如何在此循環中找到迭代索引

我正在尋找一種方法來用迭代編號替換「行」。如何在此循環中找到迭代索引

directions = 'down', 'up', 'left', 'right' 
actions = 'idle', 'walk', 'fight', 'death' 
frames = range(4) 
duration = 0.2 

animations = { 
    action : { 
     direction : (
      (duration, ('row', x)) for x in frames 
     ) for direction in directions 
    } for action in actions 
}

因此，例如，在「空閒」向下將爲0，至多1，左2，右3，則在步行4，下降5等。

來源

2014-02-28 Anonymous Entity

您是否嘗試過使用'range'？嘗試'爲我在範圍內（len（方向））'？ – nish

通過'actions'，'directions'或'frames'獲取迭代編號？或所有3的產品？ –

@MartijnPieters的方向+行動 –

您可以使用enumerate()來計算你的迭代：

animations = { 
    action : { 
     direction : (
      (duration, (i * len(directions) + j, x)) for x in frames 
     ) for j, direction in enumerate(directions) 
    } for i, action in enumerate(actions) 
}

這將產生：

>>> pprint({action: {direction: [(duration, (i * len(directions) + j, x)) for x in frames] for j, direction in enumerate(directions) } for i, action in enumerate(actions)}) 
{'death': {'down': [(0.2, (12, 0)), 
        (0.2, (12, 1)), 
        (0.2, (12, 2)), 
        (0.2, (12, 3))], 
      'left': [(0.2, (14, 0)), 
        (0.2, (14, 1)), 
        (0.2, (14, 2)), 
        (0.2, (14, 3))], 
      'right': [(0.2, (15, 0)), 
        (0.2, (15, 1)), 
        (0.2, (15, 2)), 
        (0.2, (15, 3))], 
      'up': [(0.2, (13, 0)), 
        (0.2, (13, 1)), 
        (0.2, (13, 2)), 
        (0.2, (13, 3))]}, 
'fight': {'down': [(0.2, (8, 0)), 
        (0.2, (8, 1)), 
        (0.2, (8, 2)), 
        (0.2, (8, 3))], 
      'left': [(0.2, (10, 0)), 
        (0.2, (10, 1)), 
        (0.2, (10, 2)), 
        (0.2, (10, 3))], 
      'right': [(0.2, (11, 0)), 
        (0.2, (11, 1)), 
        (0.2, (11, 2)), 
        (0.2, (11, 3))], 
      'up': [(0.2, (9, 0)), (0.2, (9, 1)), (0.2, (9, 2)), (0.2, (9, 3))]}, 
'idle': {'down': [(0.2, (0, 0)), 
        (0.2, (0, 1)), 
        (0.2, (0, 2)), 
        (0.2, (0, 3))], 
      'left': [(0.2, (2, 0)), 
        (0.2, (2, 1)), 
        (0.2, (2, 2)), 
        (0.2, (2, 3))], 
      'right': [(0.2, (3, 0)), 
        (0.2, (3, 1)), 
        (0.2, (3, 2)), 
        (0.2, (3, 3))], 
      'up': [(0.2, (1, 0)), (0.2, (1, 1)), (0.2, (1, 2)), (0.2, (1, 3))]}, 
'walk': {'down': [(0.2, (4, 0)), 
        (0.2, (4, 1)), 
        (0.2, (4, 2)), 
        (0.2, (4, 3))], 
      'left': [(0.2, (6, 0)), 
        (0.2, (6, 1)), 
        (0.2, (6, 2)), 
        (0.2, (6, 3))], 
      'right': [(0.2, (7, 0)), 
        (0.2, (7, 1)), 
        (0.2, (7, 2)), 
        (0.2, (7, 3))], 
      'up': [(0.2, (5, 0)), (0.2, (5, 1)), (0.2, (5, 2)), (0.2, (5, 3))]}}

或者，使用itertools.count() object與next();如果內部循環使用基於外部循環的可變數量的元素，這很有幫助：

from itertools import count 

iter_count = count() 

animations = { 
    action : { 
     direction : (
      (duration, (count, x)) for c in (next(iter_count),) for x in frames 
     ) for direction in directions 
    } for action in actions 
}

來源

2014-02-28 19:37:42

好吧，這解決了我的問題，幫助我看到沒有必要在內部循環中包含持續時間，以及讓我質疑使用這種方式字典理解。謝謝。 –

我只想建議使用for循環作出反擊，並建立您的結構：

animations = {} 
row = 0 

for action in actions: 
    d = {} 
    for direction in directions: 
     rows = [] 
     for x in frames: 
      rows.append((duration, (row, x)) 
      row += 1 
     d[direction] = rows 
    animations[action] = d

來源

2014-02-28 19:33:33 nneonneo

如何在此循環中找到迭代索引

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