Python 如何将元组列表转换为字典列表？

我有一个列表，它是元组，看起来像：

z = [[(2, 81), (2, 52), (2, 53), (2, 86), (3, 4), (3, 5), (3, 72), (3, 9), (3, 16), (3, 17), (3, 18), (3, 84), (3, 73), (3, 41), (3, 63), (3, 30), (3, 31), (4, 48), (4, 5), (4, 38), (4, 7), (4, 8), (4, 95), (4, 11), (4, 12), (4, 16), (4, 19), (4, 53), (4, 54), (4, 57), (4, 27), (4, 30), (4, 37), (5, 51), (5, 67), (5, 6), (5, 65), (5, 11), (5, 15), (5, 19), (5, 21), (5, 24), (5, 36), (5, 71), (5, 31), (6, 83), (6, 71), (6, 76), (6, 98), (6, 78), (6, 16), (6, 19), (6, 20), (6, 22), (6, 87), (6, 52), (6, 27), (6, 86), (7, 49), (7, 33), (7, 60), (7, 96), (7, 43), (7, 12), (7, 14), (7, 15), (7, 17), (7, 46), (7, 22), (7, 28), (8, 32), (8, 34), (8, 98), (8, 40), (8, 96), (8, 10), (8, 44), (8, 13), (8, 84), (8, 21), (8, 87), (8, 28), (8, 30), (9, 65), (9, 40), (9, 14), (9, 55), (9, 25)], [(2, 32), (2, 3), (2, 39), (2, 8), (2, 9), (2, 10), (2, 43), (2, 13)], [(1, 85), (1, 22), (1, 23), (1, 56), (1, 25), (1, 70), (1, 42)], [(1, 79), (1, 17), (1, 67)], [(1, 54), (1, 6), (1, 7), (1, 92), (1, 10)], [(1, 3), (1, 4)], [(0, 91), (0, 92)], [(0, 88), (0, 89)], [(0, 70), (0, 74), (0, 83)], [(0, 66), (0, 68), (0, 69)], [(0, 26), (0, 29), (0, 33), (0, 35), (0, 38), (0, 41), (0, 45), (0, 50), (0, 51), (0, 55), (0, 57), (0, 59), (0, 65)], [(0, 24)], [(0, 14), (0, 15), (0, 20)],[(0, 13)], [(0, 9), (0, 11)], [(0, 7), (0, 8)], [(0, 5), (0, 6)], [(0, 3), (0, 4)]]

现在我真的想把这些列表整理成字典。所以我使用了集合中的defaultdict：

d = defaultdict(list)

for k,v in z[0]:
    d[k].append(v)

sorted(d.items())

我想让它把所有的清单分类成字典。我试着在z上循环，但不起作用。我用过：

for i in np.arange(len(z)):
    d = defaultdict(list)

    for k,v in z[i]:
        d[k].append(v)

sorted(d.items())

---

如何将字典列表作为一个大列表？

我将您的问题解释为，您希望为每个键对列表进行排序。在这种情况下：

from collections import defaultdict
d = defaultdict(list)

for k,v in z[0]:
    d[k].append(v)

d = {k: sorted(v) for k, v in d.items()}

因此d变成：

{2: [52, 53, 81, 86],
 3: [4, 5, 9, 16, 17, 18, 30, 31, 41, 63, 72, 73, 84],
 4: [5, 7, 8, 11, 12, 16, 19, 27, 30, 37, 38, 48, 53, 54, 57, 95],
 5: [6, 11, 15, 19, 21, 24, 31, 36, 51, 65, 67, 71],
 6: [16, 19, 20, 22, 27, 52, 71, 76, 78, 83, 86, 87, 98],
 7: [12, 14, 15, 17, 22, 28, 33, 43, 46, 49, 60, 96],
 8: [10, 13, 21, 28, 30, 32, 34, 40, 44, 84, 87, 96, 98],
 9: [14, 25, 40, 55, 65]}

根据您在下面的评论，这可以简单地扩展到列表列表：

out = []
for entry in z:
    d = defaultdict(list)
    for k,v in entry:
        d[k].append(v)
    out.append([{k: sorted(v) for k, v in d.items()}])
print(out)

