如何从numpy数组列表中“删除"一个numpy数组? [英] How do you 'remove' a numpy array from a list of numpy arrays?

问题描述

如果我有一个numpy数组列表，则使用remove方法将返回值错误.

If I have a list of numpy arrays, then using remove method returns a value error.

例如:

import numpy as np

l = [np.array([1,1,1]),np.array([2,2,2]),np.array([3,3,3])]

l.remove(np.array([2,2,2]))

请给我

ValueError:具有多个元素的数组的真值不明确.使用a.any()或a.all()

ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()

我似乎无法使all()工作，这是不可能的吗?

I can't seem to get the all() to work, is it just not possible?

推荐答案

这里的问题是，当两个numpy数组与==进行比较时，如remove()和index()方法中那样，一个numpy布尔值数组(逐个元素比较)将返回，这被解释为模棱两可.比较两个numpy数组是否相等的一种好方法是使用numpy的array_equal()函数.

The problem here is that when two numpy arrays are compared with ==, as in the remove() and index() methods, a numpy array of boolean values (the element by element comparisons) is returned which is interpretted as being ambiguous. A good way to compare two numpy arrays for equality is to use numpy's array_equal() function.

由于列表的remove()方法没有键参数(就像sort()一样)，因此我认为您需要创建自己的函数才能执行此操作.这是我做的:

Since the remove() method of lists doesn't have a key argument (like sort() does), I think that you need to make your own function to do this. Here's one that I made:

def removearray(L,arr):
    ind = 0
    size = len(L)
    while ind != size and not np.array_equal(L[ind],arr):
        ind += 1
    if ind != size:
        L.pop(ind)
    else:
        raise ValueError('array not found in list.')

如果您需要更快的速度，则可以对其进行Cython化.

If you need it to be faster then you could Cython-ize it.

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