Question

我有一个大型数据集，我想用其他人使用lapply函数对1列进行求和。但我有问题，其他专栏消失了......我想保留它们。

我有一个例子：）

例如：

我有数据集：

   X  Y   Z     date     columnSum
1: A  a1  z1   2018.01         4
2: A  a1  z1   2018.01         4
2: B  a2  z3   2018.02        10
2: B  a2  z5   2018.02        30
2: B  a2  z5   2018.02        10
3: C  a2  z3   2018.02        10
4: D  a3  z4   2018.03         0
4: D  a3  z6   2018.03         0

我要总结＆＃34; columnSum ＆＃34; by＆＃34; X ＆＃34;，＆＃34; Y ＆＃34;和＆＃34; 日期＆＃34;。我想保留专栏＆＃34; Z ＆＃34;

我试过了：

DT[, lapply(.SD,sum,na.rm=TRUE), .SDcols="columnSum", by=list(X,Y,date)]

今天我有结果：

   X  Y   date    columnSum
1: A  a1  2018.01         8
2: B  a2  2018.02        50
3: C  a2  2018.02        10
4: D  a3  2018.03         0

我想要结果：

   X  Y   Z     date     columnSum
1: A  a1  z1   2018.01         8
2: B  a2  z3   2018.02        50
3: B  a2  z5   2018.02        50
4: C  a2  z3   2018.02        10
5: D  a3  z4   2018.03         0
6: D  a3  z6   2018.03         0

Answer 1

df <- read.table(text = "X  Y   Z     date     columnSum
 A  a1  z1   2018.01         4
                 A  a1  z1   2018.01         4
                 B  a2  z3   2018.02        10
                 B  a2  z5   2018.02        30
                 B  a2  z5   2018.02        10
                 C  a2  z3   2018.02        10
                 D  a3  z4   2018.03         0
                  D  a3  z6   2018.03         0", 
                 header = TRUE, stringsAsFactors = FALSE)
library(data.table)

setDT(df)
df[, columnSum := sum(columnSum), by = c("X", "Y", "date")] # summing columnSum by X, Y, date and retaining column Z
df <- unique(df) # filtering duplicate records

   #    X  Y  Z    date columnSum
   # 1: A a1 z1 2018.01         8
   # 2: B a2 z3 2018.02        50
   # 3: B a2 z5 2018.02        50
   # 4: C a2 z3 2018.02        10
   # 5: D a3 z4 2018.03         0
   # 6: D a3 z6 2018.03         0

Answer 2

这有效：

# recreated your example
DT <- data.table(X = c("A", "A", "B", "B", "B", "C", "D", "D"),
                 Y = c("a1", "a1", "a2", "a2", "a2", "a2", "a3", "a3"),
                 Z = c("z1", "z1", "z3", "z5", "z5", "z3", "z4", "z6"),
                 date = c("2018.01", "2018.01", "2018.02", "2018.02", 
                          "2018.02", "2018.02", "2018.03", "2018.03"),
                 columnSum = c(4, 4, 10, 30, 10, 10, 0, 0))

sums <- DT[, sum(columnSum), .(X, Y, date)]
keep <- unique(DT[, .(X, Y, Z, date)])
merge.data.frame(keep, sums)

Answer 3

library(dplyr)
practiceData <- tibble(X=c("A","A","B","B","B","C","D","D"),
                   Y=c('a1', 'a1', 'a2','a2', 'a2', 'a2','a3','a3'),  
                   Z=c('z1','z1', 'z3', 'z5','z5','z3','z4','z6'),
                   date=c('2018.01',
                          '2018.01',
                          '2018.02',
                          '2018.02',
                          '2018.02',
                          '2018.02',
                          '2018.03',
                          '2018.03'),
                   U=c(4,4,10,30,10,10,0,0))

NEW <- practiceData %>% group_by(X,Y,date) %>% summarise(colsumnew=sum(U)) 
lol <- unique(practiceData[,c(1,2,3,4)]) %>% data.frame()
lol2 <- left_join(NEW,lol) %>% unique()

使用lapply data.table汇总列并保留其他列

3 个答案: