I have a list of dictionaries
[{'abc':1},{'abcd':2},{'ab':1}]
Based on length of key list has to sorted
[{'abcd':2},{'abc':1},{'ab':1}]
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Vivek Sable
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user3522967
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1Have you seen this: http://stackoverflow.com/questions/72899/how-do-i-sort-a-list-of-dictionaries-by-values-of-the-dictionary-in-python?rq=1 – EdChum Mar 05 '15 at 10:47
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Where is your code, and what exactly is the problem with it? – jonrsharpe Mar 05 '15 at 10:48
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Will your dictionaries have only one key always? – thefourtheye Mar 05 '15 at 10:49
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@EdChum my keys are dynamic not the way they specified in the link – user3522967 Mar 05 '15 at 10:49
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you can't sort regular dictionaries, except using ordereddict. – Marcin Mar 05 '15 at 10:50
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@thefourtheye dictionaries have only one key – user3522967 Mar 05 '15 at 10:50
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@Marcin: this is sorting a *list* **containing** dictionaries. – Martijn Pieters Mar 05 '15 at 10:57
2 Answers
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If you want to sort by the longest key:
l = [{'abc':1},{'abcd':2},{'ab':1}]
l.sort(key=lambda x: len(next(iter(x))), reverse=True)
print(l)
Martijn Pieters
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Padraic Cunningham
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`lambda x: len(x.keys()[0])` would work too as `.keys()` return list of keys. – Tanveer Alam Mar 05 '15 at 10:54
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@TanveerAlam: not in Python 3, where `dict.keys()` returns a *dictionary view*. `next(iter(x))` is better as it works in both versions. – Martijn Pieters Mar 05 '15 at 10:54
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If u like to sort by 1st key it's gonna be this:
l.sort(lambda a, b: -cmp(len(a.keys()[0]), len(b.keys()[0])))
However if you want to sort by max key, here the answer:
def maxKeyLen(d): return max(len(k) for k in d.keys())
l.sort(lambda a, b: -cmp(maxKeyLen(a), maxKeyLen(b)))
Mux
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when we haev list of dictionaries inside dictionaries [{'abc':{'count':1}},{'abcd:{'count:2'}'}], sorting this list based on 'count' , is it possible – user3522967 Mar 05 '15 at 11:22