给予

[[{2: [52, 53, 81, 86],
   3: [4, 5, 9, 16, 17, 18, 30, 31, 41, 63, 72, 73, 84],
   4: [5, 7, 8, 11, 12, 16, 19, 27, 30, 37, 38, 48, 53, 54, 57, 95],
   5: [6, 11, 15, 19, 21, 24, 31, 36, 51, 65, 67, 71],
   6: [16, 19, 20, 22, 27, 52, 71, 76, 78, 83, 86, 87, 98],
   7: [12, 14, 15, 17, 22, 28, 33, 43, 46, 49, 60, 96],
   8: [10, 13, 21, 28, 30, 32, 34, 40, 44, 84, 87, 96, 98],
   9: [14, 25, 40, 55, 65]}],
 [{2: [3, 8, 9, 10, 13, 32, 39, 43]}],
 [{1: [22, 23, 25, 42, 56, 70, 85]}],
 [{1: [17, 67, 79]}],
 [{1: [6, 7, 10, 54, 92]}],
 [{1: [3, 4]}],
 [{0: [91, 92]}],
 [{0: [88, 89]}],
 [{0: [70, 74, 83]}],
 [{0: [66, 68, 69]}],
 [{0: [26, 29, 33, 35, 38, 41, 45, 50, 51, 55, 57, 59, 65]}],
 [{0: [24]}],
 [{0: [14, 15, 20]}],
 [{0: [13]}],
 [{0: [9, 11]}],
 [{0: [7, 8]}],
 [{0: [5, 6]}],
 [{0: [3, 4]}]]

---

您可以使用

itertools.groupby

和

operator.itemgetter

对每个元组中的键和值进行分组

from itertools import groupby
from operator import itemgetter

[{k: sorted(map(itemgetter(1), v)) for k, v in groupby(l, itemgetter(0))} for l in z]

结果:

[{2: [52, 53, 81, 86], 3: [4, 5, 9, 16, 17, 18, 30, 31, 41, 63, 72, 73, 84], 4: [5, 7, 8, 11, 12, 16, 19, 27, 30, 37, 38, 48, 53, 54, 57, 95], 5: [6, 11, 15, 19, 21, 24, 31, 36, 51, 65, 67, 71], 6: [16, 19, 20, 22, 27, 52, 71, 76, 78, 83, 86, 87, 98], 7: [12, 14, 15, 17, 22, 28, 33, 43, 46, 49, 60, 96], 8: [10, 13, 21, 28, 30, 32, 34, 40, 44, 84, 87, 96, 98], 9: [14, 25, 40, 55, 65]}, {2: [3, 8, 9, 10, 13, 32, 39, 43]}, {1: [22, 23, 25, 42, 56, 70, 85]}, {1: [17, 67, 79]}, {1: [6, 7, 10, 54, 92]}, {1: [3, 4]}, {0: [91, 92]}, {0: [88, 89]}, {0: [70, 74, 83]}, {0: [66, 68, 69]}, {0: [26, 29, 33, 35, 38, 41, 45, 50, 51, 55, 57, 59, 65]}, {0: [24]}, {0: [14, 15, 20]}, {0: [13]}, {0: [9, 11]}, {0: [7, 8]}, {0: [5, 6]}, {0: [3, 4]}]

---

您也可以这样做，而无需使用任何导入进行数据操作

from pprint import pprint

l = []
for list_item in z:
    d = {}
    for index, value in list_item:
        if index not in d:
            d[index] = [value]
        else:
            d[index].append(value)
    l.append({k: sorted(v) for k, v in d.items()})

pprint(l)

输出结果如下：

[{2: [52, 53, 81, 86],
  3: [4, 5, 9, 16, 17, 18, 30, 31, 41, 63, 72, 73, 84],
  4: [5, 7, 8, 11, 12, 16, 19, 27, 30, 37, 38, 48, 53, 54, 57, 95],
  5: [6, 11, 15, 19, 21, 24, 31, 36, 51, 65, 67, 71],
  6: [16, 19, 20, 22, 27, 52, 71, 76, 78, 83, 86, 87, 98],
  7: [12, 14, 15, 17, 22, 28, 33, 43, 46, 49, 60, 96],
  8: [10, 13, 21, 28, 30, 32, 34, 40, 44, 84, 87, 96, 98],
  9: [14, 25, 40, 55, 65]},
 {2: [3, 8, 9, 10, 13, 32, 39, 43]},
 {1: [22, 23, 25, 42, 56, 70, 85]},
 {1: [17, 67, 79]},
 {1: [6, 7, 10, 54, 92]},
 {1: [3, 4]},
 {0: [91, 92]},
 {0: [88, 89]},
 {0: [70, 74, 83]},
 {0: [66, 68, 69]},
 {0: [26, 29, 33, 35, 38, 41, 45, 50, 51, 55, 57, 59, 65]},
 {0: [24]},
 {0: [14, 15, 20]},
 {0: [13]},
 {0: [9, 11]},
 {0: [7, 8]},
 {0: [5, 6]},
 {0: [3, 4]}]

---

你期望的结果是什么？你所说的“把所有的列表分类成字典”是什么意思？您期望的输出是什么？我希望列表列表中的每个列表都作为字典排序，并且所有这些字典都作为单个列表包含在大列表中。实际上，我希望列表列表中的每个列表都作为字典排序，并且所有这些字典都作为单个列表包含在大列表中。这有意义吗？