Question

我有一个大的数据集，大约1 mil行和8个cols（变量）。其中一个变量ORDER的类别从1到90.我想创建一个新的data.frame，变量ORDER（4）1,2,3 +和ALL的类别数量减少，其中ALL是总和所有类别（1-90）和3+的频率是类别的频率之和＆gt; = 3（所以3到90）。

YEAR  PROVINCE  ZONA91OK AGE5 ORDER NATIONALITY_MOTHER NATIONALITY_FATHER FREQUENCY
 1979        1      101   15     1      No computable      No computable        10
 1989        3      102   20     1      No computable      No computable        50

我对R的数据管理非常新，所以非常感谢任何有关此问题的帮助！

以下是data.frame

的示例

mydata<-structure(list(YEAR = c(1981, 1981, 1981, 1981, 1981, 1981, 1981, 
1981, 1981, 1981, 1981, 1981, 1981, 1981, 1981, 1981, 1981, 1981, 
1981, 1981, 1981), PROVINCE = c(1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1), ZONA91OK = c(101, 101, 101, 
101, 101, 101, 101, 101, 101, 101, 101, 101, 101, 101, 101, 101, 
101, 101, 101, 101, 101), AGE5 = c(15, 20, 20, 25, 25, 25, 25, 
30, 30, 30, 30, 30, 35, 35, 35, 35, 35, 35, 40, 40, 40), ORDER = c(1, 
1, 2, 1, 2, 3, 4, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 12, 1, 3, 5), 
NATIONALITY_MOTHER = structure(c(9L, 9L, 9L, 9L, 9L, 9L, 
9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L
), .Label = c("España", "UE-15 y PD", "Resto Europa", "Magreb", 
"África Sub-sahariana", "Latinoamérica", "Asia", "Resto del Mundo", 
"No computable"), class = "factor"), NATIONALITY_FATHER = structure(c(9L, 
9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 
9L, 9L, 9L, 9L, 9L), .Label = c("España", "UE-15 y PD", "Resto Europa", 
"Magreb", "África Sub-sahariana", "Latinoamérica", "Asia", 
"Resto del Mundo", "No computable"), class = "factor"), FREQUENCY = c(10, 
40, 20, 50, 30, 10, 1, 10, 15, 10, 1, 1, 5, 5, 5, 1, 1, 1, 
1, 1, 1)), .Names = c("YEAR", "PROVINCE", "ZONA91OK", "AGE5", 
"ORDER", "NATIONALITY_MOTHER", "NATIONALITY_FATHER", "FREQUENCY"
 ), row.names = 60175:60195, class = "data.frame")

Answer 1

如果您的数据有1M行，您可能会想要使用data.table

library(data.table)
myDT <- data.table(mydata, key="ORDER")

specialCats <- c(1, 2, 3)

rbind(
    myDT[, list(SUM_FOR="ALL", FREQ_SUM=sum(FREQUENCY))]
  , myDT[!.(specialCats), list(SUM_FOR="3+", FREQ_SUM=sum(FREQUENCY))]
)

## RESULTS: 
       SUM_FOR FREQ_SUM
1:     ALL      219
2:      3+        7

更新：评论

要将ORDER列更改为您的要求，请使用：

myDT[, order := ifelse(ORDER %in% specialCats, as.character(ORDER), "3+")]

注1：为了使3+成为值，您需要转换为字符串注意2：为"ALL"添加一行没有多大意义，因为您要为AGE，PROVINCE等添加一行？

R data.frame操作

1 个答案:

更新：评论