正文
[PY3]——求TopN/BtmN 和 排序问题的解决
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需求
K长的序列，求TopN
K长的序列，求BtmN
排序问题
解决
-
heap.nlargest(）、heap.nsmallest( )
-
sorted( )+切片
-
max( )、min( )
总结和比较
1）在Top N问题中，如果 N=1，则直接用max(iterable)/min(iterable) 即可（效率最高）。
2）如果N很大，接近集合元素，则为了提高效率，采用 sort+切片 的效率会更高，如：
求最大的N个元素：sorted(iterable, key=key, reverse=True)[:N]
求最小的N个元素：sorted(iterable, key=key)[:N]
3）当要查找的元素个数相对比较小的时候，使用 nlargest() 和 nsmallest() 是很合适的
详解max( )/min( )函数用法
-
求简单的序列TopN/BtmN（N=1）问题
lst=[1,2,3,4,5]
print(max(lst))
5
-
通过key属性的使用，设置函数条件为判断的标准
a=[-9,-8,1,3,-4,6]
print(max(a,key=lambda x:abs(x)))
-9
-
找出字典中值最大的那组数据
prices = {
'A':123,
'B':450.1,
'C':12,
'E':444,
}
//在对字典进行数据操作的时候，默认只会处理key，而不是value
//先使用zip把字典的keys和values翻转过来，再用max取出值最大的那组数据
max_prices=max(zip(prices.values(),prices.keys()))
print(max_prices)
(450.1, 'B')
nlargest( )/nsmallest( )详解
-
nlargest(n,iterable) 求序列iterable中的TopN | nsmallest(n,iterable) 求序列iterable中的BtmN
import heapq
nums=[16,7,3,20,17,8,-1]
print(heapq.nlargest(3,nums))
print(heapq.nsmallest(3,nums))
[20, 17, 16]
[-1, 3, 7]
-
nlargest(n, iterable, key=lambda) | nsmallest(n, iterable, key=lambda) key接受关键字参数，用于更复杂的数据结构中
def print_price(dirt):
for i in dirt:
for x,y in i.items():
if x=='price':
print(x,y)
portfolio = [
{'name': 'IBM', 'shares': 100, 'price': 91.1},
{'name': 'AAPL', 'shares': 50, 'price': 543.22},
{'name': 'FB', 'shares': 200, 'price': 21.09},
{'name': 'HPQ', 'shares': 35, 'price': 31.75},
{'name': 'YHOO', 'shares': 45, 'price': 16.35},
{'name': 'ACME', 'shares': 75, 'price': 115.65}
]
cheap=heapq.nsmallest(3,portfolio,key=lambda x:x['price'])
expensive=heapq.nlargest(3,portfolio,key=lambda y:y['price'])
print_price(cheap)
print_price(expensive)
price 16.35
price 21.09
price 31.75
price 543.22
price 115.65
price 91.1
sorted( )详解
-
sorted(iterable, key=None, reverse=False)
-
reverse=True 逆序
nums=[16,7,3,20,17,8,-1]
print(sorted(nums))
print(sorted(nums,reverse=True))
[-1, 3, 7, 8, 16, 17, 20]
[20, 17, 16, 8, 7, 3, -1]
str=['b','a','A','s']
print(sorted(str))
print(sorted(str,reverse=True))
['A', 'a', 'b', 's']
['s', 'b', 'a', 'A']
-
key接受一个函数，且是个只接受一个元素的函数
-
多条件的key应该怎么写？
//按长度排序
L = [{1:5,3:4},{1:3,6:3},{1:1,2:4,5:6},{1:9}]
print(sorted(L,key=lambda x: len(x)))
[{1: 9}, {1: 5, 3: 4}, {1: 3, 6: 3}, {1: 1, 2: 4, 5: 6}]
//根据指定的值来排序(例如字典中的某个key)
L = [
('john', 'A', 15),
('jane', 'B', 10),
('dave', 'B', 12),
]
print(sorted(L,key=lambda x:x[2],reverse=True))
[('john', 'A', 15), ('dave', 'B', 12), ('jane', 'B', 10)]
portfolio = [
{'name': 'IBM', 'shares': 100, 'price': 91.1},
{'name': 'AAPL', 'shares': 50, 'price': 543.22},
{'name': 'FB', 'shares': 200, 'price': 21.09},
{'name': 'HPQ', 'shares': 35, 'price': 31.75},
