正则表达式 (RegEx) 和 dplyr::filter() [英] Regular expressions (RegEx) and dplyr::filter()

问题描述

我有一个简单的数据框，如下所示:

I have a simple data frame that looks like this:

x <- c("aa", "aa", "aa", "bb", "cc", "cc", "cc")
y <- c(101, 102, 113, 201, 202, 344, 407)
df = data.frame(x, y)    

    x   y
1   aa  101
2   aa  102
3   aa  113
4   bb  201
5   cc  202
6   cc  344
7   cc  407

我想使用 dplyr::filter() 和 RegEx 来过滤掉所有以数字 1

I would like to use a dplyr::filter() and a RegEx to filter out all the y observations that start with the number 1

我想象代码看起来像这样:

I'm imagining that the code will look something like this:

df %>%
  filter(y != grep("^1")) 

但是我收到一个 Error in grep("^1") :缺少参数x"，没有默认值

推荐答案

您需要仔细检查 grepl 和 filter 的文档.

You need to double check the documentations for grepl and filter.

对于 grep/grepl，您还必须提供要检入的向量(在本例中为 y)并且 filter 需要一个逻辑向量(即你需要使用 grepl).如果你想提供一个索引向量(来自 grep)，你可以使用 slice 代替.

For grep/grepl you have to also supply the vector that you want to check in (y in this case) and filter takes a logical vector (i.e. you need to use grepl). If you want to supply an index vector (from grep) you can use slice instead.

df %>% filter(!grepl("^1", y))

或者使用从 grep 派生的索引:

Or with an index derived from grep:

df %>% slice(grep("^1", y, invert = TRUE))

但是你也可以只使用 substr 因为你只对第一个字符感兴趣:

But you can also just use substr because you are only interested in the first character:

df %>% filter(substr(y, 1, 1) != 1)

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