Question

x = int(input("Pick one small number: "))
y = int(input("And a bigger number: "))

if x > y:
    print ("Doesn't work. ")

elif x < y:

    for i in range(x,y):
        if i%7 == 0 and i%5 !=0:
            z = sum(i)
            print (z)

Answer 1

您可能想一次将i求和：

z = 0
for n in range(x,y):
    if n%7 == 0 and n%5 !=0:
        n += i
        print("Running total:", z)
print("Final total:", z)

如果您想使用sum，则必须将其应用于列表：

filtered_list = []
for n in range(x,y):
    if n % 7 == 0 and n % 5 !=0:
        filtered_list.append(n)
print("Final total:", sum(filtered_list))

或使用生成器表达式：

print(sum(n for n in range(x, y) if n%7 == 0 and n%5 !=0))

还有一种快捷方式来获取range(x, y)中7的所有倍数：

multiples = range(x + 7 - x % 7, y, 7)

然后，您只需要检查第二个条件（不能被5整除）

print(sum(n for n in multiples if n%5 !=0))

您还可以创建2套并计算差值：

def multiples(start, stop, d):
    """ This function returns the set of all multiples of d between start and stop """
    return set(range(start + divisor - start % divisor, stop, divisor))

print(sum(multiples(x, y, 7) - multiples(x, y, 5))

最后是一种数学方法。您的if条件是选择7的倍数并过滤掉5和7的倍数的数字。这等效于减去5和7的最小公倍数（即35）的倍数之和。）从范围内7的倍数之和...

# we need a function from the math module that calculates the greatest common denominator
# we will use this to help calculate the least common multiple
from math import gcd 

def sum_of_multiples(start, stop, mult):
    """ Calculate sum of multiples of mult that lie in the range start, stop """
    start //= mult
    stop //= mult
    return mult * (stop - start) * (stop + start + 1) / 2

lcm = 5 * 7 // gcd(5, 7)   # The least common multiple of 5 and 7 = 35
print("Total:", sum_of_multiples(x, y, 7) - sum_of_multiples(x, y, lcm))

（当然，在上面您可以只写35而不是lcm。如果您要使用其他数字应用此方法，我将给出计算结果。）

Answer 2

您之所以遇到该异常，是因为sum期望一个可以被迭代的对象，或者换句话说，可以返回一个迭代器，例如列表或元组。

一种适当的方法是利用range函数返回的迭代器对象来获取x和y之间所有数字的累加值，这些数字是7的因数，而不是5的因数。

x = int(input("Pick one small number: "))
y = int(input("And a bigger number: "))

if x > y:
    print ("Doesn't work. ")

elif x < y:
    print(sum([i for i in range(x, y) if i%7 == 0 and i%5 != 0]))

该代码将在条件后输出i，但不会输出i的总和。错误指出“ int”对象不可迭代

2 个答案: