sorted

python排序之sort,sorted用法

生来就可爱ヽ(ⅴ<●) 提交于 2019-12-22 18:37:00
【推荐】2019 Java 开发者跳槽指南.pdf(吐血整理) >>> python内置排序方式有两种,一种是针对列表的.sort()方法,另外一种是sorted()函数,他们方法类似,区别有两点 1、list.sort()方法只适合列表对象,sorted()函数适合任何可迭代对象。 2、list.sort()用于改变当前列表,sorted()返回一个新的序列。 用法示例: >>>help(sorted) Help on built-in function sorted in module builtins: sorted(...) sorted(iterable, key=None, reverse=False) --> new sorted list >>>ls = [5,1,9,7] >>> sorted(ls) [1, 5, 7, 9] >>>ls.sort() >>>ls [1, 5, 7, 9] #逆序 >>ls.sort(reverse = True) >>>ls [9,7,5,1] 通过key函数指定排序方法:(因为sort()和sorted()功能类似,之后都使用sorted()示例) #假定有个学生年级、年龄名单 >>>students = {'mhye':(2,18),'Jackey':(3,15),'Lucy':(1,16)} #如果想要按年级排序 >>

Convert Sorted Array to Binary Search Tree (Picturing the Recursion)

人走茶凉 提交于 2019-12-22 15:09:18
问题 I want to convert a sorted integer array into a binary search tree. I believe I understand how to do this. I have posted my pseudo-code below. What I cannot picture is how the recursion actually works. So if my array is 1, 2, 3, 4, 5... I first make 3 the root of my BST. Then I make 2 the left-child node of 3. Then do I make 1 the left-child node of 2, and come back to 2? I don't get how the recursion steps through the whole process... Thanks in advance, and apologies if this question is

Collection modify item

微笑、不失礼 提交于 2019-12-22 07:08:13
问题 I have read tons of articles on choosing the correct collection for a specific implementation, and I understand that in the end it will come down to benchmarking real data, but while I'm busy doing that: What sorted collection in c# allows the modification of an item contained?I can't seem to find any? Is this because a modify would probably be implemented as a removal then re-insertion, thus making an explicit 'Modify' function pointless? I am in need of a collection (custom or standard

efficient sorted Cartesian product of 2 sorted array of integers

∥☆過路亽.° 提交于 2019-12-22 04:50:48
问题 Need Hints to design an efficient algorithm that takes the following input and spits out the following output. Input: two sorted arrays of integers A and B, each of length n Output: One sorted array that consists of Cartesian product of arrays A and B. For Example: Input: A is 1, 3, 5 B is 4, 8, 10 here n is 3. Output: 4, 8, 10, 12, 20, 24, 30, 40, 50 Here are my attempts at solving this problem. 1) Given that output is n^2, Efficient algorithm can't do any better than O(n^2) time complexity.

python3 sort()与sorted()的区别

余生长醉 提交于 2019-12-21 23:52:20
1, sort()只用于list,sorted()可以对所有的迭代对象进行排序 2, sort()原地排序返回None, sorted()返回一个新的list 3, 调用方式: listname.sort(reverse=False) sorted(iterable,key=None,reverse=False) iterable---可迭代对象 key---用来进行比较的元素,只有一个参数,具体的函数参数就是取自于可迭代对象中,指定可迭代对象中的一个元素来进行排序 4, sorted()可用于dict的排序: dict1 = {"a":12,"5":98,"e":80} print(sorted(dict1.items(),key=lambda x:x[1])) 来源: CSDN 作者: 奈之若何℡ 链接: https://blog.csdn.net/qq_42036970/article/details/103647977

Sorted hash table (map, dictionary) data structure design

强颜欢笑 提交于 2019-12-18 11:42:38
问题 Here's a description of the data structure: It operates like a regular map with get , put , and remove methods, but has a sort method that can be called to sorts the map. However, the map remembers its sorted structure, so subsequent calls to sort can be much quicker (if the structure doesn't change too much between calls to sort ). For example: I call the put method 1,000,000 times. I call the sort method. I call the put method 100 more times. I call the sort method. The second time I call

Automatically sorted by values map in Java

馋奶兔 提交于 2019-12-17 10:57:33
问题 I need to have an automatically sorted-by-values map in Java - so that It keeps being sorted at any time while I'm adding new key-value pairs or update the value of an existing key-value pair, or even delete some entry. Please also have in mind that this map is going to be really big (100's of thousands, or even 10's of millions of entries in size). So basically I'm looking for the following functionality: Supposed that we had a class 'SortedByValuesMap' that implements the aforementioned

Real time sorted by value, auto-discarding, bounded collection ?

蹲街弑〆低调 提交于 2019-12-14 03:26:45
问题 I spent some time to try to make a collection that: 1) is sorted by value (not by key) 2) is sorted each time an element is added or modified 3) is fixed size and discard automatically smallest/biggest element depending of the sort way 4) is safe thread So 3) and 4) I think it is quite ok. For 1) and 2) it was a bit more tricky. I spent quite a long time on this thread, experimenting the different sample, but one big issue is that the collection are sorted only once when object are inserted.

python dict 排序

隐身守侯 提交于 2019-12-13 21:14:29
python dict按照value 排序 我们知道Python的内置dictionary数据类型是无序的,通过key来获取对应的value。可是有时我们需要对dictionary中 的item进行排序输出,可能根据key,也可能根据value来排。到底有多少种方法可以实现对dictionary的内容进行排序输出呢?下面摘取了 一些精彩的解决办法。 #最简单的方法,这个是按照key值排序: def sortedDictValues1(adict): items = adict.items() items.sort() return [value for key, value in items] #又一个按照key值排序,貌似比上一个速度要快点 def sortedDictValues2(adict): keys = adict.keys() keys.sort() return [dict[key] for key in keys] #还是按key值排序,据说更快。。。而且当key为tuple的时候照样适用 def sortedDictValues3(adict): keys = adict.keys() keys.sort() return map(adict.get, keys) #一行语句搞定: [(k,di[k]) for k in sorted(di.keys())]

Better way to find matches in two sorted lists than using for loops? (Java)

烂漫一生 提交于 2019-12-13 13:11:53
问题 I have two sorted lists, both in non-decreasing order. For example, I have one sorted linked list with elements [2,3,4,5,6,7...] and the other one with elements [5,6,7,8,9...] . I need to find all common elements in both lists. I know I can use a for loop and a nested loop to iterate all matches to find the same two elements. However, is there another way to do this that has running time less than O(n^2) ? 回答1: You can do it in O(n) time. Pseudocode: a = list1.first b = list2.first repeat: if