PYTHON

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)

python - sort dictionary by value

d = {'one':1,'three':3,'five':5,'two':2,'four':4}

# Sort
a = sorted(d.items(), key=lambda x: x[1])

# Reverse sort
r = sorted(d.items(), key=lambda x: x[1], reverse=True)

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))

how to sort dictionary in python by value

# your dictionary
a = {'a':4, 'c':5, 'b':3, 'd':0}

# sort x by keys
a_keys = dict(sorted(a.items(),key=lambda x:x[0],reverse = False)) # ascending order
# output: {'a': 4, 'b': 3, 'c': 5, 'd': 0}

# # sort x by values
a_values = dict(sorted(a.items(),key=lambda x:x[1],reverse = False)) # ascending order
# output: {'d': 0, 'b': 3, 'a': 4, 'c': 5}

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}

sort dict by value

dict(sorted(x.items(), key=lambda item: item[1]))

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}

sort dictionary by value and then key python

sorted(y.items(), key=lambda x: (x[1],x[0]))

python sort the values in a dictionaryi

from operator import itemgetter
new_dict = sorted(data.items(), key=itemgetter(1))

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))

python sort dictionary by value

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

sort a dict by values

{k: v for k, v in sorted(dic.items(), key=lambda item: item[1])}

sort dict by values

>>> 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}

sort a dictionary by value then key

d = {'apple': 2, 'banana': 3, 'almond':2 , 'beetroot': 3, 'peach': 4}
[v[0] for v in sorted(d.items(), key=lambda kv: (-kv[1], kv[0]))]

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)]

sort dict by value

for w in sorted(d, key=d.get, reverse=True):
    print(w, d[w])

python Sort the dictionary based on values

dt = {5:4, 1:6, 6:3}

sorted_dt = {key: value for key, value in sorted(dt.items(), key=lambda item: item[1])}

print(sorted_dt)

python Sort the dictionary based on values

dt = {5:4, 1:6, 6:3}

sorted_dt_value = sorted(dt.values())
print(sorted_dt_value)

sort dict by value

d = {'one':1,'three':3,'five':5,'two':2,'four':4}
a = sorted(d.items(), key=lambda x: x[1])    
print(a)

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

{12:2, 9:1,  14:2}
{100:1, 90:4, 99:3, 92:1, 101:1}

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)]