Как найти сумму квадратов в python

What about :

sum(map(lambda x:x*x,l))

we also use reduce:

print reduce(lambda x,y: x+y*y,l) # as pointed by @espang reduce(lambda x,y: x+y*y,l) is only ok, when the first value is 1 (because 1*1 == 1). The first value is not squared

We can take the first element, get its square, then add it to the head of the list so we can make sure that it’s squared. Then we continue using reduce. It isn’t worth all that work, as we have better alternatives.

reduce(lambda x,y: x+y*y,[l[:1][0]**2]+l[1:])

Just out of curiosity, I tried to compare the three solutions to sum the squares of 10000 numbers generated by range, and compute the execution time of every operation.

l=range(10000) 
from datetime import datetime
start_time = datetime.now()
print reduce(lambda x,y: x+y*y,l)
print('using Reduce numbers: {}'.format(datetime.now() - start_time))

from datetime import datetime
start_time = datetime.now()
print sum(map(lambda x:x*x,l))
print('Sum after map square operation: {}'.format(datetime.now() - start_time))

from datetime import datetime
start_time = datetime.now()
print sum( i*i for i in l)
print('using list comprehension to sum: {}'.format(datetime.now() - start_time))

Output:

Using list comprehension is faster

333283335000
using Reduce numbers: 0:00:00.003371
333283335000
Sum after map square operation: 0:00:00.002044
333283335000
using list comprehension to sum: 0:00:00.000916

Python being the language of magicians can be used to perform many tedious and repetitive tasks in a easy and concise manner and having the knowledge to utilize this tool to the fullest is always useful. One such small application can be finding sum of squares of list in just one line. Let’s discuss certain ways in which this can be performed. 

Method #1 : Using reduce() + lambda The power of lambda functions to perform lengthy tasks in just one line, allows it combined with reduce which is used to accumulate the subproblem, to perform this task as well. Works with only Python 2. 

The original list is : [3, 5, 7, 9, 11]
The sum of squares of list is : 285

Time Complexity: The reduce function in this code has a time complexity of O(n), where n is the number of elements in the input list “test_list”. The lambda function used inside the reduce function takes constant time to execute, so the overall time complexity is O(n).
Auxiliary Space: The reduce function in this code has an auxiliary space complexity of O(1), as it only requires a single accumulator variable to store the intermediate and final results. The lambda function used inside the reduce function takes constant space, so the overall auxiliary space complexity is O(1).

Method #2 : Using map() + sum() The similar solution can also be obtained using the map function to integrate and sum function to perform the summation of the squared number. 

The original list is : [3, 5, 7, 9, 11]
The sum of squares of list is : 285

Time Complexity: O(n)
Auxiliary Space: O(1)

Method #3 : Using for loop and math.pow()

The original list is : [3, 5, 7, 9, 11]
The sum of squares of list is : 285

Time Complexity: O(n), where n is the number of elements in the input list.
Auxiliary Space: O(1), as we are not using any additional data structure or creating any new list to store intermediate results. 

Method #4 : Using simple list comprehension

Here is an approach using a list comprehension:

The original list is : [3, 5, 7, 9, 11]
The sum of squares of list is : 285

Time Complexity: O(n), where n is the length of the input list, as the list comprehension iterates over each element in the input list once.
Auxiliary Space: O(1), as the space used by the algorithm is constant and does not depend on the size of the input.

Method #5: Using numpy

Output:

The original list is : [3, 5, 7, 9, 11]
The sum of squares of list is : 285

Time Complexity: O(N)
Auxiliary Space : O(1)

Approach using reduce() and operator.mul():

The reduce() function from the functools module is used to reduce the list to a single value by applying a given function to the elements of the list. The operator.mul() function from the operator module is used to multiply the squared element of the list. In this approach, we use reduce() and operator.mul() to calculate the sum of the squares of the elements in the input list.

Algorithm:

Import reduce() and operator.mul() functions from the respective modules.
Take the input list as input_list.
Use reduce() function with operator.mul() and map() function to calculate the sum of squares of input_list.
Print the result.

The original list is :  [3, 5, 7, 9, 11]
The sum of squares of list is :  285

Time complexity: The time complexity of this approach is O(n) because map() and reduce() function iterate over the input list once.
Auxiliary space: The space complexity of this approach is O(1) because we are not using any additional data structure or creating any new list to store intermediate results.

Last Updated :
02 May, 2023

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Введение

В этом разделе мы изучим код Python для вычисления суммы квадратов n натуральных чисел, предоставленных пользователем.

Математическое выражение: (n * (n + 1) * (2 * n + 1)) // 6

суррогатного материнства

def sum_square(n): sum = 0 # Математическое выражение sum = (n*(n+1) * (2*n + 1))//6 print(sum) n = int(input("Пожалуйста, введите число: ")) сумма_квадрат(n)

Результат

объяснение

В приведенной выше программе мы создали переменную n для хранения пользовательского ввода. Функция sum_square() вычисляет сумму квадратов n натуральных чисел с помощью математического выражения, а вывод отображается с помощью функции print().