根据动态模式过滤行答案

【问题标题】：Filter rows based on dynamic pattern根据动态模式过滤行
【发布时间】：2021-06-16 13:11:18
【问题描述】：

我在数据框dfin 列Orthographic 中有语音数据：

df <- data.frame(
  Orthographic = c("this is it at least probably",
                   "well not probably it's not intuitive",
                   "sure no it's I mean it's very intuitive",
                   "I don't mean to be rude but it's anything but you know",
                   "well okay maybe"),
  Repeat = c(NA, "probably", "it's,intuitive", "I,mean,it's", NA),
  Repeat_pattern = c(NA, "\\b(probably)\\b", "\\b(it's|intuitive)\\b", "\\b(I,mean|it's)\\b", 
                     NA))

我想filter基于一个动态模式的行，即no、never、not作为单词的出现或n't之前 Repeat 列中列出的任何单词。但是，使用模式\\b(no|never|not)\\b|n't\\b\\s 与列Repeat_pattern 中的交替模式，我得到这个错误：

df %>%
   filter(grepl(paste0("\\b(no|never|not)\\b|n't\\b\\s", Repeat_pattern), Orthographic))
                             Orthographic         Repeat         Repeat_pattern
1    well not probably it's not intuitive       probably       \\b(probably)\\b
2 sure no it's I mean it's very intuitive it's,intuitive \\b(it's|intuitive)\\b
Warning message:
In grepl(paste0("\\b(no|never|not)\\b|n't\\b\\s", Repeat_pattern),  :
  argument 'pattern' has length > 1 and only the first element will be used

我不知道为什么“只使用第一个元素”，因为两个模式组件似乎连接得很好：

paste0("\\b(no|never|not)\\b|n't\\b\\s", df$Repeat_pattern)
[1] "\\b(no|never|not)\\b|n't\\b\\sNA"                     "\\b(no|never|not)\\b|n't\\b\\s\\b(probably)\\b"      
[3] "\\b(no|never|not)\\b|n't\\b\\s\\b(it's|intuitive)\\b" "\\b(no|never|not)\\b|n't\\b\\s\\b(I,mean|it's)\\b"   
[5] "\\b(no|never|not)\\b|n't\\b\\sNA"

预期的输出是这样的：

2                   well not probably it's not intuitive       probably       \\b(probably)\\b
3                sure no it's I mean it's very intuitive it's,intuitive \\b(it's|intuitive)\\b
4 I don't mean to be rude but it's anything but you know    I,mean,it's    \\b(I,mean|it's)\\b

【问题讨论】：

标签： r regex dplyr

【解决方案1】：

这里看起来像是矢量化问题，您需要在这里使用stringr::str_detect 而不是grepl。

另外，您没有很好地对否定词替代品进行分组，所有这些都必须位于一个组中，并且您的 n't 现在是字符串中的必填项。

另外，NA 值被强制转换为文本并添加到正则表达式模式中，而您似乎想要丢弃 Repeat_pattern 为 NA 的项目。

您可以通过使用来修复您的代码

df %>%
    filter(ifelse(is.na(Repeat_pattern), FALSE, str_detect(Orthographic, paste0("(?:\\bno|\\bnever|\\bnot|n't)\\b.*", Repeat_pattern))))

输出：

                                            Orthographic         Repeat         Repeat_pattern
1                   well not probably it's not intuitive       probably       \\b(probably)\\b
2                sure no it's I mean it's very intuitive it's,intuitive \\b(it's|intuitive)\\b
3 I don't mean to be rude but it's anything but you know    I,mean,it's    \\b(I|mean|it's)\\b

我也认为最后一个模式一定是\\b(I|mean|it's)\\b，而不是\\b(I,mean|it's)\\b。

如果“否”字与Repeat 列中的字之间只能有空格，请在我的模式中将.* 替换为\\s+。我使用.*\b 来确保在“否”字右侧的任何地方都有匹配项。

【讨论】：