我有一list本字典python。我想按keys进行排序，即，首先是 by chain_id，然后是 by start.residue_number，然后是 by end.residue_number。在list下面，第 4 个和第 6 个元素应该相反。我可以使用sortedwithfrom operator import itemgetter, attrgetter进行排序chain_id，但是如何使用dictionaryinside进行排序dictionary？
"mappings": [
{
"entity_id": 1,
"end": {
"residue_number": 63
},
"start": {
"residue_number": 1
},
"chain_id": "A",
"struct_asym_id": "A"
},
{
"entity_id": 1,
"end": {
"residue_number": 116
},
"start": {
"residue_number": 1
},
"chain_id": "A",
"struct_asym_id": "A"
},
{
"entity_id": 1,
"end": {
"residue_number": 124
},
"start": {
"residue_number": 1
},
"chain_id": "A",
"struct_asym_id": "A"
},
{
"entity_id": 1,
"end": {
"residue_number": 116
},
"start": {
"residue_number": 1
},
"chain_id": "B",
"struct_asym_id": "B"
},
{
"entity_id": 1,
"end": {
"residue_number": 124
},
"start": {
"residue_number": 1
},
"chain_id": "B",
"struct_asym_id": "B"
},
{
"entity_id": 1,
"end": {
"residue_number": 63
},
"start": {
"residue_number": 1
},
"chain_id": "B",
"struct_asym_id": "B"
}
]
我假设d作为你的字典，解决了sorted和lambda
# Sort with chain_id
In [93]: sort_1 = sorted(d['mappings'],key=lambda x:x['chain_id'])
# Sort with start residue number
In [94]: sort_2 = sorted(sort_1,key=lambda x:x['start']['residue_number'])
# Sort with end residue number
In [95]: sort_3 = sorted(sort_2,key=lambda x:x['end']['residue_number'])
或者你可以在一条线上做。
sorted(d['mappings'],key=lambda x:(x['chain_id'],x['start']['residue_number'],x['end']['residue_number']))
灵感来自 leotrubach 的回答。
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