linux group by date column and show count
Best way to simulate "group by" from bash?
In the movie Wonder Woman, we see Wonder Woman as a child who grows into an adult. During the same time we see the people around her - the Amazons - not seem to age at all. It's revealed that Wonder Woman is special.      She's actually a God and presumably immortal   The Amazons are, well, not quite human, but they aren't the same as WW.   It's also revealed that:      A long time ago there was a war between the Gods and all of them perished by the hands of Ares. Before Zeus died, he created Wonder Woman out of clay.   All of the Amazons have bronze aged weaponry and do not seem aware of the changes to the outside world. This suggests that they are several thousand years old.   Presumably Wonder Woman has a normal life span as a child and has just become an adult at the beginning of the film. This seems very incongruent to me. Maybe I missed something though.  So, does the film or the DC Extended Universe give any indication of how old Wonder Woman is? Or is my assumption that she's several thousand years old (based on the weaponry) correct, despite the apparent incongruity I brought up? Suppose that $y_1(x)$ and $y_2(x)$ are solutions of the differential equation $y''+py'+qy=0$ on $I$. How can I show that if $y_1$ and $y_2$ vanish at the same point then they are not linearly independent?   Here is my attempt to prove the problem. Since $y_1(x)$ and $y_2(x)$ are solutions of the differential equation $y''+py'+qy=0$ on $I$, it is enough for me to show that the Wronskian of $y_1$ and $y_2$ denoted by $W(y_1,y_2)$ is zero. Now since $y_1$ and $y_2$ vanish at the same point say $p$ we have $y_1(p)=0$ and also $y_2(p)=0$.   Solving for $W(y_1,y_2)(p)$ we have: $y_1(p)y_2'(p)-y_2(p)y_1'(p)=0$ since $y_1(p)=0$ and also $y_2(p)=0$.   Am I correct? Thanks Anyone on the Android Jelly Bean OS and not able to buy Apps/Books through wallet? I have just bought the Samsung GalaxyNote 800 and am facing this issue.   The error is (RPC:S-7:AEC-0).   What is the solution? Basic desktop actions are not available on Ubuntu 20.04 Cannot turn script back on I've got an issue. When (Input.GetMouseButtonUp (0), Speedy script is disabled, thus we cannot enable it again from inside the script like shown below. How can we go around this?  public class  Speedy : MonoBehaviour {  GameObject car;   if (speed == good)     {         GameObject car = GameObject.FindWithTag ("Car");          if (Input.GetMouseButtonUp (0))          {             car.GetComponent&lt;Speedy&gt;().enabled = false;         }          else car.GetComponent&lt;Speedy&gt;().enabled = true;     } } $\\$   A normal subgroup $N$ is a subgroup where the left cosets are the same as the right cosets. $N$ is normal $\iff $ $xnx^{-1} \in N,  \forall x\in G$.   5.) Why is it that if $[G:H]=2 \implies $ $H$ is normal subgroup?  6.) Can we say that a factor group is just a group that has left cosets of $N$ (being a normal subgroup) as its elements? So if $N$ is a normal subgroup, then the left cosets of $N$ forms a group under coset multiplication given by $aNbN = abN$.   7.) The group of left cosets of $N$ in $G$ is called the factor group, why do we denote this by $G/N$? These are the same things as the integers modulo $n$ groups? How can I relate those exactly? If $|\lbrace g \in G: \pi (g)=g^{-1} \rbrace|>\frac{3|G|}{4}$, then $G$ is an abelian group. Assume that $\pi$ is an automorphism of a finite group $G$. Let $S$ denote the set $\lbrace g \in G: \pi (g)=g^{-1} \rbrace$. Show that if $|S|&gt;\frac{3|G|}{4}$, then $G$ is an abelian group.    