【问题标题】：Convert data with one column and multiple rows into multi column multi row data将一列多行数据转换为多列多行数据
【发布时间】：2017-12-01 09:33:25
【问题描述】：

我在 R 中有一个网络抓取数据的输出，如下所示

Name1
Email: email1@xyz.com
City/Town: Location1
Name2
Email: email2@abc.com
City/Town: Location2
Name3
Email: email3@pqr.com
City/Town: Location3

某些名称可能没有电子邮件地址或位置。我想将上述数据转换为表格格式。输出应该是这样的

Name      Email           City/Town
Name1   email1@xyz.com  Location1
Name2   email2@abc.com  Location2
Name3   email3@pqr.com  Location3
Name4                   Location4
Name5   email5@abc.com

【问题讨论】：

源数据如下所示。名称1 电子邮件：email1@abc.com 城市/城镇：位置1 名称2 电子邮件：email2@xyz.com 城市/城镇：位置2 名称3 电子邮件：email3@pqr.com 城市/城镇：位置3
你能提供一个reproducible example吗？

标签： r reshape

【解决方案1】：

使用：

txt <- readLines(txt)

library(data.table)
library(zoo)

dt <- data.table(txt = txt)

dt[!grepl(':', txt), name := txt
   ][, name := na.locf(name)
     ][grepl('^Email:', txt), email := sub('Email: ','',txt)
       ][grepl('^City/Town:', txt), city_town := sub('City/Town: ','',txt)
         ][txt != name, lapply(.SD, function(x) toString(na.omit(x))), by = name, .SDcols = c('email','city_town')]

给予：

    name          email city_town
1: Name1 email1@xyz.com Location1
2: Name2 email2@abc.com Location2
3: Name3 email3@pqr.com Location3
4: Name4                Location4
5: Name5 email5@abc.com

这也适用于真实姓名。使用@uweBlock 的数据，您将获得：

                  name          email city_town
1:            John Doe email1@xyz.com Location1
2: Save the World Fund email2@abc.com Location2
3:     Best Shoes Ltd. email3@pqr.com Location3
4:              Mother                Location4
5:                Jane email5@abc.com

每个部分有多个键（同样使用@UweBlock 的数据）：

                  name                          email             city_town
1:            John Doe email1@xyz.com, email1@abc.com             Location1
2: Save the World Fund                 email2@abc.com             Location2
3:     Best Shoes Ltd.                 email3@pqr.com             Location3
4:              Mother                                Location4, everywhere
5:                Jane                 email5@abc.com

使用过的数据：

txt <- textConnection("Name1
Email: email1@xyz.com
City/Town: Location1
Name2
Email: email2@abc.com
City/Town: Location2
Name3
Email: email3@pqr.com
City/Town: Location3
Name4
City/Town: Location4
Name5
Email: email5@abc.com")

【讨论】：

【解决方案2】：

在每个名称前插入\nName:，然后使用read.dcf 读取它（如果数据来自文件，则在第一行代码中将textConnection(Lines) 替换为文件名，例如"myfile.dat"。）没有包被使用了。

L <- trimws(readLines(textConnection(Lines)))
ix <- !grepl(":", L)
L[ix] <- paste("\nName:", L[ix])
read.dcf(textConnection(L))

使用末尾注释中的输入给出以下内容：

     Name    Email            City/Town  
[1,] "Name1" "email1@xyz.com" "Location1"
[2,] "Name2" NA               "Location2"
[3,] "Name3" "email3@pqr.com" NA

注意：使用的输入。对问题稍作修改，以表明如果缺少电子邮件或城市/城镇，它可以工作：

Lines <- "Name1
Email: email1@xyz.com
City/Town: Location1
Name2
City/Town: Location2
Name3
Email: email3@pqr.com"

【讨论】：

所有其他答案都在解决 Y 问题，这个是在解决 X！
不错的答案！我猜read.dcf 不是一个众所周知的函数（尽管在基础 R 中可用）
很好地展示了学习base R的好处。如果参数all = TRUE 与read.dcf() 一起使用，则该解决方案还能够处理重复条目，例如，多个电子邮件地址，如txt2 样本数据集here 中一样。

【解决方案3】：

输入数据带来了几个挑战：

数据以直字符向量的形式给出，而不是带有预定义列的 data.frame。
行部分由键/值对组成，由": " 分隔
其他行用作节标题。下面行中的所有键/值对都属于一个部分，直到到达下一个标题。

以下代码仅依赖于两个假设：

键/值对包含一个且只有一个": "
完全没有节标题。

一个部分中的多个键，例如，具有电子邮件地址的多行通过将toString()指定为dcast()的聚合函数来处理。

library(data.table)
# coerce to data.table
data.table(text = txt)[
  # split key/value pairs in columns
  , tstrsplit(text, ": ")][
    # pick section headers and create new column 
    is.na(V2), Name := V1][
      # fill in Name into the rows below
      , Name := zoo::na.locf(Name)][
        # reshape key/value pairs from long to wide format using Name as row id
        !is.na(V2), dcast(.SD, Name ~ V1, fun = toString, value.var = "V2")]

    Name City/Town          Email
1: Name1 Location1 email1@xyz.com
2: Name2 Location2 email2@abc.com
3: Name3 Location3 email3@pqr.com
4: Name4 Location4             NA
5: Name5        NA email5@abc.com

数据

txt <- c("Name1", "Email: email1@xyz.com", "City/Town: Location1", "Name2", 
"Email: email2@abc.com", "City/Town: Location2", "Name3", "Email: email3@pqr.com", 
"City/Town: Location3", "Name4", "City/Town: Location4", "Name5", 
"Email: email5@abc.com")

或者，尝试更“真实”的名称

txt1 <- c("John Doe", "Email: email1@xyz.com", "City/Town: Location1", "Save the World Fund", 
"Email: email2@abc.com", "City/Town: Location2", "Best Shoes Ltd.", "Email: email3@pqr.com", 
"City/Town: Location3", "Mother", "City/Town: Location4", "Jane", 
"Email: email5@abc.com")

这将导致：

                  Name City/Town          Email
1:     Best Shoes Ltd. Location3 email3@pqr.com
2:                Jane        NA email5@abc.com
3:            John Doe Location1 email1@xyz.com
4:              Mother Location4             NA
5: Save the World Fund Location2 email2@abc.com

或者，每个部分有多个键

txt2 <- c("John Doe", "Email: email1@xyz.com", "Email: email1@abc.com", "City/Town: Location1", "Save the World Fund", 
"Email: email2@abc.com", "City/Town: Location2", "Best Shoes Ltd.", "Email: email3@pqr.com", 
"City/Town: Location3", "Mother", "City/Town: Location4", "City/Town: everywhere", "Jane", 
"Email: email5@abc.com")

                  Name             City/Town                          Email
1:     Best Shoes Ltd.             Location3                 email3@pqr.com
2:                Jane                                       email5@abc.com
3:            John Doe             Location1 email1@xyz.com, email1@abc.com
4:              Mother Location4, everywhere                               
5: Save the World Fund             Location2                 email2@abc.com

【讨论】：

【解决方案4】：

使用 dplyr 和 tidyr，在 @Jaap txt 和 @UweBlock txt1 提供的两个数据上进行测试：

library(dplyr)
library(tidyr)

# data_frame(txt = txt1) %>%     
data_frame(txt = txt) %>% 
  mutate(txt = if_else(grepl(":", txt), txt, paste("Name:", txt)),
         rn = row_number()) %>% 
  separate(txt, into = c("mytype", "mytext"), sep = ":") %>% 
  spread(key = mytype, value = mytext) %>% 
  select(-rn) %>% 
  fill(Name) %>% 
  group_by(Name) %>% 
  fill(1:2, .direction = "down") %>% 
  fill(1:2, .direction = "up") %>% 
  unique() %>% 
  ungroup() %>% 
  select(3:1)

# # A tibble: 5 x 3
#     Name           Email `City/Town`
#    <chr>           <chr>       <chr>
# 1  Name1  email1@xyz.com   Location1
# 2  Name2  email2@abc.com   Location2
# 3  Name3  email3@pqr.com   Location3
# 4  Name4            <NA>   Location4
# 5  Name5  email5@abc.com        <NA>

