CODEWITHSUNDEEP
rfor-loopdplyr
nigel griffin
nigel griffin

Reputation: 1

changing the status value to 1 when the diff is <0 in r

I am trying to update the Status column in my data set when the Diff value is < 0, so therefore when the Diff value is a -neg number I want to make the Status column = 1 to represent a change. below is sample code of my data set which is all numeric:

df <- as.data.frame(list(diff=c("4","0","0","0","0","0","0","0","0","0",
                       "0","-30","0","0","0","0","0","0","0","0",
                       "14","0","0","0","0","0","0","-55","0","0",
                       "0","0","0","0","0","0","0","0","0","0",
                       "0","0","0","-40","0","0","0","0","0","0",
                       "0","0","0","0","0","0","0","0","0","0",
                       "0","-30","0","0","0","0","0","0","0","0",
                       "0","0","0","0","0"),
             status=c("0","0","0","0","0","0","0","0","0","0",
                      "0","0","0","0","0","0","0","0","0","0",
                      "0","0","0","0","0","0","0","0","0","0",
                      "0","0","0","0","0","0","0","0","0","0",
                      "0","0","0","0","0","0","0","0","0","0",
                      "0","0","0","0","0","0","0","0","0","0",
                      "0","0","0","0","0","0","0","0","0","0",
                      "0","0","0","0","0")))

Upvotes: 0

Views: 90

Answers (2)

IRTFM
IRTFM

Reputation: 263441

You've created a difficult problem by giving a set of character values to the data.frame function. By default that will create factor variables unless you set stringsAsFactors=FALSE. Two thing are consequential: the comparisons to 0 will fail and assignment to a value that doesn't exist in the df$status column will likewise fail.

df <- data.frame(diff=c("4","0","0","0","0","0","0","0","0","0",
                       "0","-30","0","0","0","0","0","0","0","0",
                       "14","0","0","0","0","0","0","-55","0","0",
                       "0","0","0","0","0","0","0","0","0","0",
                       "0","0","0","-40","0","0","0","0","0","0",
                       "0","0","0","0","0","0","0","0","0","0",
                       "0","-30","0","0","0","0","0","0","0","0",
                       "0","0","0","0","0"),
             status=c("0","0","0","0","0","0","0","0","0","0",
                      "0","0","0","0","0","0","0","0","0","0",
                      "0","0","0","0","0","0","0","0","0","0",
                      "0","0","0","0","0","0","0","0","0","0",
                      "0","0","0","0","0","0","0","0","0","0",
                      "0","0","0","0","0","0","0","0","0","0",
                      "0","0","0","0","0","0","0","0","0","0",
                      "0","0","0","0","0"), stringsAsFactors=FALSE)

Once those issues are put aside you can get success with a logical indexed assignment like:

df$status[as.numeric(df$diff) < 0 ] <- 1

Upvotes: 1

Joseph Noirre
Joseph Noirre

Reputation: 387

library(dplyr)    
df <- df %>%
      mutate(status = ifelse(diff < 0, 1, 0))

Upvotes: 0

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