I have the following table:     source_id bands duration     24 L BAND 0.215447   24 L BAND 0.00017   24 S BAND 0.00017   88 L BAND 0.215447   88 L BAND 0.00017   72 L BAND 0.00017     For every combination of &quot;source_id&quot; and &quot;bands&quot; values, I would like to output the entry with the longest duration, such that the output would be:     source_id bands duration     24 L BAND 0.215447   24 S BAND 0.00017   88 L BAND 0.215447   72 L BAND 0.00017     What would be the easiest and/or best way to achieve this?
My table is:     id home datetime player resource     1 10 04/03/2009 john 399   2 11 04/03/2009 juliet 244   5 12 04/03/2009 borat 555   3 10 03/03/2009 john 300   4 11 03/03/2009 juliet 200   6 12 03/03/2009 borat 500   7 13 24/12/2008 borat 600   8 13 01/01/2009 borat 700     I need to select each distinct home holding the maximum value of datetime. Result would be:     id home datetime player resource     1 10 04/03/2009 john 399   2 11 04/03/2009 juliet 244   5 12 04/03/2009 borat 555   8 13 01/01/2009 borat 700     I have tried: -- 1 ..by the MySQL manual:   SELECT DISTINCT   home,   id,   datetime AS dt,   player,   resource FROM topten t1 WHERE datetime = (SELECT   MAX(t2.datetime) FROM topten t2 GROUP BY home) GROUP BY datetime ORDER BY datetime DESC  Doesn't work. Result-set has 130 rows although database holds 187. Result includes some duplicates of home. -- 2 ..join  SELECT   s1.id,   s1.home,   s1.datetime,   s1.player,   s1.resource FROM topten s1 JOIN (SELECT   id,   MAX(datetime) AS dt FROM topten GROUP BY id) AS s2   ON s1.id = s2.id ORDER BY datetime   Nope. Gives all the records. -- 3 ..something exotic:   With various results.
Hide textfield blinking cursor On the number divisors of a number Let $d(n)$ be the number divisors of a natural number $n$. Now let $m$ be a natural number. Find least natural number $n$ such that $d(n)=m$.  Thank you How can I add some html to my Mail.app signature? I want to add my company logo and and some juice to my contact information in the signature part in Mail.app. How can I do that with little or no efforts? I finally got dragonrend and defeated Alduin, but I'm still too weak for fighting ancient dragons. Any certain weapon for killing them.  (I've got lvl 20 one handed weapons and lvl 46 for archery, so the Dragonbane is out of the question) How do I format my code blocks? Do 'learn by heart' & 'learn by rote' mean the same? Here in India, both the phrases learning by heart and learning by rote are taken to have the same meaning, i.e., blind memorisation without true understanding.    However, some sources say that to learn something by heart implies that one knows something so well that it has been thoroughly internalised and imbibed.    So is it wrong to conflate the two? Assume $R=(\mathcal{U}_n)_n$ is a sequence of distinct ultrafilters on some set $X$. Since every Hausdorff space has an infinite discrete subspace, there is a subsequence $R=(\mathcal{V}_n)_n$ of $R=(\mathcal{U}_n)_n$ such that $\{\mathcal{V}_n|n \in \mathbb{N} \}$ is a discrete subspace of $\beta X$. In particular, there is a sequence $(A_n)_n$ of sets with $A_n \in \mathcal{V}_m$ if and only if $m =n$. If we let $B_n=A_n - (A_0 \cup...\cup A_{n−1})$, then the sequence $(B_n)_n$ is pairwise disjoint and $B_n \in \mathcal{V}_m$  iff $m=n$. If we let $B =\cup_n B_{2n}$, then $B \in \mathcal{V}_m$ if and only if $m$ is even. In other words, if $B =\{ \mathcal{V} \in βX|B \in \mathcal{V}\}$, then $B$ is a clopen set with $\mathcal{V}_m \in B$ iff $m$ is even. Therefore, we conclude that the sequence $(\mathcal{V}_m)_m$ cannot converge to any point, so the sequence $(\mathcal{U}_n)_n$ cannot converge to any point either.     I would like to know that:      (1) Why "$B_n \in \mathcal{V}_m$  iff $m=n$"? And " $B =\cup_n B_{2n}$, then $B \in \mathcal{V}_m$ if and only if $m$ is even "?      (2)$B$ is a clopen set with $\mathcal{V}_m \in B$ iff $m$ is even This or That person Reset password in Drupal 6? How many associative binary operations for a set of two elements? Suppose we have a set  $S=\{a,b\}.$ Obviously, the total number of binary operations on $S$ is the number of all the pairs such that $$*(a,b)=a \, \text{or}  \, b,$$ which gives  $ 2 \cdot 2 \cdot 2 \cdot 2=2^4$ possible ones. Is there a way to find which of these $16$ binary operations are associative without writing all the operation matrices and checking it by hand? Overflow to left instead of right I have a div with overflow:hidden, inside which I show a phone number as the user types it. The text inside the div is aligned to right and incoming characters are added to right as the text grows to left.  But once the text is big enough not to fit in the div, last characters of the number is automatically cropped and the user cannot see the new characters she types.  What I want to do is crop the left characters, like the div is showing the rightmost of its content and overflowing to the left side. How can I create this effect? Selenium python drop down menu select Theorem 13.9 in these  is the polynomial spectral mapping theorem:     Theorem 13.9 For a polynomial $p$ we have $\sigma(p(T)) = p(\sigma(T))$.     My questions:   First direction: Why would $q(T) (p(T) - p(\lambda) I)$ be an inverse for $T - \lambda I$? Update: @RobertIsrael said that instead $(p(T) - p(\lambda) I)^{-1} q(T)$ should be $(p(T) - p(\lambda) I)^{-1}$ be? According to the lines in the proof above, we have $$ (p(T) - p(\lambda) I)^{-1} = ((T - \lambda I) q(T))^{-1} \overset{?!}{=} q(T)^{-1} (T - \lambda I)^{-1}, $$ but as $\lambda \in \sigma(T)$ by assumption, $(T - \lambda I)^{-1}$ doesn't exist, right? For the second direction: Why does $x_i \in \sigma(T)$ imply that $p(x_i) - \mu = 0$? I thought that $x_i \in \sigma(T)$ only implies that $T - x_i I$ is not boundedly invertible. Since $\mu \in \sigma(p(T))$ we have that $p(T) - \mu I = c \prod_{k = 1}^{n} (T - x_i I)$ is not boundedly invertible but how does that help?   In both statements there seem to be a notion of "a product of operators is not invertible iff all factors aren't invertible", but how can that be true? Consider i.e. shift operators on a sequence spaces such that $ST = I$, i.e. the left and right shift operators, then the above mentioned notion doesn't hold, right? 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