我有2本字典
gen1 = {0: 'RR', 1: 'RR', 2: 'rR', 3: 'rR', 4: 'RR', 5: 'RR', 6: 'rR', 7: 'rR', 8: 'Rr', 9: 'Rr', 10: 'rr', 11: 'rr', 12: 'Rr', 13: 'Rr', 14: 'rr', 15: 'rr'}
和
gen2 = {0: 'PP', 1: 'PP', 2: 'PP', 3: 'PP', 4: 'PP', 5: 'PP', 6: 'PP', 7: 'PP', 8: 'PP', 9: 'PP', 10: 'PP', 11: 'PP', 12: 'PP', 13: 'PP', 14: 'PP', 15: 'PP'}
我想用相同的键将它们连在一起
gen3 = {0: 'RRPP', 1: 'RRPP', 2: 'rRPP', 3: 'rRPP', 4: 'RRPP', 5: 'RRPP', 6: 'rRPP', 7: 'rRPP', 8: 'RrPP', 9: 'RrPP', 10: 'rrPP', 11: 'rrPP', 12: 'RrPP', 13: 'RrPP', 14: 'rrPP', 15: 'rrPP'}
您可以使用字典理解:
{k:f'{v}{gen2.get(k, "")}' for k,v in gen1.items()}
{0: 'RRPP', 1: 'RRPP', 2: 'rRPP', 3: 'rRPP', 4: 'RRPP', 5: 'RRPP', 6: 'rRPP',
7: 'rRPP', 8: 'RrPP', 9: 'RrPP', 10: 'rrPP', 11: 'rrPP', 12: 'RrPP', 13: 'RrPP',
14: 'rrPP', 15: 'rrPP'}
dict理解-与@yatu的回答略有不同。这将对gen1和gen2的所有唯一键都起作用-如果某个键仅存在于一个词典中,则所得字典将仅具有相应gen字典中的值。
>>> gen1 = {0: 'RR', 1: 'RR', 2: 'rR', 3: 'rR', 4: 'RR', 5: 'RR', 6: 'rR', 7: 'rR', 8: 'Rr', 9: 'Rr', 10: 'rr', 11: 'rr', 12: 'Rr', 13: 'Rr', 14: 'rr', 15: 'rr'}
>>> gen2 = {0: 'PP', 1: 'PP', 2: 'PP', 3: 'PP', 4: 'PP', 5: 'PP', 6: 'PP', 7: 'PP', 8: 'PP', 9: 'PP', 10: 'PP', 11: 'PP', 12: 'PP', 13: 'PP', 14: 'PP', 15: 'PP'}
>>> {k: gen1.get(k) + gen2.get(k) for k in set(list(gen1.keys()) + list(gen2.keys()))}
{0: 'RRPP',
1: 'RRPP',
2: 'rRPP',
3: 'rRPP',
4: 'RRPP',
5: 'RRPP',
6: 'rRPP',
7: 'rRPP',
8: 'RrPP',
9: 'RrPP',
10: 'rrPP',
11: 'rrPP',
12: 'RrPP',
13: 'RrPP',
14: 'rrPP',
15: 'rrPP'}
如果您确定gen1
和gen2
具有完全相同的键,这应该起作用:
gen3 = { k: gen1[k] + gen2[k] for k in gen1 }