回答編集履歴
2
追記
answer
CHANGED
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@@ -1,20 +1,29 @@
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1
1
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素直な方法。
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2
2
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```Python
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3
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-
arr = [["A","B","C"],["D","E","F"],["G","H","I"]]
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3
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+
arr = [["A", "B", "C"], ["D", "E", "F"], ["G", "H", "I"]]
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4
4
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5
5
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for y, row in enumerate(arr):
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6
6
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try:
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7
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-
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7
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pos = (y, row.index("G"))
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8
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-
pos = (y, x)
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9
8
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break
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10
9
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except ValueError:
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11
10
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pass
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12
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-
else:
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13
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-
pos = None
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14
11
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15
12
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print(pos)
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16
13
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```
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17
14
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15
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+
ちょっと捻った方法。
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16
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```Python
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17
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from itertools import chain
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18
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19
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arr = [["A", "B", "C"], ["D", "E", "F"], ["G", "H", "I"]]
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20
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21
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pos = divmod(
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22
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list(chain(*arr)).index("G"), len(arr[0])
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23
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)
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24
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print(pos)
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25
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```
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26
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+
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18
27
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内包表記を使った方法。
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19
28
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```Python
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20
29
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arr = [["A", "B", "C"], ["D", "E", "F"], ["G", "H", "I"]]
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1
追記
answer
CHANGED
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@@ -1,3 +1,30 @@
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1
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+
素直な方法。
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2
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+
```Python
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3
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+
arr = [["A","B","C"],["D","E","F"],["G","H","I"]]
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4
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+
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5
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+
for y, row in enumerate(arr):
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6
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try:
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7
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x = row.index("G")
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8
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pos = (y, x)
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9
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break
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10
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except ValueError:
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11
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pass
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12
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else:
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13
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pos = None
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14
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+
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15
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print(pos)
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16
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```
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17
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+
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18
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+
内包表記を使った方法。
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19
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+
```Python
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20
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arr = [["A", "B", "C"], ["D", "E", "F"], ["G", "H", "I"]]
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21
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22
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pos = [
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23
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(x, y) for y, row in enumerate(arr) for x, elem in enumerate(row) if arr[x][y] == "G"
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24
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][0]
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25
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print(pos)
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26
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```
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27
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1
28
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NumPyを使うのもアリかもしれませんね。
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2
29
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```Python
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3
30
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import numpy as np
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@@ -6,5 +33,5 @@
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6
33
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)
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7
34
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8
35
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pos = tuple(nd[0] for nd in np.where(arr == "G"))
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9
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-
print(pos)
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36
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+
print(pos)
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10
37
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```
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