my_list = [[1, 2, 3], [['a', 'b', 'c'], 5, 6]] print(my_list[1][0][1])

print(my_list3[1][0][1]) Output:- b

This is a very difficult topic. The presentation was so good that I understood it so easily.

these videos rock. thank you !

great, i was having issue in understanding Lists. You explained it very well. Thanks

My_list3[1][0][1]='b'

li1 = [[1,2,3],[['a','b','c'],[5,6]]] print(li1[1][0][1])

how do I access an individual char from a multi-dimensional list? so if I want to access the N in Neso? is my_list [1] [1] ?

l1=[[1,2,3],['A','B','C'],5,6] print(l1[1][1]) we can use this type

16:40 my_list3 = [[1, 2, 3], [['a', 'b', 'c'], 5, 6]] for accessing 'b' item from my_list3 print(my_list3[1][0][1])

# Homework-Problem # Q -> Access value "b"? # Code: my_list3 = [[1, 2, 3], ['a', 'b', 'c'], [5, 6]] value_b = my_list3[1][1] print(value_b)

The answer is: my_list2[1][0][1]

I have a question Jaspreet Sir! Why list's index starts from 0 and not 1?

my_list=[[1,2,3],"abc",[4,5,6]] print(my_list[1][1]) Output b

my_list3=[1][0][1] the output is b

Answer to the Homework Question: print(my_list3[1][1])

List =[[1,2,3],[['a','b'],[5,6]]] Print(list[1][0][1])

I was here to comment the correct answer but already everyone has posted it.

Mylist_3 [1] [1]

Nice

list3 = [[1,2,3],['a','b','c'], [5,6]] print (list3 [1][1])

print(my_list3[1][0][1]) giving error of indexError: string index out of range. the correct answer is [1][1]

my_list3[1][0][1] = 'b'

my_list=[[1,2,3],['a','b','c'],[3,4,5]] print(my_list[1][1])

my_list =[[1,2,3],['a','b','c'],5,6] k= my_list[1][1] print(k)

my_list = [1,2,3],['a','b','c'], [5,6] print(my_list[1][1])

Answer of HW -- print(my_list3[1][0][-2])😅Sorry for taking negative indexing only i like that

my_list3 = [[1,2,3],[["a","b","c"],5,6]] print(my_list3[1][0][1])

print(my_list[1][0][1])

my_list3=[[1,2,3],[['a','b','c'],5,6]] print(my_list3[1][0][1]) output # b

❤

my_list = [[1,2,3],[["a","b","c"]'5'6]] Print (my_list[1][0][1])

print(my_list[1][1])

Answer to homework question : print(my_list3[1][0][1])

Answer : my_list3[1][0][1]

print(my_list3[1][1])

Print(my_list[2])

my_list = [[1,2,3], [['a','b','c'], 5,6]] print(my_list[1][0][1]) ans:- b ;)

My_list 3[2]

my_list3 = [[1,2,3],[['a','b','c'],5,6]] list3 = my_list3[1][0][1] print(list3) thank you for support, good luck

print(my_list3[1][1]) homework b

b = [["a","b","c"],5,6]

my_list3[1][0][1]=b

mylist3[1][1]

My_list[1][0][1]

my_list [1] [0] [1]

My list3[1][0]

My list3[1][0][1]

[1][1]

my_list3 [1][1]

lists=[[1,2,3],["a","b","c",5,6]] print (lists[1][1])

my_list = [[1, 2, 3], ['a', 'b', 'c'], [5, 6]] print(my_list[1][1]) print(my_list[-2][-2]) print(my_list[1][-2]) print(my_list[-2][1])

Sure, I can explain it simply! First, let's break down the code: mylist = [[1, 2, 3], [5, 6, ['a', 'b', 'c']]] This line creates a list called `mylist` containing two lists inside it. Each of these inner lists contains numbers and another list. Now, let's break down the indexing part: print(mylist[1][2][1]) - `mylist[1]` refers to the second element of `mylist`, which is `[5, 6, ['a', 'b', 'c']]`. - Then, `mylist[1][2]` refers to the third element of that second list, which is `['a', 'b', 'c']`. - Finally, `mylist[1][2][1]` refers to the second element of that inner list, which is `'b'`. So, `print(mylist[1][2][1])` will output `'b'`

my_list3 [1] [0] [1]='b'

my_list3 = [[1, 2, 3], [['a', 'b', 'c'], 5, 6]] print(my_list3[1][0][1])

my_list = [[1, 2, 3], [['a', 'b', 'c'], [5, 6]]] print(my_list[1][0][1])

❤

my_list3[1][0][1]=b

my_list = [[1, 2, 3], [['a', 'b', 'c'], 5, 6]] print(my_list[1][0][1])