Python 如何创建元组的合并列表

我有两个元组列表：

leftlist = [(1.52, u'testSphere.vtx[299]'), (1.4450000000000001, u'testSphere.vtx[298]'), (1.0, u'testSphere.vtx[277]')   ]

rightlist = [(1.52, u'testSphere.vtx[289]'), (1.4450000000000001, u'testSphere.vtx[290]'), (1.1, u'testSphere.vtx[299]')  ]

我想从每个元组中提取第二个元素，并将其放入一个新的元组列表中，如下所示：

merged_list = [('testSphere.vtx[299]', 'testSphere.vtx[289]'), ('testSphere.vtx[298]', 'testSphere.vtx[290]'), ('testSphere.vtx[277]', 'testSphere.vtx[299]') 

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最好使用列表理解还是for循环

使用zip和地图/列表理解：

zip((x[1] for x in leftlist),(y[1] for y in rightlist))
zip(map(lambda x : x[1],leftlist),map(lambda y: y[1],rightlist))

---

两者最终都会得到相同的结果

使用zip和地图/列表理解：

zip((x[1] for x in leftlist),(y[1] for y in rightlist))
zip(map(lambda x : x[1],leftlist),map(lambda y: y[1],rightlist))

---

两者最终都会得到相同的结果

下面是一个列表：

>>> leftlist = [(1.52, u'testSphere.vtx[299]'), (1.4450000000000001, u'testSphere.vtx[298]'), (1.0, u'testSphere.vtx[277]')   ]
>>> rightlist = [(1.52, u'testSphere.vtx[289]'), (1.4450000000000001, u'testSphere.vtx[290]'), (1.1, u'testSphere.vtx[299]')  ]
>>> result = [(x[1], y[1]) for x, y in zip(leftlist, rightlist)]
>>> result
[(u'testSphere.vtx[299]', u'testSphere.vtx[289]'), (u'testSphere.vtx[298]', u'testSphere.vtx[290]'), (u'testSphere.vtx[277]', u'testSphere.vtx[299]')]

只是我注意到这和一小时前删除的答案是一样的。如果@struchtmaster重新给出了答案，请将此归功于@struchtmaster

---

后来：

如果您希望代码速度更快（即使不那么直观），这很好：

result = [(x, y) for (_, x), (_, y) in zip(leftlist, rightlist)]

如果您喜欢这种微优化，您可以在此处看到计时结果：

In [4]: %timeit [(leftlist[i][1],j[1]) for i,j in enumerate(rightlist)]
1000000 loops, best of 3: 680 ns per loop

In [5]: %timeit [(x[1], y[1]) for x, y in zip(leftlist, rightlist)]
1000000 loops, best of 3: 787 ns per loop

In [6]: %timeit [(x, y) for (_, x), (_, y) in zip(leftlist, rightlist)]
1000000 loops, best of 3: 630 ns per loop

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以下是一份清单：

>>> leftlist = [(1.52, u'testSphere.vtx[299]'), (1.4450000000000001, u'testSphere.vtx[298]'), (1.0, u'testSphere.vtx[277]')   ]
>>> rightlist = [(1.52, u'testSphere.vtx[289]'), (1.4450000000000001, u'testSphere.vtx[290]'), (1.1, u'testSphere.vtx[299]')  ]
>>> result = [(x[1], y[1]) for x, y in zip(leftlist, rightlist)]
>>> result
[(u'testSphere.vtx[299]', u'testSphere.vtx[289]'), (u'testSphere.vtx[298]', u'testSphere.vtx[290]'), (u'testSphere.vtx[277]', u'testSphere.vtx[299]')]

只是我注意到这和一小时前删除的答案是一样的。如果@struchtmaster重新给出了答案，请将此归功于@struchtmaster

---

后来：

如果您希望代码速度更快（即使不那么直观），这很好：

result = [(x, y) for (_, x), (_, y) in zip(leftlist, rightlist)]

如果您喜欢这种微优化，您可以在此处看到计时结果：

In [4]: %timeit [(leftlist[i][1],j[1]) for i,j in enumerate(rightlist)]
1000000 loops, best of 3: 680 ns per loop

In [5]: %timeit [(x[1], y[1]) for x, y in zip(leftlist, rightlist)]
1000000 loops, best of 3: 787 ns per loop

In [6]: %timeit [(x, y) for (_, x), (_, y) in zip(leftlist, rightlist)]
1000000 loops, best of 3: 630 ns per loop

---

您还可以使用enumerate，它似乎是最有效的

leftlist = [(1.52, u'testSphere.vtx[299]'), (1.4450000000000001, u'testSphere.vtx[298]'), (1.0, u'testSphere.vtx[277]')   ]

rightlist = [(1.52, u'testSphere.vtx[289]'), (1.4450000000000001, u'testSphere.vtx[290]'), (1.1, u'testSphere.vtx[299]')  ]

In [12]: [(leftlist[i][1],j[1]) for i,j in enumerate(rightlist)]
Out[12]: 
[(u'testSphere.vtx[299]', u'testSphere.vtx[289]'(u'testSphere.vtx[298]',u'testSphere.vtx[290]'),(u'testSphere.vtx[277]', u'testSphere.vtx[299]')]





In [7]: %timeit [(leftlist[i][1],j[1]) for i,j in enumerate(rightlist)
]1000000 loops, best of 3: 849 ns per loop

In [8]: %timeit  [(x[1], y[1]) for x, y in zip(leftlist, rightlist)]
1000000 loops, best of 3: 951 ns per loop

In [9]: %timeit zip(map(lambda x : x[1],leftlist),map(lambda y: y[1],rightlist))
100000 loops, best of 3: 1.86 µs per loop

In [10]: %timeit zip((x[1] for x in leftlist),(y[1] for y in rightlist))
100000 loops, best of 3: 2.11 µs per loop

---

您还可以使用enumerate，它似乎是最有效的

leftlist = [(1.52, u'testSphere.vtx[299]'), (1.4450000000000001, u'testSphere.vtx[298]'), (1.0, u'testSphere.vtx[277]')   ]

rightlist = [(1.52, u'testSphere.vtx[289]'), (1.4450000000000001, u'testSphere.vtx[290]'), (1.1, u'testSphere.vtx[299]')  ]

In [12]: [(leftlist[i][1],j[1]) for i,j in enumerate(rightlist)]
Out[12]: 
[(u'testSphere.vtx[299]', u'testSphere.vtx[289]'(u'testSphere.vtx[298]',u'testSphere.vtx[290]'),(u'testSphere.vtx[277]', u'testSphere.vtx[299]')]





In [7]: %timeit [(leftlist[i][1],j[1]) for i,j in enumerate(rightlist)
]1000000 loops, best of 3: 849 ns per loop

In [8]: %timeit  [(x[1], y[1]) for x, y in zip(leftlist, rightlist)]
1000000 loops, best of 3: 951 ns per loop

In [9]: %timeit zip(map(lambda x : x[1],leftlist),map(lambda y: y[1],rightlist))
100000 loops, best of 3: 1.86 µs per loop

In [10]: %timeit zip((x[1] for x in leftlist),(y[1] for y in rightlist))
100000 loops, best of 3: 2.11 µs per loop

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zip（[x[1]表示左列表中的x]，[x[1]表示右列表中的x]）

？

zip（[x[1]表示左列表中的x]，[x[1]表示右列表中的x]）

？+1用于记住在第一行末尾使用生成器表达式而不是列表理解u missesd）+1用于记住在第一行末尾使用生成器表达式而不是列表理解u missesd）