合并列表和列表中的元组

Question

假设我有以下列表：

l1 = [['a','b','c'],['x','y','z'],['i','j','k']]
l2 = [(0,0.1),(0,0.2),(2,0.3),(2,0.4),(1,0.5),(0,0.6)]

我想在 l2 的键和 l1 的索引上合并两个列表，这样我得到：

[[0.1,['a','b','c'],
[0.2,['a','b','c'],
[0.3,['i','j','k'],
[0.4,['i','j','k'],
[0.5,['x','y','z'],
[0.6,['a','b','c']]

我想知道这是怎么做到的？因为合并不在键上。

Answer 1

带有一些索引的列表理解会很好

l1 = [['a', 'b', 'c'], ['x', 'y', 'z'], ['i', 'j', 'k']]
l2 = [(0, 0.1), (0, 0.2), (2, 0.3), (2, 0.4), (1, 0.5), (0, 0.6)]

result = [[value, l1[idx]] for idx, value in l2]

for 循环等价物

result = []
for idx, value in l2:
    l1_item = l1[idx]
    result.append([value, l1_item])

Answer 2

简单理解即可：

[[v, l1[i]] for i, v in l2]
# [[0.1, ['a', 'b', 'c']], 
#  [0.2, ['a', 'b', 'c']], 
#  [0.3, ['i', 'j', 'k']], 
#  [0.4, ['i', 'j', 'k']], 
#  [0.5, ['x', 'y', 'z']], 
#  [0.6, ['a', 'b', 'c']]]

Answer 3

l1 = [['a','b','c'],['x','y','z'],['i','j','k']]
l2 = [(0,0.1),(0,0.2),(2,0.3),(2,0.4),(1,0.5),(0,0.6)]

print([[v,l1[k]] for k,v in l2])

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