Question

我想知道是否有一种快速方法可以找到两个文本字符串之间的有向交集，例如

 t1 <- "I have achieved my goals over the past 20 years and look forward for my next chalanges"
 t2 <- " have achieved goals and look my chalanges some other words bla bla"

t1 isContainedIn t2将返回7，因为在t1中出现的7个单词也在t2中出现。此外，t1和t2是数据帧中的2列，因此我需要在整个数据帧上应用该函数，并将结果列附加到原始数据帧。这就是我的数据框'data.selected'的样子：

        keywords                                         title
1  Samsung UN48H6350 48" Samsung UN48H6350 48" Full 1080p Smart HDTV 120Hz with Wi-Fi +$50 Visa Gift Card
2  Samsung UN48H6350 48"     Samsung UN48H6350 48" Full HD Smart LED TV -Bundle- (See Below for Contents)
3  Samsung UN48H6350 48"      Samsung UN48H6350 48" Class Full HD Smart LED TV -BUNDLE- See below Details
4  Samsung UN48H6350 48"     Samsung UN48H6350 48" Full HD Smart LED TV With BD-H5100 Blu-ray Disc Player
5  Samsung UN48H6350 48"                 Samsung UN48H6350 48" Smart 1080p Clear Motion Rate 240 LED HDTV
6  Samsung UN48H6350 48"            Samsung UN48H6350 - 48-Inch Full HD 1080p Smart HDTV 120Hz with Wi-Fi
7  Samsung UN48H6350 48"               Samsung 6350 Series UN48H6350 48" 1080p HD LED LCD Internet TV NEW
8  Samsung UN48H6350 48"  Samsung Un48h6350af 75" 1080p Led-lcd Tv - 16:9 - Hdtv 1080p - (un75h6350afxza)
9  Samsung UN48H6350 48"                         Samsung UN48H6350 - 48" HD 1080p Smart HDTV 120Hz Bundle
10 Samsung UN48H6350 48"   Samsung UN48H6350 - 48-Inch Full HD 1080p Smart HDTV 120Hz with Wi-Fi, (R#416)

Answer 1

我想另一种类似的方法就是使用简单的match

string <- strsplit(c(t1, t2), "\\s+") # similar to @Richard
length(na.omit(match(string[[2]], string[[1]])))
## [1] 7

或者lapply

length(unlist(lapply(string[[2]], intersect, string[[1]])))
## [1] 7

Answer 2

我对方向问题的含义并不十分清楚。除非您更改数据，否则交叉点的长度不应更改。这可能就是你要找的东西。

length(Reduce(intersect, strsplit(c(t1, t2), "\\s+")))
# [1] 7

如果您将c(t1, t2)切换为c(t2, t1)，则可以看到Reduce输出中的差异。但正如我所说，长度仍将是相同的。它只是集合的顺序不同。

R文本挖掘 - 文本字段之间的交集

2 个答案: