Question

如何对每次k迭代获得的X的所有值求和？

print('Enter value for N between 1 and 50:') #User inputs number for N
N=int(input())
k=1

if N < 1:
    print('N must be between 1 and 50')

while k <= N:
    if N < 50:
        x=(1/(2**k))    #Value X is calculated for each value of N
        print('x=', x) 
        k=k + 1       
    else:
        print('N must be between 1 and 50')
        break

Answer 1

只需创建一个名为sum的新变量，该变量从0开始。每当您拥有x的新值时，就将x的值累加到其中。

print('Enter value for N between 1 and 50:') #User inputs number for N
N=int(input())
k=1
sum=0

if N < 1 or N > 50:
    print('N must be between 1 and 50')
    exit()

while k <= N:
    x=(1/(2**k))    #Value X is calculated for each value of N
    sum=sum+x
    print('x=', x) 
    k=k + 1       

print("Sum = ", sum)

请注意，我还在循环之前将N的验证转移了，因为您只需要检查一次即可。

Answer 2

您可以尝试变得更加精确吗？是否要像当前一样在每次迭代中显示x，并在末尾显示一个全局计数，将所有x相加？

在这种情况下，您可以引入一个额外的变量，如下所示

print('Enter value for N between 1 and 50:') #User inputs number for N
N=int(input())
k=1

if N < 1:
    print('N must be between 1 and 50')

total_count = 0  # new variable
while k <= N:
    if N < 50:
        x=(1/(2**k))    #Value X is calculated for each value of N
        print('x=', x) 
        total_count += x # this updates the total count
        k=k + 1       
    else:
        print('N must be between 1 and 50')
        break

Answer 3

即使在多达50次的操作中（通常）不需要，我也会使用numpy来向量化我的答案。

import numpy as np

N=int(input('Enter value for N between 1 and 50: '))

if N < 1 or N > 50:
    print('N must be between 1 and 50')
else:
    x = 1/(2**np.arange(1, N+1, dtype=float))
    print(x)  # in case you want to know the value at each "iteration"
    sum = np.sum(x)
    print(sum)

Answer 4

为此做一个证明，但是1 - (1/2**N)也会给你答案。

如何求和不同的迭代值？

4 个答案: