将具有一列和多行的数据转换为多列多行数据

时间：2017-06-28 06:45:36

标签： r reshape

我在R中输出了网页报废数据，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或位置。我想将上面的数据转换为表格格式。输出应该看起来像

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

5 个答案:

答案 0 :(得分：4)

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')]

给出：

    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（再次使用@ UweBlock＆＃39;）

                  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")

答案 1 :(得分：4)

在每个名称前面插入\nName:，然后使用read.dcf将其读取（如果数据来自文件，则使用文件名替换textConnection(Lines)，例如"myfile.dat"，第一行代码。）没有使用包。

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

使用最后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。这个问题略有修改，表明如果缺少电子邮件或城市/城镇，它可以正常工作：

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

答案 2 :(得分：3)

输入数据提出了一些挑战：

  • 数据以直线字符向量的形式给出，而不是具有预定义列的data.frame。
  • 行部分由键/值对组成，由": "
    • 分隔
  • 其他行充当节标题。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：

  1. 键/值对包含一个且仅包含一个toString()
  2. 部分标题完全没有。

    3. 通过将dcast()指定为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")] 的聚合函数来处理部分中的多个密钥，例如，具有电子邮件地址的多个行。

        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

    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")

    数据

    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")

    或者，尝试更多＆＃34;现实＆＃34;名称

                      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

    将导致：

    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")

    或者，每个部分有多个键

                      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

    {{1}}

答案 3 :(得分：3)

使用 dplyr 和 tidyr ，对@Jaap txt和@UweBlock txt1提供的数据进行了测试：

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：

  • 请参阅here我们需要rn的原因。
  • 希望有人会仅使用 tidyverse 建议更好/更简单的代码。

答案 4 :(得分：2)

基准：

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
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