字符串拆分data.table列生成NA
发布时间：2020-12-14 18:26:56 所属栏目：资源 来源：网络整理
导读：这是关于SO的第一个问题,请告诉我是否可以改进.我正在研究R中的自然语言处理项目,并且正在尝试构建包含测试用例的data.table.在这里,我构建了一个简化的示例： texts.dt - data.table(string = c("one","two words","three words here","four useless words
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这是关于SO的第一个问题,请告诉我是否可以改进.我正在研究R中的自然语言处理项目,并且正在尝试构建包含测试用例的data.table.在这里,我构建了一个简化的示例：
texts.dt <- data.table(string = c("one","two words","three words here","four useless words here","five useless meaningless words here","six useless meaningless words here just","seven useless meaningless words here just to","eigth useless meaningless words here just to fill","nine useless meaningless words here just to fill up","ten useless meaningless words here just to fill up space"),word.count = 1:10,stop.at.word = c(0,1,2,4,3,6,7,5))
这将返回我们将要处理的data.table： string word.count stop.at.word 1: one 1 0 2: two words 2 1 3: three words here 3 2 4: four useless words here 4 2 5: five useless meaningless words here 5 4 6: six useless meaningless words here just 6 3 7: seven useless meaningless words here just to 7 3 8: eigth useless meaningless words here just to fill 8 6 9: nine useless meaningless words here just to fill up 9 7 10: ten useless meaningless words here just to fill up space 10 5 在实际应用程序中,stop.at.word列中的值是随机确定的(上限= word.count – 1).此外,字符串不按长度排序,但不应该有所不同. 代码应该添加两列input和output,其中input包含从位置1到stop.at.word的子字符串,输出包含后面的单词(单个单词),如下所示： >desired_result
string word.count stop.at.word input
1: one 1 0
2: two words 2 1 two
3: three words here 3 2 three words
4: four useless words here 4 2 four useless
5: five useless meaningless words here 5 4 five useless meaningless words
6: six useless meaningless words here just 6 2 six useless
7: seven useless meaningless words here just to 7 3 seven useless meaningless
8: eigth useless meaningless words here just to fill 8 6 eigth useless meaningless words here just
9: nine useless meaningless words here just to fill up 9 7 nine useless meaningless words here just to
10: ten useless meaningless words here just to fill up space 10 5 ten useless meaningless words here
output
1:
2: words
3: here
4: words
5: here
6: meaningless
7: words
8: to
9: fill
10: just
不幸的是,我得到的是： string word.count stop.at.word input output 1: one 1 0 2: two words 2 1 NA NA 3: three words here 3 2 NA NA 4: four useless words here 4 2 NA NA 5: five useless meaningless words here 5 4 NA NA 6: six useless meaningless words here just 6 3 NA NA 7: seven useless meaningless words here just to 7 3 NA NA 8: eigth useless meaningless words here just to fill 8 6 NA NA 9: nine useless meaningless words here just to fill up 9 7 NA NA 10: ten useless meaningless words here just to fill up space 10 5 ten NA 注意结果不一致,第1行为空字符串,第10行返回“10”. 这是我正在使用的代码： texts.dt[,c("input","output") := .(
substr(string,sapply(gregexpr(" ",string),"[",stop.at.word) - 1),substr(string,stop.at.word),stop.at.word + 1) - 1)
)]
我运行了很多测试,当我在控制台中尝试单个字符串时,substr指令运行良好,但在应用于data.table时失败. 我非常感谢一些帮助. 解决方法
我可能会这样做
texts.dt[stop.at.word > 0,"output") := {
sp = strsplit(string," ")
list(
mapply(function(p,n) paste(p[seq_len(n)],collapse = " "),sp,mapply(`[`,stop.at.word+1L)
)
}]
# partial result
head(texts.dt,4)
string word.count stop.at.word input output
1: one 1 0 NA NA
2: two words 2 1 two words
3: three words here 3 2 three words here
4: four useless words here 4 2 four useless words
交替： library(stringi)
texts.dt[stop.at.word > 0,"output") := {
patt = paste0("((w+ ){",stop.at.word-1,"}w+) (.*)")
m = stri_match(string,regex = patt)
list(m[,2],m[,4])
}]
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