I want to run a function that will create a file with the name of a variable and write in it the content of that variable.  list_A = 'one two three'.split()  def write_list(list_to_write):      flux = open('%s.list' %list_to_write, 'w')     flux.write('\n'.join(list_to_write))     flux.close()  write_list (list_A)   This creates a file called ['one', 'two', 'three'].list   I would like it to create a file called liste_A.list instead.  PS: I found many similar posts, but none that seemed to address my issue.
Say I have a variable named choice it is equal to 2. How would I access the name of the variable? Something equivalent to  In [53]: namestr(choice) Out[53]: 'choice'   for use in making a dictionary. There's a good way to do this and I'm just missing it.  EDIT:  The reason to do this is thus. I am running some data analysis stuff where I call the program with multiple parameters that I would like to tweak, or not tweak, at runtime. I read in the parameters I used in the last run from a .config file formated as  filename no_sig_resonance.dat  mass_peak 700  choice 1,2,3   When prompted for values, the previously used is displayed and an empty string input will use the previously used value.  My question comes about because when it comes to writing the dictionary that these values have been scanned into. If a parameter is needed I run get_param which accesses the file and finds the parameter.   I think I will avoid the problem all together by reading the .config file once and producing a dictionary from that. I avoided that originally for... reasons I no longer remember. Perfect situation to update my code!
Decidability of a predicate. I have the following problem:  Let the following predicates be: $P1, P2, Q1, Q1 : R \to \{0,1\} $ .  It is given that $P1 \lor P2$ and $Q1 \lor Q2$ are semidecidable and $P2 = \lnot Q2$  What can it be said about $P1 \lor Q1$?  I have searched all over the internet and I simply cannot understand the concept of decidability of a predicate, let alone a logic operation with predicates. If a ring element is right-invertible, but not left-invertible, then it has infinitely many right-inverses. How to remove URI encoding from file names? I realise this might be quite a simple question, but I'm still quite new to the command line and only have a grasp of basic commands.  I've downloaded some lecture presentations (~25 or so) from my University however on doing so they've been named things like...  L2%20Development%20of%20immune%20system.pptx L4%20Molecular%20Recognition.pdf   As you can see they've downloaded with the URL encoding %20 instead of a space.  My question is how to batch rename all of these files so that the %20 is removed and replaced with a space instead? Novel about interstellar war: Both sides win, both sides think they lost Divergence of $\sum_{n=1}^{\infty} \frac{1}{2+i^n}$? I notice that,  $i^n = i, \quad if \text{ } n \equiv \text{ } 1 \text{ } mod \text{ } 4$  $i^n = -1, \quad if \text{ }n \equiv \text{ } 2 \text{ } mod \text{ } 4$  $i^n = -i, \quad if \text{ }n \equiv \text{ } 3 \text{ } mod \text{ } 4$  $i^n = 1, \quad if \text{ } n \equiv \text{ } 0 \text{ } mod \text{ } 4$  so the series $\sum_{n=1}^{\infty} \frac{1}{2+i^n}$ seems to be an alternating series that does not converge to any point in the complex plane. Is there a rigorous way to prove this divergence? How would you write a proof to this divergence or convergence (if I am wrong)?  Thanks! How do I change Wi-Fi Hotspot settings in Ubuntu with GNOME 3? How do I manage my hotspot connected devices or change the hotspot password? I can only see the Network Name, Security type and the Password but I can't seem to manage users or the network properties. How far can you go with rm -rf / as root Madurai to Germany via Dubai - SpiceJet I am an Indian citizen with a German residency permit. I am travelling from Germany to Dubai by Pegasus Airlines and then from Dubai to Madurai by SpiceJet separately, in both directions. I need to transfer myself in Dubai. I have found that transit is possible in Dubai airport via transit desk without UAE transit visa.  But, I have read on SpiceJet's website that Indian citizens travelling from India to Dubai need "Ok to Go" stamp in their passports. It is not clearly explained for transit people like me with Schengen visa.  Q1: Will SpiceJet in Madurai check "Ok to Go" and reject me (Indian citizen with Schengen residence permit) in Madurai airport during my return journey?  If yes; Q2: Is there is any other possibility to solve the above problem?  If "Ok to Go" is ultimately needed for SpiceJet in Madurai; Q3: How can I get it from Germany and how much does it cost? $d_1(x_1,y_1)$ and  $d_2(x_2,y_2)$ are metric on $X$ and $d(x,y)$ is defined as: $$d(x,y)= \sqrt{d_1(x_1,y_1)^2+d_2(x_2,y_2)^2}.$$ I am trying to show this is a metric. Can you give me some clue about proving the triangle inequality for $d$ ? Thank you for your help. Unary intersection of the empty set Intertext vertical spacing How do I reduce the vertical spacing of \intertext? Gauss law question with regard to this example I am really confused in Gauss law.  Why do E3 and E2 pointing up? and also E1 pointing down?  The lecture note said infer from symmetry and you will get the following but I dont really understand. Could anyone explain to me? I am new to gauss law. Procedure cache (plan cache) size Can I kill an indestructible creature with -X/-X and sufficient damage marked on it? Can I hit a 4/4 indestructible creature with a  (3 damage) and give it -1/-1 with  to kill it? (Does it matter which way around I cast those?)  I read  here and it says, damage is removed in cleanup step, so it should be on the creature still after both spells.     702.12. Indestructible   702.12a Indestructible is a static ability.   702.12b A permanent with indestructible can’t be destroyed. Such permanents aren’t destroyed by lethal damage, and they ignore the state-based action that checks for lethal damage (see rule 704.5g).   702.12c Multiple instances of indestructible on the same permanent are redundant. bash regex with quotes? The following code  number=1 if [[ $number =~ [0-9] ]] then   echo matched fi   works. If I try to use quotes in the regex, however, it stops:  number=1 if [[ $number =~ "[0-9]" ]] then   echo matched fi   I tried "\[0-9\]", too. What am I missing?  Funnily enough,  suggests this should work.  Bash version 3.2.39. I have some questions regarding suspended users:   Why do suspended users have a reputation of one? According to the default reputation of a new user, he/she will have reputation of one and he/she has the privilege of posting questions. When comparing this, this is getting violated when compared to suspended users' reputation. Even though they have one reputation point, they cannot post questions. As suspended users do not have privilege of posting questions, there is no use of making "Ask Question" visible. This is an unnecessary server hit. Am I right? As suspended users do not have the privilege of answering questions, there is no use of making "Answer Question" visible. I don't have a minecraft folder. I've looked through forums and videos, followed all the steps. I just don't have one. I have windows 7, and I have "view hidden files" turned on. I have look in the roaming folder of my app data folder, with hidden files turned on. I want to find my folder so I can install a mod, could someone please tell me why my folder isn't showing up? Hair Paricle System not rendering properly on plane I'm running macOS 10.15 Catalina Beta.  When I try to install Spotify I see the following error:    Is this a Spotify issue or a Catalina issue?  Thanks. Can a full version of Ubuntu be Installed on a USB Drive?