PYTHON

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)

dictionary sort python

d={
  3: 4,
  1: 1,
  0: 0,
  4: 3,
  2: 1
}
y=dict(sorted(d.items(), key=lambda item: item[1]))
print(y) # {0: 0, 2: 1, 1: 2, 4: 3, 3: 4}

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 dictionary

#for dictionary d
sorted(d.items(), key=lambda x: x[1]) #for inceasing order
sorted(d.items(), key=lambda x: x[1], reverse=True) # for decreasing order
#it will return list of key value pair tuples

sort the dictionary in python

d = {2: 3, 1: 89, 4: 5, 3: 0}
od = sorted(d.items())
print(od)

python sort dict by key

A={1:2, -1:4, 4:-20}
{k:A[k] for k in sorted(A)}

output:
{-1: 4, 1: 2, 4: -20}

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)])

python dictionary sort

# empty dictionary
dictionary = {}
# lists
list_1 = [1, 2, 3, 4, 5]
list_2 = ["e", "d", "c", "b", "a"]
# populate a dictionary.
for key, value in zip(list_1, list_2):
    dictionary[key] = value
# original
print(f"Original dictionary: {dictionary}")

# Sort dictionary based on value
dictionary_sorted = dict(sorted(dictionary.items(), key=lambda value: value[1]))
print(f"Sort dictionary by value: {dictionary_sorted}")

# Sort dictionary based on key
dictionary_sorted = dict(sorted(dictionary.items(), key=lambda key: key[0]))
print(f"Sort dictionary by key: {dictionary_sorted}")

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 dict sortieren

# dict has no order - use a list to store
l = {1: 40, 2: 60, 3: 50, 4: 30, 5: 20}
d1 = list(sorted(l.items(),key=lambda x:x[1],reverse=True))

sort dictionary by key

dictionary_items = a_dictionary.items()
sorted_items = sorted(dictionary_items)

python sort dictionary by key

def sort_dict(dictionary, rev = True):
    l = list(dictionary.items())
    l.sort(reverse = rev)
    a = [item[1] for item in l]    
    z = ''    
    for x in a:
        z = z + str(x)
    return(z)

sort dictionary by key python

for key in sorted(a_dictionary):
        print ("{}: {}".format(key, a_dictionary[key]))

dict sort

[(k,di[k]) for k in sorted(di.keys())] 

sorting dictionary in python

Sorting a dictionary in python