Sometimes, while working with data, we can have a problem in which we require to gather a data that is of the form of sequence of increasing element tuple with each tuple containing the element N times. Let’s discuss certain ways in which this task can be performed.
Method #1 : Using generator expression + tuple() The combination of above functions can be used to perform this task. In this, we need to iterate through N using generator expression and construction of tuple using tuple().
Python3
# Python3 code to demonstrate working of# N element incremental tuples# Using generator expression + tuple# initialize NN = 3# printing Nprint("Number of times to repeat : " + str(N))# N element incremental tuples# Using generator expression + tupleres = tuple((ele, ) * N for ele in range(1, 6))# printing resultprint("Tuple sequence : " + str(res)) |
Number of times to repeat : 3 Tuple sequence : ((1, 1, 1), (2, 2, 2), (3, 3, 3), (4, 4, 4), (5, 5, 5))
Method #2 : Using repeat() + list comprehension This task can also be performed using combination of above functions. In this, we use repeat() to repeat elements N times. And iteration is handled using list comprehension.
Python3
# Python3 code to demonstrate working of# N element incremental tuples# Using generator expression + tuplefrom itertools import repeat# initialize NN = 3# printing Nprint("Number of times to repeat : " + str(N))# N element incremental tuples# Using generator expression + tupleres = tuple(tuple(repeat(ele, N)) for ele in range(1, 6))# printing resultprint("Tuple sequence : " + str(res)) |
Number of times to repeat : 3 Tuple sequence : ((1, 1, 1), (2, 2, 2), (3, 3, 3), (4, 4, 4), (5, 5, 5))
Method #3 : Using for loop, while loop and tuple() method
Python3
# Python3 code to demonstrate working of# N element incremental tuples# Using generator expression + tuple# initialize NN = 3# printing Nprint("Number of times to repeat : " + str(N))# N element incremental tuplesres=[]for i in range(1,6): x=[] j=0 while(j<N): x.append(i) j+=1 res.append(tuple(x))res=tuple(res)# printing resultprint("Tuple sequence : " + str(res)) |
Number of times to repeat : 3 Tuple sequence : ((1, 1, 1), (2, 2, 2), (3, 3, 3), (4, 4, 4), (5, 5, 5))
Method #4 : Using * operator and tuple() method
Python3
# Python3 code to demonstrate working of# N element incremental tuples# initialize NN = 3# printing Nprint("Number of times to repeat : " + str(N))# N element incremental tuplesres=[]for i in range(1,6): a=[i]*N res.append(tuple(a))res=tuple(res)# printing resultprint("Tuple sequence : " + str(res)) |
Number of times to repeat : 3 Tuple sequence : ((1, 1, 1), (2, 2, 2), (3, 3, 3), (4, 4, 4), (5, 5, 5))
Time complexity: O(n) where n is the number of elements in the range from 1 to 6, as the loop iterates n times and each iteration takes constant time.
Auxiliary space: O(n) to store the list of tuples where n is the number of tuples in the list.
Method 5: Using numpy library
Note: Install numpy module using command “pip install numpy”
Python3
import numpy as np# initialize NN = 3# printing Nprint("Number of times to repeat : " + str(N))# N element incremental tuples using numpyres = [tuple(np.full(N, i)) for i in range(1, 6)]# printing resultprint("Tuple sequence : " + str(res)) |
Output:
Number of times to repeat : 3 Tuple sequence : [(1, 1, 1), (2, 2, 2), (3, 3, 3), (4, 4, 4), (5, 5, 5)]
Time Complexity: O(n^2), where n is the number of elements in the result.
Auxiliary Space: O(n^2)
Method 6: using itertools module
Step-by-step algorithm for implementing the approach
- Initialize the variable N to the given value.
- Define the range of numbers to choose from, using the range() function.
- Use the product() function from the itertools library to generate all possible tuples of length N whose elements are chosen from the numbers iterable.
- Filter the resulting list of tuples to only those where all elements are the same, using a list comprehension and the all() function.
- Print the value of N to the console.
- Print the resulting list of tuples to the console.
- Return the list of tuples.
Python3
from itertools import product# Set the number of times to repeat each elementN = 3# Define the range of numbers to choose fromnumbers = range(1, 6)# Generate all possible tuples of length N whose elements are chosen from the numbers iterableres = list(product(numbers, repeat=N))# Filter the tuples to only those where all elements are the sameres = [t for t in res if all(x == t[0] for x in t)]# Print the number of times to repeat each elementprint("Number of times to repeat : " + str(N))# Print the resulting tuple sequenceprint("Tuple sequence : " + str(res)) |
Number of times to repeat : 3 Tuple sequence : [(1, 1, 1), (2, 2, 2), (3, 3, 3), (4, 4, 4), (5, 5, 5)]
Time complexity:
- The product() function generates all possible tuples of length N whose elements are chosen from the numbers iterable. The number of possible tuples is len(numbers)^N.
- The filter operation uses a list comprehension and the all() function to filter the tuples where all elements are the same. This operation takes O(N) time for each tuple.
- Therefore, the overall time complexity of the algorithm is O(N*len(numbers)^N).
Auxiliary space:
- The product() function uses O(N*len(numbers)) auxiliary space to generate all possible tuples.
- The list comprehension used to filter the tuples uses O(N) auxiliary space for each tuple.
- Therefore, the overall auxiliary space complexity of the algorithm is O(N*len(numbers)).
