我认为这应该是一个相对简单的任务,但我似乎无法弄清楚。
a = {1: 'the last bus', 2: 'you take', 4: 'train I try', 6: 'see its'}
b = ['I miss', 'nan', 'nan', 'the next', 'nan', 'but you', 'nan', 'hard to explain']
期望的输出:
['I miss', 'the last bus', 'you take', 'the next', 'train I try', 'but you', 'see its', 'hard to explain']
只需循环dict
,然后将list
中的每个键重新分配给关联的值:
for i, value in a.items():
b[i] = value
这会修改b
的位置;如果您希望不修改它,最简单的解决方案是预先复制它:
c = b[:]
for i, value in a.items():
c[i] = value