Numpy实践_集合操作

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一、构造集合

找出数组中的唯一值并返回已排序的结果：

import numpy as np

x = np.unique([1, 1, 3, 2, 3, 3])
print(x)  # [1 2 3]

x = sorted(set([1, 1, 3, 2, 3, 3]))
print(x)  # [1, 2, 3]

x = np.array([[1, 1], [2, 3]])
u = np.unique(x)
print(u)  # [1 2 3]

x = np.array([[1, 0, 0], [1, 0, 0], [2, 3, 4]])
y = np.unique(x, axis=0)
print(y)
# [[1 0 0]
#  [2 3 4]]

x = np.array(['a', 'b', 'b', 'c', 'a'])
u, index = np.unique(x, return_index=True)
print(u)  # ['a' 'b' 'c']
print(index)  # [0 1 3]
print(x[index])  # ['a' 'b' 'c']

x = np.array([1, 2, 6, 4, 2, 3, 2])
u, index = np.unique(x, return_inverse=True)
print(u)  # [1 2 3 4 6]
print(index)  # [0 1 4 3 1 2 1]
print(u[index])  # [1 2 6 4 2 3 2]

u, count = np.unique(x, return_counts=True)
print(u)  # [1 2 3 4 6]
print(count)  # [1 3 1 1 1]

二、布尔运算

前面的数组是否包含于后面的数组，返回布尔值。返回的值是针对第一个参数的数组的，所以维数和第一个参数一致，布尔值与数组的元素位置也一一对应

import numpy as np

test = np.array([0, 1, 2, 5, 0])
states = [0, 2]
mask = np.in1d(test, states)
print(mask)  # [ True False  True False  True]
print(test[mask])  # [0 2 0]

mask = np.in1d(test, states, invert=True)
print(mask)  # [False  True False  True False]
print(test[mask])  # [1 5]

三、求两个集合的交集

求两个数组的唯一化+求交集+排序函数

import numpy as np
from functools import reduce

x = np.intersect1d([1, 3, 4, 3], [3, 1, 2, 1])
print(x)  # [1 3]

x = np.array([1, 1, 2, 3, 4])
y = np.array([2, 1, 4, 6])
xy, x_ind, y_ind = np.intersect1d(x, y, return_indices=True)
print(x_ind)  # [0 2 4]
print(y_ind)  # [1 0 2]
print(xy)  # [1 2 4]
print(x[x_ind])  # [1 2 4]
print(y[y_ind])  # [1 2 4]

x = reduce(np.intersect1d, ([1, 3, 4, 3], [3, 1, 2, 1], [6, 3, 4, 2]))
print(x)  # [3]

四、求两个集合的并集

计算两个集合的并集，唯一化并排序

import numpy as np
from functools import reduce

x = np.union1d([-1, 0, 1], [-2, 0, 2])
print(x)  # [-2 -1  0  1  2]
x = reduce(np.union1d, ([1, 3, 4, 3], [3, 1, 2, 1], [6, 3, 4, 2]))
print(x)  # [1 2 3 4 6]

五、求两个集合的差集

集合的差，即元素存在于第一个函数不存在于第二个函数中

import numpy as np

a = np.array([1, 2, 3, 2, 4, 1])
b = np.array([3, 4, 5, 6])
x = np.setdiff1d(a, b)
print(x)  # [1 2]

六、求两个集合的异或

集合的对称差，即两个集合的交集的补集。简言之，就是两个数组中各自独自拥有的元素的集合

import numpy as np

a = np.array([1, 2, 3, 2, 4, 1])
b = np.array([3, 4, 5, 6])
x = np.setxor1d(a, b)
print(x)  # [1 2 5 6]

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参考：阿里云天池