DekGenius.com
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 values in python from dictionary to list
x = {'a':1, 'b':2, 'c':3, 'd':4, 'e':5}
accending_x_keys = sorted(x, key=x.get)
descending_x_keys = sorted(x, key=x.get, reverse=True)
for k in accending_x_keys:
print(k, x.get(k), end=" ")
#output: a 1 b 2 c 3 d 4 e 5
for v in descending_x_keys:
print(v, x.get(v), end=" ")
#output: e 5 d 4 c 3 b 2 a 1
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}
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}
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 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)
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}
python sort array of dictionary by value
my_list = sorted(my_list, key=lambda k: k['name'])
sort dictionary by value and then key python
sorted(y.items(), key=lambda x: (x[1],x[0]))
python sort the values in a dictionary
from operator import itemgetter
new_dict = sorted(data.items(), key=itemgetter(1))
python sort dict by value
A={1:2, -1:4, 4:-20}
{k:A[k] for k in sorted(A, key=A.get)}
output:
{4: -20, 1: 2, -1: 4}
how to sort dict by value
dict1 = {1: 1, 2: 9, 3: 4}
sorted_dict = {}
sorted_keys = sorted(dict1, key=dict1.get) # [1, 3, 2]
for w in sorted_keys:
sorted_dict[w] = dict1[w]
print(sorted_dict) # {1: 1, 3: 4, 2: 9}
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 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])}
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 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 dict by values
{k: v for k, v in sorted(dic.items(), key=lambda item: item[1])}
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]))]
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)
How to sort a Python dict by value
# How to sort a Python dict by value
# (== get a representation sorted by value)
>>> xs = {'a': 4, 'b': 3, 'c': 2, 'd': 1}
>>> sorted(xs.items(), key=lambda x: x[1])
[('d', 1), ('c', 2), ('b', 3), ('a', 4)]
# Or:
>>> import operator
>>> sorted(xs.items(), key=operator.itemgetter(1))
[('d', 1), ('c', 2), ('b', 3), ('a', 4)]
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)
python order list of dictionaries by value
newlist = sorted(list_to_be_sorted, key=lambda d: d['name'])
© 2022 Copyright:
DekGenius.com