注意事项：

请参阅 here 为什么我们需要 rn。
希望有人建议仅使用 tidyverse 的更好/更简单的代码。

【讨论】：

更简单：data_frame(text = txt) %>% separate_rows(text, sep = '\n') %>% separate(text, c('var', 'val'), sep = ': ', fill = 'left') %>% mutate(entry = cumsum(is.na(var)), var = coalesce(var, 'Name')) %>% spread(var, val) %>% select(4:2) 其中txt 是字符向量或路径。或者使用data_frame(text = read_lines(txt)) 而不是separate_rows。
@alistaire 无法使用 txt2，重复行错误。也许添加 rn？另外，也许添加为新答案或者我可以添加到我的？
您可以使用toString进行分组和汇总，例如data_frame(text = txt2) %>% #separate_rows(text, sep = '\n') %>% separate(text, c('var', 'val'), sep = ': ', fill = 'left') %>% mutate(entry = cumsum(is.na(var)), var = coalesce(var, 'Name')) %>% group_by(entry, var) %>% summarise(val = toString(val)) %>% spread(var, val) %>% ungroup() %>% select(4:2) 或为键添加索引，但我不太喜欢这两种选择。如果您愿意，请继续添加。

【解决方案5】：

基准测试：

代码：

txt2 <- c("John Doe", "Email: email1@xyz.com", "Email: email1@abc.com", "City/Town: Location1", "Save the World Fund", 
          "Email: email2@abc.com", "City/Town: Location2", "Best Shoes Ltd.", "Email: email3@pqr.com", 
          "City/Town: Location3", "Mother", "City/Town: Location4", "City/Town: everywhere", "Jane", 
          "Email: email5@abc.com")

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

microbenchmark(ans.uwe = data.table(text = txt2)[, tstrsplit(text, ": ")
                                                 ][is.na(V2), Name := V1
                                                   ][, Name := zoo::na.locf(Name)
                                                     ][!is.na(V2), dcast(.SD, Name ~ V1, fun = toString, value.var = "V2")],
               ans.zx8754 = data_frame(txt = txt2) %>% 
                 mutate(txt = ifelse(grepl(":", txt), txt, paste("Name:", txt)),
                        rn = row_number()) %>% 
                 separate(txt, into = c("mytype", "mytext"), sep = ":") %>% 
                 spread(key = mytype, value = mytext) %>% 
                 select(-rn) %>% 
                 fill(Name) %>% 
                 group_by(Name) %>% 
                 fill(1:2, .direction = "down") %>% 
                 fill(1:2, .direction = "up") %>% 
                 unique() %>% 
                 ungroup() %>% 
                 select(3:1),
               ans.jaap = data.table(txt = txt2)[!grepl(':', txt), name := txt
                                                 ][, name := zoo::na.locf(name)
                                                   ][grepl('^Email:', txt), email := sub('Email: ','',txt)
                                                     ][grepl('^City/Town:', txt), city_town := sub('City/Town: ','',txt)
                                                       ][txt != name, lapply(.SD, function(x) toString(na.omit(x))), by = name, .SDcols = c('email','city_town')],
               ans.G.Grothendieck = {
                 L <- trimws(readLines(textConnection(txt2)))
                 ix <- !grepl(":", L)
                 L[ix] <- paste("\nName:", L[ix])
                 read.dcf(textConnection(L))},
               times = 1000)

结果：

Unit: microseconds
               expr       min         lq       mean     median        uq        max neval  cld
            ans.uwe  4243.754  4885.4765  5305.8688  5139.0580  5390.360  92604.820  1000   c 
         ans.zx8754 39683.911 41771.2925 43940.7646 43168.4870 45291.504 130965.088  1000    d
           ans.jaap  2153.521  2488.0665  2788.8250  2640.1580  2773.150  91862.177  1000  b  
 ans.G.Grothendieck   266.268   304.0415   332.6255   331.8375   349.797    721.261  1000 a

【讨论】：