在二维数组中删除的行具有相同的值 [英] Removing rows in a 2D array that have the same value

问题描述

我在寻找一个快速的方法来删除重复值present在二维阵列先到先得。我知道一种方法，如果他们是相同的删除行，但没有如果只有一个值是present。

I'm looking for a quick way to remove duplicate values present in a 2D array on a first come first serve basis. I know a way to remove the rows if they are identical, but not if only one of the values is present.

a = array([[0, 1],
           [3, 4],
           [3, 5],
           [2, 5],
           [1, 2]])

由于3是present在[1]和[2]我想删除任何未来价值的发生。同样在2 [3]和[4]所以输出将是：

As 3 is present in a[1] and a[2] I would like to delete any future occurrence of a value. Similarly with 2 in a[3] and a[4] So the output would be:

a = array([[0, 1],
           [3, 4],
           [2, 5]])

如可以看到的，所用的值5重叠的任何建议是AP preciated

As can be seen, there is overlap with the value 5. Any suggestions are appreciated.

推荐答案

一个纯Python方式使用起来会设定一个列表-COM prehension：

A pure-Python way will be to use set with a list-comprehension:

>>> seen = set()
>>> np.array([x for x in a if seen.isdisjoint(x) and not seen.update(x)])
array([[0, 1],
       [3, 4],
       [2, 5]])

一班轮只是滥用的事实， set.update 回报无，所以当的 seen.isdisjoint（X） 是真我们就可以更新看到使用不seen.update（X）。

The one-liner simply abuses the fact that set.update returns None, so when seen.isdisjoint(x) is True we can update the seen set using not seen.update(x).

我们也可以写上code为：

We can also write the above code as:

seen = set()
out = []
for item in a:
    # if none of items in current sub-array are present in seen set
    # then add current sub-array to our list. Plus update the seen
    # set with the items from current sub-array
    if seen.isdisjoint(item):
        out.append(item)
        seen.update(item)
...         
>>> out
[array([0, 1]), array([3, 4]), array([2, 5])]
>>> np.array(out)
array([[0, 1],
       [3, 4],
       [2, 5]])

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