How to perform a slope analysis in QGIS? I have contour information in the form of line features (see picture below) and I want to do a slope analysis in QGIS. Is it possible?
Calculating slope in percentage with QGIS I'm working with QGIS, v. 2.0.1. I have several soil types. They are (amongst others) described with slope indications. I already calculated the slope from the DTM in degrees, but as description of the soil types is in percentage, I need to calculate the slope in percentage. I tried two things:  I tried the analysis tool from the raster section, but it didn't work. The result was just black with values from 0 to 0. The same happened when I tried to calculate the percentage in the raster calculator but the result was the same.  Maybe I need to say that the terrain is pretty plain. What else can I do? Update I tried the raster calculator and did what @radouxju suggested. It gave me a black picture with values from 1756.64 to 1756.64. But as soon as I click in the picture the values are &quot;normal&quot;, say 5.88 or 13.224 or something like that. The same happens when I try the gdaldem tool but with values from 0 to 0. I also tried to change the contrast enhancement (like stretch to min/max) but it didn't change anything. Update 2 I just  that maybe I have to set the scale to 111120  but that doesn't work neither (result: black, values 0-0).
Meaning of hash symbol in pin name I'm reading the datasheet of a PSRAM (available ). In the top-right hand corner all the pins are listed. Some of the pin names end with a hash, e.g. LB#, OE#.  What does the # notation mean for pins? The problem is, it seems when I enable a disabled module, the module's schema version in the system table is automatically updated to the latest version supplied by hook_update_N.  So my current module with the schema version -1 is modified to 7106 when I enable the module, and no hook_update_N functions are run.   So when I come to the next step of running database updates, no hook_update_N functions are run because the module thinks it's already the correct version.  The only method I can see is to manually set the module version in hook_enable, however I'll then need to update the module code to remove the hook_enable to make sure it doesn't run if the module is ever re-enabled in the future, so obviously this approach is a bit dirty.  I'm hoping there's a better solution? Can US stocks list on one stock exchange but trade on other US stock exchanges? I am trying to understand how the US stock market works. From my understanding, there is one stock exchange where a company is &quot;listed&quot;. This stock exchange listing requires the company to adhere to the regulations of the stock exchange where it is listed. Once a company is &quot;listed&quot;, its shares can start to trade on any stock exchange. For example, once a company lists on NASDAQ, its shares can be traded on NASDAQ, NYSE, Boston Stock Exchange, Philadelphia Stock Exchange, National Stock Exchange, EDGX Exchange, EDGA Exchange, NYSE Arca, etc. Is my understanding correct? Is it true that once a stock is listed on a US stock exchange, the stock can then trade on any US stock exchange? Consider the Level 2 quotes for Microsoft (NASDAQ: MSFT). Microsoft is only listed on NASDAQ, but it also trades on the NYSE and other stock exchanges:  My questions are:  Does this mean that companies only need to be listed on one stock exchange in order for their stock to trade on other stock exchanges?  If so, do companies need to adhere to the &quot;listing requirements&quot; of all the stock exchanges where their stock is traded, even though they are not &quot;listed&quot; on those stock exchanges? Does Microsoft have to follow the listing requirements of the NYSE even though it is not listed there?  If stocks can trade on any US stock exchange regardless of which US stock exchange the stock is listed on, why don't NYSE-listed companies reduce their yearly listing fees by listing on NASDAQ instead? Extrusions causing problems in edit mode When I extruded vertices in edit mode, for some reason one of the sides of my 3-D model (the textured-looking side to the right) is creating a lot of errors. First off, it automatically sets extrusions on the x-axis. Secondly, it doesn't allow me to extrude planes into 3-D shapes. Thirdly, I can't increase the height of vertices without it creating a new face. Finally, it renders those automatic textures to create a grainy black texture in render mode. I attached an image with the solid object (top) and the rendered object (bottom). Note: I did not set any textures or alter the lighting on the model. Also, the left gray side is working perfectly fine without errors. This is a variation of the  questioned some time ago.  