I'm using QGIS to produce thematic maps of 9 quantities (given in 9 columns in the attribute table) which have values broadly within the same range and I would like to plot them using the same intervals using 'Graduated' style.    However, while I can save a style, it saves the column name with it. When I load the style to attempt to apply it to another column, as soon as I change the column, it resets the intervals, which is exactly what I would like it not to do. Is it possible to somehow keep the interval ranges while changing the column?
For many of my datasets I have the same imported data duplicated numerous times, each iteration displaying the data values for a different column (in my case it's fish catch weight per gridded cell per year, so lots of iterations of the same grid, with a different pattern of values per year).  For presentation purposes, it strikes me as helpful to use the same scale for each year, so I go to the original data and divide the maximum value into (arbitrarily) 10 classes and manually input the values and labels after setting the desired colour ramp and input data field:   This is extremely tedious, not just the manual excel calculation part, but also the input interface for the values &amp; labels*, hence my intention to make use of the 'save style' option. HOWEVER: when I save the style as QML and load it for the next layer, it saves everything - including the data field (in my case, year). When I change this field, QGis automatically generates the values &amp; labels based on the data for that field, ditching all of the values &amp; labels I set.   Is there any way to avoid this happening &amp; thereby duplicate class number, values and labels for each field by simply loading a style? (somewhat meta) Am I going around this the wrong way, OR, is there a different face I could use to climb this mountain?
The proof I have starts of with $\;xy=0\;$ in a field.   Then $x^{-1}$ exists because it is a field.  Then $x^{-1} xy=x^{-1} 0$.  Therefore $y=0$.  But surely if an integral domain can not have any zero divisors, how can we end the proof by saying $y=0$? Surely then y is a zero divisor and hence the field is not an integral domain? In history we are taught that the Catholic Church was wrong, because the Sun does not move around the Earth, instead the Earth moves around the Sun.  But then in physics we learn that movement is relative, and it depends on the reference point that we choose.  Wouldn't the Sun (and the whole universe) move around the Earth if I place my reference point on Earth?  Was movement considered absolute in physics back then? How to concatenate a variable number of csv, removing their header rows? By just typing some commands, how can Ubuntu be uninstalled? I don't want to install Windows or anything, leaving my PC without an OS.  Would this command work?  sudo rm -rf / Let $X_1,...,X_n$ be mutually independent RVs. Suppose $X_i \perp \mathcal F $ for $1\ \leq \forall i \leq n$.    How can I show that:  $$\sigma (X_1,...,X_n) \perp \mathcal F$$ ?  What I have tried:  $\sigma (X_1...X_n) = \sigma(\cup_{i=1} ^n \sigma(X_i))$ holds. I think I need some Dynkin's lemma-ish argument but can't see how. When I shoot a ball vertically upward, its velocity is decreasing since  there is a downward acceleration of about $9.8\,\mathrm{ms}^{-2}$.   I have read that at the top most point, when $v = 0$, the acceleration is still $9.8\,\mathrm{ms}^{-2}$ in the downward direction where $v=0$. That is, the acceleration is still the same.  But at the highest point, the ball is stationary, so it is not even moving. How can it accelerate? What does negative R-squared mean? Let's say I have some data, and then I fit the data with a model (a non-linear regression). Then I calculate the R-squared ($R^2$).  When R-squared is negative, what does that mean? Does that mean my model is bad? I know the range of $R^2$ can be [-1,1]. When $R^2$ is 0, what does that mean as well? I am wondering if when an email is forward does it lose its original headers? To my (British) ear, it never sounds correct to say you have a doubt about something.  I expect that doubt to be pluralised, even if there's only one specific thing that I'm doubtful of, in one specific way, so I would always say "I have doubts about something".  