在 python 中切片类似于 MATLAB

Question

In Matlab, slice can be a vector:

a = {'a','b','c','d','e','f','g'}; % cell array
b = a([1:3,5,7]);

How can I do the same thing in python?

a = ['a','b','c','d','e','f','g']
b = [a[i] for i in [0,1,2,4,6]]

but when 1:3 becomes 1:100, this will not work. Using range(2),4,6 returns ([0,1,2],4,6), not (0,1,2,4,6). Is there a fast and "pythonic" way?

Answer 1

If you want to do things that are similar to Matlab in Python, NumPy should always be your first guess. In this case, you need numpy.r_ :

from numpy import array, r_
a = array(['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h'])
print a[r_[1:3, 5, 7]]

['b' 'c' 'f' 'h']

Answer 2

One way is using itertools.chain :

>>> b = [a[i] for i in itertools.chain(range(2), [5, 6])]
>>> b
['b', 'c', 'f', 'g']

Notes:

1. Ranges adapted from Matlab (1-based indexing) to Python (0-based indexing)
2. You may gain by changing range to xrange if you have Python 2.x, to avoid creating the whole range list on the fly. I don't think it will make a big performance difference, but it's nice to know about.

Answer 3

Try

[a[i] for i in range(2) + [4, 6]]

If you use NumPy, then you have some more options:

import numpy as N
a = N.array(['a', 'b', 'c', 'd', 'e', 'f', 'g'])
b = a[range(2) + [4, 6]]
c = a.take(range(2) + [4, 6])