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二次元配列で重複部分を検索する方法を考えています。

以下の配列を例にとります。

 1    2    3    4    5
 6    7    8    9   10
11   12   13   14   15
16   17   18   19   20
21   22   23   24   25
26    *8    9*   29   30
31   *13   14   15*   35
17   *18   19*   39   40
41   *23   24*   44   45
46   47   48   49   50

重複した領域を自動的に検索して座標を保存できる方法はありますか?

4

4 に答える 4

1 
>>> l=[[1,    2,    3,    4,    5],
... [6,    7,    8,    9,   10],
... [11,   12,   13,   14,   15],
... [16,   17,   18,   19,   20],
... [21,   22,   23,   24,   25],
... [26,    8,    9,   29,   30],
... [31,   13,   14,   15,   35],
... [17,   18,   19,   39,   40],
... [41,   23,   24,   44,   45],
... [46,   47,   48,   49,   50]]
>>> seen = set()
>>> dupes = {}
>>> for i_index, i in enumerate(l):
...     for j_index, j in enumerate(i):
...         if j in seen:
...             dupes[(i_index, j_index)] = j
...         seen.add(j)
...
>>> for coord, num in dupes.iteritems():
...     print "%s: %s" % (coord, num)
...
(7, 0): 17
(8, 2): 24
(7, 1): 18
(8, 1): 23
(6, 1): 13
(6, 3): 15
(6, 2): 14
(5, 1): 8
(5, 2): 9
(7, 2): 19
于 2013-04-29T01:41:05.907 に答える
0

あなたの質問を正しく理解できれば、単一の重複した値だけでなく、一連の値を探す必要があります。つまり、 in say[1,2,3,4]の重複が見つかります。[2,3,4][39,87,2,3,4]

インポートとテスト値

import itertools,pprint
from collections import defaultdict
l = ((1, 2, 3, 4, 5),
 (6, 7, 8, 9, 10),
 (11, 12, 13, 14, 15),
 (16, 17, 18, 19, 20),
 (21, 22, 23, 24, 25),
 (26, 8, 9, 29, 30),
 (31, 13, 14, 15, 35),
 (17, 18, 19, 39, 40),
 (41, 23, 24, 44, 45),
 (46, 47, 48, 49, 50))

メインコード:

seen = defaultdict(dict)
for y,row in enumerate(l):
        rowlen = len(row)
        values = [ [ (row[i:k+1]) for (i,k) in zip(range(rowlen),range(e,rowlen,1))] for e in range(rowlen) ]
        for valueGroup in values:
            for x,value in enumerate(valueGroup):
                seen[value]['count'] = seen[value].get('count',0) + 1
                seen[value]['x-coOrd'] = x
                seen[("R",y)][value] = True

for y in range(len(l)):
    my_rows_vals = seen[("R",y)].keys()
    for value in my_rows_vals:
        if seen[value]['count'] > 1:
            print "{0} repeated at ({1},{2})".format(value,seen[value]['x-coOrd'],y)

サンプルとして出力します（さらに出力がありました）：

(13, 14) repeated at (1,6)
(14, 15) repeated at (2,6)
(13,) repeated at (1,6)
(13, 14, 15) repeated at (1,6)
(14,) repeated at (2,6)
(17, 18) repeated at (0,7)
(18, 19) repeated at (1,7)
(17,) repeated at (0,7)
(18,) repeated at (1,7)
(19,) repeated at (2,7)
(17, 18, 19) repeated at (0,7)
(23,) repeated at (1,8)
(24,) repeated at (2,8)
(23, 24) repeated at (1,8)

リストの理解ロジックは、この例に基づいて推論されました

 l = [1,2,3,4]
 len = 4
 i:k
 0:1 1:2 2:3 3:4  i = 0,1,2,len-e  k = e,e+1,e+2,e+3    e = 0
 0:2 1:3 2:4      i = 0,1,len-e    k = e,e+1,e+2        e = 1
 0:3 1:4          i = 0,len-e      k = e,e+1            e = 2
 0:4              i = len-e        k = e                e = 3

この方法は、個々の数字と一連の数字の両方をチェックし、一致に関与する両方の当事者を強調するため、他の回答とは異なります。

于 2013-04-29T05:21:43.460 に答える
0

キーが番号であるを使用しdict、その座標をリスト内に格納します。

In [171]: lis
Out[171]: 
[[1, 2, 3, 4, 5],
 [6, 7, 8, 9, 10],
 [11, 12, 13, 14, 15],
 [16, 17, 18, 19, 20],
 [21, 22, 23, 24, 25],
 [26, 8, 9, 29, 30],
 [31, 13, 14, 15, 35],
 [17, 18, 19, 39, 40],
 [41, 23, 24, 44, 45],
 [46, 47, 48, 49, 50]]

In [172]: from collections import defaultdict

In [173]: dic=defaultdict(list)

In [174]: for i,x in enumerate(lis):
    for j,y in enumerate(x):
        dic[y].append((i,j))
   .....:         

In [175]: for num,coords in dic.items():
    if len(coords)>1:
        print "{0} was repeated at coordinates {1}".format(num,
                                             " ".join(str(x) for x in coords))
   .....:         
8 was repeated at coordinates (1, 2) (5, 1)
9 was repeated at coordinates (1, 3) (5, 2)
13 was repeated at coordinates (2, 2) (6, 1)
14 was repeated at coordinates (2, 3) (6, 2)
15 was repeated at coordinates (2, 4) (6, 3)
17 was repeated at coordinates (3, 1) (7, 0)
18 was repeated at coordinates (3, 2) (7, 1)
19 was repeated at coordinates (3, 3) (7, 2)
23 was repeated at coordinates (4, 2) (8, 1)
24 was repeated at coordinates (4, 3) (8, 2)
于 2013-04-29T01:42:41.363 に答える