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Python | Program to count number of lists in a list of lists

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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




# Python3 program to Count number
# of lists in a list of lists
 
 
def countList(lst):
    return len(lst)
 
 
# Driver code
lst = [[1, 2, 3], [4, 5], [6, 7, 8, 9]]
print(countList(lst))


Output:

3

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




# Python3 program to Count number
# of lists in a list of lists
 
 
def countList(lst):
    count = 0
    for el in lst:
        if type(el) == type([]):
            count += 1
 
    return count
 
 
# Driver code
lst = [[1, 2, 3], [4, 5], [6, 7, 8, 9]]
print(countList(lst))


Output:

3

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)


Method #3 : Using isinstance() method 

Python3




# Python3 program to Count number
# of lists in a list of lists
 
def countList(lst):
    return sum(isinstance(i, list) for i in lst)
     
# Driver code
lst = [[1, 2, 3], [4, 5], [6, 7, 8, 9]]
print(countList(lst))


Output:

3

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))


Output

3

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))


Output

3

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))


Output

3

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))


Output

3

Method #8: Using eval() 

Python3




lst = ["[1, 2, 3]", "[4, 5]", "[6, 7, 8, 9]"]
x=list(map(eval,lst) )
print(len(x))


Output

3

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))
#This code is contributed by Edula Vinay Kumar Reddy


Output

3

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))


Output

3

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




# Python3 program to Count number
# of lists in a list of lists
from functools import reduce
 
 
# Function which count the number of list in list
def countList(lst):
    return reduce(lambda a, b : a + 1 if isinstance(b, list) else a, lst, 0)
     
# Driver code
lst = [[1, 2, 3], [4, 5], [6, 7, 8, 9]]
print(countList(lst))


Output

3

Time complexity: O(N), where N is the length of the list. 
Auxiliary space: O(1), No extra space is used. 



Last Updated : 28 Apr, 2023
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