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Martin Hope
Yang Yang
Asked: 2024-11-21 05:01:04 +0800 CST
  • 8

我正在对我的流量数据进行对数皮尔逊 III 分布拟合。拟合后，我想绘制观测值和拟合分布。

然而，这个数字并不是我所期望的：

1. x 轴标签没有显示 0.99 0.02 和 0.01，尽管我已将其设置breaks为 c(0.99、0.9、0.5、0.2、0.1、0.02、0.01)。

2.如何在图中绘制拟合的 Log Pearson III 分布？我提供了一个示例图供参考，谢谢您的帮助。

library(fitdistrplus)
library(PearsonDS)
library(ggplot2)

low_flows <- data.frame(Year = 1981:2010, Flow = c(0.03357143, 0.01328571, 0.02285714, 0.02657143, 0.04957143, 0.04085714, 0.16900000, 0.01057143, 
                                                   0.04128571, 0.10871429, 0.08771429, 0.09585714, 0.22057143, 0.11571428, 0.08300000, 0.11257143, 
                                                   0.13614286, 0.07742857, 0.09785714, 0.04728571, 0.04300000, 0.08385714, 0.02828571, 0.07271429, 
                                                   0.21428571, 0.10142857, 0.04400000, 0.10928571, 0.12471429, 0.31500000))


Log_mydata <- log10(low_flows$Flow)

m <- mean(Log_mydata)
v <- stats::var(Log_mydata)
g <- e1071::skewness(Log_mydata, type = 2)
my_shape <- (2/g)^2
my_scale <- sqrt(v)/sqrt(my_shape) * sign(g)
my_location <- m - my_scale * my_shape
start <- list(shape = my_shape, location = my_location, scale = my_scale)

dPIII <<- function(x, shape, location, scale) PearsonDS::dpearsonIII(x, 
                                                                     shape, location, scale, log = FALSE)
pPIII <<- function(q, shape, location, scale) PearsonDS::ppearsonIII(q, 
                                                                     shape, location, scale, lower.tail = TRUE, log.p = FALSE)
qPIII <<- function(p, shape, location, scale) PearsonDS::qpearsonIII(p, 
                                                                     shape, location, scale, lower.tail = TRUE, log.p = FALSE)

fit_lp3 <- fitdist(Log_mydata, distr = "PIII", start = start)

plotdata = low_flows
plotdata <- plotdata[order(plotdata$Flow), ]
plotdata$prob <- seq_len(length(plotdata$Flow))/(length(plotdata$Flow) + 1)

prob_scale_points = c(0.99, 0.9, 0.5, 0.2, 0.1, 0.02, 0.01)

ggplot(data = plotdata, aes(x = prob, y = Flow)) + 
  geom_point(size = 2) + 
  xlab("Probability") + 
  scale_x_continuous(transform = scales::probability_trans("norm", lower.tail = FALSE), 
                     breaks = prob_scale_points, 
                     sec.axis = dup_axis(name = "Return Period", 
                                         labels = function(x) {ifelse(1/x < 2, round(1/x, 2), round(1/x, 0))})) +
  theme_bw()  + 
  theme(panel.border = element_rect(colour = "black", fill = NA, linewidth = 1), 
        panel.grid = element_line(linewidth = 0.2), 
        axis.title = element_text(size = 12), axis.text = element_text(size = 10), 
        axis.title.x.top = element_text(size = 12), 
        legend.text = element_text(size = 10), legend.title = element_text(size = 10))

以下是我从上面的代码中得到的结果： 在此处输入图片描述

我的目标如下：

在此处输入图片描述

按照 Allan 的建议，Log Pearson III 可以正确绘制了！为了简化我的原始代码，我尝试使用scale_x_log10而不是scale_x_continuous。但是，Log Pearson III 的拟合线没有正确绘制。我想知道如何解决这个问题？

ggplot(data = plotdata, aes(x = prob, y = Flow)) + 
  geom_point(size = 2, color="blue") + 
  geom_function(fun = ~ 10^(qpearsonIII(.x, params = fit_lp3$estimate)),
                color = "black", linewidth = 1.5, alpha = 0.7) + 
  scale_x_log10(name = "Probability", 
                trans = "reverse", 
                breaks = prob_scale_points, 
                limits = rev(range(prob_scale_points)), 
                sec.axis = dup_axis(name = "Return Period", 
                                    labels = function(x) {ifelse(1/x < 2, round(1/x, 2), round(1/x, 0))})) + 
  scale_y_log10(breaks = seq(0.05, 0.4, 0.05)) + 
  theme_bw()  + 
  theme(panel.border = element_rect(colour = "black", fill = NA, linewidth = 1), 
        panel.grid = element_line(linewidth = 0.2), 
        axis.title = element_text(size = 12), axis.text = element_text(size = 10), 
        axis.title.x.top = element_text(size = 12), 
        legend.text = element_text(size = 10), legend.title = element_text(size = 10))

这是图。我们可以看到拟合的线不正确。

在此处输入图片描述

Martin Hope
Yang Yang
Asked: 2023-12-02 08:14:18 +0800 CST
  • 3

我是 Python 新手，在使用min().

import pandas as pd
a = [1,2,3,4,5]
print(min(20, pd.array(a)**2))

该代码应返回 [1, 4, 9, 16, 20]，但是，它返回 [1, 4, 9, 16, 25]。似乎min()没有按预期工作。原因是什么以及如何解决？谢谢。

python