Question

我有一个帧数据“testData”如下：

id     content
 1     I came from China
 2     I came from America
 3     I came from Canada
 4     I came from Japan
 5     I came from Mars

我还有另一个帧数据“addr”如下：

id   addr
 1   America
 2   Canada
 3   China
 4   Japan

那么如何在R中使用grepl，sapply或任何其他有用的函数来生成数据，如下所示：

id   content               addr
 1   I came from China     China
 2   I came from America   America
 3   I came from Canada    Canada
 4   I came from Japan     Japan
 5   I came from Mars      Mars

Answer 1

看起来您只想复制该列并删除＆＃34;我来自＆＃34;

testData$addr <- gsub("I came from ", testData$content)

Answer 2

这就是诀窍：

vec = addr$addr

testData$addr = apply(testData, 1, function(u){
    bool = sapply(vec, function(x) grepl(x, u[['content']]))
    if(any(bool)) vec[bool] else NA
})

Answer 3

这是使用某些tidyverse函数的粗略解决方案：

df1 <- read.table(text = "id     content
 1     'it is China'
 2     'She is in America now'
 3     'Canada is over there'
 4     'He comes from Japan'
 5     'I came from Mars'", header = TRUE, stringsAsFactors = FALSE)

df2 = read.table(text = "id   addr
 1   America
 2   Canada
 3   China
 4   Japan
 5   Mars", header = TRUE, stringsAsFactors = FALSE)

library(tidyverse)
crossing(df1, df2 %>% select(addr)) %>% # this creates a data frame of every possible content and add combination
  rowwise() %>% 
  filter(str_detect(content, add)) # str_detect is the same as grepl, though the arguments are reversed. This filters to only observations where addr is in content.

# A tibble: 5 x 3
     id content               addr   
  <int> <chr>                 <chr>  
1     1 it is China           China  
2     2 She is in America now America
3     3 Canada is over there  Canada 
4     4 He comes from Japan   Japan  
5     5 I came from Mars      Mars

在R中使用grepl来匹配字符串

3 个答案: