根据特定条件删除 numpy 数组的行

Question

我有一个四行的数组A = array([[-1, -1, -1, -1], [-1, -1, 1, 2], [-1, -1, 1, 1], [2, 1, -1, 2]]).每行都有4数字。如何删除row#3和row#4？在row#3和row#4中，1和2分别出现不止一次。

有没有更快的方法来为任意数量的行和列做到这一点？主要目的是删除那些多次出现非负数的行。

Answer 1

你可以使用这样的东西：首先使用 np.unique 创建子数组中每个值的出现字典，并且只保留多次出现没有正数的数组。

A = np.array([[-1, -1, -1, -1], [-1, -1, 1, 2], [-1, -1, 1, 1], [2, 1, -1, 2]])
new_array = []
# loop through each array
for array in A:
# Get a dictionary of the counts of each value
unique, counts = np.unique(array, return_counts=True)
counts = dict(zip(unique, counts))
# Find the number of occurences of postive numbers
positive_occurences = [value for key, value in counts.items() if key > 0]
# Append to new_array if no positive number appears more than once
if any(y > 1 for y in positive_occurences):
continue
else:
new_array.append(array)
new_array = np.array(new_array)

这将返回：

array([[-1, -1, -1, -1],
[-1, -1,  1,  2]])

Answer 2

我的完全矢量化方法：

• 对每一行进行排序
• 通过将排序后的数组向左移动 1 并与其自身进行比较来检测重复项
• 标记具有正重复项的行
• 落

import numpy as np
a = np.array([[-1, -1, -1, -1], [-1, -1, 1, 2], [-1, -1, 1, 1], [2, 1, -1, 2]])
# sort each row
b = np.sort(a)
# mark positive duplicates
drop = np.any((b[:,1:]>0) & (b[:,1:] == b[:,:-1]), axis=1)
# drop
aa = a[~drop, :]
Output:
array([[-1, -1, -1, -1],
[-1, -1,  1,  2]])

Answer 3

我也修改了以存储索引：

A = np.array([[-1, -1, -1, -1], [-1, -1, 1, 2], [-1, -1, 1, 1], [2, 1, -1, 2]])
new_array = []
**indiceStore = np.array([])**
# loop through each array
for array in A:
# Get a dictionary of the counts of each value
unique, counts = np.unique(array, return_counts=True)
counts = dict(zip(unique, counts))
# Find the number of occurences of postive numbers
positive_occurences = [value for key, value in counts.items() if key > 0]
# Append to new_array if no positive number appears more than once
if any(y > 1 for y in positive_occurences):
**indiceStore = np.append(indiceStore, int(array))**
continue
else:
new_array.append(array)
new_array = np.array(new_array)

如果这是对的，让我抨击一下。