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  • Median

    2017-10-17 23:38:20
    Note in this problem, the median is defined as the (m/2)-th smallest number if m,the amount of the differences, is even. For example, you have to find the third smallest one in the case of m = 6. ...
  • median

    2017-10-03 19:00:55
    median的作用:求矩阵的中位数。 median(M),每一列返回一个值,为M该列的从大到小排列的中位数. median(M,dim),dim为维度。 其中1表示按每列返回一个值,为该列从大到小排列的中位数, 2表示...
    median的作用:求矩阵的中位数。

    median(M),每一列返回一个值,为M该列的从大到小排列的中位数.

    median(M,dim),dim为维度。
    其中1表示按每列返回一个值,为该列从大到小排列的 中位数,
    2表示按每行返回一个值,为该行从大到小排列的 中位数.
    (和mean()一样...)
    注意:如果行或列的个数为偶数,返回中间两个值的平均值。举例如下:

    A=[1 4 5;2 8 3;9 7 6];
    >> median(A)

    ans =

            2       7       5

    >> median(A,1)

    ans =

            2       7       5

    >> median(A,2)
    ans =

            4
            3
            7
    展开全文
  • Python计算中位数 numpy.median

    万次阅读 多人点赞 2017-07-04 17:46:48
    numpy模块下的median作用为: 计算沿指定轴的中位数 返回数组元素的中位数其函数接口为:median(a, axis=None, out=None, overwrite_input=False, keepdims=False)其中各参数为: a:输入的数组; axis:...

    numpy模块下的median作用为:
    计算沿指定轴的中位数
    返回数组元素的中位数

    其函数接口为:

    median(a, 
           axis=None, 
           out=None,
           overwrite_input=False, 
           keepdims=False)
    

    其中各参数为:
    a:输入的数组;
    axis:计算哪个轴上的中位数,比如输入是二维数组,那么axis=0对应行,axis=1对应列,如果对于二维数组不指定长度,将拉伸为一唯计算中位数;
    out:用于放置求取中位数后的数组。 它必须具有与预期输出相同的形状和缓冲区长度;
    overwrite_input :一个bool型的参数,默认为Flase。如果为True那么将直接在数组内存中计算,这意味着计算之后原数组没办法保存,但是好处在于节省内存资源,Flase则相反;
    keepdims:一个bool型的参数,默认为Flase。如果为True那么求取中位数的那个轴将保留在结果中;

    >>> a = np.array([[10, 7, 4], [3, 2, 1]])
    >>> a
    array([[10,  7,  4],
           [ 3,  2,  1]])
    >>> np.median(a)
    # 1,2,3,4,7,10 
    # (3+4)/2 = 3.5
    3.5
    >>> np.median(a, axis=0)
    #(10+3)/2 = 6.5
    #(7+2)/2 = 4.5
    #(4+1)/2 = 2.5 
    array([ 6.5,  4.5,  2.5])
    >>> np.median(a, axis=1)
    #奇数个数中位数
    array([ 7.,  2.])
    >>> m = np.median(a, axis=0)
    >>> out = np.zeros_like(m)
    >>> np.median(a, axis=0, out=m)
    array([ 6.5,  4.5,  2.5])
    >>> m
    array([ 6.5,  4.5,  2.5])
    >>> b = a.copy()
    >>> np.median(b, axis=1, overwrite_input=True)
    array([ 7.,  2.])
    >>> assert not np.all(a==b)
    >>> b = a.copy()
    >>> np.median(b, axis=None, overwrite_input=True)
    3.5
    
    展开全文
  • Running Median

    万次阅读 2021-02-27 22:51:22
    For this problem, you will write a... After each odd-indexed value is read, output the median (middle value) of the elements received so far. 输入描述: The first line of input contains a single integer P

    For this problem, you will write a program that reads in a sequence of 32-bit signed integers. After each odd-indexed value is read, output the median (middle value) of the elements received so far.
    输入描述:
    The first line of input contains a single integer P([1, 1000]), which is the number of data sets that follow. The first line of each data set contains the data set number, followed by a space, followed by an odd decimal integer M([1, 9999]), giving the total number of signed integers to be processed. The remaining line(s) in the dataset consists of the values, 10 per line, separated by a single space. The last line in the dataset may contain less than 10 values.

    输出描述:
    For each data set the first line of output contains the data set number, a single space and the number of medians output (which should be one-half the number of input values plus one). The output medians will be on the following lines, 10 per line separated by a single space. The last line may have less than 10 elements, but at least 1 element. There should be no blank lines in the output.
    示例1
    输入
    3
    1 9
    1 2 3 4 5 6 7 8 9
    2 9
    9 8 7 6 5 4 3 2 1
    3 23
    23 41 13 22 -3 24 -31 -11 -8 -7
    3 5 103 211 -311 -45 -67 -73 -81 -99
    -33 24 56
    输出
    1 5
    1 2 3 4 5
    2 5
    9 8 7 6 5
    3 12
    23 23 22 22 13 3 5 5 3 -3
    -7 -3

