Question

在一个（非常害怕的）dplyr管道之后，我最终得到了这样一个数据集：

year   A    B     C  [....] Z  count.A    count.B     count.C [....] count.Z
1999  10    20    10  ...   6      3          5           67    ...      6
2000   3    5      5  ...   7      5          2            5    ...      5

要重现的一些示例数据：

df <- data.frame(year = c(1999, 2000), 
                 A = c(10, 20), 
                 B = c(3, 6), 
                 C = c(1, 2), 
                 count.A = c(1, 2), 
                 count.B = c(8, 9), 
                 count.C = c(5, 7))

我真正需要的是将每一列与其“计数”对应组合，即

weight.A = A / count.A, 
weight.B = B / count.B

我要以编程方式执行此操作，因为我有数百列。有没有办法在dplyr管道中做到这一点？

Answer 1

不要在列名中存储变量。如果您重塑数据以使其整洁，计算非常简单：

library(tidyverse)

df %>% gather(var, val, -year) %>%    # reshape to long
    separate(var, c('var', 'letter'), fill = 'left') %>%    # extract var from former col names
    mutate(var = coalesce(var, 'value')) %>%    # add name for unnamed var
    spread(var, val) %>%    # reshape back to wide
    mutate(weight = value / count)    # now this is very simple

#>   year letter count value     weight
#> 1 1999      A     1    10 10.0000000
#> 2 1999      B     8     3  0.3750000
#> 3 1999      C     5     1  0.2000000
#> 4 2000      A     2    20 10.0000000
#> 5 2000      B     9     6  0.6666667
#> 6 2000      C     7     2  0.2857143

Answer 2

如果您的列一直被命名（并且很容易检索），您可以使用lapply轻松完成此操作：

cols <- c("A","B","C")
df[,paste0("weighted.",cols)] <- lapply(cols, function(x) df[,x] / df[, paste0("count.",x)])

#  year  A B C count.A count.B count.C weighted.A weighted.B weighted.C
#1 1999 10 3 1       1       8       5         10  0.3750000  0.2000000
#2 2000 20 6 2       2       9       7         10  0.6666667  0.2857143

Answer 3

假设您可以以编程方式创建所有列名称的向量，以下是我为上述示例所做的操作

for (c.name in c("A", "B", "C")) {
    c.weight <- sprintf("weight.%s", c.name)
    c.count <- sprintf("count.%s", c.name)
    df[,c.weight] <- df[,c.name] / df[,c.count]
}

Answer 4

假设列按顺序排列，我们可以使用data.table。在.SDcols中指定感兴趣的列，并将Data.table的子集的子集除以另一半，并将其分配（:=）到新列

library(data.table)
setDT(df)[, paste0("weighted.",names(df)[1:3]) := .SD[,1:3]/.SD[,4:6], .SDcols = A:count.C]
df
#   year  A B C count.A count.B count.C weighted.year weighted.A weighted.B
#1: 1999 10 3 1       1       8       5            10  0.3750000  0.2000000
#2: 2000 20 6 2       2       9       7            10  0.6666667  0.2857143

在dplyr中将一组列除以（并命名）另一组

4 个答案: