如何删除三元组列表中每个元组的第二个元素？

Question

I have a big list like this:

list__=[('string id1', 'string id2', 'string id3'), ('string id4', 'string id5', 'string id6'), ... ,('string idn', 'string id', 'string idn-1')]

How can I drop the ids from this big tuple, for example:

[('string', 'string', 'string'), ('string', 'string', 'string'), ... ,('string', 'string', 'string')]

Any idea of how can I aproach this?. I tried with:

OutputTuple = [(a, b, d) for a, b, c, d in ListTuple]

But it just drop the second element.

Answer 1

使用列表理解：

my_list = [tuple([j.split()[0] for j in i]) for i in my_list]

Answer 2

Unpacking will be more efficient than using a double for loop:

[(a.split()[0], b.split()[0], c.split()[0]) for a, b, c in  list__ ]

You could also index to the whitespace:

  [(a[:a.index(" ")], b[:b.(" ")], c[:c.index(" ")]) for a,b,c in  list__ ]

Interestingly using str.find is the most efficient solution using python2.7.

In [41]: timeit [(a[:a.find(" ")], b[:b.find(" ")], c[:c.find(" ")]) for a,b,c in  list__ ]
100000 loops, best of 3: 2.27 µs per loop

In [42]: timeit [tuple([j.split()[0] for j in i]) for i in list__]
100000 loops, best of 3: 3.85 µs per loop

In [43]: timeit [(a.split()[0], b.split()[0], c.split()[0]) for a, b, c in  list__ ]
100000 loops, best of 3: 2.73 µs per loop