    #include<bits/stdc++.h>
    using namespace std;
    char buf[1<<20],*P1=buf,*P2=buf;
    #define gc() (P1==P2&&(P2=(P1=buf)+fread(buf,1,1<<20,stdin),P1==P2)?EOF:*P1++)
    #define TT template<class T>inline
    TT bool read(T &x)
    {
        x=0;char c=gc();bool f=0;
        while(c<48||c>57){if(c==EOF)return 0;f^=(c=='-'),c=gc();}
        while(47<c&&c<58)x=(x<<3)+(x<<1)+(c^48),c=gc();
        if(f)x=-x;return 1;
    }
    TT bool read(T&a,T&b){return read(a)&&read(b);}
    TT bool read(T&a,T&b,T&c){return read(a)&&read(b)&&read(c);}
    typedef long long ll;
    const ll MAXN=10;
    #define lowbit(x) (x&(-x))
    int main()
    {
        int t,cas;
        for(read(t);t;--t)
    	{
            priority_queue<int>h1;
            priority_queue<int,vector<int>,greater<int>>h2;
            int n;
            read(cas,n);
            printf("%d %d\n",cas,n/2+1);
            for(int i=1,x;i<=n;++i)
    		{
                read(x);
                if(h1.empty()||x<h1.top())h1.push(x);
                else h2.push(x);
                if(~i&1)continue;
                while(h1.size()>h2.size()+1)h2.push(h1.top()),h1.pop();
                while(h1.size()<h2.size()+1)h1.push(h2.top()),h2.pop();
                if(i>1&&(i-1)/2%10==0)putchar('\n');
                printf("%d ",h1.top());
            }
    		putchar('\n');
        }
        return 0;
    }
    
    展开全文
  • 1029 Median

    2021-07-12 15:52:37
    题目来源:PAT (Advanced Level) Practice Given an increasing sequence S of... For example, the median of S1 = { 11, 12, 13, 14 } is 12, and the median of S2 = { 9, 10, 15, 16, 17 } is 15. The median of t

    题目来源:PAT (Advanced Level) Practice

    Given an increasing sequence S of N integers, the median is the number at the middle position. For example, the median of S1 = { 11, 12, 13, 14 } is 12, and the median of S2 = { 9, 10, 15, 16, 17 } is 15. The median of two sequences is defined to be the median of the nondecreasing sequence which contains all the elements of both sequences. For example, the median of S1 and S2 is 13.

    Given two increasing sequences of integers, you are asked to find their median.

    Input Specification:

    Each input file contains one test case. Each case occupies 2 lines, each gives the information of a sequence. For each sequence, the first positive integer N (≤2×10​5​​) is the size of that sequence. Then N integers follow, separated by a space. It is guaranteed that all the integers are in the range of long int.

    Output Specification:

    For each test case you should output the median of the two given sequences in a line.

    Sample Input:

    4 11 12 13 14
    5 9 10 15 16 17

    Sample Output:

    13

    words:

    median 中间的,中值的        

    思路:

    1. 本题的意思是从两个数列中找出中间的那个数,首先根据两个数列的长度计算出最中间的数序号;

    2. 将两个数列归并为一个数列,其实不用真正的归并,只需要在归并时找出第median_index个数是多少就行了;

    //PAT ad 1029 Median
    #include <iostream>
    #define N 200005
    using namespace std;
    #include <cmath>
    
    int n1,n2;
    void input(int s[N],int &nn)
    {
    	int n;
    	cin>>n;
    	for(int i=0;i<n;i++)
    		cin>>s[i];
    	nn=n;
    } 
    
    int main()
    {
    	int s1[N],s2[N];
    	input(s1,n1);   //输入数列s1
    	input(s2,n2);
    	int i,j,k=0;
    	int median_index=ceil((n1+n2)/2.0);   //中间位置
    	int median;             //中间数
    	for(i=0,j=0;i<n1&&j<n2;)
    	{
    		if(s1[i]<s2[j])
    		{
    			k++;i++;
    			if(k==median_index)
    			{
    				cout<<s1[i-1]<<endl;return 0;
    			}
    				
    		}
    		else
    		{
    			k++;j++;
    			if(k==median_index)
    			{
    				cout<<s2[j-1]<<endl;return 0;
    			}
    				
    		}
    	}
    	while(i<n1)
    	{
    		k++;i++;
    		if(k==median_index)
    		{
    			cout<<s1[i-1]<<endl;return 0;
    		}			
    	}
    	while(j<n2)
    	{
    		k++;j++;
    		if(k==median_index)
    		{
    			cout<<s2[j-1]<<endl;return 0;
    		}		
    	}
    	
    	return 0;
    }

    展开全文
  • median函数.xls

    2021-09-18 20:21:21
    median函数.xls
  • numpy中的中值计算方法是使用median median的方法的,已经是我现在知道的最快的算法了。 一些是np的中值计算方法从Python世界怎么一步一步调用到C世界的: np.median() # python() np._median() # python() if ...
  • median-filter-源码

    2021-03-27 20:42:43
    median-filter
  • MedianFlow.rar

    2019-12-03 21:17:43
    TLD开源代码MedianFlow部分,改造成一个运行程序示例,包含测试主程序及测试视频序列。运行环境:vs2008 + opencv2.4.2
  • L1 Median

    2020-11-19 20:43:43
    L1 Median:定义为空间中的一点,该点到数据集中其它点的欧式距离之和最小。 https://www.cnblogs.com/yhlx125/p/5049069.html
  • median filtering

    2013-05-04 19:54:45
    matlab编写median滤波 去除椒盐噪声(给出图片实例) 最后附加自带程序medfilt2(J)
  • median filter

    2019-03-26 12:40:36
    Wiki link: https://en.wikipedia.org/wiki/Median_filter
  • PAT 1029 Median

    2020-04-02 08:43:54
    Given an increasing sequence S of N integers, the median is the number at the middle position. For example, the median of S1 = { 11, 12, 13, 14 } is 12, and the median of S2 = { 9, 10, 15, 16, 17...
  • POJ 3579 Median

    千次阅读 2020-10-07 15:16:36
    POJ 3579 Median 题目链接 Description Given N numbers, X1, X2, … , XN, let us calculate the difference of every pair of numbers: ∣Xi - Xj∣ (1 ≤ i ≤ N). We can get C(N,2) differences through this ...
  • T he median filter is a popular image processing technique for removing salt and pepper (“shot”) noise from images. With this technique, the eight direct neigh- bors and center point of a sliding 3-...

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