Question

我有一个类似于下面的numpy数组，代表每个人的邻居（这首先由igraph包生成然后转换为numpy数组

import numpy as np
import igraph
Edges = 2
NumNodes = 30
DisGraph = igraph.GraphBase.Barabasi(NumNodes, Edges)
Neighbors = map(DisGraph.neighbors, range(NumNodes))
Neighbors = np.asarray(DisNeighbors)

）：

Neighbors=[[1, 2, 3, 4, 5, 6, 8, 9, 11, 23, 24, 27]
 [0, 2, 3, 4, 9, 10, 13, 16, 17, 19, 25, 27] [0, 1, 10, 22]
 [0, 1, 5, 6, 7, 8, 12, 14, 15, 21, 22] [0, 1]
 [0, 3, 7, 11, 15, 23, 24, 25, 29] [0, 3] [3, 5, 18] [0, 3, 12, 16, 18]
 [0, 1, 13] [1, 2, 14, 20] [0, 5] [3, 8, 19] [1, 9, 21, 28]
 [3, 10, 17, 20, 26] [3, 5] [1, 8] [1, 14, 26] [7, 8] [1, 12] [10, 14, 28]
 [3, 13] [2, 3] [0, 5] [0, 5] [1, 5] [14, 17] [0, 1, 29] [13, 20] [5, 27]]

我想找到一种方法从这个阵列中获取某些数字，可能没有使用循环。例如，如果我有一个列表：

List = [0 1 2 3 4 5 6 7 8 9 10]

然后，我希望生成的Neighbors数组在List中删除这些值。

任何帮助将不胜感激。

我目前的答案如下：

for aa in List:
    i=0
    for bb in Neighbors:
        Neighbors[i] = [cc for cc in bb if cc != aa]
        i=i+1

但是我想知道是否有更有效的方法来处理这个问题，因为我正在处理数百万的数组。

Answer 1

我不知道你的意思＆＃34;将这些值移到列表中＆＃34; （你是什么意思，＆＃34;删除＆＃34;？）。但是，通常，您可以通过以下方式选择数组中的点：

import numpy as np

a = np.random.random_integers(0,10,[10,10])
b = np.random.random_integers(0,10,5)

for r in b:
    a[a==r] = -999


a
Out[12]: 
array([[   5,    1, -999,    3,    7,    5,    8,    3,    8,    4],
       [   8,    8, -999,    7, -999,    8, -999, -999,    4,    7],
       [  10, -999, -999, -999, -999,    1, -999,    7,   10, -999],
       [   3,   10,    8, -999,    8,    4, -999,    7,    4,    3],
       [   4, -999,    4, -999, -999, -999, -999, -999, -999, -999],
       [   5,    3, -999,   10,   10, -999,   10,   10,    3,    8],
       [   8,    5, -999, -999,    7, -999,    1,    8, -999,    8],
       [   4,    3,    8, -999,    3,    5,    4, -999,    4,   10],
       [   4,    3,    7,    4, -999,    7,    7,    7, -999,    8],
       [-999,   10, -999,    5,    1,    5,    1,   10,    5,    1]])

如何在没有循环的情况下摆脱Numpy数组中的值？

1 个答案: