For example if I have the lists:  x = [1,2,3,4,5,6] y = [4,3,2,5,6,7]   How do I subtract my lists in order to get:  z = "x-y" = [1-4,2-3,3-2,4-5,5-6,6-7] z = [-3,-1,1,-1,-1,-1]   ?  Meaning, as long as the numbers from my lists share the same index then subtract them.  Thanks in advance.
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Right now I have vector3 values represented as lists. is there a way to subtract 2 of these like vector3 values, like  [2,2,2] - [1,1,1] = [1,1,1]   Should I use tuples?  If none of them defines these operands on these types, can I define it instead?  If not, should I create a new vector3 class?
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How do I remove an element from a list by index in Python?  I found the list.remove method, but say I want to remove the last element, how do I do this? It seems like the default remove searches the list, but I don't want any search to be performed.
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If $\{x,\{x,y\}\} = \{z, \{z,t\}\}$, then must it be $x = z$ and $y = t$?   Can you explain, or can you give hint?
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If I have a dictionary like:  { 'a': 1, 'b': 2, 'c': 3 }   How can I convert it to this?  [ ('a', 1), ('b', 2), ('c', 3) ]   And how can I convert it to this?  [ (1, 'a'), (2, 'b'), (3, 'c') ]
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Is it possible to delete multiple elements from a list at the same time? If I want to delete elements at index 0 and 2, and try something like del somelist[0], followed by del somelist[2], the second statement will actually delete somelist[3].  I suppose I could always delete the higher numbered elements first but I'm hoping there is a better way.
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Question: Find the image of (1,1,1) by T  $$T(x,y,z) = (y,6y,x+y+z)$$   I found the standard matrix to be:   \begin{bmatrix}0&amp;1&amp;0\\0&amp;6&amp;0\\1&amp;1&amp;1\end{bmatrix}  I'm confused on how to approach this question.   In order to find the image is it possible to inverse the values of it's standard matrix?  Would I able to use the identity matrix of 3 and inverse the values of that to find the image?    Any help will be greatly appreciated.
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Why does list.index throw an exception, instead of using an arbitrary value (for example, -1)? What's the idea behind this?  To me it looks cleaner to deal with special values, rather than exceptions.  EDIT: I didn't realize -1 is a potentially valid value. Nevertheless, why not something else? How about a value of None?
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What is the difference between:  some_list1 = [] some_list1.append("something")   and  some_list2 = [] some_list2 += ["something"]
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What's the simplest, most standard, and/or most efficient way to split a List into two sub-Lists in Java? It's OK to mutate the original List, so no copying should be necessary. The method signature could be  /** Split a list into two sublists. The original list will be modified to  * have size i and will contain exactly the same elements at indices 0   * through i-1 as it had originally; the returned list will have size   * len-i (where len is the size of the original list before the call)   * and will have the same elements at indices 0 through len-(i+1) as   * the original list had at indices i through len-1.  */ &lt;T&gt; List&lt;T&gt; split(List&lt;T&gt; list, int i);   [EDIT] List.subList returns a view on the original list, which becomes invalid if the original is modified. So split can't use subList unless it also dispenses with the original reference (or, as in Marc Novakowski's answer, uses subList but immediately copies the result).
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Does $3 - 4 + 1 = 0$ or $3 - 4 + 1 = -2$? Makes sense that $(3 - 4) + 1 = 0$ and $3 - (4 + 1) = -2$, but what if there are no parenthesis? Also, if I have $4$ apples and I add $1$ more apple, then I have $5$ apples, but if I eat $3$ apples, then I have $2$ left. Any ideas?
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I want to make an enumerated list in the following format:  [1] foo foo [2] foo foo ...    How I can do that using enumerated lists? I have tried the following:  \begin{enumerate}[\left[ 1 \right]]   but it does not work.
