Question

我似乎找不到解决我问题的答案。

这是示例数据

Credit Card Type  Bank   Year   Total Balance
MASTER CARD       BOFA   2017   $100
MASTER CARD       BOFA   2017   $100
MASTER CARD       BOFA   2017   $700
VISA              Wells  2018   $60 
VISA              Wells  2018   $50
VISA              Wells  2018   $60

等

我正在尝试找出如何通过所有变量的总余额获得模式所以最终会像这样

所需的输出：

Credit Card Type  Bank   Year   Mode
MASTER CARD       BOFA   2017   $100
VISA              Wells  2018   $60

Answer 1

按照弗兰克的建议，使用stackoverflow.com/q/2547402中的Mode，使用dplyr很容易。

library(dplyr)

df %>% 
    group_by(CreditCardType, Bank, Year) %>%
    summarise(mode = Mode(TotalBalance))

df在哪里：

df <- read.table(text = 'CreditCardType  Bank   Year   TotalBalance
MASTERCARD       BOFA   2017   $100
MASTERCARD       BOFA   2017   $100
MASTERCARD       BOFA   2017   $700
VISA              Wells  2018   $60 
VISA              Wells  2018   $50
VISA              Wells  2018   $60', header = T, stringsAsFactors = F)

Answer 2

这个问题obtaining 3 most common elements of groups, concatenating ties, and ignoring less common values

library(plyr)
getmode<- function(origtable,groupby,columnname) {
  data <- ddply (origtable, groupby, .fun = function(xx){
    c(m1 = paste(names(sort(table(xx[,columnname]),decreasing=TRUE)[1]))
    ) } ) 
  return(data)
}

getmode(df,c("CreditCardType","Bank","Year"),"TotalBalance")

df<-read.table(text="CreditCardType  Bank   Year   TotalBalance
MASTERCARD       BOFA   2017   $100
MASTERCARD       BOFA   2017   $100
MASTERCARD       BOFA   2017   $700
VISA              Wells  2018   $60 
VISA              Wells  2018   $50
VISA              Wells  2018   $60", header=T, stringsAsFactors=F)

Answer 3

另一种dplyr解决方案：

df %>%
  add_count(Credit_Card_Type, Bank, Year, Total_Balance) %>%
  filter(n == max(n)) %>%
  distinct() %>%
  select(-n)

考虑关系并选择第一个模式值：

df %>%
  add_count(Credit_Card_Type, Bank, Year, Total_Balance) %>%
  filter(n == max(n)) %>%
  distinct() %>%
  select(-n) %>%
  group_by(Credit_Card_Type, Bank, Year) %>%
  summarise(Total_Balance = first(Total_Balance))

数据：

df <- read.table(text = "Credit_Card_Type Bank Year Total_Balance
           MASTER_CARD BOFA 2017 100
           MASTER_CARD BOFA 2017 100
           MASTER_CARD BOFA 2017 700
           VISA Wells 2018 60
           VISA Wells 2018 50
           VISA Wells 2018 60", header = TRUE)

Answer 4

我找到了使用data.table和最适度的软件包的解决方案。

library(data.table)
library(modeest)
dt <- data.table("Type"=c(rep("MASTERCARD",3),rep("VISA",3)),"Bank"=c(rep("BOFA",3),rep("Wells",3)),"Year"=c(rep(2017,3),rep(2018,3)),"TotalBalance"=c(100,100,700,60,50,60))
dt[,mfv(TotalBalance)[1],by=c("Type","Bank","Year")]

          Type  Bank Year  V1
 1: MASTERCARD  BOFA 2017 100
 2:       VISA Wells 2018  60

多种模式的变量

4 个答案: