是否有更优雅的方法可以将参差不齐的数据转换为整洁的数据框答案

【问题标题】：Are there more elegant ways to transform ragged data into a tidy dataframe是否有更优雅的方法可以将参差不齐的数据转换为整洁的数据框
【发布时间】：2014-08-03 07:35:56
【问题描述】：

我有一个数据框，其中包含一列参差不齐的数据：“主题”，其中每个主题都是一串字符，相邻的主题由分隔符（在本例中为“|”）相互分隔：

library(lubridate)
events <- data.frame(
  date  =dmy(c(     "12/6/2012",           "13/7/2012",    "4/8/2012")),
  days  =    c(               1,                     6,           0.5),
  name  =    c("Intro to stats", "Stats Winter school", "TidyR tools"),
  topics=    c( "probability|R", "R|regression|ggplot", "tidyR|dplyr"),
  stringsAsFactors=FALSE
  )

events 数据框如下所示：

        date days                name              topics
1 2012-06-12  1.0      Intro to stats       probability|R
2 2012-07-13  6.0 Stats Winter school R|regression|ggplot
3 2012-08-04  0.5         TidyR tools         tidyR|dplyr

我想转换此数据框，以便每行包含一个主题，并指示在该主题上花费了多少天，假设如果在 D 天内呈现 N 个主题，则每个主题花费 D/N 天主题。

我不得不匆匆忙忙地这样做，并且这样做了：

library(dplyr)

events %>%
  # Figure out how many topics were delivered at each event
  mutate(
    ntopics=sapply(
      gregexpr("|", topics, fixed=TRUE),
      function(x)(1 + sum(attr(x, "match.length") > 0 ))
      )
    ) %>%
  # Create a data frame with one topic per row
  do(data.frame(
    date    =rep(   .$date, .$ntopics),
    days    =rep(   .$days, .$ntopics),
    name    =rep(   .$name, .$ntopics),
    ntopics =rep(.$ntopics, .$ntopics),
    topic   =unlist(strsplit(.$topics, "|", fixed=TRUE)),
    stringsAsFactors=FALSE
    )) %>%
  # Estimate roughly how many days were spent on each topic
  mutate(daysPerTopic=days/ntopics)

这给了我们

        date days                name ntopics       topic daysPerTopic
1 2012-06-12  1.0      Intro to stats       2 probability         0.50
2 2012-06-12  1.0      Intro to stats       2           R         0.50
3 2012-07-13  6.0 Stats Winter school       3           R         2.00
4 2012-07-13  6.0 Stats Winter school       3  regression         2.00
5 2012-07-13  6.0 Stats Winter school       3      ggplot         2.00
6 2012-08-04  0.5         TidyR tools       2       tidyR         0.25
7 2012-08-04  0.5         TidyR tools       2       dplyr         0.25

我很想知道如何更优雅地实现这一目标。

【问题讨论】：

你可能正在寻找的是unnest()，但我还没有实现它：github.com/hadley/tidyr/issues/3
很高兴知道 (a) 我没有错过一个明显的 dplyr/tidyr 解决方案 (b) 这个概念在 tidyr 的经典中占有一席之地。

标签： r dplyr tidyr

【解决方案1】：

你可以试试：

library(data.table)
library(devtools)
source_gist(11380733) ## 

dat <- cSplit(events, "topics", sep="|", "long")

dat1 <-  dat[, c("ntopics", "daysperTopic") := {m= length(days);list(m, days/m)},
                 by=name][,c(1:3,5,4,6),with=F]

dat1
#         date days                name ntopics      topics daysPerTopic
# 1: 2012-06-12  1.0      Intro to stats       2 probability         0.50
# 2: 2012-06-12  1.0      Intro to stats       2           R         0.50
# 3: 2012-07-13  6.0 Stats Winter school       3           R         2.00
# 4: 2012-07-13  6.0 Stats Winter school       3  regression         2.00
# 5: 2012-07-13  6.0 Stats Winter school       3      ggplot         2.00
# 6: 2012-08-04  0.5         TidyR tools       2       tidyR         0.25
# 7: 2012-08-04  0.5         TidyR tools       2       dplyr         0.25

dplyr 可以缩短

library(stringr)
library(dplyr)

res <- mutate(events %>% 
 mutate(
 ntopics = str_count(
     topics, pattern = "\\|") + 1, N = row_number()) %>% 
  do(data.frame(
        .[rep(.$N, .$ntopics), ], 
     topic = unlist(strsplit(.$topics, "|", fixed = TRUE)))), 
   daysPerTopic = days/ntopics) %>%
  select(-topics, -N)
 res
 #        date days                name ntopics       topic daysPerTopic
 #1 2012-06-12  1.0      Intro to stats       2 probability         0.50
 #2 2012-06-12  1.0      Intro to stats       2           R         0.50
 #3 2012-07-13  6.0 Stats Winter school       3           R         2.00
 #4 2012-07-13  6.0 Stats Winter school       3  regression         2.00
 #5 2012-07-13  6.0 Stats Winter school       3      ggplot         2.00
 #6 2012-08-04  0.5         TidyR tools       2       tidyR         0.25
 #7 2012-08-04  0.5         TidyR tools       2       dplyr         0.25

【讨论】：

【解决方案2】：

我想我会添加一个 base R 解决方案，尽管将其扩展为 更优雅。只是一个简单的字符串拆分和重塑

# split topics column 
events <- cbind(events, 
                read.table(text=events$topics, sep="|", fill=TRUE, 
                                                header=FALSE, na.strings=""))

# calculate statistics
events$ntopics <- rowSums(!is.na(events[paste0("V",1:3)]))
events$daysPerTopic <- events$days / events$ntopics

# reshape
na.omit(reshape(events, varying = list(paste0("V",1:3)),
                                        v.names="topics", direction="long"))

【讨论】：