[英]Python - how can i find the angle of a line at a given point?
我正在处理简单的 OHLC 时间序列数据，这是一个示例：
2021-02-26 08:00:00 51491.322786
2021-02-26 12:00:00 51373.462137
2021-02-26 16:00:00 51244.591670
2021-02-26 20:00:00 51061.134204
2021-02-27 00:00:00 50985.592434
2021-02-27 04:00:00 50923.287370
2021-02-27 08:00:00 50842.103282
2021-02-27 12:00:00 50695.160604
2021-02-27 16:00:00 50608.462150
2021-02-27 20:00:00 50455.235146
2021-02-28 00:00:00 50177.377531
2021-02-28 04:00:00 49936.652091
2021-02-28 08:00:00 49860.396537
2021-02-28 12:00:00 49651.901082
2021-02-28 16:00:00 49625.153441
2021-02-28 20:00:00 49570.275193
2021-03-01 00:00:00 49531.874272
2021-03-01 04:00:00 49510.381676
2021-03-01 08:00:00 49486.289712
2021-03-01 12:00:00 49481.496645
2021-03-01 16:00:00 49469.806692
2021-03-01 20:00:00 49471.958606
2021-03-02 00:00:00 49462.095568
2021-03-02 04:00:00 49453.473575
2021-03-02 08:00:00 49438.986536
2021-03-02 12:00:00 49409.492007
2021-03-02 16:00:00 49356.563396
2021-03-02 20:00:00 49331.037118
2021-03-03 00:00:00 49297.823947
2021-03-03 04:00:00 49322.049974
2021-03-03 08:00:00 49461.314013
2021-03-03 12:00:00 49515.137712
2021-03-03 16:00:00 49571.990877
2021-03-03 20:00:00 49592.320461
2021-03-04 00:00:00 49592.249409
2021-03-04 04:00:00 49593.938380
2021-03-04 08:00:00 49593.055971
2021-03-04 12:00:00 49592.025698
2021-03-04 16:00:00 49585.661437
2021-03-04 20:00:00 49578.693824
2021-03-05 00:00:00 49543.067346
2021-03-05 04:00:00 49540.706794
2021-03-05 08:00:00 49513.586831
2021-03-05 12:00:00 49494.990328
2021-03-05 16:00:00 49493.807248
2021-03-05 20:00:00 49461.133698
2021-03-06 00:00:00 49432.770930
2021-03-06 04:00:00 49412.087821
2021-03-06 08:00:00 49368.106499
2021-03-06 12:00:00 49290.581114
2021-03-06 16:00:00 49272.222740
2021-03-06 20:00:00 49269.814982
2021-03-07 00:00:00 49270.328825
2021-03-07 04:00:00 49293.664209
2021-03-07 08:00:00 49339.999430
2021-03-07 12:00:00 49404.798067
2021-03-07 16:00:00 49450.447631
2021-03-07 20:00:00 49528.402294
2021-03-08 00:00:00 49571.353158
2021-03-08 04:00:00 49572.687451
2021-03-08 08:00:00 49597.518988
2021-03-08 12:00:00 49648.407014
2021-03-08 16:00:00 49708.063384
2021-03-08 20:00:00 49862.237773
2021-03-09 00:00:00 50200.833030
2021-03-09 04:00:00 50446.201489
2021-03-09 08:00:00 50727.063301
2021-03-09 12:00:00 50952.697141
2021-03-09 16:00:00 51152.798741
2021-03-09 20:00:00 51392.873289
2021-03-10 00:00:00 51472.273233
2021-03-10 04:00:00 51601.351944
2021-03-10 08:00:00 51759.387477
2021-03-10 12:00:00 52053.982892
2021-03-10 16:00:00 52437.071119
2021-03-10 20:00:00 52648.225156
我试图找到一种方法来获得线在每个点的倾斜或陡峭程度。 基本上我只需要知道这条线是向上、向下还是横向以及上升了多少，所以理想的情况是得到某种系数或数字来告诉我这条线有多陡。
为了做到这一点，我有了计算斜率的想法，所以我尝试了从这里得到的以下代码：
def slope( close, length=None, as_angle=None, to_degrees=None, vertical=None, offset=None, **kwargs):
"""Indicator: Slope"""
# Validate arguments
length = int(length) if length and length > 0 else 1
as_angle = True if isinstance(as_angle, bool) else False