{'name': 'YHOO', 'shares': 45, 'price': 16.35},
{'name': 'ACME', 'shares': 75, 'price': 115.65}
]
print(sorted(portfolio,key=lambda x:x['price']))
[{'shares': 45, 'name': 'YHOO', 'price': 16.35}, {'shares': 200, 'name': 'FB', 'price': 21.09}, {'shares': 35, 'name': 'HPQ', 'price': 31.75}, {'shares': 100, 'name': 'IBM', 'price': 91.1}, {'shares': 75, 'name': 'ACME', 'price': 115.65}, {'shares': 50, 'name': 'AAPL', 'price': 543.22}]
//不规则字符串，按“小写-大写-奇数-偶数”顺序排序
s = 'asdf234GDSdsf23'
print("".join(sorted(s, key=lambda x: (x.isdigit(),x.isdigit() and int(x) % 2 == 0,x.isupper(),x))))
addffssDGS33224
//一道面试题：要求：正数在前负数在后 2.整数从小到大 3.负数从大到小
list1=[7, -8, 5, 4, 0, -2, -5]
print(sorted(list1,key=lambda x:(x<0,abs(x))))
[0, 4, 5, 7, -2, -5, -8]
-
用operator中的函数加快速度和进行多级排序
from operator import itemgetter, attrgetter
暂不讨论
比较三种方法的效率
-
只求TopN=1/BtmN=1时，比较max( )和nlargest( )两种效率
In [8]: nums=random.sample(range(1,10000),999)
In [9]: print(max(nums))
9999
In [10]: %time
CPU times: user 0 ns, sys: 0 ns, total: 0 ns
Wall time: 13.1 µs
In [11]: heapq.nlargest(1,nums)
Out[11]: [9999]
In [12]: %time
CPU times: user 0 ns, sys: 0 ns, total: 0 ns
Wall time: 14.1 µs
-
当K为10，N为9(即N无限接近K时)，比较了sorted( )+切片和nlargest( )两种方法的效率
In [23]: nums=random.sample(range(1,10000),10)
In [24]: sorted(nums,reverse=True)[:9]
Out[24]: [8814, 7551, 7318, 5597, 5257, 4437, 4211, 2776, 2440]
In [25]: %time
CPU times: user 0 ns, sys: 0 ns, total: 0 ns
Wall time: 11.4 µs
In [26]: heapq.nlargest(9,nums)
Out[26]: [8814, 7551, 7318, 5597, 5257, 4437, 4211, 2776, 2440]
In [27]: %time
CPU times: user 0 ns, sys: 0 ns, total: 0 ns
Wall time: 154 µs
-
当N较小时，比较了nlargest( )和sorted( )+切片两种方法
In [18]: nums=[16,7,3,20,17,8,-1]
In [19]: heapq.nlargest(3,nums)
Out[19]: [20, 17, 16]
In [20]: %time
CPU times: user 0 ns, sys: 0 ns, total: 0 ns
Wall time: 4.05 µs
In [21]: sorted(nums,reverse=True)[:3]
Out[21]: [20, 17, 16]
In [22]: %time
CPU times: user 0 ns, sys: 0 ns, total: 0 ns
Wall time: 5.48 µs
以上代码用到的import和show_tree( )
import math
import io
from io import StringIO
import heapq
import random
import time
from functools import wraps
def show_tree(tree, total_width=36, fill=' '):
output =io.StringIO() #创建stringio对象
last_row = -1
for i, n in enumerate(tree): #
if i:
row = int(math.floor(math.log(i+1, 2)))
else:
row = 0
if row != last_row:
output.write('\n')
columns = 2**row
col_width = int(math.floor((total_width * 1.0) / columns))
output.write(str(n).center(col_width, fill))
last_row = row
print(output.getvalue())
print('-' * total_width)
print(' ')
return
参考资料
python3-cookbook-1.4-查找最大或最小元素
详解Python中heapq模块的用法（这里有实现show_tree的函数代码）
理解堆和堆排序的文章1
理解堆和堆排序的文章2
理解堆和堆排序的文章3
python奇技淫巧——max/min函数的用法
lst=[1,2,3,4,5]
print(max(lst))
5a=[-9,-8,1,3,-4,6]
print(max(a,key=lambda x:abs(x)))
