Question

给定一个包含列的数据框：

“length1”整数作为字符
“length2”每个元素都是一串数字

我想得到length2列相对于length1列的百分比。所以像df $ length2 / df $ lenght1 * 100。请参阅以下最小示例：

> df=data.frame(length1=c("10","12","14"))
> df$length2=list("2,3,4","4,5,3","3,2,6")
> df

length1 length2
1      10   2,3,4
2      12   4,5,3
3      14   3,2,6

> dfresult=df
> dfresult$resultInPercent=list("20,30,40","33,41,25","21,14,42")
> dfresult

  length1 length2 resultInPercent
1      10   2,3,4        20,30,40
2      12   4,5,3        33,41,25
3      14   3,2,6        21,14,42

我无法让它工作，我的方法是：

dfresult=apply(df, 1, function(x) 
{

  lapply(lapply(lapply(x$length2,strsplit,split=","),as.numeric),function(y)
     {
        round(as.numeric(y)/as.numeric(x$length1)*100)
     }

  )
 } 
)

lapply中的错误（lapply（x $ length2，strsplit，split =“，”），as.numeric） :( list）对象无法强制输入'double'

我放弃了，感觉我的所作所为是复杂的。

Answer 1

另一个想法：

library(dplyr)
library(tidyr)

df %>%
  separate_rows(length2) %>% 
  mutate_all(funs(as.numeric(as.character(.)))) %>%
  group_by(length1) %>%
  summarise(l2 = list(length2), 
            l3 = list(round(100 * length2 / length1)))

给出了：

## A tibble: 3 x 3
#  length1        l2        l3
#    <dbl>    <list>    <list>
#1      10 <dbl [3]> <dbl [3]>
#2      12 <dbl [3]> <dbl [3]>
#3      14 <dbl [3]> <dbl [3]>

这会将结果存储在list中，以便于进一步操作：

#Observations: 3
#Variables: 3
#$ length1 <dbl> 10, 12, 14
#$ l2      <list> [<2, 3, 4>, <4, 5, 3>, <3, 2, 6>]
#$ l3      <list> [<20, 30, 40>, <33, 42, 25>, <21, 14, 43>]

Answer 2

这是使用data.table

的某种矢量化解决方案

library(data.table)
temp <- round(setDT(df)[, tstrsplit(length2, ",", fixed = TRUE, type.convert = TRUE)] /
              as.numeric(levels(df$length1))[df$length1] * 100)
df[, resultInPercent := do.call(paste, c(temp, sep = ","))]
df
#    length1 length2 resultInPercent
# 1:      10   2,3,4        20,30,40
# 2:      12   4,5,3        33,42,25
# 3:      14   3,2,6        21,14,43

一些基准

library(data.table)
library(microbenchmark)
library(dplyr)
library(tidyr)

set.seed(123)
bigdf <- data.frame(length1 = sample(1e4),
                    length2 = I(replicate(1e4, "2,3,4", simplify = FALSE)))
bigdf2 <- copy(bigdf)

Steve <- function(df){ # changed `list` to `toStirng` so all outputs match
  df %>%
    separate_rows(length2) %>% 
    mutate_all(funs(as.numeric(as.character(.)))) %>%
    group_by(length1) %>%
    summarise(res = toString(round(100 * length2 / length1)))
}

David <- function(df) {
  temp <- round(setDT(df)[, tstrsplit(length2, ",", fixed = TRUE, type.convert = TRUE)] /
                as.numeric(levels(df$length1))[df$length1] * 100)
  df[, resultInPercent := do.call(paste, c(temp, sep = ","))]
  df
}

akrun <- function(df) {
  df["resultInPercent "] <- 
  mapply(function(x,y) toString(round(x/y)), 
       lapply(strsplit(as.character(df$length2), ","), as.numeric),
       as.numeric(as.character(df$length1))/100)
  df
}

microbenchmark(Steve(bigdf), David(bigdf2), akrun(bigdf))    
#          expr       min        lq      mean    median       uq      max neval cld
#  Steve(bigdf) 475.62891 488.96441 501.77668 497.47626 507.9581 571.5748   100   c
# David(bigdf2)  17.78974  18.16284  18.77208  18.36107  18.6625  29.8744   100 a  
#  akrun(bigdf) 145.98749 149.93839 154.36653 151.82216 154.4384 218.4145   100  b

Answer 3

由于列为factor类，我们将＆＃39; length2＆＃39;在通过分隔符character转换为,类后，将list中的元素转换为numeric，使用mapply来划分list的元素使用＆＃39; length1＆＃39;，vector的相应round元素输出并转换为单个字符串（toString是paste(., collapse=", ")的包装器<） / p>

mapply(function(x,y) toString(round(x/y)), 
    lapply(strsplit(as.character(df$length2), ","), as.numeric),
      as.numeric(as.character(df$length1))/100)
#[1] "20, 30, 40" "33, 42, 25" "21, 14, 43"

两列之间的数据帧操作，并将结果添加为第三列

3 个答案: