How to get a list excluding elements whose index is i ...?

I'm curious about the fastest way to get a list without * changes * to the list ʻitems * without ʻitems [i], or a reasonably fast and smart way to write (sometimes excluding the i-th in math). , Because I want something like that).

I measured it with % timeit of Jupyter.

If you don't use numpy

items = list(range(1000000))
i = 17 #suitable

Combine slices

%timeit items[:i] + items[i+1:]
11.2 ms ± 349 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

Included notation (using ʻenumerate`)

%timeit [item for k, item in enumerate(items) if k != i]
65.1 ms ± 212 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

Comprehension notation

%timeit [items[k] for k in range(len(items)) if k != i]
75.5 ms ± 800 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

It seems that slicing is overwhelmingly faster. By the way, I was a little surprised that using the second ʻenumerate` is faster than the third example.

When using numpy

@ antimon2's Maru Park, but there are the following methods.

import numpy as np
items = np.arange(1000000)
i = 17

np.hstack

%timeit np.hstack([items[:i], items[i+1:]])
584 µs ± 6.32 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

np.delete

%timeit np.delete(items, i)
585 µs ± 3.73 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

Boolean index

%timeit items[np.arange(items.size) != i]
1.9 ms ± 5.82 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

Comprehension notation

%timeit np.array([items[k] for k in range(len(items)) if k != i])
182 ms ± 3.85 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

Again, the slices combine very quickly. Considering readability, is np.delete the best because the speed does not change so much?

If you have any other good ones, please let me know.

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