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I have the following data structure (a list of lists)  [  ['4', '21', '1', '14', '2008-10-24 15:42:58'],   ['3', '22', '4', '2somename', '2008-10-24 15:22:03'],   ['5', '21', '3', '19', '2008-10-24 15:45:45'],   ['6', '21', '1', '1somename', '2008-10-24 15:45:49'],   ['7', '22', '3', '2somename', '2008-10-24 15:45:51'] ]   I would like to be able to   Use a function to reorder the list so that I can group by each item in the list. For example I'd like to be able to group by the second column (so that all the 21's are together)  Use a function to only display certain values from each inner list. For example i'd like to reduce this list to only contain the 4th field value of '2somename'    so the list would look like this   [      ['3', '22', '4', '2somename', '2008-10-24 15:22:03'],       ['7', '22', '3', '2somename', '2008-10-24 15:45:51'] ]
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This is a continued discussion from () by return count only.     Problem      Given four lists A, B, C, D of integer values, compute how many tuples   (i, j, k, l) there are such that A[i] + B[j] + C[k] + D[l] is zero.      To make problem a bit easier, all A, B, C, D have same length of N   where \$0 \le N \le 500\$. All integers are in the range of \$-2^{28}\$ to \$2^{28} - 1\$   and the result is guaranteed to be at most \$2^{31} - 1\$.      Example:      Input:  A = [ 1, 2] B = [-2,-1] C = [-1, 2] D = [ 0, 2]       Output:  2       Explanation:      The two tuples are:         (0, 0, 0, 1) -> A[0] + B[0] + C[0] + D[1] = 1 + (-2) + (-1) + 2 = 0   (1, 1, 0, 0) -> A[1] + B[1] + C[0] + D[0] = 2 + (-1) + (-1) + 0 = 0      I'm wondering if there are any ideas to have a solution less than \$O(n^2)\$ time complexity.  Source code in Python 2.7,  from collections import defaultdict def four_sum(A, B, C, D):     sum_map = defaultdict(int)     result = 0     for i in A:         for j in B:             sum_map[i+j] += 1     for i in C:         for j in D:             if -(i+j) in sum_map:                 result += sum_map[-(i+j)]      return result  if __name__ == "__main__":     A = [1, 2]     B = [-2, -1]     C = [-1, 2]     D = [0, 2]     print four_sum(A,B,C,D)
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Is there is any difference between these two.  [[ $a == z* ]]   and   [ $a == z* ]    Can I have an example where they would have different outputs?  Furthermore, how does the working of [[ ]] differs from [ ]?
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In Python, I can do:  &gt;&gt;&gt; list = ['a', 'b', 'c'] &gt;&gt;&gt; ', '.join(list) 'a, b, c'   Is there any easy way to do the same when I have a list of objects?  &gt;&gt;&gt; class Obj: ...     def __str__(self): ...         return 'name' ... &gt;&gt;&gt; list = [Obj(), Obj(), Obj()] &gt;&gt;&gt; ', '.join(list) Traceback (most recent call last):   File "&lt;stdin&gt;", line 1, in &lt;module&gt; TypeError: sequence item 0: expected string, instance found   Or do I have to resort to a for loop?
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given a polynomial $p(x,y)$ from $\mathbb Z[x,y]$. I want to substitute $x=1$ leaving $y$ as it is. The command Value(p,[1,y]) does not work. Can you give me a hint?  After that it gets a little more complicated. Let  $$p(x,y)=x^6+y^6$$ After substituting $x=1$ we have $p(1,y)=1+y^6$. Given a natural number $j$. I want to find all expressions $y^j$ in $p$ and substitute these $y^j$ by $1$ and the others by $0$. In this case (and for $j=6$) we would get $1+1=2$. If $$p(x,y)=x^4y^2+x^2y^6$$ we should get $0+1=1$. Is there any way to do this?  Thanks for reading! edit: Well, it turns out I used the function Value wrong. The right code for my problem would be Value(p,[x,y],[1,y]). So my first question is solved.
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I have two models - we'll say a sphere and hollow cylinder - that I'm working with.  Effectively, I want to 'subtract' the hollow-cylinder shape from the sphere.  I know from certain orientations (the top of the sphere) this is fairly easily done, just a matter of deleting and creating new faces.  However, from other orientations the edges/geometry doesn't line up so easily.  Is there an automatic (or at least, not terribly difficult/manual) way to 'subtract' one mesh from another?