Python 打印 1-99 奇数的最有效代码
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most efficient code for printing odd numbers from 1-99
提问by neenertronics
The task is to print the odd numbers from 1-99 on separate lines.
任务是在不同的行上打印 1-99 的奇数。
Codeeval deemed this code partially correct (98 out of 100): (edited)
Codeeval 认为此代码部分正确（100 个中有 98 个）：（已编辑）
liszt = (i for i in range(1,100) if i%2!=0)
for i in liszt:
print i
Codeeval deemed the below code completely correct:
Codeeval 认为以下代码完全正确：
liszt = range(1,100)
for i in liszt:
if i%2!=0:
print i
New to Python, so just looking to understand why one method might be considered better than the other. Is the second method more efficient?
Python 新手，所以只是想了解为什么一种方法可能被认为比另一种更好。第二种方法更有效吗？
Thanks for the help!
谢谢您的帮助！
采纳答案by Ankur Ankan
In the first code you are iterating over two generators first range(1, 100)and then over lisztwhereas in the second case the iteration is only over liszt. Other than that the operation is same in both the cases, so the second method is more efficient.
在第一个代码中，您首先迭代两个生成器range(1, 100)，然后再迭代，liszt而在第二种情况下，迭代仅结束liszt。除此之外，两种情况下的操作相同，因此第二种方法更有效。
Since every second number after 1 would be odd, a better solution could be:
由于 1 之后的每个第二个数字都是奇数，因此更好的解决方案可能是：
for i in range(1, 100, 2):
print(i)
回答by Hugh Bothwell
print("\n".join(str(i) for i in range(1,100,2)))
