how can I sort a dictionary in python according to its values?
s = {1: 1, 7: 2, 4: 2, 3: 1, 8: 1}
k = dict(sorted(s.items(),key=lambda x:x[0],reverse = True))
print(k)how can I sort a dictionary in python according to its values?
s = {1: 1, 7: 2, 4: 2, 3: 1, 8: 1}
k = dict(sorted(s.items(),key=lambda x:x[0],reverse = True))
print(k)python sort dict by key
A={1:2, -1:4, 4:-20}
{k:A[k] for k in sorted(A)}
output:
{-1: 4, 1: 2, 4: -20}sort dictionary
#for dictionary d
sorted(d.items(), key=lambda x: x[1]) #for inceasing order
sorted(d.items(), key=lambda x: x[1], reverse=True) # for decreasing order
#it will return list of key value pair tuplespython sort dictionary by key
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)])python sort dictionary by key
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)Copyright © 2021 Codeinu
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