Question

这是我的杰森:

{"data": [{"L": "Leinster"}, {"RN": "Roscommon"}, {"G": "Galway"}, {"LS": "Laois"}, {"LD": "Longford"}, {"OY": "Offaly"}, {"KK": "Kilkenny"}, {"SO": "Sligo"}, {"C": "Connaught"}, {"CO": "Cork"}, {"M": "Munster"}, {"WD": "Waterford"}, {"CE": "Clare"}]}

我想排序这样的话, 判令会按值的 abc 排序, 例如 :

  {"data": [ {"CE": "Clare"},{"C": "Connaught"}, {"CO": "Cork"}, {"G": "Galway"},{"KK": "Kilkenny"}, {"LS": "Laois"},{"L": "Leinster"}, {"LD": "Longford"}, {"M": "Munster"},{"OY": "Offaly"}, {"RN": "Roscommon"}, {"SO": "Sligo"}, {"WD": "Waterford"} ]}

some ideas? tried this Python: sort this dictionary (dict in dict) but this is sorting values. not the values in the dict.

谢谢!

Answer 1

Python 2 用于 Python 2

>>> d = { data : [{ L :  Leinster }, { RN :  Roscommon }, { G :  Galway }, { LS :  Laois }, { LD :  Longford }, { OY :  Offaly }, { KK :  Kilkenny }, { SO :  Sligo }, { C :  Connaught }, { CO :  Cork }, { M :  Munster }, { WD :  Waterford }, { CE :  Clare }]}
>>> dict((k,sorted(v,key=dict.values)) for k,v in d.iteritems())
{ data : [{ CE :  Clare }, { C :  Connaught }, { CO :  Cork }, { G :  Galway }, { KK :  Kilkenny }, { LS :  Laois }, { L :  Leinster }, { LD :  Longford }, { M :  Munster }, { OY :  Offaly }, { RN :  Roscommon }, { SO :  Sligo }, { WD :  Waterford }]}

这是Python 3的解决方案, 最好能知道是否有比这更好的东西, 因为我不喜欢它。

>>> {k:sorted(v,key=lambda x: tuple(x.values())) for k,v in d.items()}
{ data : [{ CE :  Clare }, { C :  Connaught }, { CO :  Cork }, { G :  Galway }, { KK :  Kilkenny }, { LS :  Laois }, { L :  Leinster }, { LD :  Longford }, { M :  Munster }, { OY :  Offaly }, { RN :  Roscommon }, { SO :  Sligo }, { WD :  Waterford }]}

Answer 2

或者你可以简单地做

按值排序 < 强度 > 更新。

In [41]: r={"data": [{"L": "Leinster"}, {"RN": "Roscommon"}, {"G": "Galway"}, {"LS": "Laois"}, {"LD": "Longford"}, {"OY": "Offaly"}, {"KK": "Kilkenny"}, {"SO": "Sligo"}, {"C": "Connaught"}, {"CO": "Cork"}, {"M": "Munster"}, {"WD": "Waterford"}, {"CE": "Clare"}]}

In [42]: for v in r.itervalues():
....:     v.sort(key=dict.values)
....: 

In [43]: r
Out[43]:  
{ data : [{ CE :  Clare },
      { C :  Connaught },
      { CO :  Cork },
      { G :  Galway },
      { KK :  Kilkenny },
      { LS :  Laois },
      { L :  Leinster },
      { LD :  Longford },
      { M :  Munster },
      { OY :  Offaly },
      { RN :  Roscommon },
      { SO :  Sligo },
      { WD :  Waterford }]}

Answer 3

另一种解决办法,因为越多越好!

>>> r={"data": [{"L": "Leinster"}, {"RN": "Roscommon"}, {"G": "Galway"}, {"LS": "Laois"}, {"LD": "Longford"}, {"OY": "Offaly"}, {"KK": "Kilkenny"}, {"SO": "Sligo"}, {"C": "Connaught"}, {"CO": "Cork"}, {"M": "Munster"}, {"WD": "Waterford"}, {"CE": "Clare"}]}
>>>
>>> map(functools.partial(list.sort, key=dict.values), r.itervalues())
[None]
>>>
>>> r
{ data : [{ CE :  Clare }, { C :  Connaught }, { CO :  Cork }, { G :  Galway }, { KK :  Kilkenny }, { LS :  Laois }, { L :  Leinster }, { LD :  Longford }, { M :  Munster }, { OY :  Offaly }, { RN :  Roscommon }, { SO :  Sligo }, { WD :  Waterford }]}

友情链接