axis

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: This question in regarding formatting plots produced using ggplot2 + ggExtra and is not related to any bug. require(ggplot2) #> Loading required package: ggplot2 require(ggExtra) #> Loading required package: ggExtra p1 <- ggplot(data = mpg,aes(x = cty,y = cty)) + geom_point()+ xlab("City driving (miles/gallon)") + ylab("City driving (miles/gallon)") ggMarginal(p = p1,type= "boxplot") The y-axis marginal plot in this chart is usually not similar to the x-axis marginal plot i.e. the width of the 2 boxplots are not similar. This problem become

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: I have two vector pairs (before and after rotation). before rotation: [x1,y1,z1] [x2,y2,z2] after rotation: [x1',y1',z1'] [x2',y2',z2'] How to create a quaternion representing this rotation? 回答1: In most cases there is no rotation which transforms 2 vectors into 2 other vectors. Here is a simple way to visualize why: a rotation does not change the angle between vectors. If the angle between the 2 vectors before the rotation is different from the angle between the 2 vectors after the rotation, then there is no rotation which meets your

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: Is it possible to use numpy.nanargmin , so that it returns numpy.nan , on columns where there are only nans in them. Right now, it raises a ValueError , when that happens. And i cant use numpy.argmin , since that will fail when there are only a few nans in the column. http://docs.scipy.org/doc/numpy/reference/generated/numpy.nanargmin.html says that the ValueError is raised for all-nan slices. In that case, i want it to return numpy.nan (just to further mask the "non-data" with nans) this next bit does this, but is super-slow and not really

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: I create a web service and it works fine when the return is List like this: @WebService public class IndicatorWSImpl extends SpringBeanAutowiringSupport implements IndicatorWS { @Autowired private HomeIndicadoresReportService homeIndicadoresReportService; @Override public List<String> entidades() { homeIndicadoresReportService.buildValuesToGraph(resultMesAtual, selectedEntity, selectedBS, selectedTS, selectedFunctionality, filter, false); List<String> lista = new ArrayList<String>(); for (EntityIndicatorVO entityIndicatorVO : resultMesAtual)

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: Instead of the default "boxed" axis style I want to have only the left and bottom axis, i.e.: +------+ | | | | | | ---> | | | | +------+ +------- This should be easy, but I can't find the necessary options in the docs. 回答1: This is the suggested Matplotlib 2.0 solution from the official website HERE : import numpy as np import matplotlib.pyplot as plt x = np.linspace(0, 2*np.pi, 100) y = np.sin(x) ax = plt.subplot(111) ax.plot(x, y) # Hide the right and top spines ax.spines['right'].set_visible(False) ax.spines['top'].set_visible(False) #

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: I am plotting number of residents against the dorm room numbers (4 digits). The room numbers are supposed to be strings. But when I used as.character(RmNum), the axis still shows as numeric. meanResidents = c(3, 4, 3, 2, 4, 5) rmNumber = c(2034, 3043, 4012, 2035, 2022, 3013) plot(as.character(rmNumber), meanResidents, xlab = as.character(rmNumber)) I would want the dorm numbers showing vertically in the axis. Can someone help me with that? 回答1: With the function axis you can specify the position of the axis, where to put the tick marks ( at

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: Is there a way to calculate many histograms along an axis of an nD-array? The method I currently have uses a for loop to iterate over all other axes and calculate a numpy.histogram() for each resulting 1D array: import numpy import itertools data = numpy.random.rand(4, 5, 6) # axis=-1, place `200001` and `[slice(None)]` on any other position to process along other axes out = numpy.zeros((4, 5, 200001), dtype="int64") indices = [ numpy.arange(4), numpy.arange(5), [slice(None)] ] # Iterate over all axes, calculate histogram for each cell for

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: Using iPython Notebook, I have been able to bring up a globe of the Earth with code like: from mayavi import mlab from mayavi.sources.builtin_surface import BuiltinSurface ocean_blue = (0.4, 0.5, 1.0) r = 6371 # km sphere = mlab.points3d(0, 0, 0, name='Globe', scale_mode='none', scale_factor=r * 2.0, color=ocean_blue, resolution=50) sphere.actor.property.specular = 0.20 sphere.actor.property.specular_power = 10 continents_src = BuiltinSurface(source='earth', name='Continents') continents_src.data_source.on_ratio = 1 # detail level continents

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: I have a 2D matrix and I want to take norm of each row. But when I use numpy.linalg.norm(X) directly, it takes the norm of the whole matrix. I can take norm of each row by using a for loop and then taking norm of each X[i] , but it takes a huge time since I have 30k rows. Any suggestions to find a quicker way? Or is it possible to apply np.linalg.norm to each row of a matrix? 回答1: Note that, as perimosocordiae shows , as of NumPy version 1.9, np.linalg.norm(x, axis=1) is the fastest way to compute the L2-norm. If you are computing an L2-norm