Question

如果没有<条码>x,我就想制造一系列的编号: 1至<条码>。

is there a "prettier" way to do it instead of [i for i in range(n) if i != x]? thanks!

Answer 1

使用高级索引np.r_。

np.arange(n)[np.r_[0:x, x+1:n]]

def myrange(n, exclude):
    return np.arange(n)[np.r_[0:exclude, exclude+1:n]]

>>> myrange(10, exclude=3)
array([0, 1, 2, 4, 5, 6, 7, 8, 9])

%timeit myrange(10000000, 7008)
1 loop, best of 3: 79.1 ms per loop

%timeit other(10000000, 7008)
1 loop, best of 3: 952 ms per loop

地点

def myrange(n, exclude):
    return np.arange(n)[np.r_[0:exclude, exclude+1:n]]

def other(n, exclude):
    return [i for i in range(n) if i != x]

Answer 2

我建议使用itertools。

f或 a in itertools.chain(range(x), range(x+1, n)):
    print(a)

或

list(itertools.chain(range(x), range(x+1, n)))

I don t know why, but [itertools.chain(range(x), range(x+1, n))] doesn t w或k though.

EDIT: Thanks to @rafaelc f或 how to make it w或k on square brackets.

[*itertools.chain(range(x), range(x+1, n))]

Answer 3

您可以确定两个范围:

np.concatenate((np.arange(x), np.arange(x + 1, n)))

也可删除一个项目:

np.delete(np.arange(n), x)

你可以掩盖:

mask = np.ones(n, dtype=bool)
mask[x] = False
np.arange(n)[mask]

你们甚至可以使用一片面罩,视你的申请而定:

a = np.ma.array(np.arange(n), mask=np.zeross(n, dtypebool))
a.mask[x] = True