Question

我有一个元组列表和带有相关值的列表：

unsorted = [(1,3), (None,4), (19,31), (20,32), (8,None), (9,14)]
dists =  [3,None,4,5,None,9]

我想对“未排序”的元组列表进行排序，看起来像这样：

sorted = [(1,3), (None,4), (8,None), (9,14), (19,31), (20,32)]

如果元组中的一个元素为None，则取第二个元素（它们总是递增，因此不存在[(8,12),(9,11)]）

的情况

同时我想对相关的dists进行排序：

dists_sorted = [3, None, None, 9, 4, 5]

我知道我可以这样做：

sorted, dists_sorted = (list(t) for t in zip(*sorted(zip(unsorted, dists))))

但我不知道如何正确地对Nones进行排序。

提前致谢！

编辑：

我在更长的列表上尝试了你的方法（它已经排序了）：

l = [(901, 1), (902, 2), (903, 3), (904, 4), (905, 5), (906, 6), (907, 7), (908, 8), (909, 9), (910, 10), (911, 11), (912, 12), (913, 13), (914, 14), (915, 15), (916, None), (None, 16), (None, 17), (917, 18), (918, 19), (None, 20), (919, 21), (920, 22), (921, 23), (922, 24), (923, 25), (924, 26), (925, 27), (926, 28), (927, 29), (928, 30), (929, 31), (930, 32), (931, 33), (932, None), (933, None), (934, None), (935, None), (None, 34), (None, 35), (None, 36), (None, 37), (None, 38), (None, 39), (936, 40), (937, None), (938, None), (939, 41), (940, 42), (941, 43), (942, 44), (943, 45), (944, 46), (945, 47), (946, 48), (947, 49), (948, 50), (949, 51), (950, 52), (951, 53), (952, 54), (953, 55), (954, 56), (955, 57), (956, 58), (957, 59), (958, 60), (959, 61), (960, 62), (961, 63), (962, 64), (963, 65), (964, 66), (965, 67), (966, 68), (967, 69), (968, 70), (969, 71), (970, 72), (971, 73), (972, 74), (973, None), (974, None), (975, None), (None, 75), (None, 76), (None, 77)]

我得到了这个：

out = [(None, 16), (None, 17), (None, 20), (None, 34), (None, 35), (None, 36), (None, 37), (None, 38), (None, 39), (None, 75), (None, 76), (None, 77), (901, 1), (902, 2), (903, 3), (904, 4), (905, 5), (906, 6), (907, 7), (908, 8), (909, 9), (910, 10), (911, 11), (912, 12), (913, 13), (914, 14), (915, 15), (916, None), (917, 18), (918, 19), (919, 21), (920, 22), (921, 23), (922, 24), (923, 25), (924, 26), (925, 27), (926, 28), (927, 29), (928, 30), (929, 31), (930, 32), (931, 33), (932, None), (933, None), (934, None), (935, None), (936, 40), (937, None), (938, None), (939, 41), (940, 42), (941, 43), (942, 44), (943, 45), (944, 46), (945, 47), (946, 48), (947, 49), (948, 50), (949, 51), (950, 52), (951, 53), (952, 54), (953, 55), (954, 56), (955, 57), (956, 58), (957, 59), (958, 60), (959, 61), (960, 62), (961, 63), (962, 64), (963, 65), (964, 66), (965, 67), (966, 68), (967, 69), (968, 70), (969, 71), (970, 72), (971, 73), (972, 74), (973, None), (974, None), (975, None)]

所以一开始就崩溃了。你有什么想法吗？

Answer 1

key参数可以帮助您解决它。无论如何看看sorted mini HOWTO。是一个非常好的资源。

>>> unsorted = [(1,3), (None,4), (19,31), (20,32), (8,None), (9,14)]
>>> sorted(unsorted, key=lambda x:x[0] if x[0] is not None else x[1])
[(1, 3), (None, 4), (8, None), (9, 14), (19, 31), (20, 32)]

<强> [编辑] 根据以上评论，您希望(a,b)的规则少于(c,d)当且仅为：

如果c且d不是None c<d
如果c或d为无a<b

......但这不是一个排序规则。在示例的最后，您说l已排序，但您可以找到(916, None)<(None, 16)<(932, None)：没有规则可以执行此操作。

Answer 2

您可以使用适当的key排序：

>>> sorted(unsorted,key= lambda x: x[0] if x[0] else x[1])
[(1, 3), (None, 4), (8, None), (9, 14), (19, 31), (20, 32)]

如评论中所述，如果您有0，请使用is not None检查None：

>>> sorted(unsorted,key= lambda x: x[0] if x[0] is not None else x[1])
[(1, 3), (None, 4), (8, None), (9, 14), (19, 31), (20, 32)]

Answer 3

一种方式（更容易理解）将是：

def f(e):
    if e[0] is None:
        return e[1]
    return e[0]

print sorted(unsorted, key = f)

使用lambda的简化方法是：

print sorted(unsorted, lambda e: e[1] if e[0] is None else e[0])

两者都会导致相同的结果。

修改

为了与您所做的相匹配，您可以对函数f进行一些修改：

def f(e): if e[0][0] is None: return e[0][1] return e[0][0] sorted1, dists_sorted = (list(t) for t in zip(*sorted(zip(unsorted, dists), key = f))) >>> [(1, 3), (None, 4), (8, None), (9, 14), (19, 31), (20, 32)] [3, None, None, 9, 4, 5]

请注意，e[0]将是unsorted的项目，e[1]中的dists对应{。}}。

使用Nones对元组列表进行排序，同时对相关列表进行排序

3 个答案: