want to have button for apply transformations, but when I tried this code, it just doesn't work :(  layout.operator("object.transform_apply(location=False, rotation=True, scale=True)"), text="Reset Trasforms")   can someone please help me?
Basically, I want to have a single operator which can do multiple things, based on some attributes values.  class MyPanel(bpy.types.Panel):     # (...)     layout = self.layout     row = layout.row()     row.operator("my.button", text="Button text")  class MY_BUTTON_OT_Button(bpy.types.Operator):     bl_idname = "my.button"     bl_description = "Button description"     bl_label = "Button"     foo = bpy.props.IntProperty()     bar = bpy.props.BoolProperty()      if bar:         # Do something with foo     else:         # Do something else :P   I know how to set one attribute, like so:      row.operator("my.button", text="Button text").foo=5   But what about multiple arguments? I tried several things that didn't worked, like (e.g.):   With setattr(): row.operator("my.button", text="Button text").setattr(foo=5,bar=True)   With a custom method setVal() in the operator class: row.operator("my.button", text="Button text").setVal(foo=5,bar=True)   Maybe I must use a single EnumProperty? I hope to avoid this, so my question is: Is it possible to pass multiple custom arguments to set attributes in an operator class? And if yes, how ?  Thanks for your help.
Gravitational waves and time Which Standard Visitor visa should I apply for? Why is "idea" sometimes pronounced as "idear"? I know that idea is pronounced as /aɪˈdiə/, but I've meet several people in real life who put an 'r' at the end of the word. How come? I'd like to do it in Python. What I'd like to do in this example in C: #include &lt;stdio.h&gt;  int main() {     int i;     for (i=0; i&lt;10; i++) printf(&quot;.&quot;);     return 0; }  Output: ..........  In Python: &gt;&gt;&gt; for i in range(10): print('.') . . . . . . . . . . &gt;&gt;&gt; print('.', '.', '.', '.', '.', '.', '.', '.', '.', '.') . . . . . . . . . .  In Python print will add a \n or space, how can I avoid that? Now, it's just an example, don't tell me I can first build a string then print it. I'd like to know how to &quot;append&quot; strings to stdout. What off-camera flash solution should I use with an entry-level Nikon? Real Size Image Printing Play audio with Python Let the search find questions by ID How to deserialize a JSON String to Apex Recently I came across a debate about whether it's fair to call RAW 'unprocessed' or 'unaltered' sensor data of a DSLR. As far as I know, the analog sensor data is processed into a digital RAW file using the ISO setting, a process during which the Bayer-sensor data is demosaiced as well. However, those are just the necessary steps for saving a digital file while maintaining maximum quality.  Are those assumptions correct? If so, are there any additional steps that need to be performed between the sensor receiving light and the saving of the RAW file? And can those be reasonably be described as an 'alteration' of the sensor data, or is it fair to say that a RAW file holds the unaltered sensor data? If I use  \usepackage[colorlinks]{hyperref}   I get colored links for \cite (not good for printing or even viewing) but table of content looks fine.  If I use  \usepackage[colorlinks=false]{hyperref}   I get \cite  links in black with a colored frame (this looks very good) but the table of content looks strange with extra long boxes.  What are my other options? Can you have two styles in different parts of a document?   How do you make the page numbers in TOC to be the links instead of the section headers? Visual Studio C# statement collapsing When editing really long code blocks (which should definitely be refactored anyway, but that's beyond the scope of this question), I often long for the ability to collapse statement blocks like one can collapse function blocks.  That is to say, it would be great if the minus icon appeared on the code outline for everything enclosed in braces.  It seems to appear for functions, classes, regions, namespaces, usings, but not for conditional or iterative blocks.  It would be fantastic if I could collapse things like ifs, switches, foreaches, that kind of thing!  Googling into that a bit, I discovered that apparently C++ outlining in VS allows this but C# outlining in VS does not.  I don't really get why.  Even notepad++ will so these collapses if I select the C# formatting, so I don't get why Visual Studio doesn't.  Does anyone know of a VS2008 add-in that will enable this behavior?  Or some sort of hidden setting for it?  Edited to add: inserting regions is of course an option and it did already occur to me, but quite frankly, I shouldn't have to wrap things in a region that are already wrapped in braces... if I was going to edit the existing code, I would just refactor it to have better separation of concern anyway. ("wrapping" with new methods instead of regions ;) Will I get OTA updates after installing Android L Developer preview? Proving $\sum\limits_{l=1}^n \sum\limits _{k=1}^{n-1}\tan \frac {lk\pi }{2n+1}\tan \frac {l(k+1)\pi }{2n+1}=0$ Prove that $$\sum _{l=1}^{n}\sum _{k=1}^{n-1}\tan \frac {lk\pi } {2n+1}\tan \frac {l( k+1) \pi } {2n+1}=0$$     It is very easy to prove this identity for each fixed $n$ . For example let $n = 6$; writing out all terms in a $5 \times 6$ matrix, we obtain:  $\begin{matrix}  \tan \dfrac {\pi } {13}\tan \dfrac {2\pi } {13}  &amp;  \tan \dfrac {2\pi } {13}\tan \dfrac {3\pi } {13} &amp; \tan \dfrac {3\pi } {13}\tan \dfrac {4\pi } {13} &amp; \tan \dfrac {4\pi } {13}\tan \dfrac {5\pi } {13} &amp; \tan \dfrac {5\pi } {13}\tan \dfrac {6\pi } {13}  \\  \tan \dfrac {2\pi } {13}\tan \dfrac {4\pi } {13}  &amp;  \tan \dfrac {4\pi } {13}\tan \dfrac {6\pi } {13} &amp; \tan \dfrac {6\pi } {13}\tan \dfrac {8\pi } {13} &amp; \tan \dfrac {8\pi } {13}\tan \dfrac {10\pi } {13} &amp; \tan \dfrac {10\pi } {13}\tan \dfrac {12\pi } {13} \\ \tan \dfrac {3\pi } {13}\tan \dfrac {6\pi } {13} &amp; \tan \dfrac {6\pi } {13}\tan \dfrac {9\pi } {13} &amp; \tan \dfrac {9\pi } {13}\tan \dfrac {12\pi } {13} &amp; \tan \dfrac {12\pi } {13}\tan \dfrac {15\pi } {13} &amp; \tan \dfrac {15\pi } {13}\tan \dfrac {18\pi } {13} \\ \tan \dfrac {4\pi } {13}\tan \dfrac {8\pi } {13} &amp; \tan \dfrac {8\pi } {13}\tan \dfrac {12\pi } {13} &amp; \tan \dfrac {12\pi } {13}\tan \dfrac {16\pi } {13} &amp; \tan \dfrac {16\pi } {13}\tan \dfrac {20\pi } {13} &amp; \tan \dfrac {20\pi } {13}\tan \dfrac {24\pi } {13} \\ \tan \dfrac {5\pi } {13}\tan \dfrac {10\pi } {13} &amp; \tan \dfrac {10\pi } {13}\tan \dfrac {15\pi } {13} &amp; \tan \dfrac {15\pi } {13}\tan \dfrac {20\pi } {13} &amp; \tan \dfrac {20\pi } {13}\tan \dfrac {25\pi } {13} &amp; \tan \dfrac {25\pi } {13}\tan \dfrac {30\pi } {13} \\ \tan \dfrac {6\pi } {13}\tan \dfrac {12\pi } {13} &amp; \tan \dfrac {12\pi } {13}\tan \dfrac {18\pi } {13} &amp; \tan \dfrac {18\pi } {13}\tan \dfrac {24\pi } {13} &amp; \tan \dfrac {24\pi } {13}\tan \dfrac {30\pi } {13} &amp; \tan \dfrac {30\pi } {13}\tan \dfrac {36\pi } {13}  \end{matrix}$  one can notice then, that the first column vanish the fourth one :  $\tan \dfrac {\pi } {13}\tan \dfrac {2\pi } {13}=-\tan \dfrac {12\pi } {13}\tan \dfrac {15\pi } {13}$  $\tan \dfrac {2\pi } {13}\tan \dfrac {4\pi } {13}=-\tan \dfrac {24\pi } {13}\tan \dfrac {30\pi } {13}$  $\tan \dfrac {3\pi } {13}\tan \dfrac {6\pi } {13}=-\tan \dfrac {16\pi } {13}\tan \dfrac {20\pi } {13}$  $\tan \dfrac {4\pi } {13}\tan \dfrac {8\pi } {13}=-\tan \dfrac {4\pi } {13}\tan \dfrac {5\pi } {13}$  $\tan \dfrac {5\pi } {13}\tan \dfrac {10\pi } {13}=-\tan \dfrac {8\pi } {13}\tan \dfrac {10\pi } {13}$  $\tan \dfrac {6\pi } {13}\tan \dfrac {12\pi } {13}=-\tan \dfrac {20\pi } {13}\tan \dfrac {25\pi } {13}$  and the third column vanish the fifth one :  $\tan \dfrac {3\pi } {13}\tan \dfrac {4\pi } {13}=-\tan \dfrac {30\pi } {13}\tan \dfrac {36\pi } {13}$  $\tan \dfrac {6\pi } {13}\tan \dfrac {8\pi } {13}=-\tan \dfrac {5\pi } {13}\tan \dfrac {6\pi } {13}$  $\tan \dfrac {9\pi } {13}\tan \dfrac {12\pi } {13}=-\tan \dfrac {25\pi } {13}\tan \dfrac {30\pi } {13}$  $\tan \dfrac {12\pi } {13}\tan \dfrac {16\pi } {13}=-\tan \dfrac {10\pi } {13}\tan \dfrac {12\pi } {13}$  $\tan \dfrac {15\pi } {13}\tan \dfrac {20\pi } {13}=-\tan \dfrac {20\pi } {13}\tan \dfrac {24\pi } {13}$  $\tan \dfrac {18\pi } {13}\tan \dfrac {24\pi } {13}=-\tan \dfrac {15\pi } {13}\tan \dfrac {18\pi } {13}$  while the second column is self-vanishing:  $\tan \dfrac {2\pi } {13}\tan \dfrac {3\pi } {13}=-\tan \dfrac {10\pi } {13}\tan \dfrac {15\pi } {13}$  $\tan \dfrac {4\pi } {13}\tan \dfrac {6\pi } {13}=-\tan \dfrac {6\pi } {13}\tan \dfrac {9\pi } {13}$  $\tan \dfrac {8\pi } {13}\tan \dfrac {12\pi } {13}=-\tan \dfrac {12\pi } {13}\tan \dfrac {18\pi } {13}$ .  So the equality occurs. But how to generalize the proof? Avoiding jumping frames in beamer I'm trying to create a command, which calls on an environment, which defines a custom column definition with one argument using \newcolumntype, and defines a second command, containing parameters for a table column.  Then, still within the first command, I begin the tabular environment and call the custom column, with the second command as its argument. I always get an undefined control sequence error, though. Any thoughts on why that might be?  I'm sure that was clear as mud, but I've cut it down to a minimal example, which follows:  \documentclass{article} \usepackage{array} \usepackage[table,dvipsnames]{xcolor} \usepackage{colortbl}  \newenvironment{tablecolor}{     \newcommand{\tlight}{\columncolor{SkyBlue!30}\color{black}}     \newcolumntype{C}[1]{&gt;{##1}c} }{}  \newcommand{\tablegoals}{ \begin{tablecolor} \begin{tabular}{l|C{\tlight}} Goal 7  &amp; \\ \end{tabular} \end{tablecolor} }  \begin{document}     \tablegoals \end{document}     Here's my second attempt. Different approach, same undefined control sequence error with \columncolor.  \documentclass{article} \usepackage{array} \usepackage[table,dvipsnames]{xcolor} \usepackage{colortbl} \usepackage{ifthen}  \newenvironment{tablecolor}{     \newcolumntype{C}[1]{&gt;{         \ifthenelse{\equal{dark}{##1}}{             \columncolor{Cyan}\color{white}             }{             \columncolor{SkyBlue!30}\color{black}         }     }c} }{ }  \newcommand{\tablegoals}[1]{     \begin{tablecolor}         \begin{tabular}{l|C{dark}}             Goal 7  &amp;   #1         \end{tabular}     \end{tablecolor} }  \begin{document}     \tablegoals{Check mark} \end{document} What happened to Control.InvokeRequired in WPF? How do you clone a Git repository into a specific folder? Executing the command git clone git@github.com:whatever creates a directory in my current folder named whatever, and drops the contents of the Git repository into that folder:  /httpdocs/whatever/public   My problem is that I need the contents of the Git repository cloned into my current directory so that they appear in the proper location for the web server:  /httpdocs/public   I know how to move the files after I've cloned the repository, but this seems to break Git, and I'd like to be able to update just by calling git pull. How can I do this? I ran dolphin like this:  dolphin . &amp;   I see messages in the console like these:  QPixmap::scaled: Pixmap is a null pixmap   How can I keep the console clean or silent? Show $1/(1+ x^2)$ is uniformly continuous on $\Bbb R$.