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 list of dictionaries python by value
newlist = sorted(list_to_be_sorted, key=lambda k: k['name'])how to sort a dictionary by value in python
import operator
x = {1: 2, 3: 4, 4: 3, 2: 1, 0: 0}
sorted_x = sorted(x.items(), key=operator.itemgetter(1))
# Sort by key
import operator
x = {1: 2, 3: 4, 4: 3, 2: 1, 0: 0}
sorted_x = sorted(x.items(), key=operator.itemgetter(0))sort dict by value
x = {1: 2, 3: 4, 4: 3, 2: 1, 0: 0}
{k: v for k, v in sorted(x.items(), key=lambda item: item[1])}
{0: 0, 2: 1, 1: 2, 4: 3, 3: 4}sorting-a-dictionary-by-value-then-by-key
In [62]: y={100:1, 90:4, 99:3, 92:1, 101:1}
In [63]: sorted(y.items(), key=lambda x: (x[1],x[0]), reverse=True)
Out[63]: [(90, 4), (99, 3), (101, 1), (100, 1), (92, 1)]Copyright © 2021 Codeinu
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