使用 lapply 基于旧列创建新列

Question

我的数据如下：

DF <- structure(list(No_Adjusted_Gross_Income = c(183454, 241199, 249506
), NoR_from_1_to_5000 = c(1035373, 4272260, 1124098), NoR_from_5000_to_10000 = c(319540, 
4826042, 1959866)), row.names = c(NA, -3L), class = c("data.table", 
"data.frame"))
val <- c(2500.5, 7500)
vn <- c("AGI_from_1_to_5000", "AGI_from_5000_to_10000")

   No_Adjusted_Gross_Income NoR_from_1_to_5000 NoR_from_5000_to_10000
1:                   183454            1035373                 319540
2:                   241199            4272260                4826042
3:                   249506            1124098                1959866

我想创建新列，基于第 2 列和第 3 列，乘以 val 中的值，使用 vn 中的名称。我尝试这样做：

DF[,2:3] <- lapply(DF[,2:3], vn := val*DF[,2:3])

但这不起作用..

期望的输出：

DF <- setDT(DF)[, vn[1]:=val[1]*DF[,2]]
DF <- setDT(DF)[, vn[2]:=val[2]*DF[,3]]

DFout <- structure(list(No_Adjusted_Gross_Income = c(183454, 241199, 249506
), NoR_from_1_to_5000 = c(1035373, 4272260, 1124098), NoR_from_5000_to_10000 = c(319540, 
4826042, 1959866), AGI_from_1_to_5000 = c(2588950186.5, 10682786130, 
2810807049), AGI_from_5000_to_10000 = c(2396550000, 36195315000, 
14698995000)), row.names = c(NA, -3L), class = c("data.table", 
"data.frame"))

   No_Adjusted_Gross_Income NoR_from_1_to_5000 NoR_from_5000_to_10000 AGI_from_1_to_5000 AGI_from_5000_to_10000
1:                   183454            1035373                 319540         2588950187             2396550000
2:                   241199            4272260                4826042        10682786130            36195315000
3:                   249506            1124098                1959866         2810807049            14698995000

Answer 1

这应该可行.. 不需要lapply()

library( data.table )
setDT( DF )
DF[, (var) := as.data.table ( t( t( DF[, 2:3] ) * val ) ) ][]


#    No_Adjusted_Gross_Income NoR_from_1_to_5000 NoR_from_5000_to_10000 AGI_from_1_to_5000 AGI_from_5000_to_10000
# 1:                   183454            1035373                 319540         2588950187             2396550000
# 2:                   241199            4272260                4826042        10682786130            36195315000
# 3:                   249506            1124098                1959866         2810807049            14698995000

Answer 2

您可以使用apply 来获取您的值，然后如果您想与您的原始DF 结合使用cbind

t(apply(DF[,2:3],1, function(x) x*val ))

 NoR_from_1_to_5000 NoR_from_5000_to_10000
[1,]         2588950187             2396550000
[2,]        10682786130            36195315000
[3,]         2810807049            14698995000

Answer 3

OP 已在 comment 中要求分组变量。

虽然接受的答案显然符合 OP 最初的要求，但我想提出一种完全不同的方法，以整齐（长）格式存储和处理数据。恕我直言，以长格式处理数据更加直接和灵活（包括聚合和分组）。

为此，数据集从宽的 Excel 样式格式重塑为长的 SQL 样式格式

library(data.table)
col <- "NoR"
long <- melt(DF, measure.vars = patterns(col), value.name = col, variable.name = "range")
long[, range := stringr::str_remove(range, paste0(col, "_"))]
long

   No_Adjusted_Gross_Income              range     NoR
1:                   183454     from_1_to_5000 1035373
2:                   241199     from_1_to_5000 4272260
3:                   249506     from_1_to_5000 1124098
4:                   183454 from_5000_to_10000  319540
5:                   241199 from_5000_to_10000 4826042
6:                   249506 from_5000_to_10000 1959866

在整洁（长）格式中，每个观察值一行，每个变量一列（请参阅Chapter 12.2 of Hadley Wickham's book R for Data Science。

乘数向量val也需要从宽格式改成长格式：

valDF <- long[, .(range = unique(range), val)]
valDF

                range    val
1:     from_1_to_5000 2500.5
2: from_5000_to_10000 7500.0

现在，valDF 的格式也很整洁，因为每个 range 都有一行。

最后，我们可以通过 update join 将新列 AGI 添加到 DF：

long[valDF, on = "range", AGI := val * NoR][]

   No_Adjusted_Gross_Income              range     NoR         AGI
1:                   183454     from_1_to_5000 1035373  2588950187
2:                   241199     from_1_to_5000 4272260 10682786130
3:                   249506     from_1_to_5000 1124098  2810807049
4:                   183454 from_5000_to_10000  319540  2396550000
5:                   241199 from_5000_to_10000 4826042 36195315000
6:                   249506 from_5000_to_10000 1959866 14698995000

如果需要演示，可以将数据集从长格式重新调整为宽格式：

dcast(long, No_Adjusted_Gross_Income ~ range, value.var = c("NoR", "AGI"))

   No_Adjusted_Gross_Income NoR_from_1_to_5000 NoR_from_5000_to_10000 AGI_from_1_to_5000 AGI_from_5000_to_10000
1:                   183454            1035373                 319540         2588950187             2396550000
2:                   241199            4272260                4826042        10682786130            36195315000
3:                   249506            1124098                1959866         2810807049            14698995000

再现了 OP 的预期结果。请注意，变量名称vn 是自动创建的。

---

可以在整形的同时进行聚合和分组

dcast(long, No_Adjusted_Gross_Income ~ range, sum, value.var = c("NoR", "AGI"))

   No_Adjusted_Gross_Income NoR_from_1_to_5000 NoR_from_5000_to_10000 AGI_from_1_to_5000 AGI_from_5000_to_10000
1:                   183454            1035373                 319540         2588950187             2396550000
2:                   241199            4272260                4826042        10682786130            36195315000
3:                   249506            1124098                1959866         2810807049            14698995000

或

dcast(long, No_Adjusted_Gross_Income ~ ., sum, value.var = c("NoR", "AGI"))

   No_Adjusted_Gross_Income     NoR         AGI
1:                   183454 1354913  4985500187
2:                   241199 9098302 46878101130
3:                   249506 3083964 17509802049

或者，可以以长格式执行聚合和分组：

long[, lapply(.SD, sum), .SDcols = c("NoR", "AGI"), by = No_Adjusted_Gross_Income]

   No_Adjusted_Gross_Income     NoR         AGI
1:                   183454 1354913  4985500187
2:                   241199 9098302 46878101130
3:                   249506 3083964 17509802049