使用聚合函數來計算數據框架中的輸出

Question

我一直在嘗試自己並在網絡上搜索了一段時間，並且沒有成功。我有一個數據框，它是應用條件和選擇投影的子集，但無法檢索彙總輸出。

數據幀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") 

假設給定的數據幀的一部分，與返回：

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

它會返回一個數據幀，例如：

salary name 
9 4000 Bartosz 
10 7600 Damian 

現在我需要計算從給定的工資a sum,avg並選擇與來自上述數據幀，優選地使用一襯墊通過修改上面的代碼至少工資僱員（我猜測這是可能的），以便它返回：

爲sum：11600

爲avg ：5800

爲至少4000巴爾託什

---

我試過的東西：（1）

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

或（2）

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

和許多組合，其僅產生錯誤說摘要函數僅與所有變量是一個數據幀上定義的數字（2）或相同的輸出作爲第一代碼，而無需sum（1）

Answer 1

的問題可能是你的數據框對象實際上包含了一堆名單。所以，如果你把

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

那麼任何後續的工作需要與as.numeric調用來充塞你的列表轉化爲向量。這些會給你答案：

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 

另外，可以定義mydf爲向量的數據幀，而不是列表的數據幀：

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

您的mydf很奇怪，所以我做了我自己的。我將mydf拆分爲city，然後從每個子組上運行必要的操作（平均值，總和等）中獲得必要的數據。

#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

您的數據框在數據框中非常規地列爲列表，這可能會造成您的問題。這裏是一個dplyr溶液（現在編輯找到個elowest工資）

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 4

我覺得dplyr是你可能是什麼尋找：

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 

有一個在這個環節鏈接一個很好的教程[https://rpubs.com/justmarkham/dplyr-tutorial]

Answer 5

有使用data.table

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 

數據集的簡單和快速的解決方案：

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