对于循环：logit模型中一个变量的不同值的预测

发布于 2025-02-13 07:54:20 字数 1478 浏览 0 评论 0原文

我估计足球门票销售的需求功能（Logit：1 =售票/盖帽，0 =票务/块未出售）。

model_Gegner <- Ausverkauft_2 ~ Preis + Block + Gegner + Spieltagsklasse + Leistung_Heim + 
  Leistung_Auswärts + Temperatur + Unsicherheit + Wochentag+ Saison

Logit_Gegner <- glm(model_Gegner, data = train, family = "binomial")

在下一步中，我想看看俱乐部可能还会收取的最高价格来出售票。为了找到这个问题，我设计了一个用于循环的：

Preis <- c(1:100)
Preis

#test object: 
SpielRevenue <- subset(test, Datum == "2018-12-02" & Kategorie == "4" & Block == "15E")

#### for in loop 
for(p in Preis ){
  SpielRevenue$Preis <- p
  print(predict(Logit_Gegner,SpielRevenue, type="response")) 
}

而不是以不同的价格出售票的概率，而是得到以下内容：

由于我只将价格限制在100下，所以我不明白440的来源。有100倍的440，因此您可以为每个440分配不同的价格。但这将是一项巨大的计数工作。

我想获得以下结果：

如果可能的话，我还想将结果保存在数据范围内。

贡献朝着：但是，我还没有找到任何帮助我的贡献。如果有一个，我很抱歉重复。

谢谢你！

原文

I have estimated a demand function for football ticket sales (logit: 1 = ticket/block was sold, 0= ticket/block was not sold).

model_Gegner <- Ausverkauft_2 ~ Preis + Block + Gegner + Spieltagsklasse + Leistung_Heim + 
  Leistung_Auswärts + Temperatur + Unsicherheit + Wochentag+ Saison

Logit_Gegner <- glm(model_Gegner, data = train, family = "binomial")

As a next step, I would like to see what would be the highest possible price the club could charge to still sell the ticket.
To find this out I have designed a for in loop:

Preis <- c(1:100)
Preis

#test object: 
SpielRevenue <- subset(test, Datum == "2018-12-02" & Kategorie == "4" & Block == "15E")

#### for in loop 
for(p in Preis ){
  SpielRevenue$Preis <- p
  print(predict(Logit_Gegner,SpielRevenue, type="response")) 
}

Instead of the probabilities of selling a ticket at the different prices, I get the following:

Since I have only limited the prices to 100, I don't understand where the 440 comes from. There are 100x the 440 so you could assign the different prices to each 440. But this would be a big counting job.

I would like to have following result:

I would also like to save the result in a dataframe if its possible.

The contribution goes in the direction of:
However, I haven't found any contribution that helps me with mine. If there is one, I am sorry for the duplication.

Thank you!

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猫瑾少女 2025-02-20 07:54:21

找到了我的问题的答案。如果其他人有同样的问题，我将在此处发布它，而不是删除我的问题：

Block15E <- matrix(nrow=100, ncol=1)

#### in for loop 
for(p in Preis ){
  SpielRevenue$Preis <- p
  Block15E[p,] <- (predict(Logit_Gegner,SpielRevenue, type="response"))
}

Block15E <- as.data.frame(Block15E)
colnames(Block15E) <- "Verkaufswahrscheinlichkeit"
Block15E$Preis <- seq.int(nrow(Block15E))
Block15E <- Block15E[,c(2,1)]

440是行号。
为了逃脱这个数字，我在循环前创建了一个矩阵，并将索引作为第二步的价格。

看起来真的不好，但是有效:)

Have found an answer to my question. If others have the same problem, I'll post it here instead of deleting my question:

Block15E <- matrix(nrow=100, ncol=1)

#### in for loop 
for(p in Preis ){
  SpielRevenue$Preis <- p
  Block15E[p,] <- (predict(Logit_Gegner,SpielRevenue, type="response"))
}

Block15E <- as.data.frame(Block15E)
colnames(Block15E) <- "Verkaufswahrscheinlichkeit"
Block15E$Preis <- seq.int(nrow(Block15E))
Block15E <- Block15E[,c(2,1)]

440 was the row number.
To escape the number, I created a matrix before the loop and took the index as the price in the second step.

Doesn't look really nice, but works :)

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