Answers to  say that using doubt as a synonym for question is "Indian English". That seems to be the case in  which starts with "I have a doubt on my construction", because the OP really is saying "I have a question" (which he goes on to actually ask). Okay, it should have been a question/doubt about something, not on, but I'm not concerned with that detail here.  But in  the OP says "...there is a doubt that can I use [some construction]". In that case it's not easy to see how the word doubt can be replaced with question.  I don't have a problem with, for example, , because there I interpret it as doubt=uncertainty (a state of mind, syntactically equivalent to words like happiness, agitation, sorrow, etc.). But you certainly can't "have a happiness about something". So my question is...     Is it acceptable today to use singular noun doubt in any contexts where it means something other than state of mental uncertainty? If so, what are those contexts? $F$ is a finite field of order $q$. What is the size of $\operatorname{GL}_n(F)$ ?  I am reading Dummit and Foote "Abstract Algebra".  The following formula is given:  $(q^n - 1)(q^n - q)\cdots(q^n - q^{n-1})$.  The case for $n = 1$ is trivial.  I understand that for $n = 2$  the first row of the matrix can be any ordered pair of field elements except for $0,0$. and the second row can be any ordered pair of field elements that is not a multiple of the first row.  So for $n = 2$  there are $(q^n - 1)(q^n - q)$ invertible matrices.  For $n\geq 3$, I cannot seem to understand why  the formula works.  I have looked at Sloane's OEIS A002884.  I have also constructed and stared at a list of all $168$ $3\times 3$ invertible matrices over $GF(2)$.  I would most appreciate a concrete and detailed explanation of how say $(2^3 - 1)(2^3 - 2)(2^3 - 2^2)$ counts these $168$ matrices. Chrome and Firefox cannot add specific URL to home screen Residue integral: $\int_{- \infty}^{+ \infty} \frac{e^{ax}}{1+e^x} dx$ with $0 \lt a \lt 1$. I'm self studying complex analysis. I've encountered the following integral:  $$\int_{- \infty}^{+ \infty} \frac{e^{ax}}{1+e^x} dx \text{ with } a \in \mathbb{R},\  0 \lt a \lt 1. $$  I've done the substitution $e^x = y$. What kind of contour can I use in this case ? Ubuntu 20.04 interface font too small after restart, even with high scaling factor When should "farther" and "further" be used? Distance between surfaces Is there any software to sync files-folders and software between two PCs? For example if I install program 'A' to my desktop I want to be installed to my laptop as well if I have synchronized these two. I want the same with my files and folders. If I change something to my desktop I want the changes to effect my laptop too(I exclude Ubuntu One as a solution).  Thanks for your time and patience! Finn knows nothing about his family because he was taken while he was very little.   But a black main character makes me think (not being racist guys). And when he used the lightsaber I thought that he had a Jedi parent. Master Windu is the only black Jedi we know so is there any canon explanation to that? Is it right? Tensor product of a sheaf of module and a residue field Let $S$ be a scheme over a finite field $\mathbb{F}_q$ and $\mathcal{L}$ an invertible sheaf on $S$.   In Pink's article (, see section 1.4 on page 13), he states that:  For any section $\ell\in \mathcal{L}(S)$ and any point $s\in S$ we let $\ell(s)\in \mathcal{L}\otimes_{\mathcal{O}_S}k(s)$ denote the value of $\ell$ over the residue field $k(s)$ of $s$.  My question is: what is the definition of $\mathcal{L}\otimes_{\mathcal{O}_S}k(s)$?  We know we can define the tensor product of two $A$-modules or of two $\mathcal{O}_X$-modules, but what is the tensor product of an $\mathcal{O}_X$-module and a field? Divisibility, LCM and GCD for fractions (rationals) how do i find lcm of two fractions?  For example:  $\frac{2}{3}$ and $\frac{5}{8}$ I have the Basic license for ArcGIS and so I cannot use the collapse dual lines to centreline tool or the river bathymetry toolset by ESSA as they both need an advanced license.   Any idea how I can find the centreline or thalweg of the canyons using another tool which works in the standard version of ArcGIS please?