-9prices = {
'A':123,
'B':450.1,
'C':12,
'E':444,
}
//在对字典进行数据操作的时候，默认只会处理key，而不是value
//先使用zip把字典的keys和values翻转过来，再用max取出值最大的那组数据
max_prices=max(zip(prices.values(),prices.keys()))
print(max_prices)
(450.1, 'B')- nlargest(n,iterable) 求序列iterable中的TopN | nsmallest(n,iterable) 求序列iterable中的BtmN
import heapq
nums=[16,7,3,20,17,8,-1]
print(heapq.nlargest(3,nums))
print(heapq.nsmallest(3,nums))
[20, 17, 16]
[-1, 3, 7]- nlargest(n, iterable, key=lambda) | nsmallest(n, iterable, key=lambda) key接受关键字参数，用于更复杂的数据结构中
def print_price(dirt):
for i in dirt:
for x,y in i.items():
if x=='price':
print(x,y)
portfolio = [
{'name': 'IBM', 'shares': 100, 'price': 91.1},
{'name': 'AAPL', 'shares': 50, 'price': 543.22},
{'name': 'FB', 'shares': 200, 'price': 21.09},
{'name': 'HPQ', 'shares': 35, 'price': 31.75},
{'name': 'YHOO', 'shares': 45, 'price': 16.35},
{'name': 'ACME', 'shares': 75, 'price': 115.65}
]
cheap=heapq.nsmallest(3,portfolio,key=lambda x:x['price'])
expensive=heapq.nlargest(3,portfolio,key=lambda y:y['price'])
print_price(cheap)
print_price(expensive)
price 16.35
price 21.09
price 31.75
price 543.22
price 115.65
price 91.1
sorted( )详解
-
sorted(iterable, key=None, reverse=False)
-
reverse=True 逆序
nums=[16,7,3,20,17,8,-1]
print(sorted(nums))
print(sorted(nums,reverse=True))
[-1, 3, 7, 8, 16, 17, 20]
[20, 17, 16, 8, 7, 3, -1]
str=['b','a','A','s']
print(sorted(str))
print(sorted(str,reverse=True))
['A', 'a', 'b', 's']
['s', 'b', 'a', 'A']
-
key接受一个函数，且是个只接受一个元素的函数
-
多条件的key应该怎么写？
//按长度排序
L = [{1:5,3:4},{1:3,6:3},{1:1,2:4,5:6},{1:9}]
print(sorted(L,key=lambda x: len(x)))
[{1: 9}, {1: 5, 3: 4}, {1: 3, 6: 3}, {1: 1, 2: 4, 5: 6}]
//根据指定的值来排序(例如字典中的某个key)
L = [
('john', 'A', 15),
('jane', 'B', 10),
('dave', 'B', 12),
]
print(sorted(L,key=lambda x:x[2],reverse=True))
[('john', 'A', 15), ('dave', 'B', 12), ('jane', 'B', 10)]
portfolio = [
{'name': 'IBM', 'shares': 100, 'price': 91.1},
{'name': 'AAPL', 'shares': 50, 'price': 543.22},
{'name': 'FB', 'shares': 200, 'price': 21.09},
{'name': 'HPQ', 'shares': 35, 'price': 31.75},
{'name': 'YHOO', 'shares': 45, 'price': 16.35},
{'name': 'ACME', 'shares': 75, 'price': 115.65}
]
print(sorted(portfolio,key=lambda x:x['price']))
[{'shares': 45, 'name': 'YHOO', 'price': 16.35}, {'shares': 200, 'name': 'FB', 'price': 21.09}, {'shares': 35, 'name': 'HPQ', 'price': 31.75}, {'shares': 100, 'name': 'IBM', 'price': 91.1}, {'shares': 75, 'name': 'ACME', 'price': 115.65}, {'shares': 50, 'name': 'AAPL', 'price': 543.22}]
//不规则字符串，按“小写-大写-奇数-偶数”顺序排序
s = 'asdf234GDSdsf23'
print("".join(sorted(s, key=lambda x: (x.isdigit(),x.isdigit() and int(x) % 2 == 0,x.isupper(),x))))
addffssDGS33224
//一道面试题：要求：正数在前负数在后 2.整数从小到大 3.负数从大到小
