Question

我试图通过模式映射提取所需的单词。

以下是对象表中的示例数据


+-----------+-------------------------------------------------------------------------------------------------+
| Unique_Id |                                               Text                                              |
+-----------+-------------------------------------------------------------------------------------------------+
| Ax23z12   | Tool generated code 2015-8134 upon further validation, the tool confirmed the code as 2015-8134 |
+-----------+-------------------------------------------------------------------------------------------------+

使用以下代码

regmatches(table[1,2],gregexpr("2000-\\d{4}",table[1,2]))

能够将输出提取为

[[1]]
[1] "2000-0511" "2000-0511"

然而，我正在寻找的输出如下


+-----------+---------------------------------------------------------------------------+-----------+-----------+
| Unique_Id |                                    Text                                   |  Column1  |  Column2  |
+-----------+---------------------------------------------------------------------------+-----------+-----------+
| Ax23z12   | Tool generated code 2015-8134 upon further validation, the tool confirmed | 2015-8134 | 2015-8134 |
|           |   the code as 2015-8134                                                   |           |           |
+-----------+---------------------------------------------------------------------------+-----------+-----------+

文本列下的数据包含此数字多次（最多7次），因此寻找动态解决方案

非常感谢

Answer 1

这是一种方法。我使用了以下示例数据，称为foo。

#     id                                                                     text
#  <int>                                                                    <chr>
#1     1                Here is my code, 2015-8134. Here is your code, 2015-1111.
#2     2 His code is 2016-8888, her code is 2016-7777, and your code is 2016-6666

我首先使用stri_extract_all_regex()为text提取了数字。这将返回一个矩阵，因此我将其转换为数据框。然后，我使用bind_cols()将其与原始数据集合并。最后一项工作是修改列名。我使用X

中的Column替换了列名中的gsub()

library(dplyr)
library(stringi)

out <- stri_extract_all_regex(str = foo$text, pattern = "\\d+-\\d+", simplify = TRUE) %>%
                              data.frame(stringsAsFactors = FALSE) %>%
       bind_cols(foo,. )

names(out) <- names(out) %>%
              gsub(pattern = "X", replacement = "Column")

#     id                                                                     text   Column1   Column2   Column3
#  <int>                                                                    <chr>     <chr>     <chr>     <chr>
#1     1                Here is my code, 2015-8134. Here is your code, 2015-1111. 2015-8134 2015-1111          
#2     2 His code is 2016-8888, her code is 2016-7777, and your code is 2016-6666 2016-8888 2016-7777 2016-6666

DATA

foo <- structure(list(id = 1:2, text = c("Here is my code, 2015-8134. Here is your code, 2015-1111.", 
"His code is 2016-8888, her code is 2016-7777, and your code is 2016-6666"
)), .Names = c("id", "text"), class = c("tbl_df", "tbl", "data.frame"
), row.names = c(NA, -2L))

Answer 2

使用stringr和data.table：

1）使用str_match_all提取所有匹配的模式;

2）使用transpose将提取的模式转换为列;

3）通过将提取的列与原始列组合来构建新的数据帧;

library(stringr)
library(data.table)

lst = transpose(str_match_all(df$Text, "2015-\\d{4}"))
data.frame(df, setNames(lst, paste0("Column", seq_along(lst))))
#  Unique_Id                                                                                            Text   Column1   Column2
#1   Ax23z12 Tool generated code 2015-8134 upon further validation, the tool confirmed the code as 2015-8134 2015-8134 2015-8134
#2   By56m22                                           Tool generated code 2015-8134 upon further validation 2015-8134      <NA>

Answer 3

这样的事可能对你有用

df[apply(df, 1, function(x) any(grepl("2000-\\d{4}", x))), ]

请参阅此可重现的示例

iris[apply(iris, 1, function(x) any(grepl("set", x))), ]

   # Sepal.Length Sepal.Width Petal.Length Petal.Width Species
# 1           5.1         3.5          1.4         0.2  setosa
# 2           4.9         3.0          1.4         0.2  setosa
# 3           4.7         3.2          1.3         0.2  setosa
# 4           4.6         3.1          1.5         0.2  setosa
# 5           5.0         3.6          1.4         0.2  setosa
# 6           5.4         3.9          1.7         0.4  setosa
# etc

使用模式R捕获文本 - 正则表达式

3 个答案: