我有一个我认为是圆锥截面的椭圆形分布。我想旋转这些点，使分布变成圆形，就好像我在直接看圆锥顶部的圆锥截面一样。
这是一些示例数据(使用函数 here 生成)
X_df <-
structure(list(x = c(550.685479223573, 411.808342674744, 125.337513241526,
-46.6813176776531, 54.1090479024869, 335.045593380922, 538.806846993829,
476.123346783785, 207.359201714354, -23.3704356149293, -1.06902389582398,
252.471032092766, 502.461757269191, 522.09464005165, 290.954504794153,
22.4116013413886, -37.4399705577234, 166.122770874069, 446.874779008936,
547.271442128866, 372.271299246978, 84.7905677157295, -50.074206596271,
90.757431249567, 378.201298931732, 547.145608993239, 443.947162007208,
161.476775837252, -38.5517112166543, 25.2627436553199, 296.503160027896,
524.775126009974, 500.784559912938, 245.712512645379, -4.31860487373257,
-21.9661658669887, 211.218663607589, 479.198761786515, 535.314989389215,
330.010941011427, 50.0215808044216, -46.3823119064223, 130.383487121344,
416.170638042649, 549.268852072098, 407.813005658263, 119.940919946473,
-46.978590114418, 57.7409750579334, 340.505587064792, 539.650771180236,
472.254339573695, 201.890629521184, -25.163386210777, 1.25193046435474,
256.776302252232, 506.305676724803, 520.004964534048, 284.257495593069,
18.8183745840118, -35.9075114459174, 172.662124500953, 452.343060560759,
546.468842888411, 367.193523099128, 81.9151159445705, -49.726312730029,
95.1883131124973, 382.503271032958, 548.377552998115, 439.474201456606,
157.248088356873, -39.8634174011649, 28.665422852919, 301.243788141946,
526.815879166266, 497.683488701185, 240.939374274905, -7.78381612220116,
-19.3411744866129, 217.640180353188, 483.134755325255, 534.947529479343,
324.801587123232, 45.3957868762181, -45.0069945691924, 134.781896592204,
420.833721926428, 550.278658272823, 403.464000037755, 116.273973349216,
-48.5483252399878, 62.3918399072614, 345.924165684106, 540.282415561272,
468.621672005007, 195.304995872248, -28.2738679786754, 4.25351768918281,
262.272866287766, 509.296144374104), y = c(150.522375543584,
317.792592638159, 332.783726315973, 177.890614907595, -1.30774215535761,
-41.9959828735621, 94.0557252742373, 281.491416261009, 347.931229803675,
232.411150772918, 41.2498141860971, -50.240758928064, 42.4333078345691,
233.202857825371, 347.930304275537, 279.801242902484, 93.7702821671593,
-42.9605564915062, -0.128821245055916, 179.228629398932, 331.853832274807,
317.39347890031, 147.731418574477, -19.2449921421865, -31.3298018420377,
123.222814177221, 301.499351173211, 340.662811705721, 204.270915882133,
17.5570334546183, -47.8527634953491, 69.1925023774197, 260.846781070028,
350.702299892942, 255.267980339802, 64.3048063206447, -47.7550214881633,
21.0575085383169, 209.247480999408, 342.716023607699, 298.578792586917,
118.053124236616, -33.3190320226709, -16.5618829502486, 154.097078032767,
319.940208485997, 330.141085013461, 175.063055837321, -4.17911647131882,
-40.77087978947, 96.7492568480405, 284.629419943293, 347.18869555741,
228.951107374031, 37.6230640656836, -49.9549156886126, 45.2896584936418,
236.96620488459, 349.2183672397, 277.259838492706, 88.5019845874813,
-43.1419505604449, 1.98249146234145, 183.766146834555, 334.721418603224,
314.869389642466, 144.917094343997, -21.0615467657252, -29.137753346726,
126.928148173889, 305.196445845808, 339.338402720207, 201.609402013064,
15.0779976117978, -47.1046400880924, 71.7191693530443, 263.543031729657,
350.145628648054, 252.792513384296, 61.2531942049295, -49.0697852698499,
24.2068045114243, 212.793083656477, 343.533262533366, 295.969945687212,
115.135250419503, -34.4145910896749, -14.7817206225652, 156.282366465729,
322.116360452784, 328.626788731125, 171.323808201914, -6.54021515590322,
-40.1360134761796, 101.492490333309, 286.854230399582, 347.010792855229,
226.829162028581, 35.3880363162166, -50.8418314561365, 48.893760376765
)), .Names = c("x", "y"), row.names = c(NA, -101L), class = "data.frame")
绘制椭圆，这类似于我的实际数据:
ggplot(X_df, aes(x, y)) +
geom_point() +
coord_equal()
我想对 xy 坐标应用一个函数以围绕其长轴旋转椭圆，将椭圆的顶部移向我，将椭圆的底部移离我，以获得类似这样的结果(具有相同的维度像上面的椭圆一样沿 x 轴，但沿 y 轴的维度不同):
我想旋转我的数据，就像这个椭圆被旋转以产生圆形，在那里我看到形状顶部和底部之间的最大距离(即垂直于长轴的最长可能轴)，作为形状绕着它的长轴旋转。
我该怎么做？
最佳答案
您的意思似乎是，如果我从 x=-1000，y=0 处查看“圆”边，我会看到一条逆时针旋转的线(超出 xy 平面)远离y轴。目标是将圆旋转回 y 轴。
旋转角度为 acos(1/ratio)(在本例中为 0.839 弧度或 48.06 度)，其中 ratio 为 diff(range(X_df$ x))/diff(range(X_df$y))(假设旋转轴在 xy 平面内并平行于 x 轴，并且您的数据包括 x 轴两侧的点圆圈)。
实际上，要将圆旋转回 xy 平面，您只需将 y 点乘以 ratio，然后为了保持相同的中心，减去 (ratio - 1) * mean(X_df$y)(我假设数据点均匀分布在圆周上)。
换句话说(或者实际上是代码):
ratio = diff(range(X_df$x))/diff(range(X_df$y))
X_df$ynew = ratio * X_df$y - (ratio - 1) * mean(X_df$y)
ggplot(X_df, aes(x, ynew)) +
geom_point() +
coord_equal()
比较原始与“旋转”:
关于旋转椭圆的标绘点得到一个圆，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/37193380/