For a complex Hilbert space $H$ let $T: H \rightarrow H$ be a bounded operator. We call that $T$ is a trace-class operator if the following sum  $$\sum_{i}{\langle |T|e_{i}, e_{i} \rangle} &lt; \infty$$  converges, where $|T| = (T T^{*})^{\frac{1}{2}}$ is the absolute value of the operator.  Assume that  $$\sum_{i}{\langle Te_{i}, e_{i} \rangle}$$  converges for any basis in the space $H$. How to prove that if the aformentioned property holds then the operator is a trace class operator?  The progress on the problem is the following: Given an arbitrary bounded operator $T: H \rightarrow H$, one can use the following decomposition $$T = \big( \frac{T + T^{*}}{2} \big) ^{*} + i \big( \frac{T^{*} - T}{2i} \big) ^{*}$$  The latter line gives the decomposition $$T = A + i B$$ where $A, B$ are normal operators.  For the normal operators we can apply the spectral theorem that proposes that $T$ is unitary equivalent to  $$(UT U^{-1})(f(x)) = g(x) f(x)$$  where  $$U: H \rightarrow L^{2}(X, \mu)$$ $$g \in L^{\infty}(X, \mu)$$  Though this decomposition classify the operator in a broad sence, i see no direct way to conclude the statement. Are the any hints that may extend the previous argument? If not, are there any ways to conclude the statement? “between” vs. “among” Today I was cut off in the middle of the following sentence:     Between Cook, Strauss, and Pietersen—   My friend said I was wrong. He said that for more than two entities, among/amongst are used, and that between is only for two entities.   I vaguely remember some rule like this, but I asked him to hear the whole sentence, which was:     Between Cook, Strauss, and Pietersen, they’ve notched up 56 centuries in Test cricket.   Should I have used among here? Seems correct to me to use between, but my friend was certain.   Another friend pointed out the following sentence:     I’m still choosing between Harvard, Yale, and MIT.    Certainly among doesn’t seem to fit here, and she, to me, was correct in using between. The observable universe is approximately 13.7 billion years old. But yet it is 80 billion light years across. Isn't this a contradiction? 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? The literature defines multinomial coefficients (or extended bnomial coefficients) as   $$ \binom{n}{r_1,r_2,\cdots,r_l} = \frac{n!}{r_1!r_2!\cdots r_l!}$$ where $$ r_1+r_2+\cdots+r_l = n$$  Which is the coefficient of $x_1^{r_1}x_2^{r_2}\cdots x_l^{r_l}$ in expansion of $(x_1+x_2+\cdots+x_l)^n$.  Question: How do we call the coefficients of $x^r$ in expansion of $(x+x^2+\cdots+x^l)^n$. Is it bad to name them multinomial coefficients?  In case of binomial coefficients the two are same in expansion of $(x_1+x_2)^n$ and the expansion of $(x+x^2)^n$. Obviously the two are different for multinomial expansion.   Example: As $(x_1+x_2+x_3)^2= x_1^2+2 x_2 x_1+2 x_3 x_1+x_2^2+x_3^2+2 x_2 x_3$. The coefficients are $\{ 1,2,2,1,1,2 \}$. Are these MULTI-NOMIAL?  $ As (x+x^2+x^3)^2=x^2 + 2 x^3 + 3 x^4 + 2 x^5 + x^6$. The coefficients are $\{1,2,3,2,1\}$. Are these too the MULTI-NOMIAL?  I prefer different terminologies to avoid confusion. Therefore want to know how these are distinguished in existing literature. How to determine if a string is a valid variable name? What's the mathematics definition of the F value in Analysis of Variance Table? Suppose I have a linear regression model Y ~ b1*x1 + b2*x2 with the sample data:          x1      x2        y 0  -0.7251 -0.6033  10.2508 1  -1.3148  3.7067  10.3054 2  -1.2818 -5.8136   7.2010 3   0.6154  3.6336  11.2807 4   1.4887 -2.8337  10.6299 5  -2.0224 -0.3931   7.4502 6   0.5419 -1.7807  10.7339 7  -1.0745 -5.0296   5.8734 8   3.2116 -0.4681  11.7184 9   1.6767 -4.6040   9.0145 10 -1.3355 -2.6874   6.5014 11  0.8296  2.6793  11.8295 12 -2.8647  4.2916   7.8414 13  1.8445 -3.3924   9.6520 14 -0.1036  2.4273  11.9766 15  1.0886 -3.5636  10.6023 16 -3.3026 -6.9837   4.9962 17  1.4050 -3.8453  10.4630 18 -0.3111 -1.4278   9.2283 19 -3.5454  2.4960   7.3720   And using the following R code to get the Analysis of Variance Table:  &gt; df &lt;- read.csv('f:/data.csv', header=T, sep=",") &gt; model_All &lt;- lm(y ~ x1 + x2, data=df) &gt; anova(model_All) Analysis of Variance Table  Response: y           Df Sum Sq Mean Sq F value    Pr(&gt;F)     x1         1 44.440  44.440  58.277 6.865e-07 *** x2         1 26.640  26.640  34.934 1.711e-05 *** Residuals 17 12.964   0.763                       --- Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1   