Question

FAMILY<- c('FAMILYA', 'FAMILYA', 'FAMILYA', 'FAMILYA', 'FAMILYA', 'FAMILYB', 'FAMILYB', 'FAMILYB', 'FAMILYB', 'FAMILYB', 'FAMILYC', 'FAMILYC', 'FAMILYC', 'FAMILYC', 'FAMILYC')

CHILDREN<-c('JAKE', 'PETE', 'JASON', 'KEVIN', 'ALFRED','DALE', 'STEVE', 'MELISSA', 'DAN', 'THOMAS', 'CAIT', 'BRANDON', 'DEAN', 'ADAM', 'KELSEY')

CHANGE<-c(1000, -1000, 2000, 3000, 5000, 100, 300, 1234, -1022, -1111, -1112, 1000, 1002, 2131, 1231)

df1<-data.frame(FAMILY, CHILDREN, CHANGE)

df1

    FAMILY CHILDREN CHANGE
1  FAMILYA     JAKE   1000
2  FAMILYA     PETE  -1000
3  FAMILYA    JASON   2000
4  FAMILYA    KEVIN   3000
5  FAMILYA   ALFRED   5000
6  FAMILYB     DALE    100
7  FAMILYB    STEVE    300
8  FAMILYB  MELISSA   1234
9  FAMILYB      DAN  -1022
10 FAMILYB   THOMAS  -1111
11 FAMILYC     CAIT  -1112
12 FAMILYC  BRANDON   1000
13 FAMILYC     DEAN   1002
14 FAMILYC     ADAM   2131
15 FAMILYC   KELSEY   1231

我希望将此数据帧转换为4个新的额外列：前两个显示1）最大值的Child，2）第二大值Child，最后两列显示3）最小值Child，4）2nd最小值子。

我也希望它旁边的变化是相应的孩子。

最终格式应如下所示：

 FAMILY  TOTAL CHANGE       INCREASE #1      INCREASE #2       DECREASE #1      DECREASE #2
FAMILYA         10000      ALFRED: 5000      KEVIN: 3000       PETE: -1000       JAKE: 1000 
FAMILYB          -499     MELISSA: 1234       STEVE: 300     THOMAS: -1111       DAN: -1022 
FAMILYC          4252        ADAM: 2131     KELSEY: 1231       CAIT: -1112    BRANDON: 1000

如果您认为将每个子项的值放在其旁边的单独列中会更容易，但这也是我需要帮助执行的概念。

任何帮助都会很棒，谢谢！

Answer 1

library(dplyr)
library(tidyr)

# below function helps to get the second max  or second min 
myfun <-  function(x, y) {
  u <- unique(x)
  u <- sort(u, decreasing = TRUE)
  if(y<0)
    u[length(x)-1]
  else
    u[y]
}

df2 <- df1 %>% group_by(FAMILY) %>% 
      summarise(a1=CHILDREN[which(CHANGE == max(CHANGE))]   , a2 = max(CHANGE), 
                b2 = myfun(CHANGE, 2)          , b1=CHILDREN[which(CHANGE == b2)] ,  
                c1=CHILDREN[which(CHANGE == min(CHANGE))]   , c2 = min(CHANGE), 
                d2 = myfun(CHANGE,-2)          , d1=CHILDREN[which(CHANGE == d2)]) 
#df2
#   FAMILY      a1    a2      b1    b2     c1    c2     d1    d2
#   <fctr>  <fctr> <dbl>  <fctr> <dbl> <fctr> <dbl> <fctr> <dbl>
#1 FAMILYA  ALFRED  5000  3000  KEVIN   PETE -1000  1000    JAKE
#2 FAMILYB MELISSA  1234   300  STEVE THOMAS -1111 -1022     DAN
#3 FAMILYC    ADAM  2131  1231 KELSEY   CAIT -1112  1000 BRANDON

# little clumpsy here... would like if someone could suggest a better way of uniting efficiently
df3 <- unite(df2, "A1", 2,3,sep = ":")
df4 <- unite(df3, "B1", 4,3,sep = ":")
df5 <- unite(df4, "c1", 4,5,sep = ":")
df6 <- unite(df5, "c1", 6,5,sep = ":")

#df6
#   FAMILY           A1           B1           c1         c1
#   <fctr>        <chr>        <chr>        <chr>      <chr>
#1 FAMILYA  ALFRED:5000  KEVIN:3000   PETE:-1000    JAKE:1000
#2 FAMILYB MELISSA:1234   STEVE:300 THOMAS:-1111    DAN:-1022
#3 FAMILYC    ADAM:2131 KELSEY:1231   CAIT:-1112 BRANDON:1000

注意：忘记添加TOTAL_CHANGE列在TOTAL CHANGE = sum(CHANGE)中添加summarise()，在unite（）列索引中添加+1

Answer 2

以下是使用自定义函数和do（来自dplyr）将其应用于每个家庭组的方法。自定义函数也使用dplyr。

首先，自定义函数生成（并排序）有序更改。然后，它返回总变化（总和）以及顺序中的第一个和最后两个变化。它必须以data.frame返回才能与do一起正常使用。

myFamFunction <- function(CHILDREN, CHANGE){
  toOut <-
    paste(CHILDREN, CHANGE, sep = ": ")[order(CHANGE, decreasing = TRUE)]

  c(sum(CHANGE)
    , head(toOut, 2)
    , tail(toOut, 2)) %>%
    rbind() %>%
    data.frame(stringsAsFactors = FALSE) %>%
    setNames(c("Total Change"
               , "Biggest Change"
               , "Second Biggest Change"
               , "Second Smallest Change"
               , "Smallest Change"))
}

请注意，如果少于2个孩子，这可能会引发错误（但是，如果小于4，则结果已经被怀疑）。如果你有更复杂的实际数据，告诉我们你想要发生什么，可以防止这些边缘情况。

然后，只需group_by，将您想要的列传递给函数，瞧：

df1 %>%
  group_by(FAMILY) %>%
  do(myFamFunction(.$CHILDREN, .$CHANGE))

返回：

   FAMILY `Total Change` `Biggest Change` `Second Biggest Change` `Second Smallest Change` `Smallest Change`
   <fctr>          <chr>            <chr>                   <chr>                    <chr>             <chr>
1 FAMILYA          10000     ALFRED: 5000             KEVIN: 3000               JAKE: 1000       PETE: -1000
2 FAMILYB           -499    MELISSA: 1234              STEVE: 300               DAN: -1022     THOMAS: -1111
3 FAMILYC           4252       ADAM: 2131            KELSEY: 1231            BRANDON: 1000       CAIT: -1112

重塑Dataframe以获得前2个值

2 个答案: