I want to remove property 'b' from all objects how i remove???   let result=[     {        'id':'1',        'b':'asd'     },     {        'id':'2',        'b':'asd'     },     ...     ,     {         'id':'2000',        'b':'asd'     },  ]     // delete object.b; its only for object and want to remove from whole     //  array 2000 records     Using foreach loop is it correct way to delete key + value pair from array of objects
Say I create an object as follows: let myObject = {   &quot;ircEvent&quot;: &quot;PRIVMSG&quot;,   &quot;method&quot;: &quot;newURI&quot;,   &quot;regex&quot;: &quot;^http://.*&quot; };  How should I remove the property regex to end up with new myObject as follows? let myObject = {   &quot;ircEvent&quot;: &quot;PRIVMSG&quot;,   &quot;method&quot;: &quot;newURI&quot; };
SQL-Server: The backup set holds a backup of a database other than the existing Suppose $n$ is a non negative integer $\geq 4$ and $\sigma\in S_n$ a permutation. Conjugacy classes in $S_n$ are completley determined by the cycle structure of $\sigma$. If we let the alternating group $A_n$.act on $S_n$ by conjugation, the orbits coincide with the conjugacy classes in $S_n$ except for those (even) permutations whose cycle structure is composed of even cycles (which corresponds to cycles of uneven length) of distinct order (fixed points are treated as cycles of length $1$.) See   For instance a $3$-cycle in $A_4$ has cycle structure $[3,1]$ which fits the bill, and a $5$-cycle in $A_5$ or $A_6$ works too, since the cycle structure is $[5]$ and $[5,1]$ respectively. On the other hand, a $3$-cycle in $A_5$ will does not satisfy the criteria as it has cycle structure $[3,1,1]$ with two fixed points.  In this case, the conjugacy class of $\sigma$ inside $S_n$ splits in $2$ orbits of the conjugation action of $A_n$, and if $\tau$ is any odd permutation, for instance $(12)$, then we have $$\mathrm{Conj}_{S_n}(\sigma)=\mathcal{O}(\sigma)\coprod\mathcal{O}(\tau\sigma\tau^{-1})$$  My question is How can we differentiate between the two orbits? Does it have geometric meaning?    I can see that this information can be useful when looking at representations of the alternating groups, since there are as many irreducible representations as conjugacy classes. As for some geometric interpretation, I ask because there is one for the three cycles in $A_4$. If $S_4$ is understood as the group of symmetries of a regular tetrahedron, then $A_4$ is the subgroup of direct symmetries, and a $3$-cycle corresponds to a rotation of angle $\frac{2\pi}{3}$ or $\frac{4\pi}{3}$ with axis passing through one of the vertices. Conjugacy classes inside $A_n$ preserve the angle, and that's how we can tell them apart. In this case the computations are so simple we don't really need this picture, but I think it is a nice way to understand the fact $3$-cycles in are not all conjugate inside $A_4$... PHP script not executing on Apache server How can I revert a config file back to the originally installed version after I have edited it? Automatically rename files when they are placed in a specific directory Is it possible to automatically rename a file when it's placed in a specific directory?  For example I have a directory named "dir0".I move or copy a file named "file1" to "dir0".then "file1" should rename to "file1_{current timestamp}" Merging banking and casual style in moderncv (2) Should black holes be considered two-dimensional or three-dimensional? Question about "surely" and "certainly" Today I saw a sentence:  She surely can do it.  What does that mean? How does it differ from:  She certainly can do it. Surely she can do it.  This slight difference in meaning of those sentences I am not able to notice. Could you please help me? Thanks in advance! mysql doesn't ask for root password when installing How to backup an Android device? What can primes, except 2, 3, and 5, be congruent to $\pmod {30}$? After some trials, I found out that a prime $p \gt 5$ is congruent to $q\pmod{30}$, where $q$ is also a prime, and $1 \le q \lt 30 \;$ (i.e.  $p \equiv q\pmod{30}.$  Is there a way to write a formal proof to show this?  Is that an acceptable answer. Let $f(x)$ and $g(x)$ be two continuous functions on $[0,1]$ and  $$\int_{0}^{1}f(x) dx= \int_{0}^{1}g(x)dx = 1$$   Show that there exist  $[a,b]\subset [0,1]$, such that  $$\int_{a}^{b}f(x) dx= \int_{a}^{b}g(x)dx = \frac{1}{2} $$  The question can be solved by considering the fundamental group of $S^1$, now I am wondering if we can solve it by real-analysis.  Here is the topological solution:  Assume that for any $[a,b]\subset[0,1]$, we always have  $$(\int_{a}^{b}f(x) dx\neq \frac{1}{2}) \quad \vee \quad(\int_{a}^{b}g(x)dx \neq \frac{1}{2}) $$ Consider mapping $$\phi:D\rightarrow\mathbb{E}^2\backslash\{(\frac{1}{2},\frac{1}{2})\},\,\,(x,y)\mapsto(\int_{y}^{x}f(t) dt,\int_{y}^{x}g(t) dt)$$ where $D=\{(x,y)|0\leq x\leq y\leq 1\}$.  Let $a$ be path from $(0,0)$ to $(0,1) $in $D$ and $b$ be path from $(0,1)$ to $(1,1) $in $D$, then $ab$ is a path from $(0,0)$ to $(1,1) $in $D$ and $$\phi\circ (ab):[0,1]\rightarrow\mathbb{E}^2\backslash\{(\frac{1}{2},\frac{1}{2})\} $$ Notice that $$(\phi\circ (ab))(0)=(\phi\circ (ab))(1)=(0,0)$$  so $\phi\circ (ab)$ is a loop based on $(0,0)$ in $\mathbb{E}^2\backslash\{(\frac{1}{2},\frac{1}{2})\}$.  For any $t\in [0,\frac{1}{2}]$, it's not hard to get that $$(\phi\circ (ab))(t+\frac{1}{2})=(1,1)-(\phi\circ (ab))(t)$$ is equivalent to $$(\phi\circ (ab))(t+\frac{1}{2})-(\frac{1}{2},\frac{1}{2})=-((\phi\circ (ab))(t)-(\frac{1}{2},\frac{1}{2}))$$  Define retraction  $$r:\mathbb{E}^2\backslash\{(\frac{1}{2},\frac{1}{2})\}\rightarrow S^1,\quad (x,y)\rightarrow\ \frac{(x,y)-(\frac{1}{2},\frac{1}{2})}{||(x,y)-(\frac{1}{2},\frac{1}{2})||}$$  Then $r\circ \phi\circ (ab)$ is a loop on $S^1$, such that $$(r\circ \phi\circ (ab))(t+\frac{1}{2})=-(r\circ \phi\circ (ab))(t),\quad\forall t\in [0,\frac{1}{2}]$$  So $&lt;r\circ \phi\circ (ab)&gt;$ is not trivial in $\pi_1(S^1)$.   However, $ab$ is path homotopic to $c$ in $D$, where $c$ is the path between $(0,0)$ and $(1,1)$ in $D$. In this case, $r\circ \phi\circ (ab)$ is a point-path in $S^1$ by $$(\phi\circ c)(t)=\phi(t,t)\equiv (0,0)$$  which leads to contradiction. Generating list of manually installed packages and querying individual packages When to spell out numbers Let $A$ and $B$ be $n\times n$ matrices such that $ABAB=0_{n\times n}$.  Can we conclude that  $BABA$ must be  $0_{n\times n}$?   I fail to give a counterexample when $n=2$， so I guess the answer is "yes" Proving $|P(A\cap B)-P(A)P(B)|\leq \frac{1}4$ Let $A$ and $B$ be two events of a probability space. Prove that  $\displaystyle|P(A\cap B)-P(A)P(B)|\leq \frac{1}4$   I think it's a very challenging problem, and I've made no progress so far ...  Can someone give me a hint ? Polygonizing a raster yields lots of little squares? I have a pdf with the area of a county that I want to separate into multiple shapefiles (polygons) based on color.   As shown in the first picture there are 4 colors so I would need to separate based on color between those four. But first, I need to convert it into a vector (unless there is an easier way).   Doing so, I get the second image as an output by using the polygonize tool which when zoomed in very close, has a lot of tiny squares where color should be different. I was wondering if there is a way to break a raster up by color even though the image isn't smart. $[S]$ means linear span of S. Then $[S] = [[S]]$ My definition of span:   Suppose a vector space $(V,+,\cdot)$, and       $$S = \{u_1,\cdots,u_n\}$$      (and $S$ is a subset of $V$, not a subspace)      $$[S]=:\bigcap_{W\subset V, W\supseteq S} W$$      In other words, $[S]$ is, by definition, the intersection of all $W$,   such that $W$ is a subspace of $V$ and $W$ contains $S$.   Well, then I need to prove that   $$[S] = [[S]]$$  My strategy:  $[[S]]$ means the intersection of all subspaces that contain $[S]$. If all the subspaces contains $[S]$, so does their intersection. Therefore, $[S]\subset[[S]]$  However, I'm having trouble proving the converse:  $$[[S]]\subset[S]$$  Any hints? Let $S = \bigoplus_{n\ge 0} S_n$ be a graded commutative ring. Let $f$ be a homogeneous element of $S$ of degree $&gt; 0$. Let $S_{(f)}$ be the degree $0$ part of the graded ring $S_f$, where $S_f$ is the localization with respect to the multiplicative set $\{1, f, f^2,\dots\}$. Suppose $S$ is finitely generated algebra over $S_0$. Then $S_{(f)}$ is a finitely generated algebra over $S_0$. Grant the OP a binding close as duplicate vote Since the  to the close as duplicate mechanism it is a lot more visible to the OP that there is/might be a duplicate to their question. Empirically, I've noticed a lot more people agreeing, in the comments under their question, that their question is a dupe.   For instance, :        I'm aware that allowing  is  (back in the stone age) but now Stack Exchange is so much larger, and Stack Overflow is groaning under the weight of questions, I think this should be partially revisited.  Based on the principle that the simplest way to keep a site clean is to have as many people as possible nibbling away, I propose that the OP gets granted a binding close-as-duplicate vote.  It should simply be an additional line in the banner, which is . It might be something like this:     This question may already have an answer here:   &nbsp; "" 3 answers   &nbsp; Does this fully answer your question?  Yes / No   I'm not wedded to the Yes / No thing I just can't think of a better way at the moment.  Clicking on Yes would take the OP to the close as duplicate dialogue with that question highlighted. If there are multiple duplicate questions you'd need to ask the OP multiple times. A No could be used to track the relative success of the project, i.e.   Whether the question subsequently gets closed. Whether the question gets reopened again. How many times the OP actually clicks. How many times the OP disagrees with the analysis of the community.   Potentially, a No could be used to open a comment dialogue asking the OP to explain why they think their question is not a duplicate thereby stopping them simply putting "this is not a duplicate" in the question and formalising this process somewhat. Not sure about this idea...  I do not think this should be restricted by reputation, anyone should be able to close their own question as a duplicate. It gives people with lower reputation a more positive experience of the closing process and emphasises the importance of duplicate finding.  Most importantly, self-closes should not, initially, count towards the question ban. We can help new users learn how to use a lot more of the site than they might otherwise see and encourage them to search for duplicates more thoroughly. If the OP then goes on to do the same thing 10 times in a row then you can start using it toward the ban as they're obviously using the users as a personalised search service but at least the first 3-4 times this should not have any effect.  This question is, deliberately, only requesting 20% of the linked question (from 2009) and then only based on the new UI. The linked question asks for binding migration votes etc, which I do not want to see and I am not requesting a binding close-as-duplicate for the OP until someone else has already voted to close (though I think if the OP has 3k rep this might be allowable). This, I hope, addresses the concern's raised in the linked question and explains why this is not a duplicate of that question.