list1=[7, -8, 5, 4, 0, -2, -5]
print(sorted(list1,key=lambda x:(x<0,abs(x))))
[0, 4, 5, 7, -2, -5, -8]
-
用operator中的函数加快速度和进行多级排序
from operator import itemgetter, attrgetter
暂不讨论
比较三种方法的效率
-
只求TopN=1/BtmN=1时，比较max( )和nlargest( )两种效率
In [8]: nums=random.sample(range(1,10000),999)
In [9]: print(max(nums))
9999
In [10]: %time
CPU times: user 0 ns, sys: 0 ns, total: 0 ns
Wall time: 13.1 µs
In [11]: heapq.nlargest(1,nums)
Out[11]: [9999]
In [12]: %time
CPU times: user 0 ns, sys: 0 ns, total: 0 ns
Wall time: 14.1 µs
-
当K为10，N为9(即N无限接近K时)，比较了sorted( )+切片和nlargest( )两种方法的效率
In [23]: nums=random.sample(range(1,10000),10)
In [24]: sorted(nums,reverse=True)[:9]
Out[24]: [8814, 7551, 7318, 5597, 5257, 4437, 4211, 2776, 2440]
In [25]: %time
CPU times: user 0 ns, sys: 0 ns, total: 0 ns
Wall time: 11.4 µs
In [26]: heapq.nlargest(9,nums)
Out[26]: [8814, 7551, 7318, 5597, 5257, 4437, 4211, 2776, 2440]
In [27]: %time
CPU times: user 0 ns, sys: 0 ns, total: 0 ns
Wall time: 154 µs
-
当N较小时，比较了nlargest( )和sorted( )+切片两种方法
In [18]: nums=[16,7,3,20,17,8,-1]
In [19]: heapq.nlargest(3,nums)
Out[19]: [20, 17, 16]
In [20]: %time
CPU times: user 0 ns, sys: 0 ns, total: 0 ns
Wall time: 4.05 µs
In [21]: sorted(nums,reverse=True)[:3]
Out[21]: [20, 17, 16]
In [22]: %time
CPU times: user 0 ns, sys: 0 ns, total: 0 ns
Wall time: 5.48 µs
以上代码用到的import和show_tree( )
import math
import io
from io import StringIO
import heapq
import random
import time
from functools import wraps
def show_tree(tree, total_width=36, fill=' '):
output =io.StringIO() #创建stringio对象
last_row = -1
for i, n in enumerate(tree): #
if i:
row = int(math.floor(math.log(i+1, 2)))
else:
row = 0
if row != last_row:
output.write('\n')
columns = 2**row
col_width = int(math.floor((total_width * 1.0) / columns))
output.write(str(n).center(col_width, fill))
last_row = row
print(output.getvalue())
print('-' * total_width)
print(' ')
return
参考资料
python3-cookbook-1.4-查找最大或最小元素
详解Python中heapq模块的用法（这里有实现show_tree的函数代码）
理解堆和堆排序的文章1
理解堆和堆排序的文章2
理解堆和堆排序的文章3
python奇技淫巧——max/min函数的用法
nums=[16,7,3,20,17,8,-1]
print(sorted(nums))
print(sorted(nums,reverse=True))
[-1, 3, 7, 8, 16, 17, 20]
[20, 17, 16, 8, 7, 3, -1]
str=['b','a','A','s']
print(sorted(str))
print(sorted(str,reverse=True))
['A', 'a', 'b', 's']
['s', 'b', 'a', 'A']//按长度排序
L = [{1:5,3:4},{1:3,6:3},{1:1,2:4,5:6},{1:9}]
print(sorted(L,key=lambda x: len(x)))
[{1: 9}, {1: 5, 3: 4}, {1: 3, 6: 3}, {1: 1, 2: 4, 5: 6}]//根据指定的值来排序(例如字典中的某个key)
L = [
('john', 'A', 15),
('jane', 'B', 10),
('dave', 'B', 12),
]
print(sorted(L,key=lambda x:x[2],reverse=True))
[('john', 'A', 15), ('dave', 'B', 12), ('jane', 'B', 10)]
portfolio = [
{'name': 'IBM', 'shares': 100, 'price': 91.1},
{'name': 'AAPL', 'shares': 50, 'price': 543.22},
