I need to fill two dimensional list in python 3. I wrote some example code but it doesn't work. For some reason, each new line is overwritten. Why is this happening? code: testA = [0]*3 tmpA = [0]*3  for n in range(0, 3, 1):             for x in range(0, 3, 1):         tmpA[x] = x*n+1      testA[n] = tmpA     print(&quot;-&quot;, testA)     print(&quot;--&quot;, tmpA)  output: - [[1, 1, 1], 0, 0] -- [1, 1, 1] - [[1, 2, 3], [1, 2, 3], 0] -- [1, 2, 3] - [[1, 3, 5], [1, 3, 5], [1, 3, 5]] -- [1, 3, 5]
I needed to create a list of lists in Python, so I typed the following: my_list = [[1] * 4] * 3  The list looked like this: [[1, 1, 1, 1], [1, 1, 1, 1], [1, 1, 1, 1]]    Then I changed one of the innermost values: my_list[0][0] = 5  Now my list looks like this: [[5, 1, 1, 1], [5, 1, 1, 1], [5, 1, 1, 1]]    which is not what I wanted or expected. Can someone please explain what's going on, and how to get around it?
If I have $f:\mathbb{R}\rightarrow\mathbb{R}$ such that $f$ is onto and $f\circ f\circ f = f$, how can I prove that $f$ is bijective? I know that I only have to prove that it is 1-to-1 because I'm given the fact that it's onto, but how can I use the fact that $f\circ f\circ f=f$ to prove that it is 1-to-1?  Thanks! I have the following code:  \documentclass[a4paper,12pt]{article} \usepackage[margin=2cm]{geometry} %showframe \begin{document} \tableofcontents \newpage \section*{Abstract} \section{Introduction} \newpage \bibliographystyle{apacite} \bibliography{references} \end{document}   And when I create the table of contents it adds the "References" in the TOC, how can I remove this to not be included in TOC? So far I have tried something like   \bibliography*{references} Reverse (parse the output) of Arrays.toString(int[]) Is it same number of candies for any stage of evolution while transferring pokemon to professor? I was just wondering if the number of candies we get while transferring the pokemon to professor is same for any stage of evolution ?   For example, you would get a pidgey candy for transferring a pidgey. Is it same number of candies for transferring pidgey, pidgeotto or pidgeot ? This afternoon I set up a Win Server 2008 R2 system for use with Team Foundation source code control. In order to give my team-mates the right to connect, I had to create a user account for each person, 10 in total. (I can't, and I don't want to create a domain).  Now, after doing this, all user accounts are displayed in the login screen.  What can I do so that all the accounts are not displayed there? Differentiable but not continuously differentiable. Given $f: \mathbb{R} \rightarrow \mathbb{R}$ defined as $$f(x)=\left\{\begin{array}{cc}x^2\sin\left(\frac{1}{x}\right)&amp;,x\neq 0\\ 0&amp;,x=0\end{array}\right\}.$$ I am trying to prove $f$ is differentiable at $x=0$ but not continuously differentiable there. Ways to get free and legal access to research papers as a researcher I’ve discovered that website like ResearchGate offer the possibility to ask the paper editor to provide his paper for free. Many are collaborating.  Are there any other ways for researchers to obtain free access to papers? How to implement a single instance Java application? Can a character use the Misty Step spell to escape from the Iron Bands of Bilarro? Can a character escape the  (DMG p. 177, listed in the SRD as the ) with the  spell? To clarify, I am not asking if the bands prevent teleportation. I am asking whether the bands are teleported with the character as he teleports using misty step. Let $G$ be a group. Show that a nonempty subset $H$ is a subgroup of $G$ if any only if $ab^{-1} \in H $ for any $a,b \in H$.  The forward direction is quite easy. Suppose $H$ is a subgroup. Then by closure, $ab \in H$ for any $a,b \in H$. Every element has an inverse. Hence, if $b \in H $, then $b^{-1} \in H$. Hence, by closure again, $ab^{-1} \in H$.   Backward direction, suppose $ab^{-1} \in H $ for any $a,b \in H$. Let $a=b$. Then we have $bb^{-1}=1_H\in H$. Let $a=1_H$ , we have $1_Hb^{-1}=b^{-1} \in H$. I don't know how to show closure.  Reference: Fraleigh p. 58  Question 5.45 in A First Course in Abstract Algebra A cop vs a superhuman criminal I read the story over ten years ago while I was still trying to learn English, so I can only vaguely remember that the story is about a cop trying to track down a criminal who had superhuman abilities. Some of his abilities were pyrokinesis, using his mind to form a golem from earth and roam around while he stays in the room and stopping time. Eventually the cop was able to catch the criminal and shot him dead.   If anyone could identify the book I would be very grateful. I'd love to read it for nostalgic purposes. Any way to do DISTINCT ON? I'm trying to create a query like this:  SELECT DISTINCT ON (field1, field2) field1, field2, field3 FROM table1   But can't figure out how to get DISTINCT ON narrowed down to a few fields in Drupal like that. Calling $query-&gt;distinct() just seems to structure the query like this:  SELECT DISTINCT field1, field2, field3 FROM table1   I did figure out a hacky way to add what I want by adding an expression before I add any fields like this:  $query-&gt;addExpression('DISTINCT ON (field1, field2) field1', 'field1');   Which creates:  SELECT DISTINCT ON (field1, field2) field1 as field1, field2, field3 FROM table1   I'm looking for a better way to this ? /oracle/GR1/121/bin/sqlplus / as sysdba &lt;&lt;EOF &gt;&gt; $LOGFILE whenever sqlerror exit sql.sqlcode; set echo on; set serveroutput on; STARTUP; EXIT EOF   Already tried it with the path, but it hasn't worked. Is there a package that automatically inserts the ToC level of the reference I inserted?  