Python | Program to count number of lists in a list of lists
Last Updated :
28 Apr, 2023
Given a list of lists, write a Python program to count the number of lists contained within the list of lists.
Examples:
Input : [[1, 2, 3], [4, 5], [6, 7, 8, 9]]
Output : 3
Input : [[1], ['Bob'], ['Delhi'], ['x', 'y']]
Output : 4
Method #1 : Using len()
Python3
def countList(lst):
return len (lst)
lst = [[ 1 , 2 , 3 ], [ 4 , 5 ], [ 6 , 7 , 8 , 9 ]]
print (countList(lst))
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Time Complexity: O(n)
Auxiliary Space: O(1)
Method #2: Using type()
Use a for loop and in every iteration to check if the type of the current item is a list or not, and accordingly increment ‘count’ variable. This method has a benefit over approach #1, as it works well for a list of heterogeneous elements.
Python3
def countList(lst):
count = 0
for el in lst:
if type (el) = = type ([]):
count + = 1
return count
lst = [[ 1 , 2 , 3 ], [ 4 , 5 ], [ 6 , 7 , 8 , 9 ]]
print (countList(lst))
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Time complexity: O(n), where n is the number of elements in the list.
Auxiliary space: O(1), as we are only using a single variable (count) to store the count of lists.
A one-liner alternative approach for the above code is given below:
Python3
def countList(lst):
return sum ( type (el) = = type ([]) for el in lst)
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Method #3 : Using isinstance() method
Python3
def countList(lst):
return sum ( isinstance (i, list ) for i in lst)
lst = [[ 1 , 2 , 3 ], [ 4 , 5 ], [ 6 , 7 , 8 , 9 ]]
print (countList(lst))
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Time complexity: O(n), where n is the size of the set s.
Auxiliary space: O(n), as we are creating a list x with n elements, where n is the size of the set s.
Method#4: Using the list comprehension
Python3
lst = [[ 1 , 2 , 3 ], [ 4 , 5 ], [ 6 , 7 , 8 , 9 ]]
x = [i for i in lst]
print ( len (x))
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Time complexity: O(n), where n is the total number of elements in the list of lists.
Auxiliary space: O(n)
Method #5: Using enumerate function
Python3
lst = [ "[1, 2, 3]" , "[4, 5]" , "[6, 7, 8, 9]" ]
x = [ list (i) for i in enumerate (lst)]
print ( len (x))
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Time complexity: O(n), where n is length of list.
Auxiliary Space: O(n), where n is length of list.
Method #6: Using lambda function
Python3
lst = [[ 1 , 2 , 3 ], [ 4 , 5 ], [ 6 , 7 , 8 , 9 ]]
x = list ( filter ( lambda i: (i),lst))
print ( len (x))
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The time complexity of this code is O(n*m), where n is the number of sublists in the list and m is the average length of the sublists.
The auxiliary space required by this code is also O(n*m), since the filter function creates a new list that contains the same elements as the original list.
Method #7: Using map()
Python3
lst = [[ 1 , 2 , 3 ], [ 4 , 5 ], [ 6 , 7 , 8 , 9 ]]
x = list ( map ( str ,lst) )
print ( len (x))
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Method #8: Using eval()
Python3
lst = [ "[1, 2, 3]" , "[4, 5]" , "[6, 7, 8, 9]" ]
x = list ( map ( eval ,lst) )
print ( len (x))
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Method #9 : Using recursion
This approach involves reducing the length of the list by 1 at each recursive step and increasing the count by 1 if the first element of the list is a list. The function returns 0 when the list is empty.
Python3
def count_lists(lst):
if lst = = []:
return 0
return 1 + count_lists(lst[ 1 :])
lst = [[ 1 , 2 , 3 ], [ 4 , 5 ], [ 6 , 7 , 8 , 9 ]]
print (count_lists(lst))
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Time complexity: O(n), where n is the total number of elements in the list of lists. This is because the function processes each element of the list exactly once.
Auxiliary space: O(n) as well, since the maximum depth of the recursion tree is n in the worst case. For example, if the list of lists is a list of n single-element lists, the recursion tree will have a maximum depth of n. At each level of the tree, a new frame is added to the call stack, which takes up space in memory.
Method #10: Using ast.literal_eval()
Step-by-step approach:
- Import the ast module.
- Define the list of strings, lst.
- Create an empty list, num_list, to store the converted lists.
- Loop through each string in lst.
- Use ast.literal_eval() to convert the string to a list.
- Append the resulting list to num_list.
- Calculate the length of num_list using len().
- Print the length of num_list.
Python3
import ast
lst = [ "[1, 2, 3]" , "[4, 5]" , "[6, 7, 8, 9]" ]
num_list = []
for s in lst:
num_list.append(ast.literal_eval(s))
print ( len (num_list))
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Time complexity: O(n), where n is the length of lst. This is because we loop through each string in lst and apply ast.literal_eval() once for each string.
Auxiliary space: O(n), where n is the length of lst. This is because we create a list of the converted lists, which takes up space in memory.
Method #11: Using functools.reduce()
Step-by-step approach:
- Initialize a list of lists.
- Use reduce method on the list.
- reduce method takes a function, list, and initial value as parameters.
- reduce iterate over each item of the list and apply a function to it and return value.
- Function checks are an item an instance of the list or not, if it is then count 1 to the initial value else do nothing.
Python
from functools import reduce
def countList(lst):
return reduce ( lambda a, b : a + 1 if isinstance (b, list ) else a, lst, 0 )
lst = [[ 1 , 2 , 3 ], [ 4 , 5 ], [ 6 , 7 , 8 , 9 ]]
print (countList(lst))
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Time complexity: O(N), where N is the length of the list.
Auxiliary space: O(1), No extra space is used.
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