to_degrees = True if isinstance(to_degrees, bool) else False
close = verify_series(close, length)
offset = get_offset(offset)
if close is None: return
# Calculate Result
slope = close.diff(length) / length
if as_angle:
slope = slope.apply(npAtan)
if to_degrees:
slope *= 180 / npPi
# Offset
if offset != 0:
slope = slope.shift(offset)
# Handle fills
if "fillna" in kwargs:
slope.fillna(kwargs["fillna"], inplace=True)
if "fill_method" in kwargs:
slope.fillna(method=kwargs["fill_method"], inplace=True)
# Name and Categorize it
slope.name = f"SLOPE_{length}" if not as_angle else f"ANGLE{'d' if to_degrees else 'r'}_{length}"
slope.category = "momentum"
return slope
这是 output 的示例：
2021-02-26 08:00:00 51491.322786 -110.850644
2021-02-26 12:00:00 51373.462137 -117.860648
2021-02-26 16:00:00 51244.591670 -128.870468
2021-02-26 20:00:00 51061.134204 -183.457466
2021-02-27 00:00:00 50985.592434 -75.541770
2021-02-27 04:00:00 50923.287370 -62.305064
2021-02-27 08:00:00 50842.103282 -81.184088
2021-02-27 12:00:00 50695.160604 -146.942678
2021-02-27 16:00:00 50608.462150 -86.698454
2021-02-27 20:00:00 50455.235146 -153.227004
2021-02-28 00:00:00 50177.377531 -277.857615
2021-02-28 04:00:00 49936.652091 -240.725440
2021-02-28 08:00:00 49860.396537 -76.255553
2021-02-28 12:00:00 49651.901082 -208.495455
2021-02-28 16:00:00 49625.153441 -26.747641
2021-02-28 20:00:00 49570.275193 -54.878249
2021-03-01 00:00:00 49531.874272 -38.400921
2021-03-01 04:00:00 49510.381676 -21.492596
2021-03-01 08:00:00 49486.289712 -24.091964
2021-03-01 12:00:00 49481.496645 -4.793067
2021-03-01 16:00:00 49469.806692 -11.689953
2021-03-01 20:00:00 49471.958606 2.151914
2021-03-02 00:00:00 49462.095568 -9.863038
2021-03-02 04:00:00 49453.473575 -8.621994
2021-03-02 08:00:00 49438.986536 -14.487039
2021-03-02 12:00:00 49409.492007 -29.494528
2021-03-02 16:00:00 49356.563396 -52.928611
2021-03-02 20:00:00 49331.037118 -25.526278
2021-03-03 00:00:00 49297.823947 -33.213171
2021-03-03 04:00:00 49322.049974 24.226027
2021-03-03 08:00:00 49461.314013 139.264040
2021-03-03 12:00:00 49515.137712 53.823699
2021-03-03 16:00:00 49571.990877 56.853165
2021-03-03 20:00:00 49592.320461 20.329584
2021-03-04 00:00:00 49592.249409 -0.071052
2021-03-04 04:00:00 49593.938380 1.688971
2021-03-04 08:00:00 49593.055971 -0.882409
2021-03-04 12:00:00 49592.025698 -1.030273
2021-03-04 16:00:00 49585.661437 -6.364260
2021-03-04 20:00:00 49578.693824 -6.967614
2021-03-05 00:00:00 49543.067346 -35.626478
2021-03-05 04:00:00 49540.706794 -2.360551
2021-03-05 08:00:00 49513.586831 -27.119963
2021-03-05 12:00:00 49494.990328 -18.596504
2021-03-05 16:00:00 49493.807248 -1.183080
2021-03-05 20:00:00 49461.133698 -32.673550
2021-03-06 00:00:00 49432.770930 -28.362769
2021-03-06 04:00:00 49412.087821 -20.683109
2021-03-06 08:00:00 49368.106499 -43.981322
2021-03-06 12:00:00 49290.581114 -77.525385
2021-03-06 16:00:00 49272.222740 -18.358373
2021-03-06 20:00:00 49269.814982 -2.407758
2021-03-07 00:00:00 49270.328825 0.513843
2021-03-07 04:00:00 49293.664209 23.335384
