将列列表除以第二列列表 [英] Divide list of columns by a second list of columns

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问题描述

我有一个脚本,该脚本会生成一个 data.table ,其中包含我要除以其他某些列的某些列,并将结果存储在新列中。这是一个例子。

I have script that generates a data.table with some columns I want to divide by some other columns and store the results in new columns. Here's an example.

library(data.table)
dt <- data.table(V1 = c( 5.553465,  4.989168,  2.563682,  6.987971, 19.220936),
                 V2 = c(4.248335, 19.768138,  3.840026, 17.411003, 17.939368),
                 V3 = c(9.683953, 15.344424, 11.729091,  7.534210,  5.404000),
                 V4 = c(5.949093,  4.553023,  9.765656, 11.211069,  4.085964),
                 V5 = c(11.814671,  5.460138,  2.492230,  1.48792,  8.164280))

list1 <- list(c("V1", "V2", "V3"))
list2 <- list(c("V2", "V4", "V5"))
listRatio <- list(c("rat1","rat2","rat3"))

我尝试了各种方法无法成功地将list1元素中的值除以list2元素中的值。下面有两个;

I have tried a variety of approaches to dividing the values in the list1 elements by the values in the list2 elements, unsuccessfully. Two are below; neither works.

dt[, (listRatio) := list1/list2]
dt[, c("rat1","rat2","rat3") := mapply(dt, function(x,y) x / y, x = c(V1, V2, V3),  y = c(V2, V4, V5))]


推荐答案

我们需要转换 list 转换为 vector ,方法是使用 [[,然后获取 list mget ,使用 Map 进行除法( / )每个 list 值的对应列,并将其分配给向量( listRatio [[1 ]] )。

We need to convert the list to vector by using [[ and then get the values of each vector in a list with mget, use Map to divide (/) the corresponding columns of each of the list values and assign it to the vector (listRatio[[1]]).

dt[, (listRatio[[1]]) := Map(`/`, mget(list1[[1]]), mget(list2[[1]]))]
dt
#          V1        V2        V3        V4        V5      rat1      rat2      rat3
#1:  5.553465  4.248335  9.683953  5.949093 11.814671 1.3072098 0.7141147 0.8196549
#2:  4.989168 19.768138 15.344424  4.553023  5.460138 0.2523843 4.3417611 2.8102630
#3:  2.563682  3.840026 11.729091  9.765656  2.492230 0.6676210 0.3932174 4.7062635
#4:  6.987971 17.411003  7.534210 11.211069  1.487920 0.4013537 1.5530190 5.0635854
#5: 19.220936 17.939368  5.404000  4.085964  8.164280 1.0714389 4.3904861 0.6619077

注:如评论中的@Frank所述,最好创建变量的向量名称而不是列表

NOTE: As @Frank mentioned in the comments, it is better to create a vector of variables names and not a list.

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