Question

所以我有一个奇怪的问题......

我想处理一个数组并从中自然舍入索引。

例如，如果你有数组...

pies = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

我想把它的每一个1.1元素自然地舍入。所以在这种情况下：

pies[0::round(1.1x)]

输出：

[0, 1, 2, 3, 4, 6, 7, 8, 9]

因为它会取出以下四舍五入的位置：

pies[0]->pies[0]
pies[1.1]->pies[1]
pies[2.2]->pies[2]
pies[3.3]->pies[3]
pies[4.4]->pies[4]
pies[5.5]->pies[6]
pies[6.6]->pies[7]
pies[7.7]->pies[8]
pies[8.8]->pies[9]

请注意，我们并未对内容进行四舍五入 - 我们正在对索引进行四舍五入。

作为另一个例子，请考虑：

pies = [0, 2, 5, 1, 3, 9, 2, 12, 33, 45]
pies[0::round(1.1x)]

输出：

[0, 2, 5, 1, 3, 2, 12, 33, 45]

我想知道你怎么能以最'pythonic'的方式做到这一点。

谢谢！

Answer 1

In [7]: pies = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

In [8]: stride=1.1

In [9]: [pies[x] for x in range(len(pies)) for x in [int(round(x*stride))] if x < len(pies)]
Out[9]: [0, 1, 2, 3, 4, 6, 7, 8, 9]

另一方面，对于生成器函数来说这似乎是一个很好的工作：

def rounding_iterator(seq, stride):
    try:
        i = 0
        while True:
            yield seq[int(round(i*stride))]
            i += 1
    except IndexError:
        pass


pies = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

print list(rounding_iterator(pies, 1))
print list(rounding_iterator(pies, 1.1))
print list(rounding_iterator(pies, .9))

结果：

[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
[0, 1, 2, 3, 4, 6, 7, 8, 9]
[0, 1, 2, 3, 4, 5, 5, 6, 7, 8, 9]

Answer 2

我已对此进行了编辑，因此您需要使用1.1的增量索引并对其进行舍入。正确的吗？

如果是这样，它应该是这样的：

>>> pies = [0, 1, 2, 3, 3, 2, 1, 77, 88, 99]
>>> print [ pies[int(round(x*0.1+x))] for x in xrange(len(pies)-1) ]
[0, 1, 2, 3, 3, 1, 77, 88, 99]

>>> pies = [0, 2, 5, 1, 3, 9, 2, 12, 33, 45]
>>> print [ pies[int(round(x*0.1+x))] for x in xrange(len(pies)-1) ]
[0, 2, 5, 1, 3, 2, 12, 33, 45]
>>>

这是你想要的吗？

与numpy相同：

>>> import numpy as np
>>> pies = [0, 2, 5, 1, 3, 9, 2, 12, 33, 45]
>>> print [ pies[int(round(x))] for x in np.arange(0,len(pies)-1,1.1) ]
[0, 2, 5, 1, 3, 2, 12, 33, 45]

使用numpy，您可以更改乘数：

>>> print [ pies[int(round(x))] for x in np.arange(0,len(pies)/1.1,1.1) ]
[0, 2, 5, 1, 3, 2, 12, 33, 45]
>>> print [ pies[int(round(x))] for x in np.arange(0,len(pies)/3.3,3.3) ]
[0]
>>> print [ pies[int(round(x))] for x in np.arange(0,len(pies)/2.2,2.2) ]
[0, 5, 3]
>>>

Answer 3

如何将列表中的值乘以1.1，将其四舍五入，并将其转换为int以删除小数：

pies = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
for pie in pies:
    print int(round(pie * 1.1))

Answer 4

如果它不需要一行......

def rounded_indices(ls):
    result = []
    original_index = 0
    new_index = 0
    while new_index < len(ls):
        result.append(ls[new_index])
        original_index += 1
        new_index = int(round(1.1 * original_index))

    return result

>>> pies = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
>>> rounded_indices(pies)
[0, 1, 2, 3, 4, 6, 7, 8, 9]

Python数组索引 - 每个X元素？

4 个答案: