I am a beginner programmer, while doing some work in Python 2.7, I ran into an issue I can't seem to pass. I am trying to find all permutations of all pairs of digits; out of an array that has 4 digits. Example: array = ["a", "b", "c", "d"] and I would like to see the permutations like this: ab, ac, ad, ba, cd, da... ect... Here is my code so far, I can't figure out the next step:   from itertools import permutations array = ["a", "b", "c", "d"] for p in permutations(array):     print(p)   I would appreciate any help I can get, thank you.
How do you generate all the permutations of a list in Python, independently of the type of elements in that list?  For example:  permutations([]) []  permutations([1]) [1]  permutations([1, 2]) [1, 2] [2, 1]  permutations([1, 2, 3]) [1, 2, 3] [1, 3, 2] [2, 1, 3] [2, 3, 1] [3, 1, 2] [3, 2, 1]
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Scorched3D will not install and work after updating.  ron@RonsComputer:~$ sudo objdump -p $(type -p scorched3d) | grep NEEDED  NEEDED libwx_gtk2u_adv-3.0.so.0  NEEDED libwx_gtk2u_core-3.0.so.0 NEEDED libwx_baseu-3.0.so.0 NEEDED libfreetype.so.6 NEEDED libz.so.1 NEEDED libSDL-1.2.so.0 NEEDED libSDL_net-1.2.so.0 NEEDED libexpat.so.1  NEEDED libstdc++.so.6 NEEDED libm.so.6 NEEDED libgcc_s.so.1 NEEDED libpthread.so.0 NEEDED libc.so.6   ron@ron-desktop ~ $ objdump -p $(type -p scorched3d) | grep NEEDED | awk '{print $2}' | xargs -n 1 dpkg -S  libwxgtk3.0-0v5:amd64: /usr/lib/x86_64-linux-gnu/libwx_gtk2u_adv-3.0.so.0.2.0 libwxgtk3.0-0v5:amd64: /usr/lib/x86_64-linux-gnu/libwx_gtk2u_adv-3.0.so.0 libwxgtk3.0-0v5:amd64: /usr/lib/x86_64-linux-gnu/libwx_gtk2u_core-3.0.so.0.2.0 libwxgtk3.0-0v5:amd64: /usr/lib/x86_64-linux-gnu/libwx_gtk2u_core-3.0.so.0 libwxbase3.0-0v5:amd64: /usr/lib/x86_64-linux-gnu/libwx_baseu-3.0.so.0.2.0 libwxbase3.0-0v5:amd64: /usr/lib/x86_64-linux-gnu/libwx_baseu-3.0.so.0 libfreetype6:amd64: /usr/lib/x86_64-linux-gnu/libfreetype.so.6.12.1 libfreetype6:i386: /usr/lib/i386-linux-gnu/libfreetype.so.6 libfreetype6:amd64: /usr/lib/x86_64-linux-gnu/libfreetype.so.6 libfreetype6:i386: /usr/lib/i386-linux-gnu/libfreetype.so.6.12.1 zlib1g:i386: /lib/i386-linux-gnu/libz.so.1.2.8 zlib1g:amd64: /lib/x86_64-linux-gnu/libz.so.1 zlib1g:i386: /lib/i386-linux-gnu/libz.so.1 zlib1g:amd64: /lib/x86_64-linux-gnu/libz.so.1.2.8 libsdl1.2debian:amd64: /usr/lib/x86_64-linux-gnu/libSDL-1.2.so.0.11.4 libsdl1.2debian:i386: /usr/lib/i386-linux-gnu/libSDL-1.2.so.0 libsdl1.2debian:i386: /usr/lib/i386-linux-gnu/libSDL-1.2.so.0.11.4 libsdl1.2debian:amd64: /usr/lib/x86_64-linux-gnu/libSDL-1.2.so.0 libsdl-net1.2:amd64: /usr/lib/x86_64-linux-gnu/libSDL_net-1.2.so.0 libsdl-net1.2:i386: /usr/lib/i386-linux-gnu/libSDL_net-1.2.so.0 libsdl-net1.2:amd64: /usr/lib/x86_64-linux-gnu/libSDL_net-1.2.so.0.8.0 libsdl-net1.2:i386: /usr/lib/i386-linux-gnu/libSDL_net-1.2.so.0.8.0 libexpat1:i386: /lib/i386-linux-gnu/libexpat.so.1.6.0 libexpat1:amd64: /lib/x86_64-linux-gnu/libexpat.so.1 libexpat1:amd64: /lib/x86_64-linux-gnu/libexpat.so.1.6.0 libexpat1:i386: /lib/i386-linux-gnu/libexpat.so.1 libstdc++6:amd64: /usr/share/gdb/auto-load/usr/lib/x86_64-linux-gnu/libstdc++.so.6.0.21-gdb.py libstdc++6:i386: /usr/lib/i386-linux-gnu/libstdc++.so.6 libstdc++6:amd64: /usr/lib/x86_64-linux-gnu/libstdc++.so.6 libstdc++6:amd64: /usr/lib/x86_64-linux-gnu/libstdc++.so.6.0.21 libstdc++6:i386: /usr/lib/i386-linux-gnu/libstdc++.so.6.0.21 libstdc++6:i386: /usr/share/gdb/auto-load/usr/lib/i386-linux-gnu/libstdc++.so.6.0.21-gdb.py