如何在数据框其他列的一列中搜索字符串 [英] How to search for a string in one column in other columns of a data frame

问题描述

我有一个表，称其为df，有3列，第一个是产品的标题，第二个是产品的说明，第三个是一个单词字符串。我需要做的是在整个表上运行一个操作，创建2个新列（将它们称为 exists_in_title和 exists_in_description），其值为1或0，指示第一列或第二列中是否存在第三列。我需要将其简单地做为1：1运算，例如，调用第1行 A，我需要检查单元格A3是否存在于A1中，并使用该数据创建列
exist_in_title，并且然后检查A2中是否存在A3，并使用该数据创建一列exist_in_description。然后继续前进到B行并执行相同的操作。我有成千上万的数据行，因此一次以1的方式执行这些操作是不现实的，为每行编写单独的函数，肯定需要一个函数或方法来一次性遍历表中的每一行。

I have a table, call it df, with 3 columns, the 1st is the title of a product, the 2nd is the description of a product, and the third is a one word string. What I need to do is run an operation on the entire table, creating 2 new columns (call them 'exists_in_title' and 'exists_in_description') that have either a 1 or 0 indicating if the 3rd column exists in either the 1st or 2nd column. I need it to simply be a 1:1 operation, so for example, calling row 1 'A', I need to check if the cell A3, exists in A1, and use that data to create column exists_in_title, and then check if A3 exists in A2, and use that data to create the column exists_in_description. Then move on to row B and go through the same operation. I have thousands of rows of data so it's not realistic to do these in a 1 at a time fashion, writing individual functions for each row, definitely need a function or method that will run through every row in the table in one shot.

我玩过grepl，pmatch，str_count，但似乎没有一个能真正满足我的需要。我认为grepl可能是最接近我需要的代码，这是我编写的两行代码的示例，这些代码在逻辑上可以执行我希望它们执行的操作，但似乎没有用：

I've played around with grepl, pmatch, str_count but none seem to really do what I need. I think grepl is probably the closest to what I need, here's an example of 2 lines of code I wrote that logically do what I would want them to, but didn't seem to work:

df$exists_in_title <- grepl(df$A3, df$A1)

df$exists_in_description <- grepl(df$A3, df$A2)

但是，当我运行这些命令时，我得到以下消息，这使我相信不能正常工作：参数'pattern'的长度> 1，并且将仅使用第一个元素。

However when I run those I get the following message, which leads me to believe it did not work properly: "argument 'pattern' has length > 1 and only the first element will be used"

任何有关如何执行此操作的帮助将不胜感激。谢谢！

Any help on how to do this would be greatly appreciated. Thanks!

推荐答案

grepl 将与一起使用mapply ：

示例数据框：

title <- c('eggs and bacon','sausage biscuit','pancakes')
description <- c('scrambled eggs and thickcut bacon','homemade biscuit with breakfast pattie', 'stack of sourdough pancakes')
keyword <- c('bacon','sausage','sourdough')
df <- data.frame(title, description, keyword, stringsAsFactors=FALSE)

使用 grepl 搜索匹配项：

df$exists_in_title <- mapply(grepl, pattern=df$keyword, x=df$title)
df$exists_in_description <- mapply(grepl, pattern=df$keyword, x=df$description)

结果：

            title                            description   keyword exists_in_title exists_in_description
1  eggs and bacon      scrambled eggs and thickcut bacon     bacon            TRUE                  TRUE
2 sausage biscuit homemade biscuit with breakfast pattie   sausage            TRUE                 FALSE
3        pancakes            stack of sourdough pancakes sourdough           FALSE                  TRUE

更新I

您也可以使用 dplyr 和 stringr ：

library(dplyr)
df %>% 
  rowwise() %>% 
  mutate(exists_in_title = grepl(keyword, title),
         exists_in_description = grepl(keyword, description))

library(stringr)
df %>% 
  rowwise() %>% 
  mutate(exists_in_title = str_detect(title, keyword),
         exists_in_description = str_detect(description, keyword))   

更新II

地图也是一个选项，或者使用 tidyverse 中的更多选项，另一个选项可以是 pu rrr 与 stringr ：

Update II

Mapis also an option, or using more from tidyverse another option could be purrr with stringr:

library(tidyverse)
df %>%
  mutate(exists_in_title = unlist(Map(function(x, y) grepl(x, y), keyword, title))) %>% 
  mutate(exists_in_description = map2_lgl(description, keyword,  str_detect))

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