Question

我有一个包含库存数据的数据框：

stock,value
stock1,$30
stock2,£31
stock3,$50

如何将列值溢出以使两列具有货币和值。输出的一个例子：

stock,currency,value
stock1,$,30
stock2,£,31
stock3,$,50

Answer 1

尝试

library(data.table)#v1.9.5+
setDT(df1)[, c('currency', 'value') :=tstrsplit(value,
             '(?<=[^0-9])', perl=TRUE, type.convert=TRUE)]
df1
#    stock value currency
#1: stock1    30        $
#2: stock2    31        £
#3: stock3    50        $

或者

library(tidyr)
extract(df1, value, into=c('currency', 'value'), '([^0-9]+)(.*)')
#    stock currency value
#1 stock1        $    30
#2 stock2        £    31
#3 stock3        $    50

或使用base R

cbind(df1[1],read.table(text=sub('(?<=[^0-9])', ' ', df1$value, 
                perl=TRUE), col.names=c('currency', 'value')))
#    stock currency value
#1 stock1        $    30
#2 stock2        £    31
#3 stock3        $    50

数据

df1 <- structure(list(stock = c("stock1", "stock2", "stock3"), 
value = c("$30", 
"£31", "$50")), .Names = c("stock", "value"), class = "data.frame", 
row.names = c(NA, -3L))

Answer 2

Akrun打败了我，但这又是另一个：

require(stringr)
str_match(c('$30', '£500'), '^([$£])([0-9\\.]+)' )[,2:3]
     [,1] [,2] 
[1,] "$"  "30" 
[2,] "£"  "500"

你可以充实其余部分。

Answer 3

使用sqldf：

sqldf("SELECT stock, 
      SUBSTR(value, 1, 1) As currency, 
      SUBSTR(value, 2, LENGTH(value)) AS value 
      FROM df1")

或

sqldf("SELECT stock, 
      rtrim(value,'0123456789') AS currency,
      ltrim(value,'$£') AS value
      FROM df1")

输出：

   stock currency value
1 stock1        $    30
2 stock2        £    31
3 stock3        $    50

如何将货币和价值分散到不同的城市

3 个答案:

数据