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python - 在一行中打印和 for 循环

转载 作者:行者123 更新时间:2023-11-28 20:01:42 25 4
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这个简单的代码块是否有替代的一种衬里:

n = int(input())
for i in range(n):
print(i**2)

我试过:

print(i**2 for i in range(int(input())))

它需要输入,但会出现以下错误:

<generator object <genexpr> at 0x00000000032D3E60>

我的问题与 this qs 有点不同.无论我是否需要在换行符中打印它们,qs 都会帮助打印列表中的项目。

最佳答案

正如 Jon 和 Patrick 提到的,这不是错误,而是当您打印 __repr__ 时发生的情况。生成器表达式。

所以你只需要 "splat"那个发电机。 :)

print(*(i**2 for i in range(int(input()))), sep='\n')

演示输出

10
0
1
4
9
16
25
36
49
64
81

在该演示中,我在 input 提示符下输入了 10


在评论中,我想知道 i * i 的速度与 i ** 2 相比如何。在我的旧 2GHz 32 位单核机器上,运行 Python 3.6.0,i * ii ** 2 快大约 3 或 4 倍。这是一些 timeit 代码。

from timeit import Timer

commands = {'mul' : 'num * num', 'pow' : 'num ** 2'}

def time_test(num, loops, reps):
timings = []
setup = 'num = {}'.format(num)
for name, cmd in commands.items():
result = Timer(cmd, setup).repeat(reps, loops)
result.sort()
timings.append((result, name))

timings.sort()
for result, name in timings:
print(name, result)

loops, reps = 100000, 3
num = 1
for _ in range(10):
print('num =', num)
time_test(num, loops, reps)
num <<= 1

输出

num = 1
mul [0.02114695899945218, 0.02127135100090527, 0.02303983199817594]
pow [0.08504067399917403, 0.08687452600133838, 0.12349813100081519]
num = 2
mul [0.02089159800016205, 0.021789606998936506, 0.02889108999806922]
pow [0.08612996800002293, 0.09132789800059982, 0.09559987299871864]
num = 4
mul [0.021155500999157084, 0.02333696799905738, 0.028521009000542108]
pow [0.08492234799996368, 0.08499632499660947, 0.08537705599883338]
num = 8
mul [0.02173021600174252, 0.021955170999717666, 0.02823427400289802]
pow [0.08423048700205982, 0.08541251700080466, 0.08654486299928976]
num = 16
mul [0.02176373900147155, 0.02222509399871342, 0.02816650199747528]
pow [0.08528696699795546, 0.09080051600176375, 0.0968476650014054]
num = 32
mul [0.03118283900039387, 0.03388790600001812, 0.03745272100059083]
pow [0.0943321790000482, 0.09484523300125147, 0.09691544299857924]
num = 64
mul [0.030481540998152923, 0.03292956899895216, 0.03887743200175464]
pow [0.09454960600123741, 0.09569520199875114, 0.09926063899911242]
num = 128
mul [0.030935312999645248, 0.031198748001770582, 0.03733277300125337]
pow [0.09531564099961543, 0.09669112700066762, 0.09679062199938926]
num = 256
mul [0.03280377900227904, 0.03324341500047012, 0.04479783699935069]
pow [0.09439349899912486, 0.09439918999851216, 0.09548852000079933]
num = 512
mul [0.03275527599907946, 0.03428718699797173, 0.038492286003020126]
pow [0.10492119499758701, 0.10698100599984173, 0.13057717199990293]

关于python - 在一行中打印和 for 循环,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/50212167/

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