R中关键字匹配的有效方法？答案

【问题标题】：Efficient way of keyword matching in R?R中关键字匹配的有效方法？
【发布时间】：2020-06-27 18:37:17
【问题描述】：

我正在尝试匹配两个大型书目数据集（1.8M obs 和 3.9M obs）之间的关键字，它们来自记录中的各个字段：标题、作者、出版日期、出版商。

对于每个条目（1.8M），我想将字符串中的每个关键字与另一个数据集（3.9M）的每个条目中的关键字进行匹配，并返回匹配最多的行。

我想出的方法，使用来自 tidyverse 的分离（）和聚集（）函数，以及一些基本的 dplyr，似乎有效，但无法扩展到整个数据集。

有没有更有效（或完全更好）的方法来做到这一点？

三个关键字和字符串和代码的示例数据：

library(dplyr)
library(tidyverse)

df1 <- data.frame("df1.index" = c(1:3), 
                  "keywords" = c("2013 history interpretation oxford the tractatus univ wittgensteins", 
                                 "2014 baxi law of oxford pratiksha public secrets univ", 
                                 "2014 darwin flinching from looking on oxford scientific shell-shock"))


df2 <- data.frame("df2.index" = c(1:3), 
                  "keywords" = c("2014 darwin flinching from looking on oxford scientific theatricality",
                                 "2013 interpretation oxford tractatushistory univ wittgensteins", 
                                 "2014 baxi in india law of oxford pratiksha public rape secrets trials univ"))

#separate up to 10 keywords
df1_sep <- separate(df1, keywords, into = 
                      c("key1", "key2", "key3", "key4", "key5", "key6", "key7", "key8", "key9", "key10"), 
                    sep = " ", remove = FALSE)
df2_sep <- separate(df2, keywords, into = 
                      c("key1", "key2", "key3", "key4", "key5", "key6", "key7", "key8", "key9", "key10"), 
                    sep = " ", remove = FALSE)

#gather separated keywords into one column
df1_gather <- df1_sep %>% 
  gather(keys, key.match, key1:key10, factor_key = TRUE) %>% 
  distinct()
df2_gather <- df2_sep %>% 
  gather(keys, key.match, key1:key10, factor_key = TRUE) %>% 
  distinct()

#remove NAs, blanks, trim
df1_gather <- df1_gather %>% filter(!is.na(key.match))
df1_gather <- df1_gather %>% filter(key.match != "")
df1_gather$key.match <- str_trim(df1_gather$key.match)

df2_gather <- df2_gather %>% filter(!is.na(key.match))
df2_gather <- df2_gather %>% filter(key.match != "")
df2_gather$key.match <- str_trim(df2_gather$key.match)

#join, after removing some columns from df2_gather
df2_gather <- df2_gather %>% select(df2.index, key.match)

df_join <- left_join(df1_gather, df2_gather)

#remove NAs
df_join <- df_join %>% filter(!is.na(df2.index))

#tally matches for each index, then take top match
df_join <- df_join %>% group_by(df1.index, df2.index) %>% tally()
df_join <- df_join %>% group_by(df1.index) %>% top_n(1, n)

#add back keywords to review match 
df_join$df1.keywords=df1$keywords[match(df_join$df1.index, df1$df1.index)]
df_join$df2.keywords=df2$keywords[match(df_join$df2.index, df2$df2.index)]

【问题讨论】：

标签： r string dplyr tidyverse keyword

【解决方案1】：

也许这种方法对于直接使用每个关键字进行计数可能很有用。我希望这可以帮助：

library(tidytext)
#Separate
df1 %>% mutate(keywords=as.character(keywords)) %>% unnest_tokens(word,keywords) -> l1
df2 %>% mutate(keywords=as.character(keywords)) %>% unnest_tokens(word,keywords) -> l2
#Join
l1 %>% left_join(l2) -> l3
l2 %>% left_join(l1) -> l4
#Compute number of ocuurences
table(l3$df1.index,l3$df2.index,exclude=NULL)
table(l4$df1.index,l4$df2.index,exclude=NULL)

输出：

    1 2 3 <NA>
  1 1 5 2    3
  2 2 2 9    0
  3 8 1 2    2

       1 2 3
  1    1 5 2
  2    2 2 9
  3    8 1 2
  <NA> 1 1 4

【讨论】：