在 Python 中对字典进行降序排序
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Sorting dictionary descending in Python
提问by Darien
I am new to Python and I was wondering if there is a more common way of sorting a dictionary in descending order than the way I'm doing it:
我是 Python 的新手，我想知道是否有比我这样做的方式更常见的按降序对字典进行排序的方法：
sorted_dictionary = sorted(dict.items(), key=lambda x: -x[1])
采纳答案by Martijn Pieters
There is a reverseoption to sorted()you could use instead:
您可以使用一个reverse选项sorted()：
sorted(dict.items(), key=lambda kv: kv[1], reverse=True)
This produces the exact same output, and even works if the values are not numeric.
这会产生完全相同的输出，即使值不是数字也能工作。
回答by Francesco Nazzaro
Python dictionary aren't sortable. Your sorted_dictionaryoutput is not a dictionary but a list. You have to use OrderedDict
Python 字典不可排序。您的sorted_dictionary输出不是字典而是列表。你必须使用OrderedDict
from collections import OrderedDict
sorted_dictionary = OrderedDict(sorted(dict.items(), key=lambda v: v, reverse=True))
回答by Garima Chhikara
Sort in decreasing order of values in python list:
在python列表中按值的降序排序：
items = {'a':6 , 'b':3 , 'c':1 , 'd':9}
sortedItems = sorted(items , key=items.get , reverse = True)
for element in sortedItems :
print (element, items.get(element))
Output is:
输出是：
d 9
a 6
b 3
c 1
回答by Fuji Komalan
Sorting Dictionary based on values in ascending order
根据值按升序排序字典
d = {'a':5 , 'b':3 , 'd':1 , 'e':2 , 'f':4}
result = sorted(d.items() , key=lambda t : t[1])
for k,v in result:
print(k,v)
RESULT
结果
d 1
e 2
b 3
f 4
a 5
# d.items=[('a', 5), ('b', 3), ('d', 1), ('e', 2), ('f', 4)]
# lambda t = ('a',5)
# lambda t : t[1] = 5
# result = [('d', 1), ('e', 2), ('b', 3), ('f', 4), ('a', 5)]
# k ,v = ('d',1)
# k = 'd'
# v = 1
Sorting Dictionary based on values in descending order
根据值按降序排序字典
d = {'a':5 , 'b':3 , 'd':1 , 'e':2 , 'f':4}
result = sorted(d.items() , key=lambda t : t[1] , reverse=True)
for k,v in result:
print(k,v)
RESULT
结果
a 5
f 4
b 3
e 2
d 1
# d.items=[('a', 5), ('b', 3), ('d', 1), ('e', 2), ('f', 4)]
# lambda t = ('a',5)
# lambda t : t[1] = 5
# result = [('a', 5), ('f', 4), ('b', 3), ('e', 2), ('d', 1)]
# k ,v = ('a',5)
# k = 'a'
# v = 5
回答by Michael Fitzgerald
Here is code that get all the prime factors with exponents for all numbers in a range
这是获取范围内所有数字的所有质因数和指数的代码
from sympy import primefactors
for num in range (1,10001): k = num prime_factors = {}
对于范围内的 num (1,10001)：k = num prime_factors = {}
for i in primefactors(num):
count = 0
while num % i == 0:
num = num/i
count += 1
prime_factors[i] = count
result = sorted(prime_factors.items() , key=lambda t : t[0] , reverse=True)
count = 1
num = k
print '\n%d can be broken down into the following prime factors:' % num,
print '\n%d = ' % num,
for k,v in result:
if count < len(result):
print '(%d ** %d) *' % (k,v),
count += 1
else:
print '(%d ** %d)' % (k,v)
回答by thereal90
This should work
这应该工作
{k: v for k, v in sorted(dict.items(), key=lambda item: item[1], reverse = True)}
