stringr2

rm(list=ls())
library(stringr)
x <- c("sunqi", "wanglong", "lisi")
str_detect(x, "i")

## [1]  TRUE FALSE  TRUE

# str_count()：匹配字符，返回匹配的个数
str_count(x,"i")

## [1] 1 0 2

# 返回位置str_count

str_view_all(x, "qi")

# 提取匹配内容str_extract
str_extract(c("sunqi zhangsan sunlia wangwu"),pattern ="sun|zhang")

## [1] "sun"

# str_extract_all返回全部字符
str_extract_all(c("sunqi zhangsan lisi wangwu"),pattern ="sun|zhang")

## [[1]]
## [1] "sun"   "zhang"

# 替换第一个匹配值
str_replace(x, "[sn]", "-")

## [1] "-unqi"    "wa-glong" "li-i"

# 替换所有值
str_replace_all(x, "[sn]", "-")

## [1] "-u-qi"    "wa-glo-g" "li-i"

# 拆分字符串
x <- "i like apple and banana"
# 返回为一个个单词
str_split(x, " ")

## [[1]]
## [1] "i"      "like"   "apple"  "and"    "banana"

library(tidyr)
data <- tibble(
  x = c("1234", "B246", "217C", "2357f", "21WD4")
)

str_match_all(x, "\\d")

## [[1]]
##      [,1]

# 求tibble中每个单元格中数字的和
library(tidyverse)
data %>%
  # 匹配字符中的所有数字，返回列表
  mutate(num = str_match_all(x, "\\d")) %>%
  # 将返回的列表还原为数据框
  unnest(num) %>%
  # 对提取的的数字转换为数字
  mutate_at(vars(num), as.numeric) %>%
  # 按照x分组计算和
  group_by(x) %>%
  #求和
  summarise(sum = sum(num))

## # A tibble: 5 x 2
##   x       sum
##   <chr> <dbl>
## 1 1234     10
## 2 217C     10
## 3 21WD4     7
## 4 2357f    17
## 5 B246     12