libc6:amd64: /lib/x86_64-linux-gnu/libm.so.6 libc6:i386: /lib/i386-linux-gnu/libm.so.6 libgcc1:amd64: /lib/x86_64-linux-gnu/libgcc_s.so.1 libgcc1:i386: /lib/i386-linux-gnu/libgcc_s.so.1 libc6:i386: /lib/i386-linux-gnu/libpthread.so.0 libc6:amd64: /lib/x86_64-linux-gnu/libpthread.so.0 libc6:i386: /lib/i386-linux-gnu/libc.so.6 libc6:amd64: /lib/x86_64-linux-gnu/libc.so.6 ron@ron-desktop ~ $ Vectors, structs and std::find Where can I find Vagrant packages? is a tool to build and manage virtual machines for developers.  It has a  guide that contains iffy-looking suggestions like  $ sudo ln -s /usr/bin/ruby1.8 /usr/bin/ruby  # wtf???   or installing RubyGems from source rather than from Ubuntu packages, and then using gem install to install vagrant itself.  I'm not feeling comfortable just following those instructions.  Is there perhaps a PPA?  Are there, perhaps, alternative tools that are packaged for Ubuntu? How to get cross-references to multiple figures? I am currently writing an essay on artificial moral agents, and I need to explain the difference between ethics and morals. However, I am finding it really hard to find a good book or research paper which contains a citable definition.   Could someone point me in the right direction? Auto reply to text and/or iMessage? Is it possible to set up an auto-reply to an incoming text or iMessage, either on my iPhone (or the connected Message app on my Mac)?  Ideally I'd like to customize this depending upon the sender. I want to dynamically create a sequence of strings by manipulate an array of elements and create some arithmetic procedure.  for name in FIRST SECOND THIRD FOURTH FIFTH; do     $name = $(( $6 + 1 ))     $name = "${$name}q;d"     echo "${$name}"; printf "\n" done   The desire outcome would be the below for $6 equals 0.  1q;d 2q;d 3q;d 4q;d 5q;d   But I get this error  reel_first_part.sh: line 18: FIRST: command not found reel_first_part.sh: line 19: ${$name}q;d: bad substitution reel_first_part.sh: line 18: FIRST: command not found reel_first_part.sh: line 19: ${$name}q;d: bad substitution reel_first_part.sh: line 18: FIRST: command not found reel_first_part.sh: line 19: ${$name}q;d: bad substitution   I guess it's something simple. It used to work when I did something like   FIRST=$(( $6 + 1 )) FIRST="${FIRST}q;d" is incompatible with the enumerate and itemize environments. Is there a typographical reason not to wrap a list around a figure or is it a technical limitation of LaTeX? 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All I get is:  start: Job failed to start Estimating partial sums $\sum_{n = 1}^m \frac{1}{\sqrt{n}}$ Apostol's Calculus, exercise number I 4.7 13.     Prove that if $n \geq 1$, then   $$ 2(\sqrt{n+1} - \sqrt{n}) &lt; \frac{1}{\sqrt{n}} &lt; 2(\sqrt{n} - \sqrt{n-1}) $$   and use this to prove that if $m \geq 2$, then   $$ 2 \sqrt{m} - 2 &lt; \sum_{n = 1}^m \frac{1}{\sqrt{n}} &lt; 2\sqrt{m} - 1 $$   In particular, when $m = 10^6$ the sum lies between $1998$ and $1999$.   After establishing the trivial case, for $n=1$, I can't think of a way to perform the inductive step:  assuming $2(\sqrt{n+1} - \sqrt{n}) &lt; \frac{1}{\sqrt{n}} &lt; 2(\sqrt{n} - \sqrt{n-1})$ and using this, deducing the same for $n+1$ is true as well:  $2(\sqrt{n+2} - \sqrt{n+1}) &lt; \frac{1}{\sqrt{n+1}} &lt; 2(\sqrt{n+1} - \sqrt{n})$  Any help is very appreciated, thanks Is vocative comma rule vanishing? 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