python - 在字典中追加多个值，但避免python中字典值中的重复元素

Question

我有这样的列表:

[('0023_RIGHT_CC', [(2574, 2798), (1324, 1545)]),
 ('0021_LEFT_CC', [(1180, 1420), (883, 1140)]),
 ('0106_LEFT_CC', [(911, 1135), (0, 86)]),
 ('0026_LEFT_CC', [(3738, 3968), (2144, 2352)]),
 ('0021_RIGHT_CC', [(2170, 2314), (1642, 1795)]),
 ('0106_LEFT_CC', [(1679, 1833), (964, 1102)]),
 ('0106_LEFT_CC', [(1091, 1198), (65, 160)]),
 ('0021_LEFT_CC', [(1180, 1420), (883, 1140)]),
 ('0021_RIGHT_CC', [(2170, 2314), (1642, 1795)]),
 ('0106_LEFT_CC', [(911, 1135), (0, 86)]),
 ('0106_LEFT_CC', [(1679, 1833), (964, 1102)]),
 ('0106_LEFT_CC', [(1091, 1198), (65, 160)]),
 ('0018_RIGHT_CC', [(1388, 1653), (1894, 2197)]),
 ('0023_RIGHT_CC', [(2574, 2798), (1324, 1545)]),
 ('0026_LEFT_CC', [(3738, 3968), (2144, 2352)])]

从这里我想创建一个这样的字典.. 例如;

0106_LEFT_CC: [[(911, 1135), (0, 86)], [(1091, 1198), (65, 160)], [(1679, 1833), (964, 1102)]]

请注意，应避免在字典的value 中出现重复的元素。我做了如下;

from collections import defaultdict

d = defaultdict(list)
for key, value in data:
    d[key].append(value)

我得到了以下结果。

defaultdict(list,
            {'0023_RIGHT_CC': [[(2574, 2798), (1324, 1545)],
              [(2574, 2798), (1324, 1545)]],
             '0021_LEFT_CC': [[(1180, 1420), (883, 1140)],
              [(1180, 1420), (883, 1140)]],
             '0106_LEFT_CC': [[(911, 1135), (0, 86)],
              [(1679, 1833), (964, 1102)],
              [(1091, 1198), (65, 160)],
              [(911, 1135), (0, 86)],
              [(1679, 1833), (964, 1102)],
              [(1091, 1198), (65, 160)]],
             '0026_LEFT_CC': [[(3738, 3968), (2144, 2352)],
              [(3738, 3968), (2144, 2352)]],
             '0021_RIGHT_CC': [[(2170, 2314), (1642, 1795)],
              [(2170, 2314), (1642, 1795)]],
             '0018_RIGHT_CC': [[(1388, 1653), (1894, 2197)]]})

但是结果显示，例如，在 key 0106_LEFT_CC 中，一些 values 重复了。如何避免它。

Answer 1

您可以将它们转换成一个set，然后再转换成一个list。这将使您摆脱重复的元素。

from collections import defaultdict

data = [('0023_RIGHT_CC', [(2574, 2798), (1324, 1545)]),
 ('0021_LEFT_CC', [(1180, 1420), (883, 1140)]),
 ('0106_LEFT_CC', [(911, 1135), (0, 86)]),
 ('0026_LEFT_CC', [(3738, 3968), (2144, 2352)]),
 ('0021_RIGHT_CC', [(2170, 2314), (1642, 1795)]),
 ('0106_LEFT_CC', [(1679, 1833), (964, 1102)]),
 ('0106_LEFT_CC', [(1091, 1198), (65, 160)]),
 ('0021_LEFT_CC', [(1180, 1420), (883, 1140)]),
 ('0021_RIGHT_CC', [(2170, 2314), (1642, 1795)]),
 ('0106_LEFT_CC', [(911, 1135), (0, 86)]),
 ('0106_LEFT_CC', [(1679, 1833), (964, 1102)]),
 ('0106_LEFT_CC', [(1091, 1198), (65, 160)]),
 ('0018_RIGHT_CC', [(1388, 1653), (1894, 2197)]),
 ('0023_RIGHT_CC', [(2574, 2798), (1324, 1545)]),
 ('0026_LEFT_CC', [(3738, 3968), (2144, 2352)])]

d = defaultdict(list)
for key, value in data:
    d[key] = list(set(d[key] + value))

---

编辑:我稍微更改了@JST99 的答案，现在它使用set 进行in 检查，然后将值添加到列表.这在大型数据集上更快，但对于您当前的数据，@JST99 的答案更快。

from collections import defaultdict


data = [('0023_RIGHT_CC', [(2574, 2798), (1324, 1545)]),
 ('0021_LEFT_CC', [(1180, 1420), (883, 1140)]),
 ('0106_LEFT_CC', [(911, 1135), (0, 86)]),
 ('0026_LEFT_CC', [(3738, 3968), (2144, 2352)]),
 ('0021_RIGHT_CC', [(2170, 2314), (1642, 1795)]),
 ('0106_LEFT_CC', [(1679, 1833), (964, 1102)]),
 ('0106_LEFT_CC', [(1091, 1198), (65, 160)]),
 ('0021_LEFT_CC', [(1180, 1420), (883, 1140)]),
 ('0021_RIGHT_CC', [(2170, 2314), (1642, 1795)]),
 ('0106_LEFT_CC', [(911, 1135), (0, 86)]),
 ('0106_LEFT_CC', [(1679, 1833), (964, 1102)]),
 ('0106_LEFT_CC', [(1091, 1198), (65, 160)]),
 ('0018_RIGHT_CC', [(1388, 1653), (1894, 2197)]),
 ('0023_RIGHT_CC', [(2574, 2798), (1324, 1545)]),
 ('0026_LEFT_CC', [(3738, 3968), (2144, 2352)])]


d = defaultdict(list)
s = defaultdict(set)
for key, value in data:
    s_value = tuple(value) # lists are unhasable, so we will use tuple
    if s_value not in s[key]: 
        d[key].append(value)
        s[key].add(s_value)

del s # we dont need s anymore