Python 反转列表中的元组

我有以下清单：

[（3,28）、（25,126）、（25,127）、（26,59）]

我怎样才能把它变成这样：

[（28,3）、（126,25）、（127,25）、（59,26）]

我只想反转元组中的内容

>>> lst = [(3, 28), (25, 126), (25, 127), (26, 59)]
>>> [i[::-1] for i in lst]
[(28, 3), (126, 25), (127, 25), (59, 26)]

[：：-1]

使用将其前面的容器反转。注意，这仅适用于支持切片语法的容器

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[：：-1]

使用将其前面的容器反转。注意，这仅适用于支持切片语法的容器。

使用步骤为-1的切片

>>> [x[::-1] for x in [(3, 28), (25, 126), (25, 127), (26, 59)]]
[(28, 3), (126, 25), (127, 25), (59, 26)]

---

使用步长为-1的切片

>>> [x[::-1] for x in [(3, 28), (25, 126), (25, 127), (26, 59)]]
[(28, 3), (126, 25), (127, 25), (59, 26)]

---

如果您知道元组的长度仅为2：

[(b, a) for a, b in lst]

---

如果您知道元组的长度仅为2：

[(b, a) for a, b in lst]

可能只对两个元素有用

可能只对两种元素有用。

有趣的选择

>>> L = [(3, 28), (25, 126), (25, 127), (26, 59)]
>>> zip(*zip(*L)[::-1])
[(28, 3), (126, 25), (127, 25), (59, 26)]

或者对于某些邪恶的过早优化：

>>> from operator import itemgetter
>>> L = [(3, 28), (25, 126), (25, 127), (26, 59)]
>>> map(itemgetter(slice(None, None, -1)), L)
[(28, 3), (126, 25), (127, 25), (59, 26)]

时间：

python -m timeit -s "L = [(3, 28), (25, 126), (25, 127), (26, 59)
]" "[i[::-1] for i in L]"
1000000 loops, best of 3: 1.21 usec per loop   

python -m timeit -s "L = [(3, 28), (25, 126), (25, 127), (26, 59)
]" "zip(*zip(*L)[::-1])"
100000 loops, best of 3: 2.26 usec per loop   

python -m timeit -s "L = [(3, 28), (25, 126), (25, 127), (26, 59)
]; from operator import itemgetter;" "map(itemgetter(slice(None, None, -1)), L)"    
100000 loops, best of 3: 1.69 usec per loop

python -m timeit -s "L = [(3, 28), (25, 126), (25, 127), (26, 59)
]*100" "[i[::-1] for i in L]"
10000 loops, best of 3: 87.4 usec per loop

python -m timeit -s "L = [(3, 28), (25, 126), (25, 127), (26, 59)
]*100" "zip(*zip(*L)[::-1])"
10000 loops, best of 3: 67.1 usec per loop

python -m timeit -s "L = [(3, 28), (25, 126), (25, 127), (26, 59)
]*100;from operator import itemgetter;" "map(itemgetter(slice(None, None, -1)),
L)"
10000 loops, best of 3: 66.1 usec per loop

python -m timeit -s "L = [(3, 28), (25, 126), (25, 127), (26, 59)
]*100000" "[i[::-1] for i in L]"
10 loops, best of 3: 108 msec per loop

python -m timeit -s "L = [(3, 28), (25, 126), (25, 127), (26, 59)
]*100000" "zip(*zip(*L)[::-1])"
10 loops, best of 3: 109 msec per loop

python -m timeit -s "L = [(3, 28), (25, 126), (25, 127), (26, 59)
]*100000;from operator import itemgetter;" "map(itemgetter(slice(None, None, -1)
), L)"
10 loops, best of 3: 82.9 msec per loop

有趣的选择

>>> L = [(3, 28), (25, 126), (25, 127), (26, 59)]
>>> zip(*zip(*L)[::-1])
[(28, 3), (126, 25), (127, 25), (59, 26)]

或者对于某些邪恶的过早优化：

>>> from operator import itemgetter
>>> L = [(3, 28), (25, 126), (25, 127), (26, 59)]
>>> map(itemgetter(slice(None, None, -1)), L)
[(28, 3), (126, 25), (127, 25), (59, 26)]

时间：

python -m timeit -s "L = [(3, 28), (25, 126), (25, 127), (26, 59)
]" "[i[::-1] for i in L]"
1000000 loops, best of 3: 1.21 usec per loop   

python -m timeit -s "L = [(3, 28), (25, 126), (25, 127), (26, 59)
]" "zip(*zip(*L)[::-1])"
100000 loops, best of 3: 2.26 usec per loop   

python -m timeit -s "L = [(3, 28), (25, 126), (25, 127), (26, 59)
]; from operator import itemgetter;" "map(itemgetter(slice(None, None, -1)), L)"    
100000 loops, best of 3: 1.69 usec per loop

python -m timeit -s "L = [(3, 28), (25, 126), (25, 127), (26, 59)
]*100" "[i[::-1] for i in L]"
10000 loops, best of 3: 87.4 usec per loop

python -m timeit -s "L = [(3, 28), (25, 126), (25, 127), (26, 59)
]*100" "zip(*zip(*L)[::-1])"
10000 loops, best of 3: 67.1 usec per loop

python -m timeit -s "L = [(3, 28), (25, 126), (25, 127), (26, 59)
]*100;from operator import itemgetter;" "map(itemgetter(slice(None, None, -1)),
L)"
10000 loops, best of 3: 66.1 usec per loop

python -m timeit -s "L = [(3, 28), (25, 126), (25, 127), (26, 59)
]*100000" "[i[::-1] for i in L]"
10 loops, best of 3: 108 msec per loop

python -m timeit -s "L = [(3, 28), (25, 126), (25, 127), (26, 59)
]*100000" "zip(*zip(*L)[::-1])"
10 loops, best of 3: 109 msec per loop

python -m timeit -s "L = [(3, 28), (25, 126), (25, 127), (26, 59)
]*100000;from operator import itemgetter;" "map(itemgetter(slice(None, None, -1)
), L)"
10 loops, best of 3: 82.9 msec per loop

---

你打算自己写这个项目吗？或者你只是想让StackOverflow来写？你打算自己写这个项目吗？或者你只是想让StackOverflow写入？@Volatility有趣的是，它甚至不比列表comp短。然而，它产生了一些有趣的时间。。。我会很快把它们贴出来。其实这并不有趣，但我贴了一些时间安排，展示了最快的方法。双

zip

在大屏幕上快速减速data@Volatility有趣的是，它甚至不比列表中的comp短。然而，它产生了一些有趣的时间。。。我会很快把它们贴出来。其实这并不有趣，但我贴了一些时间安排，展示了最快的方法。双

zip

在巨大的数据上快速减速，比

[（b，a）对于a，b在L中]

比

[（b，a）对于a，b在L中]