计算data.frame中的行和和产品

时间:2014-11-24 10:49:17

标签: r dataframe apply

我想在R中的data.frame中添加一列包含行总和和产品的列 请考虑以下数据框

x    y     z
1    2     3
2    3     4
5    1     2

我希望得到以下内容

x    y     z    sum    prod
1    2     3    6       6  
2    3     4    9       24 
5    1     2    8       10

我试过了

 sum = apply(ages,1,add)

但它给了我一个行向量。有人可以告诉我一个有效的命令来求和和产品并将它们附加到原始数据框中,如上所示?

4 个答案:

答案 0 :(得分:19)

尝试

 transform(df, sum=rowSums(df), prod=x*y*z)
 #  x y z sum prod
 #1 1 2 3   6    6
 #2 2 3 4   9   24
 #3 5 1 2   8   10

或者

 transform(df, sum=rowSums(df), prod=Reduce(`*`, df))
 #   x y z sum prod
 #1 1 2 3   6    6
 #2 2 3 4   9   24
 #3 5 1 2   8   10

另一种选择是使用rowProds

中的matrixStats
 library(matrixStats)
 transform(df, sum=rowSums(df), prod=rowProds(as.matrix(df)))

如果您使用apply

 df[,c('sum', 'prod')] <-  t(apply(df, 1, FUN=function(x) c(sum(x), prod(x))))
 df
 #  x y z sum prod
 #1 1 2 3   6    6
 #2 2 3 4   9   24
 #3 5 1 2   8   10

答案 1 :(得分:3)

另一种方法。

require(data.table)

# Create data
dt <- data.table(x = c(1,2,5), y = c(2,3,1), z = c(3,4,2))

# Create index
dt[, i := .I]

# Compute sum and prod
dt[, sum := sum(x, y, z), by = i]
dt[, prod := prod(x, y, z), by = i]
dt


# Compute sum and prod using .SD
dt[, c("sum", "prod") := NULL]
dt
dt[, sum := sum(.SD), by = i, .SDcols = c("x", "y", "z")]
dt[, prod := prod(.SD), by = i, .SDcols = c("x", "y", "z")]
dt


# Compute sum and prod using .SD and list
dt[, c("sum", "prod") := NULL]
dt
dt[, c("sum", "prod") := list(sum(.SD), prod(.SD)), by = i,
   .SDcols = c("x", "y", "z")]
dt


# Compute sum and prod using .SD and lapply
dt[, c("sum", "prod") := NULL]
dt
dt[, c("sum", "prod") := lapply(list(sum, prod), do.call, .SD), by = i,
   .SDcols = c("x", "y", "z")]
dt

答案 2 :(得分:2)

也可以进行以下操作,但需要输入列名:

ddf$sum = with(ddf, x+y+z)
ddf$prod = with(ddf, x*y*z)
ddf
  x y z sum prod
1 1 2 3   6    6
2 2 3 4   9   24
3 5 1 2   8   10

使用data.table,另一种形式可以是:

library(data.table)    
cbind(dt, dt[,list(sum=x+y+z, product=x*y*z),])
   x y z sum product
1: 1 2 3   6       6
2: 2 3 4   9      24
3: 5 1 2   8      10

@David Arenberg在评论中提出了一个更简单的版本:

dt[, ":="(sum = x+y+z, product = x*y*z)]

答案 3 :(得分:1)

仅是部分答案,但是如果所有值均大于或等于0,则可以使用rowSums / rowsum计算乘积:

df <- data.frame(x = c(1, 2, 5), y = c(2, 3, 1), z = c(3, 4, 2))

# custom row-product-function
my_rowprod <- function(x) exp(rowSums(log(x)))

df$prod <- my_rowprod(df)
df

通用版本为(包括底片):

my_rowprod_2 <- function(x) {
  sign <- ifelse((rowSums(x < 0) %% 2) == 1, -1, 1)
  prod <- exp(rowSums(log(abs(x)))) * sign
  prod
}
df$prod <- my_rowprod_2(df)
df