same   page and could carry a coherent message back to their staffs.   Isn't staff a mass noun with no plural when referring to people? Would you take a share in Stack Overflow? Joel Spolsky has just posted  where he talks about seeking venture capital for Stack Overflow.  If instead they opted to seek shareholders from the StackOverflow community along the lines of the , but at a larger scale, would you be interested?  I'm asking this because I suspect there might be enough fanatics willing to invest $500 or so in a product they love to raise a considerable sum. People also often talk about the motivation for high rep users. It seems to me having an actual stake in the product might be a decent incentive (especially if you also get a gold badge for buying shares). I have a large external with many subfolders that I want to move all videos, pictures, and music to another hard-drive, what would be the best way to do this? Why can't I use Stack Exchange if I'm under 13 years old, or if I'm under 16 years old and from the EU? I was just reading the , and noticed something:  You must be at least 13 years old to access or use the Network or Services, including without limitation to complete a Stack Overflow account registration. By accessing or using the Services or the Network in any manner, you represent and warrant that you are at least 13 years of age. If you are under 13 years old, you may not, under any circumstances or for any reason, access or use the Services or Network in any manner, and may not provide any personal information to or on the Services or Network (including, for example, a name, address, telephone number or email address). If you are located within the European Union, you must be at least 16 years old to access or use the Network or Services, including without limitation to complete a Stack Overflow Account Registration.  I signed up for Stack Overflow when I was around 11 or 12 and I did provide my email address. Are you saying that my use of Stack Overflow was not legal? I'm 16 now (so I'm in the clear), but that just seems a little harsh. Is that possibly related to the age limit for accepting a binding contract/agreement (although I thought that was 18 in the U.S.)? Is there a way for users under the age of 13, or for users from the European Union who are under the age of 16, to use the site legally? Help Understanding How Long It Will Take To Heat My Wire How to calculate the resistance of a complicated circuit? I learned in high school the basic rules of how to compute resistance between two points, given a network of resistors.  You try to rearrange the circuit so into parallel parts and series parts, then compute the resistance of each piece individually using those rules, until you're left with just one number, and you're done.  Maybe I explained that poorly, so consider the following schematic:         &ndash; Schematic created using   Sorry it's so large, that seems to be the default.  Anyway, it's easy to compute the resistance from A to D (regardless of the values of the all the resistors)- A-B-D, A-D, and A-C-D are all series connections (easy to compute) then you combine them with the parallel formula and you're set.  But!  How do you compute the resistance from B to C?  There seem like lots of possible ways to think about it, and they all generate different answers.  I've left the specific numbers off the diagram because this isn't homework and I don't care about the answer specifically, I just want to understand the principles involved and how to solve the problem.  Any help, or links to especially readable guides, would be very much appreciated. The locker problem - why squares? There are 1000 lockers in a high school with 1000 students.  The    problem begins with the first student opening all 1000 lockers; next    the second student closes lockers 2,4,6,8,10 and so on to locker 1000;   the third student changes the state (opens lockers closed, closes    lockers open) on lockers 3,6,9,12,15 and so on; the fourth student    changes the state of lockers 4,8,12,16 and so on. This goes on until    every student has had a turn.      How many lockers will be open at the end?   I worked out this problem by doing it for 10 lockers. The pattern was obvious (1,4,9 remained open).It's easy to then say that 31 lockers remain open at the end. I don't quite understand why only square numbers remain open at the end. Is there some deeper reason? In many situations, JavaScript parsers will insert semicolons for you if you leave them out. My question is, do you leave them out?  If you're unfamiliar with the rules, there's a description of semicolon insertion on the . Here's the key point:     If the first through the nth tokens of a JavaScript program form are grammatically valid but the first through the n+1st tokens are not and there is a line break between the nth tokens and the n+1st tokens, then the parser tries to parse the program again after inserting a virtual semicolon token between the nth and the n+1st tokens.   That description may be incomplete, because it doesn't explain @Dreas's example. Anybody have a link to the complete rules, or see why the example gets a semicolon? (I tried it in JScript.NET.)   is related, but only talks about a specific scenario. Why are empty and non empty TikZ matrix cells not equally placed? I wanted to draw an almost empty matrix of nodes with TikZ similar to the left most one below. All nodes would have same size (through minimum width and minimum height) and no separation between columns and rows. But when I include a text which needs two lines I obtain the central matrix and, if the text just needs one line the right most one. I don't know what's wrong. Do you? I'm using CVS version of TikZ.  \documentclass[tikz,border=2mm]{standalone} \usetikzlibrary{matrix}  \tikzset{ proces/.style={draw, minimum height=15mm, minimum width=2cm, outer sep=0pt, text width=1.5cm, align=center} }  \begin{document} \begin{tikzpicture}  \matrix (A) [matrix of nodes,nodes=proces,nodes in empty cells] {  &amp; \\  &amp; \\ };  \begin{scope}[xshift=4.5cm] \matrix (A) [matrix of nodes,nodes=proces,nodes in empty cells] {  &amp; OOO OOO \\  &amp; \\ }; \end{scope}  \begin{scope}[xshift=9cm] \matrix (B) [matrix of nodes,nodes=proces,nodes in empty cells] {  &amp; A \\  &amp; \\ }; \end{scope}  \end{tikzpicture} \end{document} Why does local fn=$(...) mask the $? status code Two function definitions, the only difference is that the first combines the local storage keyword with the assignment, while the second separates them:  function foo {     local fn=$(mktemp -p /path/does/not/exist 2&gt;/dev/null)     echo $? }  function bar {     local fn     fn=$(mktemp -p /path/does/not/exist 2&gt;/dev/null)     echo $? }  foo bar   This echoes "0" then "1". I expect it to echo "1" then "1". It seems like the value of $? is the result of the assignment to local, rather than the result of the command substitution.  Why does bash 4.2.46(1)-release behave this way?