{'name': 'FB', 'shares': 200, 'price': 21.09},
{'name': 'HPQ', 'shares': 35, 'price': 31.75},
{'name': 'YHOO', 'shares': 45, 'price': 16.35},
{'name': 'ACME', 'shares': 75, 'price': 115.65}
]
print(sorted(portfolio,key=lambda x:x['price']))
[{'shares': 45, 'name': 'YHOO', 'price': 16.35}, {'shares': 200, 'name': 'FB', 'price': 21.09}, {'shares': 35, 'name': 'HPQ', 'price': 31.75}, {'shares': 100, 'name': 'IBM', 'price': 91.1}, {'shares': 75, 'name': 'ACME', 'price': 115.65}, {'shares': 50, 'name': 'AAPL', 'price': 543.22}]//不规则字符串，按“小写-大写-奇数-偶数”顺序排序
s = 'asdf234GDSdsf23'
print("".join(sorted(s, key=lambda x: (x.isdigit(),x.isdigit() and int(x) % 2 == 0,x.isupper(),x))))
addffssDGS33224//一道面试题：要求：正数在前负数在后 2.整数从小到大 3.负数从大到小
list1=[7, -8, 5, 4, 0, -2, -5]
print(sorted(list1,key=lambda x:(x<0,abs(x))))
[0, 4, 5, 7, -2, -5, -8]from operator import itemgetter, attrgetter
暂不讨论- 只求TopN=1/BtmN=1时，比较max( )和nlargest( )两种效率
In [8]: nums=random.sample(range(1,10000),999)
In [9]: print(max(nums))
9999
In [10]: %time
CPU times: user 0 ns, sys: 0 ns, total: 0 ns
Wall time: 13.1 µs
In [11]: heapq.nlargest(1,nums)
Out[11]: [9999]
In [12]: %time
CPU times: user 0 ns, sys: 0 ns, total: 0 ns
Wall time: 14.1 µs- 当K为10，N为9(即N无限接近K时)，比较了sorted( )+切片和nlargest( )两种方法的效率
In [23]: nums=random.sample(range(1,10000),10)
In [24]: sorted(nums,reverse=True)[:9]
Out[24]: [8814, 7551, 7318, 5597, 5257, 4437, 4211, 2776, 2440]
In [25]: %time
CPU times: user 0 ns, sys: 0 ns, total: 0 ns
Wall time: 11.4 µs
In [26]: heapq.nlargest(9,nums)
Out[26]: [8814, 7551, 7318, 5597, 5257, 4437, 4211, 2776, 2440]
In [27]: %time
CPU times: user 0 ns, sys: 0 ns, total: 0 ns
Wall time: 154 µs- 当N较小时，比较了nlargest( )和sorted( )+切片两种方法
In [18]: nums=[16,7,3,20,17,8,-1]
In [19]: heapq.nlargest(3,nums)
Out[19]: [20, 17, 16]
In [20]: %time
CPU times: user 0 ns, sys: 0 ns, total: 0 ns
Wall time: 4.05 µs
In [21]: sorted(nums,reverse=True)[:3]
Out[21]: [20, 17, 16]
In [22]: %time
CPU times: user 0 ns, sys: 0 ns, total: 0 ns
Wall time: 5.48 µs以上代码用到的import和show_tree( )
import math
import io
from io import StringIO
import heapq
import random
import time
from functools import wraps
def show_tree(tree, total_width=36, fill=' '):
output =io.StringIO() #创建stringio对象
last_row = -1
for i, n in enumerate(tree): #
if i:
row = int(math.floor(math.log(i+1, 2)))
else:
row = 0
if row != last_row:
output.write('\n')
columns = 2**row
col_width = int(math.floor((total_width * 1.0) / columns))
output.write(str(n).center(col_width, fill))
last_row = row
print(output.getvalue())
print('-' * total_width)
print(' ')
return参考资料
python3-cookbook-1.4-查找最大或最小元素
详解Python中heapq模块的用法（这里有实现show_tree的函数代码）
理解堆和堆排序的文章1
理解堆和堆排序的文章2
理解堆和堆排序的文章3
python奇技淫巧——max/min函数的用法
