使用聚合函数计算数据框中的输出

Question

我一直在尝试自己，现在在网上搜索了一段时间，stackoverflow 没有成功。我有一个数据框，我从应用条件和选择投影中提取了该数据框，但未能检索到聚合输出。

数据框mydf:

mydf = list()
mydf = cbind(mydf, 
            c("New York", "New York", "San Francisco"),
            c(4000, 7600, 2500),
            c("Bartosz", "Damian", "Maciej"))
mydf = as.data.frame(mydf)
colnames(mydf) = c("city","salary","name")

让我们假设返回的数据帧的给定部分：

subset(mydf, city == "New York", select = c(salary, name))

返回一个数据框，例如：

   salary    name
9    4000 Bartosz
10   7600  Damian

现在我需要根据给定的薪水计算sum,avg 并从上述数据框中选择薪水最低的员工，最好通过修改上述代码使用单行（我猜这是可能的） , 使其返回：

sum: 11600

avg: 5800

至少：4000 巴托兹

---

我已经尝试过 (1)

subset(mydf, city == "New York", select = sum(salary))

或 (2)

x = subset(mydf, city == "New York", select = salary)
min(x)

还有更多的组合只会产生错误，说明汇总函数仅在数据帧上定义，所有变量都是数字 (2) 或与第一个代码相同的输出，但没有 sum (1)

Answer 1

问题可能是您的数据框对象实际上包含一堆列表。所以如果你采取

ny.df = subset(mydf, city == "New York", select = c(salary, name))

那么任何后续工作都需要使用as.numeric 调用来将您的列表转换为向量。这些将为您提供答案：

sum(as.numeric(ny.df$salary)) # sum
mean(as.numeric(ny.df$salary)) # avg
ny.df[which(as.numeric(ny.df$salary) == min(as.numeric(ny.df$salary))),] # row with min salary

或者，您可以将 mydf 定义为向量数据框，而不是列表数据框：

mydf = data.frame(c("New York", "New York", "San Francisco"),
              c(4000, 7600, 2500),
              c("Bartosz", "Damian", "Maciej"))
colnames(mydf) = c("city","salary","name")

ny.df = subset(mydf, city == "New York", select = c(salary, name))
sum(ny.df$salary)
mean(ny.df$salary)
ny.df[which(ny.df$salary == min(ny.df$salary)),]

Answer 2

你的mydf 很奇怪，所以我自己做了。我将mydf 拆分为city，然后通过对每个子组运行必要的操作（均值、求和等）获得必要的数据。

#DATA
mydf = structure(list(city = structure(c(1L, 1L, 2L), .Label = c("New York", 
"San Francisco"), class = "factor"), salary = c(4000, 7600, 2500
), name = structure(1:3, .Label = c("Bartosz", "Damian", "Maciej"
), class = "factor")), .Names = c("city", "salary", "name"), row.names = c(NA, 
-3L), class = "data.frame")

do.call(rbind, lapply(split(mydf, mydf$city), function(a)
    data.frame(employee = a$name[which.min(a$salary)], #employee with least salary
               mean = mean(a$salary), #mean salary
               sum = sum(a$salary)))) #sum of salary
#              employee mean   sum
#New York       Bartosz 5800 11600
#San Francisco   Maciej 2500  2500

Answer 3

使用data.table有一个简单快速的解决方案

library(data.table) 

setDT(mydf)[, .( salary_sum = sum(salary),
                 salary_avg = mean(salary),
                 name = name[which.min(salary)]), by= city]

>             city salary_sum salary_avg    name
> 1:      New York      11600       5800 Bartosz
> 2: San Francisco       2500       2500  Maciej

你的数据集：

mydf = data.frame(city=c("New York", "New York", "San Francisco"),
                  salary=c(4000, 7600, 2500),
                  name=c("Bartosz", "Damian", "Maciej"))

Answer 4

您的数据框在数据框中的列表结构异常，这可能是您的问题。这是一个 dplyr 解决方案（现在已编辑以找到最低工资）

library(dplyr)
mydf <- data.frame(
             city = c("New York", "New York", "San Francisco"),
             salary = c(4000, 7600, 2500),
             name = c("Bartosz", "Damian", "Maciej"))

mydf %>% 
  group_by(city) %>%
  mutate(avg = mean(salary),
         sum = sum(salary)) %>%
  top_n(-1, wt = salary) 

#            city salary    name   avg   sum
#          <fctr>  <dbl>  <fctr> <dbl> <dbl>
# 1      New York   4000 Bartosz  5800 11600
# 2 San Francisco   2500  Maciej  2500  2500

Answer 5

我认为 dplyr 是您可能正在寻找的：

   library(dplyr)
   mydf %>% 
   group_by(city) %>% 
   filter (city =="New York") %>%
   summarise(mean(salary), sum(salary))

  # A tibble: 1 x 3
  #  city mean(salary) sum(salary)
  #  <fctr>        <dbl>       <dbl>
  #1 New York         5800       11600

这个链接有很好的教程链接[https://rpubs.com/justmarkham/dplyr-tutorial]