python：优雅和代码保存方式转换列表

Question

首先抱歉这个简单的问题。我有一个列表（只是一个例子）

points = [[663963.7405329756, 6178165.692240637],
 [664101.4213951868, 6177971.251818423],
 [664099.7474887948, 6177963.323432223],
 [664041.432877932, 6177903.295650704],
 [664031.8017317944, 6177895.797176996],
 [663963.7405329756, 6178165.692240637]]

我需要将其转换为以下形式

points = [(663963.7405329756, 6178165.692240637),
 (664101.4213951868, 6177971.251818423),
 (664099.7474887948, 6177963.323432223),
 (664041.432877932, 6177903.295650704),
 (664031.8017317944, 6177895.797176996),
 (663963.7405329756, 6178165.692240637)]

使用shapely module创建Polygon对象。我写了几个循环，但真的不优雅和耗时。您知道将第一个列表转换为第二个列表的最佳方法吗？

由于

Answer 1

converted = map(tuple, points) # Python 2
converted = list(map(tuple, points)) # or BlackBear's answer for Python 3
converted = [tuple(x) for x in points] # another variation of the same

Answer 2

converted = [(a,b) for a,b in points]

Answer 3

converted = [tuple(l) for l in points]

与@BlackBear给出的解决方案相比，这适用于任意大小的子列表。

Answer 4

points = [tuple(x) for x in points]