[英]How to get more than one smallest /largest values of a dictionary?
Like if i create a dictionary using the following script: 就像我使用以下脚本创建字典一样:
r_lst = {}
for z in df.index:
t = x.loc[0, 'sn']
s = df.loc[z, 'rv']
slope, intercept, r_value, p_value, std_err = stats.linregress(t,s)
r_lst.setdefault(float(z),[]).append(float(r_value))
The resultant dictionary loos like this 结果字典像这样
{4050.32: [0.29174641574734467],
4208.98: [0.20938901991887324],
4374.94: [0.2812420188097632],
4379.74: [0.28958742731611586],
4398.01: [0.3309140298947313],
4502.21: [0.28702220304639836],
4508.28: [0.2170363811575936],
4512.99: [0.29080133884942105]}
Now if i want to find the smallest or largest values of the dictionary then i just have to use this simple command min(r_lst.values())
or max(r_lst.values())
. 现在,如果我想查找字典的最小或最大值,则只需使用此简单命令min(r_lst.values())
或max(r_lst.values())
。
What if i want the three lowest values or three highest values.? 如果我要三个最低值或三个最高值怎么办? Then how can i get that? 那我该怎么办呢? I saw some questions here but none of them answers what i'm asking for. 我在这里看到了一些问题,但没有一个回答我的要求。
Just use sorted()
: 只需使用sorted()
:
sorted(r_lst.values())
Or in reverse order: 或以相反的顺序:
sorted(r_lst.values(), reverse=True)
So to get the 3 largest values: 因此,要获得3个最大值:
sorted(r_lst.values(), reverse=True)[:3]
Yields: 产量:
[[0.3309140298947313], [0.29174641574734467], [0.29080133884942105]]
You can use heapq.nsmallest
/ heapq.nlargest
: 您可以使用heapq.nsmallest
/ heapq.nlargest
:
import heapq
res = heapq.nsmallest(3, d.values())
# [[0.20938901991887324], [0.2170363811575936], [0.2812420188097632]]
To extract as a flat list, you can use itertools.chain
: 要提取为平面列表,可以使用itertools.chain
:
from itertools import chain
import heapq
res = list(chain.from_iterable(heapq.nsmallest(3, d.values())))
# [0.20938901991887324, 0.2170363811575936, 0.2812420188097632]
These heapq
solutions will have time complexity O(( n - k )*log n ) versus O( n log n ) for solutions requiring a full sort. 对于需要完整排序的解决方案,这些heapq
解决方案的时间复杂度为O(( n - k )* log n )与O( n log n )。
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