2021-03-07 08:00:00 49339.999430 46.335221
2021-03-07 12:00:00 49404.798067 64.798637
2021-03-07 16:00:00 49450.447631 45.649564
2021-03-07 20:00:00 49528.402294 77.954663
2021-03-08 00:00:00 49571.353158 42.950863
2021-03-08 04:00:00 49572.687451 1.334294
2021-03-08 08:00:00 49597.518988 24.831537
2021-03-08 12:00:00 49648.407014 50.888026
2021-03-08 16:00:00 49708.063384 59.656369
2021-03-08 20:00:00 49862.237773 154.174389
2021-03-09 00:00:00 50200.833030 338.595257
2021-03-09 04:00:00 50446.201489 245.368460
2021-03-09 08:00:00 50727.063301 280.861811
2021-03-09 12:00:00 50952.697141 225.633840
2021-03-09 16:00:00 51152.798741 200.101599
2021-03-09 20:00:00 51392.873289 240.074549
2021-03-10 00:00:00 51472.273233 79.399943
2021-03-10 04:00:00 51601.351944 129.078712
2021-03-10 08:00:00 51759.387477 158.035533
2021-03-10 12:00:00 52053.982892 294.595415
2021-03-10 16:00:00 52437.071119 383.088226
2021-03-10 20:00:00 52648.225156 211.154038
这行得通，但问题是斜率的结果很大程度上取决于我提供的数据的大小，这意味着价格越低，我将获得更低的斜率值，更高的值斜率值越高，但由于我正在执行某种分析，因此我需要一些更“通用”的东西，它可以让我知道我正在绘制的线的倾斜度，而不取决于我正在使用的数据的大小。 可能吗？ 任何形式的建议表示赞赏。
我不确定您要达到什么目标，但是可以通过以下方式找到一系列点的斜率和角度。
假设您的 dataframe 由下式给出：
Date measure
0 2021-02-26 08:00 51491.322786
1 2021-02-26 12:00 51373.462137
2 2021-02-26 16:00 51244.591670
3 2021-02-26 20:00 51061.134204
4 2021-02-27 00:00 50985.592434
.. ... ...
71 2021-03-10 04:00 51601.351944
72 2021-03-10 08:00 51759.387477
73 2021-03-10 12:00 52053.982892
74 2021-03-10 16:00 52437.071119
75 2021-03-10 20:00 52648.225156
这正是您发布的内容。 然后，您可以将 function slope_and_angle定义为
y = range(len(df['measure'])) ##Make sure you get the range of values
def slope_and_angle(df):
for i in y:
df['slope'] = (y[i-1] - y[1]) / (df['measure'].diff())
df['angle'] = np.rad2deg(np.arctan2(y[i-1] - y[1], df['measure'].diff()))
return df
返回：
Date measure slope angle
0 2021-02-26 08:00 51491.322786 NaN NaN
1 2021-02-26 12:00 51373.462137 -0.619376 148.226940
2 2021-02-26 16:00 51244.591670 -0.566460 150.470170
3 2021-02-26 20:00 51061.134204 -0.397912 158.301777
4 2021-02-27 00:00 50985.592434 -0.966353 135.980320
.. ... ... ... ...
71 2021-03-10 04:00 51601.351944 0.565546 29.490174
72 2021-03-10 08:00 51759.387477 0.461921 24.793227
73 2021-03-10 12:00 52053.982892 0.247797 13.917410
74 2021-03-10 16:00 52437.071119 0.190557 10.788745
75 2021-03-10 20:00 52648.225156 0.345719 19.071249
您在输出示例中返回的只是df['measure'].diff() 。
有些事情你可以做，但没有通用的事情总是能很好地工作——你需要了解你的数据并选择适合你情况的东西。
例如，100 个 0 到 1 之间的随机数，每 4 小时采样一次
import numpy as np
import pandas as pd
df = pd.DataFrame({
'timestamp': pd.date_range("2021-01-01", periods=100, freq="4H"),
'value': np.random.random(100)
})
# df:
# timestamp value
# 0 2021-01-01 00:00:00 0.780008
# 1 2021-01-01 04:00:00 0.689576
# 2 2021-01-01 08:00:00 0.700937
# 3 2021-01-01 12:00:00 0.756724
# 4 2021-01-01 16:00:00 0.928890
# etc
我们可以很容易地计算梯度：
differences = df.diff()
gradient = 3600 * differences.value / differences.timestamp.dt.seconds
# gradient: max value 0.1979, min value -0.2432
# 0 NaN
# 1 0.033912
# 2 0.045422
# 3 -0.001827
# 4 -0.225796
梯度是每小时值的变化，忽略任何皱纹，如缺失值、重复时间点等。
现在，正如您所观察到的，如果这些数字的大小增加，梯度就会增加。 例如，如果我将value放大 100 倍：
df['value100'] = 100 * df.value
differences = df.diff()
gradient = 3600 * differences.value100 / differences.timestamp.dt.seconds
print(gradient.max(), gradient.min())
# gradient: max value 19.79, min value -24.32
# 0 NaN
# 1 3.391221
# 2 4.542248
# 3 -0.182714
# 4 -22.579588
在这里，我们看到梯度也大了 100 倍——正如预期的那样。
这表明我们可以只除以某个数字，但问题就变成了使用什么数字？ 这就是理解数据很重要的地方。
一种方法是使用数据的范围。 这类似于您使用matplotlib绘制图表时所看到的 - y比例将适合最大值和最小值。 例如：
sf = df.value.max() - df.value.min()
sf100 = df.value100.max() - df.value100.min()
differences = df.diff()
gradient = differences.value / sf
gradient100 = differences.value100 / sf100
# gradient, gradient100
# nan, nan
# 0.1379, 0.1379
# 0.1847, 0.1847
# -0.0074, -0.0074
# -0.9184, -0.9184
如您所见，两个渐变现在相互匹配。 当存在简单的线性缩放时，这种方法效果很好。
但是，请考虑另一种情况 - 由于异常值而产生额外范围的情况。
df['value_outlier'] = df.value
df.loc[50, 'value_outlier'] = 100 # Just set the 50th value to 100
sf = df.value.max() - df.value.min()
sf_outlier = df.value_outlier.max() - df.value_outlier.min()
differences = df.diff()
gradient = differences.value / sf
gradient_outlier = differences.value_outlier / sf_outlier
# gradient, gradient_outlier
# nan, nan
# 0.1379, 0.0014
# 0.1847, 0.0018
# -0.0074, -0.0001
# -0.9184, -0.0090
# 0.5792, 0.0057
这看起来不太好。 原因是我们在没有改变大多数点之间的实际范围的情况下夸大了value_outlier的范围。
您可以解决此问题 - 一种方法是使用四分位数范围作为比例因子：
sf = df.value.quantile(0.75) - df.value.quantile(0.25)
sf100 = df.value100.quantile(0.75) - df.value100.quantile(0.25)
sf_outlier = df.value_outlier.quantile(0.75) - df.value_outlier.quantile(0.25)
differences = df.diff()
gradient = differences.value / sf
gradient100 = differences.value100 / sf100
gradient_outlier = differences.value_outlier / sf_outlier
for a, b, c in zip(gradient, gradient100, gradient_outlier):
print(f'{a:.4f}, {b:.4f}, {c:.4f}')
# gradient, gradient100, gradient_outlier
# nan, nan, nan
# 0.2953, 0.2953, 0.3202
# 0.3956, 0.3956, 0.4289
# -0.0159, -0.0159, -0.0173
# -1.9665, -1.9665, -2.1320
# 1.2403, 1.2403, 1.3447
这些值永远不会完美匹配，但它们应该大致相同。 而且，当然，离群值所在的位置会有很大的不同。
因此，关键信息是您可以做某事，但您需要确保它适合您的数据。
如何在python中给定点（x，y）和线与y轴的夹角的情况下找到线的斜率（m）？
[英]How do I find the slope (m) for a line given a point (x,y) on the line and the line's angle from the y axis in python?
如何找到一个点的坐标，使其与给定的线成 135 度角？
[英]How to find the coordinates of a point such that it makes a 135 degree angle with a given line?
给定圆上的 2 个点和它们之间的角度，如何找到中心（在 python 中）？
[英]Given 2 points on a circle and the angle between, how to find the center (in python)?
Python：使用到点 A 的距离（x0，y0）和角度找到给定点 B 的 ax,y 坐标
[英]Python: find a x,y coordinate for a given point B using the distance from the point A (x0, y0) and the angle
如何在给定线的另一个点和垂直线上的2个点的线上找到点的X坐标？
[英]How to find the X coordinate of a point on a line given another point of the line and the 2 points on a perpendicular line?
声明:本站的技术帖子网页，遵循CC BY-SA 4.0协议，如果您需要转载，请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.