Measures of Variation 方差:measures of variation or measures of spread 源于range发现range不足以评估整个set(因为只用到largest and smallest value),所以有了方差 The Sample Standard Deviation: 偏差:How far, on average, the observations are from the mean: 几何意义: 必须取绝对值,否则第二列sum=0(因

指令集总览              1.General 1).Comment:: 注释指令. 2).Image Window: 创建新的图像窗口. True: 取像之后更新图像窗口显示:False: 有新图像时不更新图像窗口.取像成功返回true. 3).Return: 流程指令”返回“.从子程序插入一个返回. 4).Subrotine: 创建一个子程序. 5).Test: 评估声明. 2.Array: Boolean(布尔数组指令)      同 "Array: Number(整型数组指令)

1. tf.nn.moments(x, axes=[0, 1, 2])  # 对前三个维度求平均值和标准差,结果为最后一个维度,即对每个feature_map求平均值和标准差 参数说明:x为输入的feature_map, axes=[0, 1, 2] 对三个维度求平均,即每一个feature_map都获得一个平均值和标准差 2.with tf.control_dependencies([train_mean, train_var]): 即执行with里面的操作时,会先执行train_mean 和

IQR(Inter-Quartile Range)在统计中叫内距.内距又称为四分位差.具体如下:内距IQR即Inter-Quartile Range, 这是统计技术上的名词.内距又称为四分位差,是两个四分位数之差,即内距IQR=高四分位数—低四分位数.标准化四分位距——对一组按顺序排列的数据,上四分位值Q3与下四分位值Q1之间的差称为四分位距(IQR),即IQR=Q3-Q1.IQR乘以因子0.7413得标准化四分位距(Norm IQR),它是稳健统计技术处理中用于表示数据分散程度的一个量,其值相

1. tf.nn.conv2d(x, w, strides=[1, 1, 1, 1], padding='SAME')  # 对数据进行卷积操作 参数说明:x表示输入数据,w表示卷积核, strides表示步长,分别表示为样本数,长,宽,通道数,padding表示补零操作 2. tf.nn.max_pool(x, ksize=[1, 2, 2, 1], strides=[1, 2, 2, 1], padding='SAME')  # 对数据进行池化操作 参数说明:x表示输入数据,ksize表示卷

一.值域(Range) Range = Max - Min 受异常值(Outliers)影响 二.四分位差(IQR) 四分位距(interquartile range, IQR),又称四分差.是描述统计学中的一种方法,以确定第三四分位数和第一四分位数的区别(即Q1~Q3 的差距). 三.异常值(Outlier) 异常值的常用定义:Outlier < Q1 - 1.5IQR OR > Q3 + 1.5IQR 可视化--Box Plots(箱线图) & Whisker(盒须图) IQR的不

Descriptive statistics tell you about the distribution of data points in data set. The most common measures are the mean, median, mode, range, and standard deviation. For all the examples explained below, we will use the following fictional set of ma

在使用 torchvision.transforms进行数据处理时我们经常进行的操作是: transforms.Normalize((0.485,0.456,0.406), (0.229,0.224,0.225)) 前面的(0.485,0.456,0.406)表示均值,分别对应的是RGB三个通道:后面的(0.229,0.224,0.225)则表示的是标准差 这上面的均值和标准差的值是ImageNet数据集计算出来的,所以很多人都使用它们 但是如果你想要计算自己的数据集的均值和标准差,让其作为你的

应用统计学 数据的概括性度量 集中趋势 Mode众数是唯一描述无序类别数据,由图可知众数便是图形中的峰. 对于类别变量,众数就是某一种类别. 中位数和平均数都可能不是样本中的值. 中位数不受极值影响,对于类别数据来说,中位数是某一类别(同mode),各变量值与中位数的离差绝对值之和最小,与均数不同. 平均数的求法,令函数等于各变量值与平均数的离差平方之和,该函数表达如下式. 对该函数求一阶导,如下式, 当一阶导为零时该函数取到最小值,此时样本均值表达式为: 各变量值与平均数的离差平方之和最小,各

    最近帮一个客户搭建跨洋的合并复制,由于数据库非常大,跨洋网络条件不稳定,因此只能通过备份初始化,在初始化完成后向海外订阅端插入数据时发现报出如下错误: Msg 548, Level 16, State 2, Line 2 The insert failed. It conflicted with an identity range check constraint in database %s, replicated table %s, column %s. If the identit

在Java位运算总结-leetcode题目博文中总结了Java提供的按位运算操作符,今天又碰到LeetCode中一道按位操作的题目 Given a range [m, n] where 0 <= m <= n <= 2147483647, return the bitwise AND of all numbers in this range, inclusive. For example, given the range [5, 7], you should return 4. 题意:

Assume you have an array of length n initialized with all 0's and are given k update operations. Each operation is represented as a triplet: [startIndex, endIndex, inc] which increments each element of subarray A[startIndex ... endIndex] (startIndex

Given an integer array nums, return the number of range sums that lie in [lower, upper] inclusive.Range sum S(i, j) is defined as the sum of the elements in nums between indices i and j (i ≤ j), inclusive. Note:A naive algorithm of O(n2) is trivial.

Given a 2D matrix matrix, find the sum of the elements inside the rectangle defined by its upper left corner (row1, col1) and lower right corner (row2, col2). The above rectangle (with the red border) is defined by (row1, col1) = (2, 1) and (row2, co

Given an integer array nums, find the sum of the elements between indices i and j (i ≤ j), inclusive. The update(i, val) function modifies nums by updating the element at index i to val. Example: Given nums = [1, 3, 5] sumRange(0, 2) -> 9 update(1, 2

Given a 2D matrix matrix, find the sum of the elements inside the rectangle defined by its upper left corner (row1, col1) and lower right corner (row2, col2). The above rectangle (with the red border) is defined by (row1, col1) = (2, 1) and (row2, co

Given an integer array nums, find the sum of the elements between indices i and j (i ≤ j), inclusive. Example: Given nums = [-2, 0, 3, -5, 2, -1] sumRange(0, 2) -> 1 sumRange(2, 5) -> -1 sumRange(0, 5) -> -3 Note: You may assume that the array do

Given a range [m, n] where 0 <= m <= n <= 2147483647, return the bitwise AND of all numbers in this range, inclusive. For example, given the range [5, 7], you should return 4. Credits:Special thanks to @amrsaqr for adding this problem and creatin

python中的range功能非常好用 for i in range(100): print(i) 现在利用C++11的基于范围的for循环特性实现C++中的range功能 class range { public: range(int end) { _begin = 0; _end = end; _step = 1; } range(int begin, int end, int step=1) { _begin = begin; _end = end; _step = step; if (!

报错: 原来用的python3.5版本后来改为2.7出现了这个错误里面的中文无法显示 UnicodeDecodeError: 'ascii' codec can't decode byte 0xe6 in position 0: ordinal not in range(128) 解决办法: 文件抬头加入 import sys reload(sys) sys.setdefaultencoding("utf-8") 原因: 提示中的“ordinal not )”,意思是,字符不在128范