how to do the calculation with different signs-CodePudding

I have a df and I want to do a simple calculation using test*factor. How could I take care of the signs in front of the value?

df<-structure(list(test = structure(c(3L, 2L, 5L, 4L, 1L), .Label = c("<=1.5", 
"<5", ">=1", ">7.8", "6"), class = "factor"), factor = c(2, 4, 
6, 9, 8)), row.names = c(NA, -5L), class = "data.frame")

CodePudding user response：

Another possible solution:

library(tidyverse)

df %>% 
  mutate(outcome = str_c(str_remove(test, "\\d "), factor *
     as.numeric(str_extract(test, "\\d "))))

#>   test factor outcome
#> 1  >=1      2     >=2
#> 2   <5      4     <20
#> 3    6      6      36
#> 4  >78      9    >702
#> 5 <=15      8   <=120

Or with parse_number:

library(tidyverse)

df %>% 
  mutate(outcome = str_c(str_remove(test, "\\d "),
    factor * parse_number(test %>% as.character())))

#>   test factor outcome
#> 1  >=1      2     >=2
#> 2   <5      4     <20
#> 3    6      6      36
#> 4  >78      9    >702
#> 5 <=15      8   <=120

With decimals:

library(tidyverse)

df %>% 
  mutate(outcome = str_c(str_remove(df$test, "\\d (\\.\\d )?"),
   factor * parse_number(test %>% as.character())))

#>    test factor outcome
#> 1   >=1      2     >=2
#> 2    <5      4     <20
#> 3     6      6      36
#> 4  >7.8      9   >70.2
#> 5 <=1.5      8    <=12

CodePudding user response：

We could use str_replace - in the pattern, match one or more digits (\\d ) and replace with a function to multiply with the factor column after converting to numeric

library(stringr)
df$outcome <- str_replace(df$test, "\\d ", function(x) as.numeric(x) * df$factor)

-output

df$outcome
[1] ">=2"   "<20"   "36"    ">702"  "<=120"

CodePudding user response：

One more. Here we use parse_number to multiply and then paste after removing the numbers from test column:

library(dplyr)
library(readr)
df %>% 
  mutate(outcome = paste0(str_replace(test, '\\d ', ''), factor*parse_number(as.character(test))
         ))

  test factor outcome
1  >=1      2     >=2
2   <5      4     <20
3    6      6      36
4  >78      9    >702
5 <=15      8   <=120

CodePudding user response：

We can try

transform(
    df,
    outcome = paste0(gsub("\\d","",test),as.numeric(gsub("\\D", "", test)) * factor)
)

which gives

  test factor outcome
1  >=1      2     >=2
2   <5      4     <20
3    6      6      36
4  >78      9    >702
5 <=15      8   <=120

CodePudding user response：

And another solution using separate:

library(tidyr)
library(stringr)
library(dplyr)
df %>%
    separate(test, 
             into = c("temp1", "temp2"), 
             sep = "(?<=\\D|^)(?=\\d )", 
             remove = FALSE) %>%
  mutate(
    temp3 = as.numeric(temp2)*factor,
    outcome = str_c(temp1, temp3)
    ) %>%
  select(-matches("temp"))
  test factor outcome
1  >=1      2     >=2
2   <5      4     <20
3    6      6      36
4  >78      9    >702
5 <=15      8   <=120