Classificar lista de dicionários por Key Python
newlist = sorted(list_to_be_sorted, key=lambda k: k['name'])
Stupid Stork
newlist = sorted(list_to_be_sorted, key=lambda k: k['name']) l = {1: 40, 2: 60, 3: 50, 4: 30, 5: 20}
d1 = dict(sorted(l.items(),key=lambda x:x[1],reverse=True))
print(d1) #output : {2: 60, 3: 50, 1: 40, 4: 30, 5: 20}
d2 = dict(sorted(l.items(),key=lambda x:x[1],reverse=False))
print(d2) #output : {5: 20, 4: 30, 1: 40, 3: 50, 2: 60}d1 = dict(sorted(d.items(), key = lambda x:x[0]))
d = {2: 3, 1: 89, 4: 5, 3: 0}
od = sorted(d.items())
print(od)In [1]: import collections
In [2]: d = {2:3, 1:89, 4:5, 3:0}
In [3]: od = collections.OrderedDict(sorted(d.items()))
In [4]: od
Out[4]: OrderedDict([(1, 89), (2, 3), (3, 0), (4, 5)])def sort_dict(dictionary, rev = True):
l = list(dictionary.items())
l.sort(reverse = rev)
a = [item[1] for item in l]
z = ''
for x in a:
z = z + str(x)
return(z)