def example():
word_ids = [2,3,4,5,6]
boundary = 2
return [(u, v, v*2) for i, u in enumerate(word_ids) for j, v in
enumerate(word_ids[max(i - boundary, 0):i boundary 1]) if u != v]
a = example()
print(a)
這個回報：
[(2, 3, 6), (2, 4, 8), (3, 2, 4), (3, 4, 8), (3, 5, 10), (4, 2, 4), (4, 3, 6), (4, 5, 10), (4, 6, 12), (5, 3, 6), (5, 4, 8), (5, 6, 12), (6, 4, 8), (6, 5, 10)]
有人可以幫助我使用上面的行內函式來實作這一點：
[(2, [3,4], [6,8]), (3, [2,4,5], [4,8,10]), (4, [2,3,5,6] [4,6,10,12]), (5, [3,4,6], [6,8,12]), (6, [4,5], [8,10])]
uj5u.com熱心網友回復：
def sample():
values = []
possible = [2,3,4,5,6]
popping = [2,3,4,5,6]
boundary = 2
for index, num in enumerate(possible):
popping.pop(index)
values.append((num,popping[max(index-2,0):index boundary 1],[x*num for x in popping[max(index-2,0):index boundary 1]]))
popping = [2,3,4,5,6]
return values
我現在正在開發一個更簡單的版本
轉載請註明出處，本文鏈接：https://www.uj5u.com/qiye/359197.html
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