polar-coordinates

问题 I am trying to create a circular stacked barplot as mentioned here (https://www.r-graph-gallery.com/299-circular-stacked-barplot.html). And I get the following error when when I get to the step of making the plot (in bold below): Error: Aesthetics must be either length 1 or the same as the data (26): hjust Run rlang::last_error() to see where the error occurred. In addition: Warning message: Removed 208 rows containing missing values (position_stack). This is what my data looks like (with 5

问题 Log polar transform is commonly used in image with phase correlation using Fourier transform for estimating rotation and translation etc. However, I am little confusing how to apply it in audio signal. I am trying to estimate time shift between two audio signal through log polar transform (LPT) and phase correlation. I applied LPT using https://en.wikipedia.org/wiki/Log-polar_coordinates in audio signal and plotter using matlab with polar(theta,rho), until here, there is no problem, problem

问题 I need a phase portrait of the following nonlinear system given in polar form... \dot{r} = 0.5*(r - r^3) \dot{\theta} = 1 I know how to do it in Mathematica... field1 = {0.5*(r - r^3), 1}; p1 = StreamPlot[Evaluate@TransformedField["Polar" -> "Cartesian", field1, {r, \[Theta]} -> {x, y}], {x, -3, 3}, {y, -3, 3}, Axes -> True, StreamStyle -> Gray, ImageSize -> Large]; Show[p1, AxesLabel->{x,y}, ImageSize -> Large] How can I do the same using pyplot.quiver in Python? 回答1: Just very naive

问题 I am still new to ggplot2. I want to plot my data from my insect poo experiment: x-value is time of day (24h) and y value is amount of poo (frass) produced as a percentage and represented as a circular plot. I added geom_ribbon to represent standard deviation which I calculated with circular. Firstly, I had issues with the 'clock' starting as 1/24 instead of 0/24: ggplot2 clock starting from 1/24 instead of 0/24: So I added the code expand_limits(x = 0, y = 0) which helped with fixing 1/24 to

问题 I want to add another legend that tells me what ring of a circular heat map represents (from outer ring to inner ring). I tried the following from another answer previously: library(reshape) library(ggplot2) library(plyr) nba <- read.csv("http://datasets.flowingdata.com/ppg2008.csv") nba$Name <- with(nba, reorder(Name, PTS)) nba.m <- melt(nba) nba.m <- ddply(nba.m, .(variable), transform, value = scale(value)) # Convert the factor levels (variables) to numeric + quanity to determine size of

问题 I want to add another legend that tells me what ring of a circular heat map represents (from outer ring to inner ring). I tried the following from another answer previously: library(reshape) library(ggplot2) library(plyr) nba <- read.csv("http://datasets.flowingdata.com/ppg2008.csv") nba$Name <- with(nba, reorder(Name, PTS)) nba.m <- melt(nba) nba.m <- ddply(nba.m, .(variable), transform, value = scale(value)) # Convert the factor levels (variables) to numeric + quanity to determine size of

问题 I wish to add a wedge outlining a group of polar data using Python's Matplotlib. I have tried using the Wedge patch artist unsuccessfully, due to unknown reasons. I wish to either understand these unknowns, or find an alternative to the patch artist approach. The primary issue is that the Wedge patch is not displaying as I am expecting it to. Given my code, I am expecting it to be oriented at an angle, and span a range of ~0.05 in radius. This places it within the sector plot here: 1 But this

问题 i have an assignment on sympy and am struggling with the following question: "Prove with the help of Sympy that 4*(x 2 + y 2 -ax) 3 = 27 a 2 (x 2 +y 2 ) 2 can be written using r = 4a*cos(theta/3) 3 ". I have tried to substitute x = r*cos(theta) and y = r*sin(theta) . Then I tried sp.solveset(eq, r) but I only got a very longset of {}, nothing like the given polar equation. Does anyone know how to do this (I can use sympy and numpy)? 回答1: The following code builds the equation from its left

问题 i have an assignment on sympy and am struggling with the following question: "Prove with the help of Sympy that 4*(x 2 + y 2 -ax) 3 = 27 a 2 (x 2 +y 2 ) 2 can be written using r = 4a*cos(theta/3) 3 ". I have tried to substitute x = r*cos(theta) and y = r*sin(theta) . Then I tried sp.solveset(eq, r) but I only got a very longset of {}, nothing like the given polar equation. Does anyone know how to do this (I can use sympy and numpy)? 回答1: The following code builds the equation from its left

问题 I've been looking for a solution to convert cartesian coordinates (lat, long) that I have to polar coordinates in order to facilitate a simulation that I want to run, but I haven't found any questions or answers here for doing this in R. There are a number of options, including the built in function cart2pol in Matlab, but all of my data are in R and I'd like to continue getting comfortable working in this framework. Question: I have lat/long coordinates from tagging data, and I want to