# Accessing values in nested list and Fixed Number of Elements List

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by Goeduhub's Expert (9.3k points)

## Accessing values in nested list-

Starting with a three-dimensional list:
alist = [[[1,2],[3,4]], [[5,6,7],[8,9,10], [12, 13, 14]]]

Accessing items in the list:

print(alist[0][0][1])

#2
#Accesses second element in the first list in the first list
print(alist[1][1][2])
#10
#Accesses the third element in the second list in the second list
Performing support operations:
alist[0][0].append(11)
print(alist[0][0][2])
#11
#Appends 11 to the end of the first list in the first list
Using nested for loops to print the list:
for row in alist:  #One way to loop through nested lists
for col in row:
print(col)
#[1, 2, 11]
#[3, 4]
#[5, 6, 7]
#[8, 9, 10]
#[12, 13, 14]
Note- that this operation can be used in a list comprehension or even as a generator to produce efficiencies,
e.g.: [col for row in alist for col in row]
#[[1, 2, 11], [3, 4], [5, 6, 7], [8, 9, 10], [12, 13, 14]]
Not all items in the outer lists have to be lists themselves:
alist[1].insert(2, 15)
#Inserts 15 into the third position in the second list
Another way to use nested for loops. The other way is better but I've needed to use this on occasion:
for row in range(len(alist)): #A less Pythonic way to loop through lists
for col in range(len(alist[row])):
print(alist[row][col])
#[1, 2, 11]
#[3, 4]
#[5, 6, 7]
#[8, 9, 10]
#15
#[12, 13, 14]

Using slices in nested list:

print(alist[1][1:])

#[[8, 9, 10], 15, [12, 13, 14]]

#Slices still work

The final list:

print(alist)

#[[[1, 2, 11], [3, 4]], [[5, 6, 7], [8, 9, 10], 15, [12, 13, 14]]]

## Initializing a List to a Fixed Number of Elements-

For immutable elements (e.g. None, string literals etc.):
my_list = [None] * 10
my_list = ['test'] * 10
For mutable elements, the same construct will result in all elements of the list referring to the same object, for
example, for a set:
>>> my_list=[{1}] * 10
[{1}, {1}, {1}, {1}, {1}, {1}, {1}, {1}, {1}, {1}]