python sort a dictionary by values
x = {1: 2, 3: 4, 4: 3, 2: 1, 0: 0}
sort_by_key = dict(sorted(x.items(),key=lambda item:item[0]))
sort_by_value = dict(sorted(x.items(), key=lambda item: item[1]))
print("sort_by_key:", sort_by_key)
print("sort_by_value:", sort_by_value)
# sort_by_key: {0: 0, 1: 2, 2: 1, 3: 4, 4: 3}
# sort_by_value: {0: 0, 2: 1, 1: 2, 4: 3, 3: 4}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)sort dictionary python
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}sort the dictionary in python
d = {2: 3, 1: 89, 4: 5, 3: 0}
od = sorted(d.items())
print(od)python sort the values in a dictionaryi
from operator import itemgetter
new_dict = sorted(data.items(), key=itemgetter(1))python 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)])Copyright © 2021 Codeinu
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