Section 1.1 \label{a} Subsection 1.1.1 \label{b} Subsection 1.1.2  \label{c}   So that the text:  Blah \superref{a} and \superref{b} and \superref{c}   becomes:  Blah Section 1.1 and Subsection 1.1.1 and Subsection 1.1.2   So that it automatically adapts if I change the level of a label? OS: UBUNTU 18.04  I had installed Oracle Java 8 earlier. But decided to switch to open JDK 11. it's successfully installed. java -version openjdk version "11.0.2" 2019-01-15 OpenJDK Runtime Environment (build 11.0.2+9-Ubuntu-3ubuntu118.04.3) OpenJDK 64-Bit Server VM (build 11.0.2+9-Ubuntu-3ubuntu118.04.3, mixed mode, sharing)  Then I set the JAVA_HOME sudo nano /etc/environment  File changes: PATH="/usr/local/sbin:/usr/local/bin:/usr/sbin:/usr/bin:/sbin:/bin:/usr/games:/$ JAVA_HOME="/usr/lib/jvm/java-11-openjdk-amd64/"  Then executed: source /etc/environment  echo $JAVA_HOME /usr/lib/jvm/java-11-openjdk-amd64/  But the moment I exit the terminal and then launch the terminal again: I get this output:  ▶ echo $JAVA_HOME /usr/lib/jvm/java-8-oracle  which you can see is not what is expected. Please help me out Why is the Night's Watch so depleted at the start of ASOIAF? At the start of AGOT, the total number of the Night's Watch is said to be less than a thousand. During Aegon's conquest, the Watch numbered ten thousand men. Out of the nineteen castles on the wall, only three are manned today. Castle black itself is once said to have housed 5000.   This rapid deterioration doesn't make sense to me from a world-building perspective. The NW have continued to maintain themselves pretty well for nigh on 8000 years (if the tales can be believed that is, Sam Tarly finds no written evidence of Lord Commanders earlier than #673, which suggests that 8000 years may be an inflated 'legendary' number).   Whereas in the real world, one of the oldest and most venerable institutions, the Catholic Church, has been around for less than 2000 years and has been administered by ~200 popes. The Night's Watch has had a few rocky years, sure, but for it to maintain itself so well for so long and then deteriorate only in the last 300 years seems strange to me.   The NW has two purposes. The ostensible one of defending against the Others, and the (arguably much more real, at least at the start of AGOT) purpose of serving as a dumping ground for the 'losers' of Westeros. By 'losers', I mean the people on the losing side of various wars and political battles, as well as criminals and n-th sons of noble houses with nothing to inherit.  I am currently reading through Fire and Blood, and was struck by one chapter, where an armed revolt of members of the Faith were defeated by Maegor, and 2000 brothers were forced to take the black. Those sorts of numbers don't exist in the recent histories it seems.   One would expect that after Robert's Rebellion, there would be a surfeit of defeated Targaryen loyalists who were forced to take the black, but apart from a smattering of named characters (including Ser Alliser Thorne), there seem to be hardly any Night's watch recruits from recent wars.  Given the way the Night's watch is continually bolstered by the losers of wars and by criminals and other people who elsewise have nothing going for them, why, at the start of AGOT, is the force so depleted? How do I design a 2A or more power supply for my consumer USB devices? I understand that many of my consumer devices charged by USB ports can charge at higher rates than 0.5A.  However, in testing, I've found they don't consume more than this. Test Setup  I have a power supply that outputs between 0 and 30 volts, and 0 to 20 amps. I set the voltage to 5 volts, then connected up a female USB to the wires correctly. When I plug in a Samsung Galaxy S4 Mini, it pulls around 0.44 amps. It is the same with a Samsung Galaxy Tab Pro. However, when I connect an iPad Air with an official lightning cable, it only pulls around 0.11 amps. Also, I tried connecting a ZGPAX S28 smartwatch and it was still about 0.44-0.45 amps.  The power supply is more than capable of outputting the maximum 2.1 amps, so why doesn't it to at least the tablets? What do I need to do in my test setup to convince the devices to consume their maximum charging current? [i] Susan married Ed. (RHD)   [ii] My sister got married last fall. (COCA)   Is there a passive implication in [ii]? That is, the families and so on arrange and perform the marriage ceremonies for two people, and through all this the two get married. Is this implied in the expression [ii], and not in [i]? How prove this $\sum_{k=1}^{2^{n-1}}\sigma{(2^n-2k+1)}\sigma{(2k-1)}=8^{n-1}$ Given the positive integer numbers $n$,prove that $$\sum_{k=1}^{2^{n-1}}\sigma{(2^n-2k+1)}\sigma{(2k-1)}=8^{n-1}$$  where $\sigma(n)$ is defined as $$\sigma{(N)}=\sum_{d|N}d$$  This problem I think is very interesting, and have nice methods?  Thank you everyone.  and I find some of this     or can see: Generating random integer from a range