Question

我想通过对具有匹配变量的列（而不是附加列）求和来将一组数据帧组合为一个数据帧。

例如，给定

df1 <- data.frame(A = c(0,0,1,1,1,2,2), B = c(1,2,1,2,3,1,5), x = c(2,3,1,5,3,7,0))
df2 <- data.frame(A = c(0,1,1,2,2,2), B = c(1,1,3,2,4,5), x = c(4,8,4,1,0,3))
df3 <- data.frame(A = c(0,1,2), B = c(5,4,2), x = c(5,3,1))

我想用"A"和"B"进行匹配，并将"x"的值相加。对于此示例，我可以得到如下所示的期望结果：

library(plyr)
library(dplyr)
# rename columns so that join_all preserves them all:
colnames(df1)[3] <- "x1"
colnames(df2)[3] <- "x2"
colnames(df3)[3] <- "x3"
# join the data frames by matching "A" and "B" values:
res <- join_all(list(df1, df2, df3), by = c("A", "B"), type = "full")
# get the sums and drop superfluous columns:
arrange(res, A, B) %>% 
  rowwise() %>% 
  mutate(x = sum(x1, x2, x3, na.rm = TRUE)) %>% 
  select(A, B, x)

结果：

       A     B     x
   <dbl> <dbl> <dbl>
 1     0     1     6
 2     0     2     3
 3     0     5     5
 4     1     1     9
 5     1     2     5
 6     1     3     7
 7     1     4     3
 8     2     1     7
 9     2     2     2
10     2     4     0
11     2     5     3

更通用的解决方案是

library(dplyr)
# function to get the desired result for two data frames:
my_merge <- function(df1, df2)
{
  m1 <- merge(df1, df2, by = c("A", "B"), all = TRUE)
  m1 <- rowwise(res) %>% 
    mutate(x = sum(x.x, x.y, na.rm = TRUE)) %>%
    select(A, B, x)
  return(m1)
}
l1 <- list(df2, df3) # omit the first data frame
res <- df1 # initial value of the result
for(df in l1) res <- my_merge(res, df) # call the function repeatedly

是否存在用于组合大量数据帧的更有效的选择？理想情况下，它应该是递归的（即，最好在计算总和之前不要将所有数据帧都合并为一个大数据帧）。

Answer 1

一个更简单的选择是绑定数据集的行，然后按感兴趣的列分组，并通过获取'x'的sum来获得汇总输出

library(tidyverse)
bind_rows(df1, df2, df3) %>% 
        group_by(A, B) %>%
        summarise(x = sum(x))
# A tibble: 11 x 3
# Groups:   A [?]
#       A     B     x
#   <dbl> <dbl> <dbl>
# 1     0     1     6
# 2     0     2     3
# 3     0     5     5
# 4     1     1     9
# 5     1     2     5
# 6     1     3     7
# 7     1     4     3
# 8     2     1     7
# 9     2     2     2
#10     2     4     0
#11     2     5     3

如果全局环境中有许多对象，其模式为"df"，后跟一些数字

mget(ls(pattern= "^df\\d+")) %>%
        bind_rows %>%
        group_by(A, B) %>% 
        summarise(x = sum(x))

正如OP提到的memory约束，如果我们先进行join，然后将rowSums或+与reduce一起使用，那将会更多高效

mget(ls(pattern= "^df\\d+")) %>% 
      reduce(full_join, by = c("A", "B")) %>%
      transmute(A, B, x = rowSums(.[3:5], na.rm = TRUE)) %>%
      arrange(A, B)
#   A B x
#1  0 1 6
#2  0 2 3
#3  0 5 5
#4  1 1 9
#5  1 2 5
#6  1 3 7
#7  1 4 3
#8  2 1 7
#9  2 2 2
#10 2 4 0
#11 2 5 3

这也可以通过data.table

完成

library(data.table)
rbindlist(mget(ls(pattern= "^df\\d+")))[, .(x = sum(x)), by = .(A, B)]

Answer 2

理想情况下，它应该是递归的（即，最好不要在计算总和之前将所有数据帧合并为一个大数据帧）。

如果您的内存有限并且愿意牺牲速度（与@akrun的data.table方法相比），请一次循环使用一个表：

library(data.table)
tabs = c("df1", "df2", "df3")

# enumerate all combos for the results table
# initializing sum to 0
res = CJ(A = 0:2, B = 1:5, x = 0)
# loop over tabs, adding on
for (i in seq_along(tabs)){
  tab = get(tabs[[i]])
  res[tab, on=.(A, B), x := x + i.x][]
  rm(tab)
}

如果需要从磁盘读取表，请将tabs更改为文件名，将get更改为fread或其他功能。

我怀疑您是否可以在内存中容纳所有表，但不能同时将它们的rbind版复制到一起。

同样（感谢@akrun的评论），成对使用他的方法：

res = data.table(get(tabs[[1]]))[0L]

for (i in seq_along(tabs)){
  tab = get(tabs[[i]])
  res = rbind(res, tab)[, .(x = sum(x)), by=.(A,B)]
  rm(tab)
}

递归求和数据帧以匹配行

2 个答案: