Answers for "ordering dictionary by value"

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 by values

from collections import OrderedDict
dd = OrderedDict(sorted(d.items(), key=lambda x: x[1]))
print(dd)

sorting values in dictionary in python

#instead of using python inbuilt function we can it compute directly.
#here iam sorting the values in descending order..
d = {1: 1, 7: 2, 4: 2, 3: 1, 8: 1}
s=[]
for i in d.items():
  s.append(i)
for i in range(0,len(s)):
  for j in range(i+1,len(s)):
    if s[i][1]<s[j][1]:
      s[i],s[j]=s[j],s[i]
print(dict(s))

sorting-a-dictionary-by-value-then-by-key

[(14,2), (12,2), (9,1)]  # output from print 
[(90,4), (99,3), (101,1), (100,1), (92,1)]