Anyone has any idea on how to solve this ? I have no idea how to start. Java Strings: "String s = new String("silly");" cgi and tomcat im trying to run a cgi script (.cgi) with tomcat. I am getting the below error and cant find out whats wrong. I know i should really use apache and mod proxy but this really isnt my area of expertise so im taking the easy way out!  Thanks for any help.   java.io.IOException: Cannot run program "perl" (in directory "C:\Java\tomcat\webapps\my_app_name\WEB-INF\cgi"): CreateProcess error=2, The system cannot find the file specified     java.lang.ProcessBuilder.start(ProcessBuilder.java:459)     java.lang.Runtime.exec(Runtime.java:593)     java.lang.Runtime.exec(Runtime.java:431)     org.apache.catalina.servlets.CGIServlet$CGIRunner.run(CGIServlet.java:1705)     org.apache.catalina.servlets.CGIServlet.doGet(CGIServlet.java:597)     javax.servlet.http.HttpServlet.service(HttpServlet.java:627)     javax.servlet.http.HttpServlet.service(HttpServlet.java:729)     org.tuckey.web.filters.urlrewrite.UrlRewriteFilter.doFilter(UrlRewriteFilter.java:738)     org.apache.struts2.dispatcher.FilterDispatcher.doFilter(FilterDispatcher.java:416) Are nuclear weapons simply non-existent in the Star Wars universe? The resistance is attacks the Starkiller Base with what appears to be pretty small time ammunition even by modern earth missiles, which requires a long running sequence of repeated flybys.  The objective, IIRC was to destroy a critical piece of machinery on the surface which would then destabilize the entire planet/weapon (feel free to correct in edit if I remember incorrectly).  OK cool, but then why the force wouldn't they just drop a nuclear bomb on it?  Have nuclear weapons been used any other time in Star Wars (including EU), or, like smart phones, have they just not been invented in that world? GNOME 3.26 - Clicking one window brings another in the foreground (Ubuntu 17.10) I'm on Ubuntu 17.10 with GNOME 3.26.  Often I have windows tiled on the left and right side of the screen. A typical setup is to have a browser and text editor on the left, a PDF reader and terminal on the right.  Something I've seen happen very often: I'm reading the PDF on the right, having the window in focus. The text editor on the left. When I click on the text editor to write something, it brings forward another window on the right (say the file browser), covering the PDF reader.  This seems to have something to do with the Alt-Tab view history.  My question is: Is it possible to remove the automatic bring-to-front of windows? I don't understand why unrelated windows are brought to the front when I click on another application. What are your thoughts on writing your text in images?  I'm not saying throughout the whole website... But in certain key locations like image carousels. And I'm not talking about the special cases where you have font types that browsers can't render. I'm talking about text that uses standard fonts and is not too big (3/4 bullets).  Should I embed the text in the images that compose the carousel? Or should I apply the text to the images as "real text" HTML?  What are the advantages/ disadvantages of using text in an image? If a normal subgroup $N$ of order $p$($p$ prime) is contained in a group $G$ of order $p^n$,then $N$ is in the center of $G$. If a normal subgroup $N$ of order $p$($p$ prime) is contained in a group $G$ of order $p^n$,then $N$ is in the center of $G$.   I want to use induction to prove this:  It is trivial when $G=p(n=1)$, assume $G=p^n$, $N$ is a normal subgroup in $G$ with order $p$. Since $G$ is p-subgroup, $G$ has a normal subgroup $H$ with order $p^{n-1}$,according to assumption, $N \subset C(H)$. But I don't know how to go on, is $C(H)=C(G)$?Can induction work in this case? What is $1^k+2^k+\cdots+ (p-1)^k$ modulo $p$? (From Ireland and Rosen). I've been working through a bit of Ireland and Rosen's Number Theory for fun. Problem 4.11 of Ireland and Rosen asks     Prove that $1^k+2^k+\cdots+(p-1)^k\equiv 0\pmod{p}$ if $p-1\nmid k$, and $-1\pmod{p}$ if $p-1\mid k$.    