Question

我有一个看起来像这样的表：

Character   Legend  Example Sample Match
\t  Tab T\t\w{2}    T     ab
\r  Carriage return character   see below   
\n  Line feed character see below   
\r\n    Line separator on Windows   AB\r\nCD    AB
    CD
\N  Perl, PCRE (C, PHP, R…): one character that is not a line break \N+ ABC
\h  Perl, PCRE (C, PHP, R…), Java: one horizontal whitespace character: tab or Unicode space separator      
\H  One character that is not a horizontal whitespace       
\v  .NET, JavaScript, Python, Ruby: vertical tab        
\v  Perl, PCRE (C, PHP, R…), Java: one vertical whitespace character: line feed, carriage return, vertical tab, form feed, paragraph or line separator      
\V  Perl, PCRE (C, PHP, R…), Java: any character that is not a vertical whitespace      
\R  Perl, PCRE (C, PHP, R…), Java: one line break (carriage return + line feed pair, and all the characters matched by \v)

（“天”列不涉及我的偏度和峰度计算，仅在我的表中）

我想要一个表，用于按年份分组计算偏度和峰度值：

start_table <- data.frame("Water_Year" =  c("1903", "1903", "1904", "1904"), "X" = c(13, 11, 12, 
15), "Day" = c(1, 2, 1, 2))

我不知道如何按年份将其分组以执行这些计算。

Answer 1

将fBasics与data.table一起使用：

library(fBasics)
library(data.table)

setDT(start_table)[, .(Skew = skewness(X), Kurtosis=kurtosis(X)), .(Water_Year)][]

#>    Water_Year Skew Kurtosis
#> 1:       1903    0    -2.75
#> 2:       1904    0    -2.75

Answer 2

您还可以定义偏度/峰度函数：

kurtosis <- function(x) {  
 m4 <- mean((x - mean(x))^4) 
 kurtosis <- m4/(sd(x)^4) - 3  
 kurtosis
}

skewness <-  function(x) {
 m3 <- mean((x - mean(x))^3)
 skewness <- m3/(sd(x)^3)
 skewness
}

然后，将其应用于base R：

aggregate(X ~ Water_Year, 
          FUN = function(x) c(kurtosis = kurtosis(x), skewness = skewness(x)),
          data = start_table)

  Water_Year X.kurtosis X.skewness
1       1903      -2.75       0.00
2       1904      -2.75       0.00

Answer 3

一个选项是# Load Pyomo from pyomo.environ import * import numpy as np # Create model m3 = ConcreteModel() ## Declare variables with initial values WITHOUT bounds m3.x1 = Var(initialize=1) m3.x2 = Var(initialize=1) ## Declare objective m3.OBJ = Objective(expr=m3.x1**3 - m3.x2 - m3.x1*m3.x2 - m3.x2**2, sense = minimize) ## Declare equality constraints m3.h = Constraint(expr= m3.x1**2 + m3.x2**2 == 1) # initial near global min theta0 = 1.0 # initialize near local min # theta0 = np.pi m3.x1 = np.sin(theta0) m3.x2 = np.cos(theta0) # Specify solver solver=SolverFactory('baron') solver.options['LocRes'] = 1 solver.options['results'] = 1 # Solve model solver.solve(m3, tee=True) ## Return the solution print("x1 = ",value(m3.x1)) print("x2 = ",value(m3.x2)) print("objective = ",value(m3.OBJ)) print("\n")

res.lst

在R中按年计算偏斜和峰度

3 个答案: