Python 总结 range() 中的所有整数
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Sum up all the integers in range()
提问by mllnd
I need to write a program that sums up all the integers which can be divided by 3 in the range of 100 to 2000. I'm not even sure where to start, so far I've got this tiny piece of code written which isn't correct.
我需要编写一个程序，将 100 到 2000 范围内可以被 3 整除的所有整数相加。我什至不知道从哪里开始，到目前为止，我已经编写了一小段代码不正确。
for x in range(100, 2001, 3):
print(x+x)
Any help is much appreciated!
任何帮助深表感谢！
采纳答案by aga
Use generator expression and sum function here:
在此处使用生成器表达式和 sum 函数：
res = sum(x for x in range(100, 2001) if x % 3 == 0)
It's pretty self-explanatory code: you're summing all the numbers from 100 to 2000, inclusive, which are divisible by three.
这是一段不言自明的代码：您要对 100 到 2000（含）之间的所有数字求和，这些数字可以被 3 整除。
回答by Deck
There is a sum function
有一个求和函数
>>> sum(filter(lambda x: x % 3 == 0, range(100, 2000)))
664650
But this is better:
但这更好：
>>> sum(x for x in range(100, 2000) if x % 3 == 0)
664650
回答by Inbar Rose
Since you know the first number in this range that is divisible by 3 is 102, you can do the following:
由于您知道此范围内第一个可被 3 整除的数字是 102，因此您可以执行以下操作：
Solution:
解决方案：
>>> sum(range(102, 2001, 3))
664650
To make it into a robust function:
使其成为一个健壮的函数：
def sum_range_divisible(start, end, divisor):
while start % divisor != 0:
start += 1
return sum(range(start, end, divisor))
Using it:
使用它：
>>> sum_range_divisible(100, 2001, 3)
664650
Note:
笔记：
The advantage here is that you do not have to check each number in the whole range, since you are jumping by 3 each time.
这里的优点是您不必检查整个范围内的每个数字，因为您每次都跳 3。
Timing:
定时：
I have timed the different solutions, mine and aga's:
我已经为不同的解决方案计时，我的和aga 的：
>>> import timeit
>>> timeit.Timer('sum(range(102, 2001, 3))').repeat()
[9.516391893850312, 9.49330620765817, 9.508695564438462]
>>> timeit.Timer('sum(x for x in range(100, 2001) if x % 3 == 0)').repeat()
[134.757627812011, 134.46399066622394, 138.34528734198346]
Conclusion:
结论：
My answer is faster by a factor of 14
我的回答快了14倍
回答by devil00
sum(filter(lambda l : l%3 ==0, range(100,2001)))
回答by Aristide
There is a closed formula for that.
有一个封闭的公式。
If (u_i) is a sequence defined by its first term u_0 and its common difference r, then the sum of the n first terms of (u_i) is:
如果 (u_i) 是由它的第一项 u_0 和它的公差 r 定义的序列，那么 (u_i) 的前 n 项之和是：
EDIT: I have made this little videoto explain it visually.
编辑：我制作了这个小视频来直观地解释它。
A popular anecdoteattributes this formula to the young Johann Carl Friedrich Gauss.
一个流行的轶事将这个公式归功于年轻的约翰·卡尔·弗里德里希·高斯。
In your case:
在你的情况下：
- u_0 = 102
- u_{n-1} = 1998
- n = (1998 - 102) / 3 + 1 = 633
- u_0 = 102
- u_{n-1} = 1998
- n = (1998 - 102) / 3 + 1 = 633
So, the sum is (633 * (102 + 1998)) / 2 = 664650.
所以，总和是 (633 * (102 + 1998)) / 2 = 664650。
As a general Python function with the usual rangearguments start, stop, step:
作为带有常用range参数start, stop,的通用 Python 函数step：
def arithmetic_series(start, stop, step):
number_of_terms = (stop - start) // step
sum_of_extrema = start + (stop - step)
return number_of_terms * sum_of_extrema // 2
In your case, the call would be:
在您的情况下，电话将是：
arithmetic_series(102, 2001, 3)
The complexity is O(1) instead of O(n), so unsurprisingly:
复杂度是 O(1) 而不是 O(n)，所以不出所料：
%timeit sum(range(102, 2001, 3))
100000 loops, best of 3: 17.7 μs per loop
%timeit arithmetic_series(102, 2001, 3)
1000000 loops, best of 3: 548 ns per loop
