英文:
Using scipy.interpolate in python, I want to erase certain values and interpolate them
问题
在我的代码中，我有一些数值，如下所示：
my_list = [725.998474, 0.0, 0.0, 0.0, 0.0, 789.507934, 792.585388, 801.612916, 799.38916, 809.280518, 809.186036, 811.899414, .... , 412.314528]
在我的代码中，我想要对列表中的0.0值进行插值，因为它们是异常值。但由于插值仅适用于空值，所以它无法正常工作。
我该如何删除这些0.0值并进行插值处理？
英文:
In my code, I have certain values like
my_list = [725.998474, 0.0, 0.0, 0.0, 0.0, 789.507934, 792.585388, 801.612916, 799.38916, 809.280518, 809.186036, 811.899414, .... , 412.314528]
In my code, I want to interpolate the points where the list is 0.0 because they are outliers. But it is not working as interpolation only works for empty values.
How can I erase those 0.0 values and do interpolation?
答案1
得分: 5
使用您提供的数据，我们可以使用numpy创建一个“清理后”的x和y数据列表。您说数值等于0的值是异常值，但是使用浮点数进行相等性检查可能会导致问题，所以我使用了np.isclose。去除异常值后，您可以对清理后的数据进行插值。
import numpy as npfrom scipy.interpolate import make_interp_splineimport matplotlib.pyplot as pltplt.close("all")y = np.array([725.998474, 0.0, 0.0, 0.0, 0.0, 789.507934, 792.585388,801.612916, 799.38916, 809.280518, 809.186036, 811.899414,412.314528])x = np.arange(len(y))outliers = np.isclose(y, 0)y_clean = y[~outliers]x_clean = x[~outliers]spline = make_interp_spline(x_clean, y_clean)y_interped = spline(x)fig, ax = plt.subplots()ax.plot(x_clean, y_clean, ".", label="cleaned", zorder=3)ax.plot(x, y_interped, label="interpolated")ax.legend()ax.set_xlabel("x")ax.set_ylabel("y")fig.show()
如果像@Reinderien建议的那样，您的实际条件是值小于100，例如，被视为异常值，那么您可以将条件更改为outliers = y < 100。
英文:
Using the data you provided, we can create a "cleaned" list of the x and y data using numpy. You said that the values equal to 0 are the outliers, but checking equality with floating point numbers can lead to issues, so I used np.isclose. With the outliers removed, you can interpolate the cleaned data.
import numpy as npfrom scipy.interpolate import make_interp_splineimport matplotlib.pyplot as pltplt.close("all")y = np.array([725.998474, 0.0, 0.0, 0.0, 0.0, 789.507934, 792.585388,801.612916, 799.38916, 809.280518, 809.186036, 811.899414,412.314528])x = np.arange(len(y))outliers = np.isclose(y, 0)y_clean = y[~outliers]x_clean = x[~outliers]spline = make_interp_spline(x_clean, y_clean)y_interped = spline(x)fig, ax = plt.subplots()ax.plot(x_clean, y_clean, ".", label="cleaned", zorder=3)ax.plot(x, y_interped, label="interpolated")ax.legend()ax.set_xlabel("x")ax.set_ylabel("y")fig.show()
If, as @Reinderien suggested, your actual condition is that values below 100, for example, are outliers, then you could change it to that condition (i.e. outliers = y < 100).
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