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 the dictionary in python
d = {2: 3, 1: 89, 4: 5, 3: 0}
od = sorted(d.items())
print(od)sort dict by value python 3
>>> d = {"aa": 3, "bb": 4, "cc": 2, "dd": 1}
>>> for k in sorted(d, key=d.get, reverse=True):
... k, d[k]
...
('bb', 4)
('aa', 3)
('cc', 2)
('dd', 1)order dictionary by value python
x = {1: 2, 3: 4, 4: 3, 2: 1, 0: 0}
sorted_dict = {k: v for k, v in sorted(x.items(), key=lambda item: item[1])}
print(sorted_dict)
#{0: 0, 2: 1, 1: 2, 4: 3, 3: 4}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
Forgot your account's password or having trouble logging into your Account? Don't worry, we'll help you to get back your account. Enter your email address and we'll send you a recovery link to reset your password. If you are experiencing problems resetting your password contact us