My work is leading where I didn't expect to go. Here I assume $p$ id odd. First, if $p-1\mid k$, then we can wrike $k=(p-1)j$ for some $j$. But then $$ 1^k+\cdots+(p-1)^k\equiv 1^{(p-1)j}+\cdots+(p-1)^{(p-1)j}\equiv \underbrace{1+\cdots+1}_{p-1\text{ times}}=\frac{(p-1)p}{2}\equiv 0\pmod{p} $$ since $p-1$ is even.   Secondly, I suppose $p-1\nmid k$. WLOG, I can assume $0&lt;k&lt;p-1$, Since if $k=(p-1)j+r$ with $0&lt;r&lt;p-1$, then  $$ a^k\equiv a^{(p-1)j+r}\equiv a^r\pmod{p}. $$  Then if $g$ is a primitive root, $$ 1^k+\cdots+(p-1)^k=\sum_{i=1}^{p-1}(g^k)^i=\frac{g^k(1-g^{k(p-1)})}{1-g^k}. $$ Since $g$ is primitive, $p\nmid 1-g^k$, but $p\mid 1-g^{k(p-1)}$, so the above sum is again congruent to $0\pmod{p}$.  So in all cases, I get the sum is congruent to $0$ regardless. Have I messed up, or is there a typo? Thanks. Passing world as parameter results nullpointer \section*[toc]{normal section title} not working for me replacement wood threshold insert for metal door I have some late 70's doors and one is missing a threshold insert (the adjustable wood kind), anybody know of where one could by just the "wood strip" that goes in and screws down (there is a metal piece that sticks up into the wood, though you can't see it)?    They have about 4 screws that go down into them? Is there a list of error codes for Minecraft? no wifi in Ubuntu 15.04 I cannot see WiFi option in my Ubuntu 15.04. What should I do? Revert asap  lspci -nnk | grep 0280 -A2 output:-  03:00.0 Network controller [0280]: Qualcomm Atheros Device [168c:0042] (rev 30) Subsystem: Foxconn International, Inc. Device [105b:e09a]   rfkill list output:-  0: hci0: Bluetooth Soft blocked: no Hard blocked: no 1: acer-wireless: Wireless LAN Soft blocked: no Hard blocked: no      dmesg | grep ath [    9.545199] ath10k_pci 0000:03:00.0: pci irq msi-x interrupts 8 irq_mode 0 reset_mode 0 [   10.029684] ath10k_pci 0000:03:00.0: Direct firmware load for ath10k/cal-pci-0000:03:00.0.bin failed with error -2 [   10.029701] ath10k_pci 0000:03:00.0: Direct firmware load for ath10k/QCA9377/hw1.0/firmware-5.bin failed with error -2 [   10.029704] ath10k_pci 0000:03:00.0: could not fetch firmware file 'ath10k/QCA9377/hw1.0/firmware-5.bin': -2 [   10.029711] ath10k_pci 0000:03:00.0: Direct firmware load for ath10k/QCA9377/hw1.0/firmware-4.bin failed with error -2 [   10.029713] ath10k_pci 0000:03:00.0: could not fetch firmware file 'ath10k/QCA9377/hw1.0/firmware-4.bin': -2 [   10.029719] ath10k_pci 0000:03:00.0: Direct firmware load for ath10k/QCA9377/hw1.0/firmware-3.bin failed with error -2 [   10.029721] ath10k_pci 0000:03:00.0: could not fetch firmware file 'ath10k/QCA9377/hw1.0/firmware-3.bin': -2 [   10.029726] ath10k_pci 0000:03:00.0: Direct firmware load for ath10k/QCA9377/hw1.0/firmware-2.bin failed with error -2 [   10.029728] ath10k_pci 0000:03:00.0: could not fetch firmware file 'ath10k/QCA9377/hw1.0/firmware-2.bin': -2 [   10.029733] ath10k_pci 0000:03:00.0: Direct firmware load for ath10k/QCA9377/hw1.0/firmware.bin failed with error -2 [   10.029735] ath10k_pci 0000:03:00.0: could not fetch firmware (-2) [   10.029737] ath10k_pci 0000:03:00.0: could not fetch firmware files (-2) [   10.029739] ath10k_pci 0000:03:00.0: could not probe fw (-2) Every proper ideal is contained in a maximal ideal, in a commutative ring with identity. The statement is:     In a commutative ring with 1, every proper ideal is contained in a maximal ideal.   and we prove it using Zorn's lemma, that is, $I$ is an ideal, $P=\{I\subset  A\mid A\text{ is an ideal}\} $, then by set inclusion, every totally ordered subset has a bound, then $P$ has a maximal element $M$.  My question is why $M$ must contain $I$?