在python中比较多维数组的行
我有两个numpy数组在python中比较多维数组的行,python,arrays,list,numpy,scipy,Python,Arrays,List,Numpy,Scipy,我有两个numpy数组 a= np.array([[2,2],[3,2],[4,2],[3,3],[5,3]]) b= np.array([[1,1],[1,3],[5,3]]) 我想将a与b进行比较并返回a-b,以便: a-b = array([[2,2], [3,2], [4,2], [3,3]]) 我尝试过: [x for x in a if x not in b] 结果是 [array([2, 2])
a= np.array([[2,2],[3,2],[4,2],[3,3],[5,3]])
b= np.array([[1,1],[1,3],[5,3]])
我想将a与b进行比较并返回a-b,以便:
a-b = array([[2,2],
[3,2],
[4,2],
[3,3]])
我尝试过:
[x for x in a if x not in b]
结果是
[array([2, 2]), array([3, 2]), array([4, 2])] # where clearly [3,3] is missing
我甚至试着比较循环中a和b的每一行,它给了我一个错误
The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()
有谁能帮我解决这个问题吗?一种方法是将numpy数组转换为元组列表,然后将
b
转换为元组集,然后执行与对它们使用的相同的列表理解。范例-
In [1]: import numpy as np
In [2]: a= np.array([[2,2],[3,2],[4,2],[3,3],[5,3]])
In [3]: b= np.array([[1,1],[1,3],[5,3]])
In [18]: alist = list(map(tuple, a))
In [19]: bset = set(map(tuple, b))
In [20]: np.array([x for x in alist if x not in bset])
Out[20]:
array([[2, 2],
[3, 2],
[4, 2],
[3, 3]])
基于小波变换的矢量化方法-
a[~((b[:,None,:] == a).all(2)).any(0)]
使用自-
样本运行-
In [89]: a
Out[89]:
array([[2, 2],
[3, 2],
[4, 2],
[3, 3],
[5, 3]])
In [90]: b
Out[90]:
array([[1, 1],
[1, 3],
[5, 3]])
In [91]: a[~((b[:,None,:] == a).all(2)).any(0)]
Out[91]:
array([[2, 2],
[3, 2],
[4, 2],
[3, 3]])
In [92]: a[~(cdist(a,b)==0).any(1)]
Out[92]:
array([[2, 2],
[3, 2],
[4, 2],
[3, 3]])
你的输出背后有什么逻辑吗?你到底想做什么样的比较?它不是减法,不是会员制,也不是索引。看起来您的输出只是
a
的前四个元素。我很难弄清楚它与b
的关系。比较'a'和'b'的行。。返回与“b”行不匹配的“a”行
In [89]: a
Out[89]:
array([[2, 2],
[3, 2],
[4, 2],
[3, 3],
[5, 3]])
In [90]: b
Out[90]:
array([[1, 1],
[1, 3],
[5, 3]])
In [91]: a[~((b[:,None,:] == a).all(2)).any(0)]
Out[91]:
array([[2, 2],
[3, 2],
[4, 2],
[3, 3]])
In [92]: a[~(cdist(a,b)==0).any(1)]
Out[92]:
array([[2, 2],
[3, 2],
[4, 2],
[3, 3]])