Find groups of Y values for unique X values, numpy structured array?-CodePudding

I have a numpy structured array:

import numpy as np
arr1 = (np.array([32, 32, 32, 32, 32, 39, 21], dtype=np.int64),np.array([449, 451, 452, 453, 454, 463, 340], dtype=np.int64))

arr1_x = arr1[0]
arr1_y = arr1[1]

arr1_struct = np.empty(arr1_x.shape[0], dtype=[('x', int), ('y', int)])

arr1_struct["x"] = arr1_x
arr1_struct["y"] = arr1_y

I would need to determine the groups of y values that correspond to unique x values.

Using the above example, I would need to know that:

Where x=32, y=449, 451, 452, 453, 454
Where x=39, y=463
Where x=21, y=340

I was looking into np.unique but that didn't give exactly what I'm after.

Any output format would be fine that provides this information. An example would be

x_unique = [32,39,21]
y_unique = [ [449,451,452,453,454] , [463] , [340] ]

Where the X value at index 0 in the x_unique array/list has the Y values at index 0 in the y_unique array/list, and so on so forth.

Is there an easy and efficient way to do this in numpy?

CodePudding user response：

Sounds like you need to be using pandas and work with a dataframe-like structure.

df = pd.DataFrame([*arr1], index=['x', 'y']).T

which gives

Then,

>>> df.groupby('x')['y'].unique().to_dict()
{21: array([340]), 32: array([449, 451, 452, 453, 454]), 39: array([463])}

Without pandas, you can refer to this thread for analogous ways to perform a groupby with numpy or pure python.