所以我有一个如下所示的数据框，我所做的是通过简单地执行income=parse_number(family_income) 创建一个新列收入，但我想要的是两个数值的平均值然后显示为收入。
第一个条目的预期输出为 (75000+99999)/2 = 87499.5
我想使用现代 R 函数，如 parse_number 或其他更好的函数来有效地执行此操作。
library(tidyr)
family_income income
<chr> <dbl>
1 $75,000 to $99,999 75000
2 $50,000 to $74,999 50000
3 $0 to $9,999 0
4 $200,000 and up 200000
5 $100,000 to $124,999 100000
【问题讨论】：
一种方法是删除逗号，然后提取数字并取mean
library(stringr)
df$income <- purrr::map_dbl(str_extract_all(str_replace_all(df$family_income,
",", ""), "\\d+"), ~mean(as.numeric(.x)))
df
# family_income income
#1 $75,000 to $99,999 87499.5
#2 $50,000 to $74,999 62499.5
#3 $0 to $9,999 4999.5
#4 $200,000 and up 200000.0
#5 $100,000 to $124,999 112499.5
使用base R，我们可以使用regmatches和gregexpr
temp <- gsub(",", "", df$family_income)
sapply(regmatches(temp, gregexpr("\\d+", temp)), function(x) mean(as.numeric(x)))
数据
df <- structure(list(family_income = structure(c(5L, 4L, 1L, 3L, 2L
), .Label = c("$0 to $9,999", "$100,000 to $124,999", "$200,000 and up",
"$50,000 to $74,999", "$75,000 to $99,999"), class = "factor")),
row.names = c(NA,-5L), class = "data.frame")
【讨论】：
parse_number 只能从字符串中提取一个（第一个）数字，而不是多个数字。
vapply(strsplit(gsub("(\\$|,| and up)", "", df$family_income), " to ", TRUE), function(x) mean(as.numeric(x)), numeric(1L))。
Base R 解决方案不如@Ronak Shah 雄辩：
df$income <- do.call("rbind",
lapply(strsplit(gsub("[[:alpha:]]|\\s+", "",
gsub(" to ", ":",
gsub("[[:punct:]]", "", df$family_income))), ":"),
function(x){mean(as.numeric(x))}))
【讨论】：
unlist 而不是使用do.call(rbind, ...)，但既然我们已经知道我们将要简化，使用sapply 或vapply 比使用unlist(lapply(...)) 更有意义。
另一种基本 R 方法可能是：
vapply(strsplit(gsub("(\\$|,| and up)", "", df$family_income), " to ", TRUE),
function(x) mean(as.integer(x)), numeric(1L))
这与 Ronak Shah 的 stringr + purrr 方法一样快，甚至比目前共享的现有基本 R 方法更快。
如果需要更高的效率，可以考虑编写如下函数：
library(data.table)
parse_income <- function(instring) {
as.data.table(instring)[
, temp := gsub("(\\$|,| and up)", "", instring)][
, c("v1", "v2") := tstrsplit(temp, " to ", fixed = TRUE, type.convert = TRUE)][
, rowMeans(.SD, na.rm = TRUE), .SDcols = c("v1", "v2")]
}
然后可以这样使用：
parse_income(df$family_income)
## [1] 87499.5 62499.5 4999.5 200000.0 112499.5
这是一个从 Ronak Shah 回答中的样本数据开始的快速基准测试。待解析的数据已扩展至10,000个值。
inc <- rep(df$family_income, 1e4/nrow(df)) # Adjust to get a sense of how each approach scales
base_am <- function(instring) {
vapply(strsplit(gsub("(\\$|,| and up)", "", instring), " to ", TRUE),
function(x) mean(as.integer(x)), numeric(1L))
}
base_rs <- function(instring) {
temp <- gsub(",", "", instring)
sapply(regmatches(temp, gregexpr("\\d+", temp)), function(x) mean(as.numeric(x)))
}
base_hf <- function(instring) {
do.call("rbind",
lapply(strsplit(gsub("[[:alpha:]]|\\s+", "",
gsub(" to ", ":",
gsub("[[:punct:]]", "", instring))), ":"),
function(x){mean(as.numeric(x))}))
}
stringi_rs <- function(instring) {
purrr::map_dbl(str_extract_all(str_replace_all(instring, ",", ""), "\\d+"), ~mean(as.numeric(.x)))
}
bench::mark(base_am(inc), base_rs(inc), base_hf(inc), stringi_rs(inc), parse_income(inc), check = FALSE)
## # A tibble: 5 x 13
## expression min median `itr/sec` mem_alloc `gc/sec` n_itr n_gc total_time result memory time gc
## <bch:expr> <bch:tm> <bch:tm> <dbl> <bch:byt> <dbl> <int> <dbl> <bch:tm> <list> <list> <list> <list>
## 1 base_am(inc) 48ms 49.5ms 20.1 312.69KB 2.01 10 1 498ms <dbl [10,000]> <df[,3] [4 × 3… <bch:… <tibble [11…
## 2 base_rs(inc) 99.4ms 127.6ms 8.38 80.04MB 2.79 3 1 358ms <dbl [10,000]> <df[,3] [20,01… <bch:… <tibble [4 …
## 3 base_hf(inc) 67.1ms 68.4ms 14.4 547.2KB 2.06 7 1 485ms <dbl[,1] [10,000… <df[,3] [7 × 3… <bch:… <tibble [8 …
## 4 stringi_rs(inc) 50.3ms 51.8ms 19.1 324.16KB 2.12 9 1 472ms <dbl [10,000]> <df[,3] [38 × … <bch:… <tibble [10…
## 5 parse_income(inc) 14.8ms 15ms 66.1 1.01MB 0 34 0 514ms <dbl [10,000]> <df[,3] [30 × … <bch:… <tibble [34…
【讨论】：