And I can calculate the F value of hypothesis test H0： b2=0 via:  &gt; model_Drop2 &lt;- lm(y ~ x1, data=df) &gt; SSR_All = sum(resid(model_All)^2) &gt; SSR_Dp2 = sum(resid(model_Drop2)^2) &gt; F_x2 = ((SSR_Dp2 - SSR_All) / 1) / (SSR_All / (20 - 3)) &gt; F_x2 [1] 34.93434   We can see that F_x2 is equal to F value of x2 in Analysis of Variance Table. However, the value F_x1 calculated as follow is not equal to F value of x1, I can't understand why:  &gt; model_Drop1 &lt;- lm(y ~ x2, data=df) &gt; SSR_Dp1 = sum(resid(model_Drop1)^2) &gt; F_x1 = ((SSR_Dp1 - SSR_All) / 1) / (SSR_All / (20 - 3)) &gt; F_x1 [1] 70.30392   So, what is the mathematics definition of the F value in table?  EDIT The result of the features reversed as commented by @Michael M:  &gt; model_All2 &lt;- lm(y ~ x2 + x1, data=df) &gt; anova(model_All2) Analysis of Variance Table  Response: y           Df Sum Sq Mean Sq F value    Pr(&gt;F)     x2         1 17.468  17.468  22.907 0.0001718 *** x1         1 53.612  53.612  70.304 1.914e-07 *** Residuals 17 12.964   0.763                       --- Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Posting and answering questions you have already found the answer to When calling super() in a derived class, can I pass in self.__class__? How to search for email of a particular date from years ago in Gmail There are similar questions and several answers. I tried to follow the methods provided in the answers.  They did not work. So I am going to ask it again with more specifics.  I only knew the date of the email in the in-box in 2012 and the content of it but did not know the exact wording of the subject. Is it possible to find this email? Which shell am I running on? Locking an iPhone if button is broken? My friend's iPhone's sleep/wake button no longer works. Is there any way to turn off the screen besides setting up auto-lock and waiting a few minutes? How to get the Input and output in a terminal to a text file? I want to create a text file which will add each command I am going to run on the terminal as a new line entry in the text file. i.e., I need to keep a record of what are the things I'm doing in the terminal. Further if I can keep a record of outputs I'm getting in the terminal, along with the inputs in the terminal, as a text file, it will be much better.   Each time I forget a command I have to search again in the Internet for the command.  Can I can make a file like above? Why isn't $\mathbb Z [\sqrt{pq}]$ a factorial domain This is exercise 8 on page 147 of Jacobson's Basic Algebra:     Let $p$ be a prime of the form $4n+1$ and let $q$ be a prime such that $ \left( \frac{q}{p} \right) =-1$. Show that $\mathbb Z[\sqrt{pq}]$ is not factorial.   Here, $ \left( \frac{q}{p} \right) =-1$ means that $\overline{q}$ is not a quadratic number in the field $\mathbb Z / (p)$.  In the special case when $p=5$ and $q=2$, $9 = 3 \cdot 3 = (\sqrt{10} +1 )(\sqrt{10 }-1)$, with $3$ and $\sqrt{10} +1$ or $\sqrt{10} - 1$ nonassociates. But I have no idea what can I do with the general case.  I know that primes of the form $4n+1$ can be written in the form $m^2 + n^2$ with $m,n \in \mathbb Z$. But when $q$ is odd and not of the form $4n+1$, $pq$ is not the sum of two quadratic numbers. How can I use the fact $\left( \frac{q}{p} \right) =-1$?  Thanks~ The shellscript is very simple:  retroarch $* -c /tmp/retroarch/Data/retroarch/gambatte.cfg   ("retroarch" is another shellscript)  If the filename passed contains spaces it fails:  RetroArch [ERROR] :: Could not read ROM file. The following is exercise 8.2 in 3rd edition (and exercise 8.19 in 2nd edition) of Goldstein's Classical Mechanics.   Adding the total time derivative of a function of $q_i$ and t to the Lagrangian will not change the the Euler-Lagrangian equation. So if we make the following change to Lagrangian,   $$L'(q,\dot{q},t) = L(q,\dot{q},t) + \frac{dF(q_1,q_2,...,q_n,t)}{dt}$$ we can get  $$ \frac{d}{dt}\frac{\partial{L'}}{\partial{\dot{q_i}}} - \frac{\partial{L'}}{\partial{q_i}} = 0 $$ from  $$ \frac{d}{dt}\frac{\partial{L}}{\partial{\dot{q_i}}} - \frac{\partial{L}}{\partial{q_i}} = 0 $$  How can we get the corresponding Hamiltonian equation part? This is to prove  $$ \dot{p'_i} = \frac{\partial{H'}}{\partial q_i} $$ $$ -\dot{q_i} = \frac{\partial{H'}}{\partial p'_i}  $$  from   $$ \dot{p_i} = \frac{\partial{H}}{\partial q_i} $$ $$ -\dot{q_i} = \frac{\partial{H}}{\partial p_i}  $$ where $p'_i = \frac{\partial L'}{\partial \dot q_i}$.  Edit  The corresponding $H'$ is $$ H' = \sum_k{p'_k \dot{q_k}} - L' $$ where $p'_k = \frac{\partial L'}{\partial \dot q_k}$.