CODEWITHSUNDEEP
rlubridatetibble
TobKel
TobKel

Reputation: 1453

ifelse() function - refer to the following day

I have a dataframe with 2 columns: the date and the return.

df <- tibble( 
date = lubridate::today() +0:9,
return= c(1,2.5,2,3,5,6.5,1,9,3,2))

And now I want to add a third column with an ifelse-condition. If the return on day t is higher than 3.5, than the retrun on the subsequent day t+1 is NA (else = the return on day t).

Here is my desired output:

date         return      retrun_subsequent_day
<date>       <dbl>       <dbl>
1 2019-03-14    1        1
2 2019-03-15    2.5      2.5
3 2019-03-16    2        2
4 2019-03-17    3        3
5 2019-03-18    5        5
6 2019-03-19    6.5      NA
7 2019-03-20    1        NA
8 2019-03-21    9        9
9 2019-03-22    3        NA
10 2019-03-23   2        2

Can someone describe me how can I formulate this condition?

Upvotes: 6

Views: 109

Answers (5)

Rushabh Patel
Rushabh Patel

Reputation: 2764

data table way, efficient for larger dataset-

  data.table::setDT(df)[,return_sbq:=ifelse(shift(return,fill=0) >= 3.5, NA, return)]

> df
          date return return_sbq
 1: 2019-03-14    1.0        1.0
 2: 2019-03-15    2.5        2.5
 3: 2019-03-16    2.0        2.0
 4: 2019-03-17    3.0        3.0
 5: 2019-03-18    5.0        5.0
 6: 2019-03-19    6.5         NA
 7: 2019-03-20    1.0         NA
 8: 2019-03-21    9.0        9.0
 9: 2019-03-22    3.0         NA
10: 2019-03-23    2.0        2.0

Upvotes: 0

C. Braun
C. Braun

Reputation: 5201

Obfuscated, but short and fun:

df$return_subsequent_day <- df$return * lag(df$return < 3.5, 1, 1)^NA

Upvotes: 1

Prince
Prince

Reputation: 336

Simple solution using ifelse

df$return_sub_day <- ifelse(dplyr::lag(df$return) > 3.5, NA ,df$return)
df$return_sub_day[1] <- df$return[1]

Upvotes: 2

f.lechleitner
f.lechleitner

Reputation: 3812

using lag and mutate from dplyr. With lag we compare the return-value of the previous row with 3.5: if it's bigger or equal we take the NA, and if it's smaller we take the return value of the current row

library(dplyr)

df <- df %>% mutate(return_subsequent_day = ifelse(lag(return, default = 0) >= 3.5, NA, return))

output:

# A tibble: 10 x 3
   date       return return_subsequent_day
   <date>      <dbl>                 <dbl>
 1 2019-03-14    1                     1  
 2 2019-03-15    2.5                   2.5
 3 2019-03-16    2                     2  
 4 2019-03-17    3                     3  
 5 2019-03-18    5                     5  
 6 2019-03-19    6.5                  NA  
 7 2019-03-20    1                    NA  
 8 2019-03-21    9                     9  
 9 2019-03-22    3                    NA  
10 2019-03-23    2                     2

Upvotes: 6

akrun
akrun

Reputation: 887128

A base R approach would be to create a copy of the 'return' as new column 'return_sub', then using the numeric index ('i1'), assign the value to NA

i1 <- which(df$return > 3.5)
df$return_subsequent_day <- df$return
df$return_subsequent_day[pmin(i1 +1, nrow(df))] <- NA
df$return_subsequent_day
#[1] 1.0 2.5 2.0 3.0 5.0  NA  NA 9.0  NA 2.0

Upvotes: 6

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