代码：-

import numpy as np

classroom1 = np.array([["Martha", "Available"], 
                       ["Max", "Anahi"], 
                       ["Available", "Available"], 
                       ["Alexis", "Rigel"]])
list_of_names=['Yash','Sam','Itachi','Ace']

seats_available=0
for i in range(len(classroom1)):
    for j in range(len(classroom1[0])):
        if classroom1[i][j]=="Available":
            seats_available+=1
#Enter the students in order right to left and starting from the last
count=0
for i in range(len(classroom1)-1,-1,-1):
    for j in range(len(classroom1[0])-1,-1,-1):
        if classroom1[i][j]=="Available":
            if seats_available and count<len(list_of_names):
                classroom1[i][j]=list_of_names[count]
                count+=1
                seats_available-=1
print(classroom1)  #as you can see Ace is not present cause there is no available seats for them when he appeared.

输出：-

[['Martha' 'Itachi']
 ['Max' 'Anahi']
 ['Sam' 'Yash']
 ['Alexis' 'Rigel']]

这是你要找的吗？

classroom1[-1,-2] = "Alex"
print(classroom1)

输出：

[['Martha' 'Available']
 ['Max' 'Anahi']
 ['Available' 'Available']
 ['Alex' 'Rigel']]

我们可以使用反向索引或负索引来解决这个问题。使用正/正索引时，你可以从头到尾遍历一个数组。但是使用反向索引时，你使用“-”号作为索引，它会反向遍历。
考虑以下示例：

x = np.array(['a','b','c','d','e','f','g'])
print(x[2])
print(x[-5])

所以这两条指令都将输出'c'需要记住的一点是，在倒排索引中，我们从-1开始，所以如果要输出'g'，必须传递x[-1]

**现在回到原来的问题：**您有一个二维数组，这意味着您必须传递两个负索引来执行所需的修改

classroom[子数组索引][子数组内元素的索引]
考虑另一个例子：

classroom1[0,1] = "Alex"
print(classroom1)
[["Martha", "Alex"], 
 ["Max", "Anahi"], 
 ["Available", "Available"], 
 ["Alexis", "Rigel"]]

假设我们想要更新索引2处数组中索引0处的座位，那么我们必须执行反向索引，如下所示：

classroom[-2][-2] = "Peter"
[["Martha", "Alex"], 
 ["Max", "Anahi"], 
 ["Peter", "Available"], 
 ["Alexis", "Rigel"]]

型